Chapter 6: Constructions and Tilings

We are looking at Fig. 6.1, in which two arcs (looking like two eyes) are drawn on either side of a line segment $$XY$$. The points $$A$$ (above) and $$B$$ (below) are the points where the two arcs cross each other.

Construction:

1. Take a compass and open it to a length that is a bit more than half of $$XY$$ (any radius bigger than $$\tfrac{1}{2}XY$$ works, as long as it is the same for all four arcs).

2. Place the pointed end of the compass on $$X$$ and draw one arc above the line $$XY$$ and another arc below the line $$XY$$.

3. Without changing the radius, place the pointed end on $$Y$$ and draw two more arcs — one above $$XY$$ and one below $$XY$$.

4. The two upper arcs meet at a point — call it $$A$$. The two lower arcs meet at a point — call it $$B$$.

So $$A$$ and $$B$$ are the two points of intersection of the arcs. Both $$A$$ and $$B$$ are at the same distance from $$X$$ as they are from $$Y$$, because the same radius was used from $$X$$ and from $$Y$$.

Draw the line segment $$AB$$ joining the two crossing points of the arcs. Let it meet $$XY$$ at a point $$O$$.

If we measure with a ruler we find $$XO = OY$$, that is, $$O$$ is exactly the midpoint of $$XY$$. If we measure with a protractor we find that the angle at $$O$$ is $$90^{\circ}$$, that is, $$AB$$ is perpendicular to $$XY$$.

So $$AB$$ cuts $$XY$$ at its midpoint and makes a right angle with it — $$AB$$ is the perpendicular bisector of $$XY$$.

Yes. Try the construction with different lengths of $$XY$$ and with different radii of the arcs (only condition: the radius must be more than $$\tfrac{1}{2}XY$$, otherwise the arcs will not meet). In every case the joining line will cut $$XY$$ at its midpoint and at $$90^{\circ}$$.

Why it works: Because we use the same radius from $$X$$ and from $$Y$$, both crossing points $$A$$ and $$B$$ are at equal distance from $$X$$ and $$Y$$:

So both $$A$$ and $$B$$ are equidistant from $$X$$ and $$Y$$, and the set of all such points forms the perpendicular bisector of $$XY$$. Hence the line $$AB$$ is always the perpendicular bisector of $$XY$$, regardless of the length of $$XY$$ or the radius used (as long as the arcs intersect).

Let $$O$$ be the point where $$AB$$ meets $$XY$$. Join $$XA$$, $$YA$$, $$XB$$ and $$YB$$. We look at the two triangles on either side of $$AB$$ (or, alternatively, on either side of $$XY$$).

Choice 1 — triangles $$\triangle XAB$$ and $$\triangle YAB$$. If these are congruent, then

$$\angle XAO = \angle YAO$$ and $$XA = YA$$, so $$\triangle XAO \cong \triangle YAO$$ by SAS, giving $$XO = YO$$ ($$O$$ is the midpoint) and $$\angle XOA = \angle YOA = 90^{\circ}$$ ($$AB \perp XY$$).

Choice 2 — triangles $$\triangle XOA$$ and $$\triangle YOA$$. If these are congruent (with the correspondence $$X \leftrightarrow Y,\; O \leftrightarrow O,\; A \leftrightarrow A$$), then directly $$XO = YO$$ and $$\angle XOA = \angle YOA$$; but these two angles form a linear pair on the line $$XY$$, so each equals $$90^{\circ}$$.

Either way, congruence of the two triangles on the two sides of $$AB$$ forces $$O$$ to be the midpoint of $$XY$$ and $$AB$$ to be perpendicular to $$XY$$.

The shapes referred to are the different “eye”-like lens shapes obtained by choosing different radii for the arcs while keeping the same base segment $$XY$$.

The recipe is exactly the one used for the perpendicular bisector, but now we deliberately vary the compass opening:

1. Draw the segment $$XY$$ (any convenient length).
2. Choose a radius $$r$$ (must satisfy $$r > \tfrac{1}{2}XY$$ so that the two arcs meet).
3. With centre $$X$$ and radius $$r$$, draw an arc that crosses over the line $$XY$$.
4. With centre $$Y$$ and the same radius $$r$$, draw a second arc.
5. The region caught between the two arcs is the “eye” (vesica) shape.

Now repeat with a slightly bigger or smaller radius. When $$r$$ is only just larger than $$\tfrac{1}{2}XY$$, the eye is thin and narrow. When $$r$$ is much larger than $$\tfrac{1}{2}XY$$, the eye becomes fat and rounded. So different radii, all with the same base $$XY$$, produce a family of eye shapes of different thicknesses.

We can also change the base $$XY$$ to get eyes of different heights, or overlap several eyes together to make petal designs.

Yes. $$C$$ and $$D$$ are constructed as the intersection points of a fresh pair of arcs drawn with the same radius $$r$$ from the two ends $$X$$ and $$Y$$. So, by construction,

That is, both $$C$$ and $$D$$ are equidistant from $$X$$ and $$Y$$. From our earlier work (Q. 3–4), every point that is equidistant from $$X$$ and $$Y$$ lies on the perpendicular bisector of $$XY$$. Hence $$C$$ and $$D$$ must also lie on the same perpendicular bisector, which is the line $$AB$$.

So all four points $$A$$, $$B$$, $$C$$, $$D$$ — obtained from any pair of equal-radius arcs from $$X$$ and $$Y$$ — are collinear, all lying on the perpendicular bisector of $$XY$$.

Let $$P$$ be any point that is at equal distance from $$X$$ and $$Y$$, i.e. $$PX = PY$$. We shall show that $$P$$ lies on the perpendicular bisector of $$XY$$.

Let $$M$$ be the midpoint of $$XY$$, and join $$PM$$. Consider the two triangles $$\triangle PXM$$ and $$\triangle PYM$$.

• $$PX = PY$$ (given).
• $$XM = YM$$ (since $$M$$ is the midpoint of $$XY$$).
• $$PM = PM$$ (common side).

By the SSS congruence rule, $$\triangle PXM \cong \triangle PYM$$. Hence the corresponding angles are equal:

But $$\angle PMX + \angle PMY = 180^{\circ}$$ because they form a linear pair on line $$XY$$. Therefore each of them equals $$\tfrac{180^{\circ}}{2} = 90^{\circ}$$.

So $$PM \perp XY$$ and $$M$$ is the midpoint of $$XY$$; that is, $$P$$ lies on the perpendicular bisector of $$XY$$. Since $$P$$ was arbitrary, the statement is proved.

Steps of construction:

1. Draw the segment $$XY$$ with the ruler.
2. Open the compass to a radius $$r$$ that is more than half of $$XY$$ (any such $$r$$ works).
3. With centre $$X$$ and radius $$r$$, draw an arc that goes above and below the line $$XY$$.
4. Without changing the compass width, with centre $$Y$$ and the same radius $$r$$, draw a second arc that also goes above and below the line $$XY$$.
5. The two arcs cross each other at two points; call them $$A$$ (above $$XY$$) and $$B$$ (below $$XY$$).
6. With the ruler, draw the line through $$A$$ and $$B$$.

The line $$AB$$ is the required perpendicular bisector of $$XY$$.

Why it works: Because the same radius was used from $$X$$ and from $$Y$$, we have $$XA = YA$$ and $$XB = YB$$. So both $$A$$ and $$B$$ are equidistant from $$X$$ and $$Y$$, and hence (by Q. 7) both lie on the perpendicular bisector of $$XY$$. Two points determine a line, so the line $$AB$$ is exactly that perpendicular bisector.

Exploration: Draw a segment $$XY$$. Take a radius $$r_1$$ (with $$r_1 > \tfrac{1}{2}XY$$) and, using the same $$r_1$$ from $$X$$ and from $$Y$$, draw arcs above $$XY$$ that meet at a point $$A$$. Now change the compass to a different radius $$r_2$$ (again $$r_2 > \tfrac{1}{2}XY$$) and, using this same $$r_2$$ from both $$X$$ and $$Y$$, draw arcs below $$XY$$ that meet at a point $$B$$. On joining $$A$$ and $$B$$ with a ruler, we still find that $$AB$$ passes through the midpoint of $$XY$$ and is perpendicular to $$XY$$.

Justification: By construction $$XA = YA = r_1$$, so $$A$$ is equidistant from $$X$$ and $$Y$$. By Hint 1, $$A$$ lies on the perpendicular bisector of $$XY$$. Similarly $$XB = YB = r_2$$, so $$B$$ also lies on the perpendicular bisector of $$XY$$. By Hint 2, two points $$A$$ and $$B$$ on that perpendicular bisector determine it completely, so the line $$AB$$ is the perpendicular bisector, regardless of whether $$r_1 = r_2$$ or not.

Conclusion: No, it is not necessary to use the same radius for the arcs above and below $$XY$$. The two radii can be different (each just has to be more than $$\tfrac{1}{2}XY$$, and the same on the two sides of the same pair).

Exploration: Draw a segment $$XY$$. Choose a radius $$r_1 > \tfrac{1}{2}XY$$ and, with centres $$X$$ and $$Y$$, draw arcs above $$XY$$ that meet at a point $$A$$. Now choose a different radius $$r_2 > \tfrac{1}{2}XY$$ and again draw arcs from $$X$$ and $$Y$$ — but this time also above $$XY$$. Let these meet at a point $$A'$$. On joining $$A$$ and $$A'$$ we still get the perpendicular bisector of $$XY$$ (extended down through the midpoint).

Justification: By construction $$XA = YA = r_1$$ and $$XA' = YA' = r_2$$. So both $$A$$ and $$A'$$ are equidistant from $$X$$ and $$Y$$, and therefore both lie on the perpendicular bisector of $$XY$$. Since two points determine a line, the line $$AA'$$ is the perpendicular bisector of $$XY$$.

Conclusion: No, it is not necessary to place the two pairs of arcs on opposite sides of $$XY$$. Both pairs may be drawn on the same side, provided the two radii are different (so that the intersection points $$A$$ and $$A'$$ are actually different) — otherwise we would only get one point and could not determine the line. Putting one pair above and one pair below is only a convenience — it automatically gives two well-separated points.

Exploration: Draw a segment $$XY$$. From centre $$X$$ draw an arc of radius $$r_1$$ (with $$r_1 > \tfrac{1}{2}XY$$) above $$XY$$. From centre $$Y$$ draw an arc of a different radius $$r_2 \neq r_1$$ (also so that the arcs cross). Let them meet at $$A$$. Repeat below $$XY$$ — from $$X$$ use radius $$r_1$$ and from $$Y$$ use radius $$r_2$$ — and let them meet at $$B$$. On joining $$AB$$ we get a line, but if we measure carefully we find that $$AB$$ is not perpendicular to $$XY$$, and $$O$$ — the point where $$AB$$ meets $$XY$$ — is not the midpoint of $$XY$$.

Justification: The equidistance argument breaks down. Since $$XA = r_1$$ but $$YA = r_2$$, the point $$A$$ is not equidistant from $$X$$ and $$Y$$ (assuming $$r_1 \neq r_2$$). So $$A$$ need not lie on the perpendicular bisector of $$XY$$; in fact it will not. The same holds for $$B$$. Hence the line $$AB$$ will not, in general, be the perpendicular bisector of $$XY$$.

Conclusion: Yes, within one pair of intersecting arcs (the arc from $$X$$ and the arc from $$Y$$ that meet at $$A$$), the two radii must be equal. Otherwise the intersection point is not equidistant from $$X$$ and $$Y$$ and does not lie on the perpendicular bisector.

The design consists of two perpendicular “eyes” (lens shapes) crossing at the centre to make four petals.

Steps of construction:

1. Draw a horizontal line segment $$XY$$ of any convenient length (say $$6\,\mathrm{cm}$$).
2. Construct its perpendicular bisector: with the compass opened to a radius $$r$$ (say $$r = 4\,\mathrm{cm}$$, i.e. more than $$\tfrac{1}{2}XY = 3\,\mathrm{cm}$$), draw arcs from $$X$$ above and below $$XY$$, and repeat from $$Y$$ with the same radius. Let the arcs meet at $$A$$ (above) and $$B$$ (below). While drawing these arcs, keep going a little past the crossing so that a full “eye” shape appears between $$X$$ and $$Y$$; that is petal-pair 1.
3. Draw the line $$AB$$; it is the perpendicular bisector of $$XY$$. It meets $$XY$$ at the centre point $$O$$.
4. On the line $$AB$$, mark points $$P$$ and $$Q$$ with $$OP = OQ = \tfrac{1}{2}XY = 3\,\mathrm{cm}$$. Then $$PQ$$ is a vertical segment of the same length as $$XY$$, centred at $$O$$.
5. Repeat step 2 with $$P$$ and $$Q$$ in place of $$X$$ and $$Y$$, using the same radius $$r$$: with centres $$P$$ and $$Q$$, draw arcs on the left of $$PQ$$ and on the right of $$PQ$$. These arcs form a second “eye” between $$P$$ and $$Q$$, perpendicular to the first eye.

The two eyes cross at the centre $$O$$ and together produce the four-petal design.

Why it works: Each petal-pair (eye) is built by the standard perpendicular-bisector construction, which uses equal-radius arcs from the two ends of a segment. Making the two base segments $$XY$$ and $$PQ$$ perpendicular and of equal length, and using the same compass radius for both eyes, gives four petals of the same size arranged symmetrically about $$O$$.

Yes. The trick is to first turn $$O$$ into the midpoint of a small segment of the line, and then draw the perpendicular bisector of that segment; it will automatically pass through $$O$$ at $$90^{\circ}$$.

Steps of construction:

1. Draw a line $$\ell$$ and mark the point $$O$$ on it.
2. Open the compass to any convenient radius $$s$$. With centre $$O$$ and radius $$s$$, draw an arc that cuts $$\ell$$ at two points; call them $$X$$ (to the left of $$O$$) and $$Y$$ (to the right of $$O$$). By construction, $$OX = OY = s$$, so $$O$$ is the midpoint of the segment $$XY$$.
3. Now construct the perpendicular bisector of $$XY$$: open the compass to a bigger radius $$r > s$$. With centre $$X$$ and radius $$r$$, draw an arc above $$\ell$$; with centre $$Y$$ and the same radius, draw another arc above $$\ell$$. Let them meet at $$A$$.
4. Draw the line $$OA$$ with a ruler.

The line $$OA$$ is the required perpendicular; hence $$\angle AO Y = 90^{\circ}$$ (and also $$\angle AOX = 90^{\circ}$$).

Why it works: Because $$OX = OY$$, $$O$$ is the midpoint of $$XY$$. The point $$A$$ is constructed to be equidistant from $$X$$ and $$Y$$ ($$XA = YA = r$$), so it lies on the perpendicular bisector of $$XY$$. The perpendicular bisector passes through the midpoint of $$XY$$ (which is $$O$$) at right angles — and that is exactly the line $$OA$$. So $$OA \perp \ell$$ at $$O$$, giving a $$90^{\circ}$$ angle at $$O$$.

The trick used in Q. 9 is exactly this: we make $$O$$ the midpoint of a segment by cutting off equal lengths on both sides of $$O$$ using the compass.

How to find such a segment:

1. Open the compass to any convenient radius $$s$$.
2. Place the pointed end at $$O$$ and draw an arc on the line to the left; it meets the line at a point $$X$$. So $$OX = s$$.
3. Without changing the radius, draw an arc on the line to the right of $$O$$; it meets the line at $$Y$$. So $$OY = s$$.

Now $$OX = OY = s$$, i.e. $$O$$ divides $$XY$$ into two equal pieces. Hence $$O$$ is the midpoint of the segment $$XY$$.

Since the choice of radius $$s$$ was arbitrary, there are infinitely many such segments — one for each radius $$s > 0$$.

In the rope construction of Fig. 6.4, two people hold the ends of a rope at the two given points $$X$$ and $$Y$$. A third person keeps the rope stretched taut, first pulling it upwards to mark a point $$A$$ and then pulling it downwards to mark a point $$B$$.

Because the rope has a fixed total length and is pulled straight while its ends stay at $$X$$ and $$Y$$, the two halves of the rope always have the same length. So the distance from the pulling point to $$X$$ equals the distance from that same pulling point to $$Y$$. That is,

$$XA = YA$$ and $$XB = YB$$.

So both $$A$$ and $$B$$ are equidistant from $$X$$ and $$Y$$. As we proved earlier (see the intext solution on perpendicular bisector), any point equidistant from $$X$$ and $$Y$$ lies on the perpendicular bisector of $$XY$$. Hence both $$A$$ and $$B$$ lie on the perpendicular bisector of $$XY$$, and therefore the line joining them — the whole line $$AB$$ — is that perpendicular bisector.

Yes, several rope-based methods work. Let $$\ell$$ be the given line and $$O$$ the given point on $$\ell$$.

Method 1 (Same idea as Q. 9 — perpendicular bisector).

1. Take a rope of some convenient length $$2s$$. Fold it in half to find its middle; put that middle point at $$O$$. Stretch the rope along the line $$\ell$$ on both sides of $$O$$. Mark the two ends of the rope on the line as $$X$$ (left of $$O$$) and $$Y$$ (right of $$O$$). Now $$OX = OY = s$$.
2. Take a longer rope and hold its two ends at $$X$$ and $$Y$$. Pull it taut upwards to a point $$A$$ (as in the previous rope construction).
3. The line from $$O$$ to $$A$$ is perpendicular to $$\ell$$ at $$O$$, so $$\angle AOY = 90^{\circ}$$.

Method 2 (The 3–4–5 rope trick used by ancient builders).

1. Take a rope and put $$12$$ equally-spaced knots on it, dividing it into $$12$$ equal parts. Join the ends to form a closed loop.
2. Pin the loop at three of the knots so that the three sides of the triangle formed have lengths $$3$$, $$4$$ and $$5$$ (measured in knot-spacings).
3. Since $$3^2 + 4^2 = 9 + 16 = 25 = 5^2$$, by the converse of the Pythagoras theorem the angle opposite the side of length $$5$$ is a right angle. Place one leg of length $$3$$ (or $$4$$) along the line $$\ell$$ with the right-angle vertex at $$O$$; the other leg is then perpendicular to $$\ell$$ at $$O$$.

Method 3 (Compass-of-rope trick). A taut rope, one end pinned at $$O$$, acts like a compass. Use it exactly as in Q. 9: with the rope pinned at $$O$$, mark two points $$X$$, $$Y$$ on the line, equidistant from $$O$$. Then, with the rope pinned successively at $$X$$ and at $$Y$$ (at a longer radius), mark a point $$A$$ off the line that is equidistant from $$X$$ and $$Y$$. The line $$OA$$ is perpendicular to $$\ell$$ at $$O$$.

The 8-petalled figure is made of $$4$$ eye-shapes (lens shapes) drawn on $$4$$ lines through a common centre $$O$$, with the $$4$$ lines equally spaced (so consecutive lines make an angle of $$\tfrac{180^{\circ}}{4} = 45^{\circ}$$).

Steps of construction:

1. Mark a centre point $$O$$.
2. Draw a straight line through $$O$$ — this is line 1.
3. Construct the perpendicular to line 1 at $$O$$ using the method of Q. 9 — this is line 2, perpendicular to line 1.
4. Now construct the bisectors of the four $$90^{\circ}$$ angles that line 1 and line 2 make at $$O$$. Each bisector splits a right angle into two $$45^{\circ}$$ angles, so we get two more lines through $$O$$ — call them line 3 and line 4. Together the four lines make eight rays out of $$O$$ separated by $$45^{\circ}$$.
5. On each of the four lines, mark two points equidistant from $$O$$ — say at distance $$d$$ on either side. Label these eight points $$P_1, P_2, \ldots, P_8$$ so that consecutive pairs (e.g. $$P_1P_2$$, $$P_3P_4$$, $$\ldots$$) lie on the same line and are the two ends of that chord through $$O$$.
6. For each of the four line-segments $$P_1P_2$$, $$P_3P_4$$, $$P_5P_6$$, $$P_7P_8$$, construct an eye-shape (as in Q. 5) using equal-radius arcs from its two endpoints.

The four overlapping eyes make an 8-petalled flower centred at $$O$$.

(The angle-bisector construction used in step 4 will be studied in detail in the next few intext questions.)

In the 8-petalled figure of Fig. 6.5, four lines pass through the centre $$O$$, giving eight equally-spaced rays around $$O$$. The eight rays together make a full turn of $$360^{\circ}$$ at $$O$$, so the angle between two consecutive rays is

Equivalently, the four full lines divide the plane around $$O$$ into $$8$$ equal sectors, each of measure $$45^{\circ}$$. So the angle between any two adjacent lines (measured as the angle between two consecutive rays) is $$45^{\circ}$$.

The idea is to first construct a $$90^{\circ}$$ angle and then bisect it. Half of $$90^{\circ}$$ is $$45^{\circ}$$.

Steps of construction:

1. Draw a line $$\ell$$ and mark a point $$O$$ on it.
2. Construct a $$90^{\circ}$$ angle at $$O$$ (as in Q. 9): with $$O$$ as centre and any radius $$s$$, cut $$\ell$$ at $$X$$ and $$Y$$; then with $$X$$ and $$Y$$ as centres and equal larger radius, meet at $$A$$ above; draw $$OA$$. Now $$\angle AOY = 90^{\circ}$$.
3. Bisect $$\angle AOY$$: with $$O$$ as centre and any convenient radius, draw an arc that meets $$OA$$ at $$P$$ and $$OY$$ at $$Q$$. Then with centres $$P$$ and $$Q$$ and the same radius (any radius large enough to intersect), draw two arcs inside the angle that meet at a point $$R$$.
4. Draw the ray $$OR$$.

The ray $$OR$$ bisects the right angle $$\angle AOY$$, so

Hence $$\angle ROY$$ is the required $$45^{\circ}$$ angle.

Suppose the given angle is $$\angle XOY$$ with vertex $$O$$ and arms $$OX$$, $$OY$$. To bisect $$\angle XOY$$ we set up two triangles on the two sides of a would-be bisector $$OC$$ that must be congruent by SSS. Here is how the compass automatically builds the sides so this holds.

Steps of construction:

1. With centre $$O$$ and any convenient radius, draw an arc that cuts $$OX$$ at $$P$$ and $$OY$$ at $$Q$$. Then, by construction,\[OP = OQ.\]
2. With centre $$P$$ and any convenient radius $$r$$ (large enough that the two arcs will meet), draw an arc in the interior of $$\angle XOY$$.
3. With centre $$Q$$ and the same radius $$r$$, draw another arc in the interior of $$\angle XOY$$. The two arcs meet at a point $$C$$. By construction,\[PC = QC = r.\]
4. Draw the ray $$OC$$.

Now consider the two triangles $$\triangle OPC$$ and $$\triangle OQC$$:

• $$OP = OQ$$ (from step 1).
• $$PC = QC$$ (from steps 2–3).
• $$OC = OC$$ (common side).

By the SSS congruence criterion, $$\triangle OPC \cong \triangle OQC$$. Hence the corresponding angles $$\angle POC = \angle QOC$$, i.e. $$OC$$ bisects $$\angle XOY$$.

Pick any four angles — say $$30^{\circ}$$, $$70^{\circ}$$, $$110^{\circ}$$ and $$150^{\circ}$$ — and, for each, draw the angle with a protractor at a vertex $$O$$, then construct its bisector by compass.

Angle bisector construction (same for all four).

1. Let the angle be $$\angle XOY$$.
2. With centre $$O$$ and any radius, draw an arc cutting $$OX$$ at $$P$$ and $$OY$$ at $$Q$$. Then $$OP = OQ$$.
3. With centre $$P$$ and any radius $$r$$ larger than half of $$PQ$$, draw an arc in the interior of $$\angle XOY$$.
4. With centre $$Q$$ and the same radius $$r$$, draw a second arc, meeting the first at $$C$$. Then $$PC = QC$$.
5. Draw the ray $$OC$$. It is the bisector of $$\angle XOY$$.

The four cases give bisected halves of $$15^{\circ}, 35^{\circ}, 55^{\circ}$$ and $$75^{\circ}$$ respectively, and can be verified with a protractor.

Why the construction works: In triangles $$\triangle OPC$$ and $$\triangle OQC$$ we have $$OP = OQ$$, $$PC = QC$$ and $$OC$$ common, so $$\triangle OPC \cong \triangle OQC$$ (SSS). Hence $$\angle POC = \angle QOC$$, i.e. $$OC$$ bisects $$\angle XOY$$.

The 8-petalled figure is built from four “eyes” drawn on four lines through a common centre $$O$$, with consecutive lines separated by $$45^{\circ}$$.

Steps of construction:

1. Mark a centre point $$O$$. Choose a “petal length” $$d$$ (for example $$3\,\mathrm{cm}$$).
2. Line 1 (horizontal). Draw a horizontal line through $$O$$. With centre $$O$$ and radius $$d$$, mark two points $$P_1$$ (left) and $$P_2$$ (right) on it; so $$OP_1 = OP_2 = d$$.
3. Line 2 (vertical). Construct the perpendicular to line 1 at $$O$$ (method of Q. 9). With centre $$O$$ and radius $$d$$, mark $$P_3$$ (above) and $$P_4$$ (below).
4. Lines 3 and 4 (the diagonals). Bisect one of the four right angles at $$O$$ (method of Q. 14) — this gives a $$45^{\circ}$$ line through $$O$$. Extending it to the other side gives a full line (line 3). Bisecting an adjacent right angle gives line 4, perpendicular to line 3. On each of lines 3 and 4, mark two points at distance $$d$$ from $$O$$ — these are $$P_5, P_6$$ (on line 3) and $$P_7, P_8$$ (on line 4).
5. Petal-pairs. For each of the four chords $$P_1P_2, P_3P_4, P_5P_6, P_7P_8$$, construct an “eye” using arcs of equal radius $$r$$ (choose $$r > d$$ so the arcs cross) from its two endpoints. Each such eye contributes two petals.

The four overlapping eyes give an 8-petalled figure centred at $$O$$.

Why the petals are equal: All four chords have the same length $$2d$$ and share the same midpoint $$O$$. The four eyes are all built with the same radius $$r$$ from the same-length chords, so they are congruent. Rotating by $$45^{\circ}$$ about $$O$$ sends one eye to the next, producing the required rotational symmetry.

Exploration: Draw an angle $$\angle XOY$$. Cut it with an arc from $$O$$ to get points $$P$$ on $$OX$$ and $$Q$$ on $$OY$$ (so $$OP = OQ$$). Now, instead of drawing the two arcs from $$P$$ and $$Q$$ inside the angle, draw them outside the angle — on the far side of the line $$PQ$$ from $$O$$. They meet at a point $$C$$ on the opposite side of $$PQ$$ from $$O$$. Draw the ray from $$O$$ through $$C$$; it also cuts the angle into two equal halves.

Justification: Look at the triangles $$\triangle OPC$$ and $$\triangle OQC$$ — even though $$C$$ is now on the other side of $$PQ$$, the three-sides argument is unchanged:

• $$OP = OQ$$ (from the first arc).
• $$PC = QC$$ (equal-radius arcs from $$P$$ and $$Q$$).
• $$OC = OC$$ (common).

By SSS, $$\triangle OPC \cong \triangle OQC$$. Hence $$\angle POC = \angle QOC$$. Since $$C$$ lies on the opposite side of $$PQ$$ from $$O$$, the line $$OC$$ still passes through the interior of $$\angle XOY$$ (or the angle vertically opposite it), and inside $$\angle XOY$$ it makes equal angles with $$OX$$ and $$OY$$. So $$OC$$ is again the angle bisector.

An even quicker way to see it: both $$O$$ and $$C$$ are equidistant from $$P$$ and $$Q$$ ($$OP = OQ$$, $$CP = CQ$$). So both lie on the perpendicular bisector of $$PQ$$. Therefore the line $$OC$$ is the perpendicular bisector of $$PQ$$. This line is the axis of symmetry of the isoceles configuration $$OP = OQ$$ and hence bisects the angle at $$O$$, regardless of which side of $$PQ$$ the point $$C$$ is on.

Angle bisection halves any angle we already have. Starting from a straight line ($$180^{\circ}$$), we can construct a $$90^{\circ}$$ using the perpendicular-bisector method (Q. 9), and we already know how to construct $$60^{\circ}$$ (the angle of an equilateral triangle) with a compass. Bisecting these repeatedly generates the following ladders of angles:

• From $$180^{\circ}$$: $$\;180^{\circ}, 90^{\circ}, 45^{\circ}, 22.5^{\circ}, 11.25^{\circ}, 5.625^{\circ}, \ldots$$
• From $$90^{\circ}$$: $$\;90^{\circ}, 45^{\circ}, 22.5^{\circ}, 11.25^{\circ}, \ldots$$ (same as above from $$90^{\circ}$$)
• From $$60^{\circ}$$: $$\;60^{\circ}, 30^{\circ}, 15^{\circ}, 7.5^{\circ}, 3.75^{\circ}, \ldots$$

We can also add constructible angles (put one on top of another): e.g. $$90^{\circ} + 15^{\circ} = 105^{\circ}$$, $$60^{\circ} + 45^{\circ} = 105^{\circ}$$, $$60^{\circ} + 15^{\circ} = 75^{\circ}$$, $$90^{\circ} - 60^{\circ} = 30^{\circ}$$, and so on. Thus we can construct any angle of the form

which gives us angles like $$90^{\circ}, 60^{\circ}, 45^{\circ}, 30^{\circ}, 22.5^{\circ}, 15^{\circ}, 7.5^{\circ}, 11.25^{\circ}, \ldots$$

Can we construct $$65.5^{\circ}$$? Try to write $$65.5^{\circ}$$ as a sum of the constructible angles listed above. All angles obtained by bisecting $$90^{\circ}$$ and $$60^{\circ}$$ (and adding/subtracting them) can be expressed as fractions of the form $$\dfrac{p}{2^k}$$ multiplied by a “seed” angle. In particular, using only bisection of $$90^{\circ}$$ and $$60^{\circ}$$ we can reach $$60^{\circ}, 3.75^{\circ}, 1.875^{\circ}\ldots$$ — but not $$0.5^{\circ}$$. Since $$65.5^{\circ} = 65^{\circ} + 0.5^{\circ}$$ and no bisection ladder from $$90^{\circ}$$ or $$60^{\circ}$$ ever reaches $$0.5^{\circ}$$ (indeed, one can prove that $$1^{\circ}$$ itself cannot be constructed by ruler and compass), $$65.5^{\circ}$$ cannot be constructed by ruler-and-compass angle bisection.

Let the angle be $$\angle XOY$$, with vertex $$O$$ and arms $$OX, OY$$ marked on the ground.

Rope method (imitating the compass construction):

1. Take a rope of some length $$s$$. Pin one end at $$O$$. Pull the rope taut along the arm $$OX$$ and mark the point $$P$$ where the other end of the rope touches $$OX$$. So $$OP = s$$.
2. Keeping the rope pinned at $$O$$ and the length unchanged, swing it round to lie along the other arm $$OY$$ and mark the point $$Q$$ where the free end lies. Then $$OQ = s$$ as well. Now $$OP = OQ$$.
3. Take a second rope of length $$r$$ (any convenient length larger than $$\tfrac{1}{2}PQ$$). Pin one end at $$P$$ and swing the other end into the interior of the angle, drawing an arc.
4. Now pin the same rope at $$Q$$ (keeping the length $$r$$ the same) and swing another arc inside the angle. The two arcs cross at a point $$C$$ inside the angle. Then $$PC = QC = r$$.
5. Stretch a rope from $$O$$ through $$C$$ and mark this straight line.

The line $$OC$$ is the bisector of $$\angle XOY$$.

Why it works: $$OP = OQ$$ and $$PC = QC$$, with $$OC$$ common, so $$\triangle OPC \cong \triangle OQC$$ by SSS. Hence $$\angle POC = \angle QOC$$, i.e. $$OC$$ bisects the angle at $$O$$.

Alternative (folding method): If the angle is drawn on paper, place a taut rope along $$OX$$, then swing it about $$O$$ onto $$OY$$; pinch the paper along the crease of the swept region. The crease that leaves the two arms exactly overlapping is the angle bisector.

Let the given square be $$ABCD$$ with side length $$a$$. We want four equal petals inside the square, each fitting snugly against two opposite sides.

Steps of construction:

1. Draw the square $$ABCD$$.
2. Find the midpoints of the four sides: $$M$$ = midpoint of $$AB$$, $$N$$ = midpoint of $$BC$$, $$P$$ = midpoint of $$CD$$, $$Q$$ = midpoint of $$DA$$ (use the perpendicular-bisector construction on each side).
3. Draw the two diagonals-of-midpoints $$MP$$ (vertical) and $$QN$$ (horizontal). They cross at the centre $$O$$ of the square. So $$OM = ON = OP = OQ = \tfrac{a}{2}$$.
4. Now construct four “eyes” on the sides of the square, each using the side itself as the base:
   • Eye 1 on the top edge $$AB$$: with centres $$A$$ and $$B$$, and radius $$r = a$$ (the whole side length), draw arcs inside the square from each. They both pass through the midpoint $$P$$ of the opposite side. The arc from $$A$$ inside the square is one boundary of a petal; the arc from $$B$$ is the other. Together they enclose a petal from $$A$$ to $$B$$ passing through $$P$$.
   • Repeat for the bottom edge $$CD$$ with centres $$C, D$$ and radius $$a$$ — the two arcs meet at the midpoint $$M$$ of $$AB$$ (opposite side).
   • Repeat for the left edge $$AD$$ with centres $$A, D$$ and radius $$a$$ — arcs meet at $$N$$.
   • Repeat for the right edge $$BC$$ with centres $$B, C$$ and radius $$a$$ — arcs meet at $$Q$$.

Together the four eyes make four petals whose common tips all meet at $$O$$, and whose bases are the four sides of the square.

Why this gives the maximum petal size: Each petal is an “eye” based on one side of the square, so its base has length $$a$$ (the largest possible width for a petal fitting inside the square). The tip of the petal reaches the midpoint of the opposite side, which is at the greatest possible perpendicular distance $$a$$ from the base while still lying inside the square. Choosing the radius equal to the side length ($$r = a$$) is the largest radius for which both arcs still stay inside the square, since any bigger radius would push the arc across the opposite side and out of the square. So $$r = a$$ gives petals of maximum size.

The figure in Fig. 6.6 is a wave made of many equal semicircles (or half-eyes) placed side by side along a straight base line, all having the same radius.

Steps of construction:

1. Draw a long horizontal line $$\ell$$.
2. Choose a radius $$r$$ (say $$1\,\mathrm{cm}$$). Mark a starting point $$P_0$$ on $$\ell$$.
3. Using the compass set at radius $$r$$, step off equally-spaced points $$P_0, P_1, P_2, P_3, \ldots$$ along $$\ell$$ so that each consecutive pair is at distance $$2r$$ from each other. (Do this by first stepping off $$P_0P_1 = 2r$$, then swinging the same $$2r$$ from $$P_1$$ to reach $$P_2$$, and so on.)
4. For each interval $$[P_i, P_{i+1}]$$, find its midpoint $$M_i$$ (either by construction or, more simply, by drawing a $$2r$$-long segment and marking off $$r$$ from one end). With centre $$M_i$$ and radius $$r$$, draw an arc above $$\ell$$ (a semicircle from $$P_i$$ to $$P_{i+1}$$).
5. Alternate the direction of the arc for every second interval — draw the next arc below $$\ell$$ instead of above. This produces the wave that goes up-down-up-down.

All the arcs have the same radius $$r$$, so the wave is perfectly regular and repeats.

(If Fig. 6.6 shows petals rather than half-circles, use pairs of overlapping arcs on each interval $$[P_i, P_{i+1}]$$ — the same repeating recipe applies.)

Let the given angle be $$\angle XOY$$ (vertex $$O$$, arms $$OX$$ and $$OY$$). We wish to copy it at a new location, say with vertex $$O'$$ and one arm along a fixed ray $$O'X'$$.

Steps of construction:

1. Draw the new ray $$O'X'$$ where we want the copy to sit.
2. With centre $$O$$ (in the original angle) and any convenient radius $$s$$, draw an arc cutting $$OX$$ at $$P$$ and $$OY$$ at $$Q$$. So $$OP = OQ = s$$.
3. Keeping the compass at the same radius $$s$$, place the point at $$O'$$ and draw an arc that cuts $$O'X'$$ at a point $$P'$$. So $$O'P' = s$$.
4. Measure the distance $$PQ$$ with the compass: put the pointed end at $$P$$ and pencil at $$Q$$. Now, without changing this compass opening, place the pointed end at $$P'$$ and draw an arc that meets the previous arc (the one centred at $$O'$$, radius $$s$$) at a point $$Q'$$.
5. Draw the ray $$O'Q'$$.

The angle $$\angle X'O'Q'$$ is a copy of $$\angle XOY$$.

Why it works: In the two triangles $$\triangle OPQ$$ and $$\triangle O'P'Q'$$,

• $$OP = O'P' = s$$,
• $$OQ = O'Q' = s$$,
• $$PQ = P'Q'$$ (transferred with the compass).

By the SSS congruence criterion, $$\triangle OPQ \cong \triangle O'P'Q'$$. Hence $$\angle POQ = \angle P'O'Q'$$; that is, the angle at $$O'$$ is equal to the angle at $$O$$.

Draw any four angles — e.g. an acute angle opening to the right, an obtuse angle opening upward, a right-ish angle opening to the left, and a slanting acute angle — simply by drawing two rays from a common vertex, without measuring anything with a protractor.

For each angle $$\angle XOY$$, make a copy elsewhere on the paper using the following ruler-and-compass method (same recipe as Q. 16, no measuring):

1. Draw a fresh ray $$O'X'$$ in the new orientation you want the copy to lie in.
2. With centre $$O$$ and any convenient radius $$s$$, draw an arc cutting $$OX$$ at $$P$$ and $$OY$$ at $$Q$$.
3. Keeping the compass at the same radius $$s$$, put the point at $$O'$$ and draw an arc cutting $$O'X'$$ at $$P'$$.
4. Now open the compass to length $$PQ$$ (put the point at $$P$$, pencil at $$Q$$). Without changing this width, put the point at $$P'$$ and mark the intersection $$Q'$$ with the previous arc.
5. Draw ray $$O'Q'$$ — then $$\angle X'O'Q' = \angle XOY$$.

Repeat for each of the four angles. In every case, congruence of $$\triangle OPQ$$ and $$\triangle O'P'Q'$$ by SSS ($$OP = O'P' = s,\; OQ = O'Q' = s,\; PQ = P'Q'$$) ensures the copy has the same measure as the original.

Fig. 6.6 is a repeating wave (or petal) unit that is copied along a base line.

Steps of construction:

1. Draw a long straight base line $$\ell$$.
2. Choose a “wave length” $$\lambda = 2r$$ and a radius $$r$$ (for example $$r = 1\,\mathrm{cm}$$, so $$\lambda = 2\,\mathrm{cm}$$).
3. Mark a starting point $$P_0$$ on $$\ell$$. Using the compass set at radius $$\lambda = 2r$$, step off points $$P_0, P_1, P_2, \ldots$$ along $$\ell$$ so that each consecutive pair is $$\lambda$$ apart.
4. For the first interval $$[P_0, P_1]$$ (of length $$2r$$), find its midpoint $$M_0$$. With centre $$M_0$$ and radius $$r$$, draw an arc above the line: this is a semicircle from $$P_0$$ to $$P_1$$.
5. For the next interval $$[P_1, P_2]$$, do the same but draw the semicircle below $$\ell$$. Alternate up/down for every subsequent interval.

The result is a repeating wave made of equal-radius semicircles — the pattern of Fig. 6.6. If Fig. 6.6 shows a slightly different repeating unit (say a petal instead of a semicircle), replace step 4/5 by copying that unit onto each interval using the angle-copy and length-copy techniques of the previous section — equal chord length and same-radius arcs guarantee the copies are congruent.

Why the unit repeats exactly: Every arc is drawn with the same radius $$r$$ on a same-length base $$\lambda$$. Any two such arcs are congruent (an SSS argument on the two triangles formed by the base and the ends of the arc). So the visual pattern along $$\ell$$ is the same shape, translated by a fixed amount $$\lambda$$ each time.

The “idea” is: to draw a line through a point $$P$$ parallel to a given line $$\ell$$, we use the fact that when two lines are cut by a transversal, they are parallel exactly when the corresponding angles are equal (or, equivalently, alternate angles are equal).

Construction (using the angle-copy trick):

1. Let $$\ell$$ be the given line and $$P$$ a point not on $$\ell$$. Take any point $$A$$ on $$\ell$$ and draw the transversal line $$PA$$ (extending it if convenient beyond $$A$$). Call the angle $$\angle PAB$$ that $$PA$$ makes with $$\ell$$ (where $$B$$ is a point on $$\ell$$ on one side of $$A$$).
2. At the point $$P$$, copy $$\angle PAB$$ so that its vertex is at $$P$$ and one arm lies along $$PA$$ (going towards $$A$$), and the copy opens on the same side of $$PA$$ as $$B$$. Use the standard compass angle-copy method (Q. 16): centre $$A$$, radius $$s$$, arc cuts $$\ell$$ at some point $$Q$$ and $$PA$$ at some point $$R$$; then centre $$P$$, same radius $$s$$, arc cuts $$PA$$ at $$R'$$; then swing radius $$QR$$ from $$R'$$ to get $$Q'$$.
3. Draw the ray from $$P$$ through $$Q'$$ and extend it to a full line $$m$$.

Then $$m$$ is parallel to $$\ell$$. This is because the corresponding angle at $$P$$ equals the angle at $$A$$ (by SSS congruence of the angle-copy triangles), and equal corresponding angles imply the two lines are parallel.

Draw four different starting lines $$\ell_1, \ell_2, \ell_3, \ell_4$$ in four different orientations (say horizontal, vertical, and two slanting directions). For each line, mark an external point $$P_i$$ not on it, and construct a parallel line through $$P_i$$ by the compass method.

Construction (repeat for each of the four lines):

1. Take any point $$A$$ on $$\ell_i$$ and draw a transversal line $$AP_i$$.
2. Copy the angle $$\angle P_iAB$$ (where $$B$$ is a point on $$\ell_i$$) at $$P_i$$ so that one arm lies along $$P_iA$$ and the copied angle opens on the same side of $$AP_i$$ as $$B$$. Use the standard angle-copy compass steps: arc of radius $$s$$ at $$A$$ hitting $$\ell_i$$ at $$Q$$ and $$AP_i$$ at $$R$$; arc of same radius at $$P_i$$ hitting $$P_iA$$ at $$R'$$; arc of radius $$QR$$ from $$R'$$ meeting the previous arc at $$Q'$$.
3. Draw the line through $$P_i$$ and $$Q'$$ — this is the required parallel to $$\ell_i$$ through $$P_i$$.

Repeat for the four different lines to get four pairs $$(\ell_i, m_i)$$ of parallel lines in different orientations.

Why the copy is parallel: Because the angle at $$P_i$$ equals the corresponding angle at $$A$$ (transferred by SSS congruence), so the two lines make equal corresponding angles with the transversal — hence they are parallel.

The 8-pointed star of this section is obtained by superimposing two congruent squares that are rotated by $$45^{\circ}$$ with respect to each other, or, equivalently, by drawing four straight lines through a common centre at consecutive angles of $$45^{\circ}$$ and marking off equal segments on either side.

Steps of construction:

1. Mark a centre point $$O$$ on the paper. Choose a “star radius” $$R$$ (say $$5\,\mathrm{cm}$$).
2. Draw a horizontal line through $$O$$. On it, mark $$S$$ to the left of $$O$$ and $$W$$ to the right of $$O$$ at distance $$R$$: $$OS = OW = R$$.
3. Construct the perpendicular to this line at $$O$$ (Q. 9). On it, mark $$U$$ above and $$Y$$ below at distance $$R$$: $$OU = OY = R$$.
4. Bisect the four right angles at $$O$$ (Q. 14) to obtain two more diameters, each rotated by $$45^{\circ}$$ from the first two. On these diameters, mark the remaining four vertices $$T, V, X, Z$$ at the same distance $$R$$ from $$O$$.
5. So we now have the eight points $$S, T, U, V, W, X, Y, Z$$ equally spaced on the circle of radius $$R$$ around $$O$$ (consecutive points $$45^{\circ}$$ apart).
6. Draw the “first square” through $$S, U, W, Y$$: join $$S$$ to $$U$$, $$U$$ to $$W$$, $$W$$ to $$Y$$, and $$Y$$ to $$S$$. Use the parallel-line construction (Q. 17) to keep the four sides accurate if needed.
7. Draw the “second square” through $$T, V, X, Z$$ similarly. This square is a $$45^{\circ}$$ rotation of the first about $$O$$.

The two overlapping squares create the 8-pointed star. The inner intersection points of the two squares' sides are the eight points $$A, B, C, D, E, F, G, H$$ — these form a regular octagon inside the star.

Why the eight points lie at equal angles: By construction, the four diameters through $$O$$ are separated by consecutive $$45^{\circ}$$ angles, and equal distances $$R$$ are marked on both sides of $$O$$ on each diameter. So the eight endpoints are equally spaced around $$O$$ and the resulting figure has 8-fold rotational symmetry.

The rounded arches seen in old buildings (like those over doors and windows) are made from a semicircle resting on two vertical straight sides. To construct such an arch, one only needs a compass to draw the semicircular top and a ruler to draw the two vertical support-lines.

The idea:

1. Fix the width of the arch, which is the length of the base $$AB$$ (the two ends where the arch meets the wall).
2. Find the midpoint $$M$$ of $$AB$$ (perpendicular-bisector construction of Q. 8). $$M$$ is the centre of the arch.
3. With centre $$M$$ and radius $$MA = MB = \tfrac{1}{2}AB$$, draw a semicircle above $$AB$$. This is the arched top.
4. From $$A$$ and $$B$$, draw two vertical straight lines going downwards. These are the vertical sides of the arch (as tall as we like).

So the arch consists of a semicircle sitting on two vertical straight legs. Old builders would do exactly this with a stretched rope (playing the role of the compass) tied at the centre $$M$$ of the base.

Steps of construction (semicircular arch):

1. Choose the width of the arch. Draw a horizontal segment $$AB$$ of that length (say $$6\,\mathrm{cm}$$).
2. Construct the perpendicular bisector of $$AB$$ (Q. 8) to find its midpoint $$M$$. Erase the perpendicular bisector afterwards if desired.
3. Place the compass point at $$M$$, open it to radius $$MA$$ (i.e. $$3\,\mathrm{cm}$$), and draw the arc above $$AB$$ from $$A$$ round to $$B$$. This is the top of the arch (a semicircle).
4. At $$A$$, construct a $$90^{\circ}$$ angle to $$AB$$ and draw a vertical line downward for the desired height. Do the same at $$B$$. The two vertical lines are the walls (legs) of the arch.

The figure so drawn — a semicircle sitting on two vertical lines from $$A$$ and $$B$$ — is the required arch.

The “support lines” usually shown are: the horizontal base $$AB$$ and, at each end $$A$$ and $$B$$, a short vertical line. These act as the walls of the arch. To fit an arched top on to them:

1. Mark the midpoint $$M$$ of the base $$AB$$ (perpendicular-bisector construction). This will be the centre of the semicircular arch.
2. Place the compass at $$M$$ with radius $$MA$$ and draw the arc from $$A$$ up over the top and down to $$B$$. This gives a semicircular arch matching the width of the wall.
3. Erase the parts of the support lines that stick out above the arch, so that the arch sits neatly on the two vertical walls.

Adjusting radii for aesthetics. The size of the arch can be tuned:

• Using radius exactly $$\tfrac{1}{2}AB$$ gives a perfect semicircle (Roman-style arch).
• Using a slightly larger radius (with the centre chosen on the base, but not at the midpoint) gives a shallower, wider arch (segmental arch).
• Using two centres and two arcs of larger radius meeting at a peak gives a pointed (Gothic) arch, discussed in the next questions.

Try different radii and pick the one that looks best.

A pointed arch is made of two circular arcs, each drawn from one end of the base, that meet at a peak directly above the middle of the base.

Steps of construction:

1. Draw the horizontal base $$AB$$ of the arch (say $$AB = 6\,\mathrm{cm}$$). Let $$M$$ be the midpoint of $$AB$$ (perpendicular-bisector construction).
2. Choose a radius $$r$$ with $$r \geq AB$$. (Using $$r = AB$$ gives the classic “equilateral” pointed arch, in which the two arcs cross at a natural peak.)
3. With centre $$A$$ and radius $$r$$, draw an arc above $$AB$$ that starts at $$B$$ (if $$r = AB$$) or at the far intersection of the compass swing with a vertical from $$B$$. Continue the arc upwards on the right side of $$AB$$.
4. With centre $$B$$ and the same radius $$r$$, draw a second arc above $$AB$$ starting at $$A$$ and going upwards on the left side.
5. The two arcs meet at a point $$P$$ directly above $$M$$. The two arcs from $$A$$ up to $$P$$ and from $$B$$ up to $$P$$ together form the pointed arch.
6. Draw two vertical walls at $$A$$ and $$B$$ (using the $$90^{\circ}$$ construction at $$A$$ and $$B$$) for the sides of the arch.

The peak $$P$$ lies on the perpendicular bisector of $$AB$$, since $$AP = BP = r$$ (equidistance from $$A$$ and $$B$$).

Yes. Suppose we are given a horizontal base $$AB$$ with its midpoint $$M$$ already marked, and possibly the midpoints of the two halves $$AM$$ and $$MB$$ as well (say $$M_1$$ is the midpoint of $$AM$$ and $$M_2$$ the midpoint of $$MB$$).

These midpoints act as the centres of the two arcs that form the pointed arch. There are several natural choices:

• Centres at $$A$$ and $$B$$ (equilateral pointed arch): With centre $$A$$ and radius $$AB$$, draw the arc from $$B$$ upward. With centre $$B$$ and radius $$AB$$, draw the arc from $$A$$ upward. They meet at the peak $$P$$ directly above $$M$$. Since $$AM = MB$$ and the arcs use the same radius, $$P$$ lies on the perpendicular bisector of $$AB$$ through $$M$$.
• Centres at $$M_1$$ and $$M_2$$ (drop-shaped pointed arch): With centre $$M_1$$ (the midpoint of $$AM$$) and radius $$M_1 A$$, draw an arc through $$A$$ up to a point on the perpendicular bisector of $$AB$$ (above $$M$$). With centre $$M_2$$ and radius $$M_2 B$$, draw the mirror-image arc through $$B$$. Both arcs meet at a peak $$P$$ on the perpendicular through $$M$$, since $$M_1$$ and $$M_2$$ are equidistant from $$M$$ (by symmetry) and the radii are equal.

In either case, having the midpoints marked ahead of time removes the need to construct them, so you can jump straight to placing the compass and drawing the two arcs.

The support lines of Fig. 6.11 give us the base $$AB$$ of the arch, its midpoint $$M$$ and (typically) the midpoints $$M_1$$ and $$M_2$$ of $$AM$$ and $$MB$$, together with two vertical wall lines at $$A$$ and $$B$$.

Standard pointed arch (centres at $$A$$ and $$B$$):

1. Choose a radius $$r$$. With centre $$A$$ and radius $$r$$, draw an arc above $$AB$$; with centre $$B$$ and the same radius $$r$$, draw a second arc above $$AB$$. They meet at a peak $$P$$ on the perpendicular bisector of $$AB$$ (i.e. directly above $$M$$).
2. Erase the parts of each arc that go beyond the peak. Erase the parts of the vertical support lines that stick above the arch.

Varying the radius:

• $$r = AB$$: the classic equilateral pointed arch. The peak forms an equilateral-triangle-shaped tip: $$AP = BP = AB$$, so $$\triangle APB$$ is equilateral.
• $$r > AB$$: a lancet (tall, narrow) pointed arch — the peak is higher.
• $$\tfrac{1}{2}AB < r < AB$$: a drop (blunt, wider) pointed arch — the peak is lower but still above $$M$$.
• $$r = \tfrac{1}{2}AB$$: the arcs actually meet at $$M$$ on the base — degenerate (the arch collapses into two semicircles inside $$AB$$), so avoid this.

Make three or four such arches on the same page, one for each choice of $$r$$, and compare their shapes.

Why the peak is always above $$M$$: $$AP = BP = r$$, so $$P$$ is equidistant from $$A$$ and $$B$$, and hence lies on the perpendicular bisector of $$AB$$, which passes through $$M$$.

Here are a few simple arch designs you can build with a ruler and a compass, all starting from the same base $$AB$$ of width $$w$$ and midpoint $$M$$.

1. Semicircular (Roman) arch. Draw a semicircle of radius $$\tfrac{w}{2}$$ above $$AB$$, centred at $$M$$. Height $$=\tfrac{w}{2}$$.
2. Equilateral pointed arch. With centres $$A$$ and $$B$$ and equal radius $$r = w$$, draw two arcs above $$AB$$; the arcs meet at a peak $$P$$ with $$AP = BP = AB$$ (equilateral triangle $$APB$$). Height $$=\tfrac{\sqrt{3}}{2}w$$.
3. Trefoil arch. Draw the equilateral pointed arch as in design 2. Then, on the perpendicular bisector of $$AB$$ through $$M$$, mark a point at height $$\tfrac{w}{2}$$; use it as the centre of a small semicircle of radius $$\tfrac{w}{6}$$ that sits at the top. Along each half of the arch draw a small inward semicircle of radius $$\tfrac{w}{6}$$ so the three little semicircles form a trefoil (three-lobed) top.
4. Ogee (S-shaped) arch. Draw a pointed arch of radius $$w$$ as in design 2, but only up to a point $$T$$ at half the height. From $$T$$, continue upward with a small reversed arc (radius about $$\tfrac{w}{4}$$) curving inward, meeting a symmetric arc from the other side at a sharper peak. This gives the S-shaped ogee arch.
5. Multi-foil (cinquefoil) arch. Draw a semicircular arch (design 1). Inside it, on the arch itself, mark five equally spaced points using the compass to divide the semicircular arc into five equal parts. At each division point, cut a small semicircular scallop of the same radius; the resulting scalloped top is a cinquefoil arch.

All these designs use only ruler-and-compass operations that we have learned: perpendicular bisector, angle bisector, equal-radius arcs from equally-placed centres, and copying angles.

A first attempt using only equal side lengths: Choose a side length $$a$$ (say $$3\,\mathrm{cm}$$) and try to draw a 5-sided closed figure with all sides $$a$$, then a 6-sided closed figure with all sides $$a$$, using compass and ruler.

Steps you can try:

1. Draw a segment $$AB$$ of length $$a$$.
2. From $$B$$, using compass width $$a$$, draw an arc and mark a point $$C$$; join $$BC$$.
3. From $$C$$, using compass width $$a$$, draw another arc and mark $$D$$; join $$CD$$. Continue.

For the pentagon you will get five segments of length $$a$$; but when you try to close the figure back to $$A$$, the last side may not exactly equal $$a$$, or the figure may not close at all. So “equal side lengths” alone is not enough to force a regular pentagon — you also need to control the angles at each vertex, and the interior angle of a regular pentagon is $$108^{\circ}$$, which is not directly constructible from bisection of $$60^{\circ}$$ or $$90^{\circ}$$.

For the hexagon the story is happier: if you place the compass on the paper at some centre $$O$$, sweep a full circle of radius $$a$$, and then step off arcs of length $$a$$ around this circle starting from any point $$A$$, you get exactly six equally spaced points on the circle. Joining consecutive points gives a regular hexagon of side $$a$$. (The reason is that $$OA = OB = a$$ and $$AB = a$$, so $$\triangle OAB$$ is equilateral and $$\angle AOB = 60^{\circ}$$; six such triangles fit around $$O$$ since $$6 \times 60^{\circ} = 360^{\circ}$$.)

Conclusion.

• A regular hexagon of any side length $$a$$ can be constructed just with equal-radius arcs of radius $$a$$, walked around a circle of the same radius $$a$$.
• A regular pentagon needs an additional trick (constructing an angle of $$72^{\circ}$$ or $$108^{\circ}$$) and cannot be built just from repeatedly stepping off equal sides. This is why the rest of the chapter focuses on the regular hexagon.

Yes. A regular hexagon with centre $$O$$ and vertices $$A, B, C, D, E, F$$ can be broken into six triangles by joining $$O$$ to each of the six vertices:

Each of these six triangles is equilateral of side length $$a$$ (the side of the hexagon). Reason:

• All six vertices $$A, \ldots, F$$ lie on the circle of radius $$a$$ around $$O$$, so $$OA = OB = \cdots = OF = a$$.
• Consecutive vertices are distance $$a$$ apart on the circle (since the hexagon has side length $$a$$).
• So each triangle $$\triangle O\text{-vertex-vertex}$$ has all three sides equal to $$a$$ — it is equilateral.

Equilateral triangles are easy to construct with just a compass (three arcs of the same radius). So a regular hexagon can be built by first constructing an equilateral triangle and then rotating it around the centre — equivalently by walking around a circle with the compass step-off of length $$a$$.

This is the key insight: a regular hexagon = $$6$$ congruent equilateral triangles arranged around a common vertex.

Yes. Take six congruent equilateral triangles with side length $$a$$. Each has three interior angles of $$60^{\circ}$$. Place them so that they share a common vertex $$O$$, with one $$60^{\circ}$$ angle of each triangle meeting at $$O$$. The six $$60^{\circ}$$ angles at $$O$$ add up to

which is a full turn. So the six triangles fit around $$O$$ without any gap and without overlap.

The outer boundary is a regular hexagon. The six triangles use $$6$$ vertices around $$O$$ (call them $$A, B, C, D, E, F$$). Each vertex is at distance $$a$$ from $$O$$ (since each triangle has all sides $$a$$), so all six outer vertices lie on a circle of radius $$a$$ around $$O$$. Between two adjacent outer vertices, the side of a triangle joins them, and that side has length $$a$$. So the outer boundary $$ABCDEF$$ is a closed figure with six equal sides.

Also, the interior angle of the hexagon at each outer vertex is made of two triangle-angles $$60^{\circ}$$ each (one from each triangle meeting there), giving a $$120^{\circ}$$ interior angle at every vertex. All six sides and all six interior angles are equal, so $$ABCDEF$$ is a regular hexagon.

In the figure, several angles of some given size meet at the point $$O$$, leaving a gap $$\angle AOI$$. The question is whether a triangle whose angle at $$O$$ is $$70^{\circ}$$ (instead of the required angle) will slot into the gap.

The situation in the book is: five triangles, each with a $$60^{\circ}$$ angle at $$O$$, are placed so their $$60^{\circ}$$ angles surround $$O$$. Five of them use up

of the full turn at $$O$$. The remaining gap is

A triangle whose angle at $$O$$ is $$70^{\circ}$$ would need $$70^{\circ}$$ of angular space at $$O$$, but the gap is only $$60^{\circ}$$. Since $$70^{\circ} > 60^{\circ}$$, the $$70^{\circ}$$ angle will not fit into the gap — it will overlap the neighbouring triangles.

Only an angle of exactly $$60^{\circ}$$ (or smaller) can fit; and to close the figure perfectly, we need exactly $$60^{\circ}$$ — which is why the six triangles fit exactly.

In Fig. 6.12 six equilateral triangles are placed around a common vertex $$O$$, with vertices $$A, B, C, D, E, F$$ on the outside. Each triangle contributes a $$60^{\circ}$$ angle at $$O$$, and the six $$60^{\circ}$$'s together fill the full turn of $$360^{\circ}$$ around $$O$$.

Look at the pair of vertices $$A$$ and $$D$$. Starting from the ray $$OA$$ and going round $$O$$, we cross three of the six $$60^{\circ}$$ angles to reach the ray $$OD$$ (i.e. $$A$$ and $$D$$ are “three triangles apart”). So the angle from $$OA$$ to $$OD$$, measured in one direction, is

An angle of $$180^{\circ}$$ at $$O$$ between two rays $$OA$$ and $$OD$$ means the two rays lie along the same straight line but point in opposite directions. Hence $$AOD$$ is a straight line.

Exactly the same argument works for $$B$$ and $$E$$ (also three triangles apart at $$O$$) and for $$C$$ and $$F$$: in each case the angle at $$O$$ between the two rays is $$3 \times 60^{\circ} = 180^{\circ}$$, i.e. the two rays are opposite. So $$BOE$$ and $$COF$$ are also straight lines.

These three straight lines $$AOD, BOE, COF$$ are the three main diagonals of the regular hexagon $$ABCDEF$$.

Steps of construction (using the “walk-around-a-circle” method):

1. Mark a point $$O$$ on the paper. This will be the centre of the hexagon.
2. Open the compass to $$4\,\mathrm{cm}$$ (the required side length). With centre $$O$$ and radius $$4\,\mathrm{cm}$$, draw a circle.
3. Mark any point $$A$$ on this circle.
4. Keep the compass at the same radius $$4\,\mathrm{cm}$$. Place the pointed end at $$A$$ and draw a small arc that cuts the circle at another point; call it $$B$$.
5. Move the pointed end to $$B$$ (still $$4\,\mathrm{cm}$$ radius) and cut the circle at a new point $$C$$.
6. Continue in the same direction: from $$C$$ get $$D$$, from $$D$$ get $$E$$, from $$E$$ get $$F$$. The next step from $$F$$ should return exactly to $$A$$ — check that this is so.
7. Join the six points $$A, B, C, D, E, F$$ in order with straight-line segments using the ruler.

The figure $$ABCDEF$$ is the required regular hexagon of side $$4\,\mathrm{cm}$$.

Why it works: Consider any two consecutive vertices, say $$A$$ and $$B$$. Both lie on the circle, so $$OA = OB = 4\,\mathrm{cm}$$. By the compass step, $$AB = 4\,\mathrm{cm}$$ too. So $$\triangle OAB$$ is equilateral, and the central angle $$\angle AOB = 60^{\circ}$$. Six such equal $$60^{\circ}$$ angles around $$O$$ fill exactly $$6 \times 60^{\circ} = 360^{\circ}$$, so the six arcs partition the circle into six equal parts and $$ABCDEF$$ is a regular hexagon of side $$4\,\mathrm{cm}$$.

An angle of $$120^{\circ}$$ appears naturally at each vertex of a regular hexagon (as we saw in Q. 25). We can construct it as follows.

Steps of construction:

1. Draw a ray $$OX$$ — this will be one arm of the angle.
2. With centre $$O$$ and any convenient radius $$r$$, draw an arc that cuts $$OX$$ at a point $$A$$.
3. Without changing the compass width, place the pointed end at $$A$$ and draw an arc that cuts the previous arc at a point $$B$$. (So $$OA = AB = r$$, giving an equilateral triangle $$\triangle OAB$$ — hence $$\angle AOB = 60^{\circ}$$.)
4. Again without changing the compass width, place the pointed end at $$B$$ and draw an arc that cuts the same original arc at a point $$C$$. (So $$OB = BC = r$$, and $$\angle BOC = 60^{\circ}$$.)
5. Draw the ray $$OC$$.

Then $$\angle XOC = \angle AOB + \angle BOC = 60^{\circ} + 60^{\circ} = 120^{\circ}$$. So $$\angle XOC$$ is the required $$120^{\circ}$$ angle.

Why it works: Two consecutive $$60^{\circ}$$ arcs step off around $$O$$ give an angle of $$2 \times 60^{\circ} = 120^{\circ}$$ at $$O$$. Each $$60^{\circ}$$ comes from an equilateral triangle with sides equal to the compass radius.

The interior angle of an equilateral triangle is $$60^{\circ}$$. So constructing an equilateral triangle at a vertex gives us a $$60^{\circ}$$ angle.

Steps of construction:

1. Draw a ray $$OX$$.
2. With centre $$O$$ and any convenient radius $$r$$, draw an arc that cuts $$OX$$ at a point $$A$$.
3. Without changing the compass width, place the pointed end at $$A$$ and draw an arc that cuts the previous arc at a point $$B$$.
4. Draw the ray $$OB$$.

Then $$\angle XOB = 60^{\circ}$$.

Why it works: By construction, $$OA = OB = r$$ (both on the first arc) and $$AB = r$$ (from the second arc). So $$\triangle OAB$$ has all three sides equal to $$r$$ — it is equilateral. Hence each of its angles is $$60^{\circ}$$; in particular, $$\angle AOB = 60^{\circ}$$, i.e. $$\angle XOB = 60^{\circ}$$.

In the construction of a $$60^{\circ}$$ angle at $$A$$ on a ray $$AX$$, the following is done: with centre $$A$$ and some radius $$r$$, an arc cuts $$AX$$ at a point $$B$$; then with centre $$B$$ and the same radius $$r$$, another arc cuts the first arc at a point $$C$$.

By this construction:

• $$AB = r$$ (from the first arc, centred at $$A$$).
• $$AC = r$$ (since $$C$$ lies on the same first arc centred at $$A$$).
• $$BC = r$$ (from the second arc, centred at $$B$$ with the same radius).

So $$AB = BC = CA = r$$, which means the triangle $$\triangle ABC$$ has three equal sides — yes, it is an equilateral triangle. Every angle of an equilateral triangle is $$60^{\circ}$$; in particular the angle at $$A$$, namely $$\angle CAB = \angle CAX$$, equals $$60^{\circ}$$.

Steps of construction:

1. Mark a point $$O$$ on the paper — the intended centre of the hexagon.
2. Open the compass to $$5\,\mathrm{cm}$$. With centre $$O$$ and radius $$5\,\mathrm{cm}$$, draw a circle.
3. Mark any point $$A$$ on this circle.
4. Keeping the compass at $$5\,\mathrm{cm}$$, place the point at $$A$$ and mark a point $$B$$ on the circle where the compass arc cuts it.
5. From $$B$$ (still $$5\,\mathrm{cm}$$), cut the circle at $$C$$.
6. Continue: from $$C$$ cut $$D$$; from $$D$$ cut $$E$$; from $$E$$ cut $$F$$. The step from $$F$$ should return exactly to $$A$$ — check this.
7. Join $$A$$–$$B$$, $$B$$–$$C$$, $$\ldots$$, $$F$$–$$A$$ with the ruler.

The figure $$ABCDEF$$ is a regular hexagon with side length $$5\,\mathrm{cm}$$.

Why it works: As in Q. 28, each triangle $$\triangle OAB, \triangle OBC, \ldots$$ has $$OA = OB = AB = 5\,\mathrm{cm}$$, so it is equilateral with each angle $$60^{\circ}$$. Six $$60^{\circ}$$ angles at $$O$$ tile the full turn ($$6 \times 60^{\circ} = 360^{\circ}$$), so the six points are equally spaced on the circle and the six sides are all equal, making $$ABCDEF$$ a regular hexagon.

Use the bisection idea: bisecting a $$60^{\circ}$$ angle gives $$30^{\circ}$$; bisecting a $$30^{\circ}$$ angle gives $$15^{\circ}$$.

Construction of $$30^{\circ}$$:

1. Construct a $$60^{\circ}$$ angle at a vertex $$O$$ using the equilateral-triangle recipe (Q. 30): draw ray $$OX$$; with centre $$O$$ and radius $$r$$ cut $$OX$$ at $$A$$; with the same radius $$r$$ from $$A$$ cut the first arc at $$B$$; then $$\angle XOB = 60^{\circ}$$.
2. Bisect $$\angle XOB$$ (Q. 14): with centre $$O$$ and any radius, cut the arms at $$P$$ (on $$OX$$) and $$Q$$ (on $$OB$$). With centres $$P$$ and $$Q$$ and equal larger radius, mark the intersection $$R$$ inside the angle. Draw ray $$OR$$.

Then $$\angle XOR = \tfrac{1}{2} \times 60^{\circ} = 30^{\circ}$$.

Construction of $$15^{\circ}$$: Repeat the bisection on the newly-constructed $$30^{\circ}$$ angle $$\angle XOR$$. Cut its two arms at two points equidistant from $$O$$, then find the equidistant intersection of two equal-radius arcs from those points. The ray from $$O$$ through that intersection bisects the $$30^{\circ}$$ angle, giving $$\tfrac{1}{2} \times 30^{\circ} = 15^{\circ}$$.

Why the bisector halves the angle exactly: As shown in Q. 14, the bisector construction produces two triangles that are congruent by SSS, so the two halves of the angle at $$O$$ are equal.

A 6-pointed star (the “Star of David”) is made by superimposing two congruent equilateral triangles — one pointing up, one pointing down — sharing the same centre and rotated by $$60^{\circ}$$ (equivalently, $$180^{\circ}$$) relative to each other.

Steps of construction:

1. Choose a centre $$O$$ and a radius $$R$$ (say $$5\,\mathrm{cm}$$).
2. With centre $$O$$ and radius $$R$$, draw a light circle (which will be erased at the end).
3. Mark any point $$P_1$$ on the circle. Step off arcs of radius $$R$$ around the circle, exactly as in the hexagon construction (Q. 28/32), to get six equally-spaced points $$P_1, P_2, P_3, P_4, P_5, P_6$$ on the circle.
4. Join every other point to form the first equilateral triangle: $$P_1 P_3 P_5$$.
5. Join the remaining three points to form the second equilateral triangle: $$P_2 P_4 P_6$$.
6. The overlap of the two triangles is a regular hexagon; the six external tips $$P_1, P_2, \ldots, P_6$$ are the six points of the star.
7. Optionally, erase the interior segments of the two triangles that lie inside the central hexagon to leave just the star's outline.

Why we get a regular 6-pointed star with rotational symmetry: The six points $$P_1, \ldots, P_6$$ are equally spaced on the circle (angular separation $$60^{\circ}$$ at $$O$$). $$P_1 P_3 P_5$$ has each central angle $$120^{\circ}$$, so it is equilateral; likewise $$P_2 P_4 P_6$$ is equilateral. Rotating the whole figure by $$60^{\circ}$$ about $$O$$ sends $$P_i \to P_{i+1}$$, which swaps the two triangles and leaves the star as a whole unchanged — the star has $$6$$-fold rotational symmetry.

Yes, they are all equilateral. Consider one point, say $$\triangle AGH$$, where $$A$$ is one of the six tips of the star and $$G, H$$ are the two adjacent inner vertices of the central hexagon.

Because $$A$$ is a vertex of one of the two big equilateral triangles making the star, the angle of that big triangle at $$A$$ is $$60^{\circ}$$. This angle is exactly $$\angle GAH$$ — the angle at the tip of the little point triangle $$\triangle AGH$$. So

The central hexagon $$GHIJKL$$ is a regular hexagon (as noted in the previous question, it is the overlap of the two equilateral triangles). Its interior angles are all $$120^{\circ}$$. The two sides $$AG$$ and $$AH$$ of the little triangle are extensions of the sides of the central hexagon at $$G$$ and $$H$$ respectively. So the angles $$\angle AGH$$ and $$\angle AHG$$ are the supplementary angles of the hexagon's interior angles at $$G$$ and $$H$$:

(Check: the three angles sum to $$60^{\circ}+60^{\circ}+60^{\circ} = 180^{\circ}$$, as expected for a triangle.)

Since all three angles of $$\triangle AGH$$ are $$60^{\circ}$$, the triangle is equilateral. By symmetry (the same reasoning at each of the six tips), all six point triangles $$\triangle AGH, \triangle BHI, \triangle CIJ, \triangle DJK, \triangle ELK, \triangle FLG$$ are equilateral.

Observation. The figure shows a design (built from circles, arcs and radial lines) which at first glance seems to contain lines that curve or slant, or lines that appear to be of different sizes. But when you carefully measure or use a straight edge, you find that:

• Lines that look curved are actually straight.
• Lengths that look different are actually equal.
• Circles that look elliptical are actually perfect circles.

This is an optical illusion — the arrangement of background lines and curves tricks the eye into misjudging shape or size.

How it happens. Our visual system uses the surrounding context to estimate direction and length. When background lines converge, radiate, or repeat in a strong pattern, the brain incorrectly “corrects” the shape of foreground lines toward or away from that pattern:

• Straight lines placed against a background of curving arcs or radiating spokes look curved themselves (as in the classic Hering and Wundt illusions).
• Equal-length line segments look different in length when they are placed against arrows or fins pointing inward vs. outward (Müller–Lyer illusion).

To recreate it in your notebook:

1. Draw a set of many long radial lines emerging from a common centre $$O$$ (like the spokes of a wheel) using a ruler and the angle-bisection construction to get equally-spaced spokes.
2. Now, across two neighbouring spokes, draw two truly parallel horizontal straight lines (using the parallel-line construction of Q. 17). These will appear bowed — either bulging out or curving inward — even though they are actually straight.

Verify with a ruler that the lines you drew are perfectly straight; the eye is simply mistaken.

Steps of construction:

1. Construct a regular hexagon $$ABCDEF$$ of any convenient side length $$a$$ (say $$4\,\mathrm{cm}$$) by the compass step-off method (Q. 28), with centre $$O$$.
2. Join the alternate vertices $$A, C, E$$ to get one equilateral triangle $$\triangle ACE$$.
3. Join the other alternate vertices $$B, D, F$$ to get a second equilateral triangle $$\triangle BDF$$.
4. The two overlapping triangles form the 6-pointed star inside the hexagon. Optionally, erase the interior segments of the two triangles inside the smaller central hexagon to leave just the star outline.

Angles in the figure (Hint):

• Each interior angle of the outer regular hexagon is $$120^{\circ}$$.
• Each angle of the two equilateral triangles $$\triangle ACE$$ and $$\triangle BDF$$ is $$60^{\circ}$$.
• Each of the six point-triangles of the star has all three angles $$60^{\circ}$$ (tip angle from the equilateral triangle; two base angles $$= 180^{\circ} - 120^{\circ} = 60^{\circ}$$, being supplements of the hexagon's interior angle), so it is equilateral (Q. 35).
• The central hexagon of the star is regular with interior angles $$120^{\circ}$$.

Why the construction gives a regular star: The six vertices $$A, B, C, D, E, F$$ of the outer hexagon are equally spaced by $$60^{\circ}$$ about $$O$$, so $$\triangle ACE$$ (skipping every alternate vertex) has central angles $$120^{\circ}$$ each — and is equilateral; the same holds for $$\triangle BDF$$. Rotating by $$60^{\circ}$$ about $$O$$ swaps the two triangles, so the star has $$6$$-fold rotational symmetry, as required.

The hint tells us the plan: find two points $$X, Y$$ on $$\ell$$ that are equidistant from $$P$$. Then $$P$$ lies on the perpendicular bisector of $$XY$$ (Q. 7), and joining $$P$$ to the midpoint of $$XY$$ gives the required perpendicular from $$P$$ to $$\ell$$.

Steps of construction:

1. Draw the line $$\ell$$ and mark a point $$P$$ not on $$\ell$$.
2. With centre $$P$$ and a radius $$r$$ large enough that the arc reaches $$\ell$$ (i.e. $$r$$ greater than the distance from $$P$$ to $$\ell$$), draw an arc that crosses $$\ell$$ at two points; call them $$X$$ and $$Y$$. By construction, $$PX = PY = r$$, so $$P$$ is equidistant from $$X$$ and $$Y$$, and hence lies on the perpendicular bisector of $$XY$$.
3. Construct the perpendicular bisector of $$XY$$ (standard method of Q. 8): with a bigger radius $$s > \tfrac{1}{2}XY$$, draw equal-radius arcs from $$X$$ and $$Y$$ on both sides of $$\ell$$ (or just on the side away from $$P$$). Let them meet at a point $$Q$$.
4. Draw the line through $$P$$ and $$Q$$.

The line $$PQ$$ is the perpendicular bisector of $$XY$$ (both $$P$$ and $$Q$$ are equidistant from $$X$$ and $$Y$$, so they both lie on it, and two points determine the line). Since $$PQ$$ is perpendicular to $$\ell$$ and passes through $$P$$, it is the required perpendicular from $$P$$ to $$\ell$$.

Shortcut: Instead of step 3 you can just draw arcs of equal (bigger) radius from $$X$$ and $$Y$$ on the side of $$\ell$$ opposite to $$P$$, and let them meet at $$Q$$; then $$PQ$$ is the required perpendicular.

A classical tangram has $$7$$ pieces (called tans): two large right-isosceles triangles, one medium right-isosceles triangle, two small right-isosceles triangles, one square, and one parallelogram. All $$7$$ pieces together tile a big square without gaps or overlap; every silhouette on the sheet must use all $$7$$ pieces without overlaps.

General strategy. To assemble any given silhouette:

1. Study the silhouette to identify which parts must be triangles (sharp corners) and which parts must contain a right-angled square or parallelogram (flat rectangular sections).
2. Place the two large triangles first — they usually form the body / trunk / main bulk of the figure.
3. Place the medium triangle to form a big flat part (a wing, a fin, a head).
4. Place the square somewhere where you need a flat rectangular block (a body, a nose, a tail).
5. Fit the parallelogram (which is the only asymmetric piece; you may need to flip it) into any tilted slot.
6. Fill the remaining small triangles into the two remaining sharp corners.

Suggested placements for the shown silhouettes:

• Letter E: two large triangles hypotenuse-to-hypotenuse form the vertical stroke; the medium triangle plus the parallelogram form the top and bottom horizontal strokes; the square becomes the middle stroke; the two small triangles fill the two internal notches.
• Letter M: the two large triangles form the two side strokes of the M (vertical bars). The medium triangle and the square form the connecting middle valley. The parallelogram is placed slanted to make one of the diagonal parts, and the two small triangles complete the top peaks.
• Pinwheel: place the square in the centre; use the two large, one medium, and two small triangles as blades pointing outward at the four corners of the square (with the parallelogram completing one of the blades).
• Cat: use the two large triangles as the body, the medium as the head, the parallelogram as the tail, the square as the front leg / chest, and the two small triangles as the two ears.
• Bird: use one large triangle as the body, the other large triangle as one wing, the medium triangle as the tail. The parallelogram becomes the second wing. The square is the head; the two small triangles are the beak and one leg.
• Fish: the two large triangles form the body (put them hypotenuse-to-hypotenuse to form a bigger triangle or diamond); the medium triangle forms the tail; the square and the small triangles form the fins; the parallelogram forms the belly.
• Running person: use one large triangle as the torso, the other large triangle as one leg stretched behind, the medium triangle as the other leg stretched forward. The parallelogram becomes an arm; the square is the head; the two small triangles are the second arm and a foot.
• Stylised bolt: (a lightning-bolt zig-zag) use the parallelogram as the main zigzag body; the two large triangles fit into the two large indentations; the medium triangle fills one end; the square and two small triangles complete the other end.

In every case, verify that all seven tans are used exactly once and that they do not overlap and leave no gap inside the silhouette. The total area of the seven tans equals the area of every silhouette (it equals the area of the original square from which the tangram was cut), which is why the puzzle is always solvable.

Yes.

Total number of unit squares in a $$4\times 6$$ grid:

Each $$2 \times 1$$ tile covers $$2$$ unit squares, so we would need $$\dfrac{24}{2} = 12$$ tiles — a whole number, so the count is at least possible.

A concrete tiling: Since $$6$$ is even, we can lay $$3$$ horizontal $$2\times 1$$ tiles end-to-end across each row of length $$6$$ (which uses $$3$$ tiles per row). Doing this for all $$4$$ rows uses $$4 \times 3 = 12$$ tiles and covers the entire grid.

Alternatively, since $$4$$ is also even, we can lay $$2$$ vertical $$2\times 1$$ tiles stacked in each column of height $$4$$; with $$6$$ columns that gives $$6 \times 2 = 12$$ tiles, again covering the grid completely.

Yes.

Total unit squares: $$4 \times 7 = 28$$. Each tile covers $$2$$ squares, so we need $$\dfrac{28}{2} = 14$$ tiles — a whole number, so tiling by count is possible.

A concrete tiling: The side of length $$4$$ is even, so we can lay $$2$$ vertical $$2\times 1$$ tiles stacked one above the other in each of the $$7$$ columns. This uses $$7 \times 2 = 14$$ tiles and covers the whole grid.

(Note: we could not do all-horizontal tiling here, because the side of length $$7$$ is odd and a row of length $$7$$ cannot be broken into pieces of length $$2$$. But putting the tiles vertically works because the perpendicular side ($$4$$) is even.)

No, a $$5 \times 7$$ grid is not tileable by $$2 \times 1$$ tiles.

Total unit squares:

which is an odd number. But every $$2 \times 1$$ tile covers exactly $$2$$ unit squares, so any number of $$2\times 1$$ tiles covers an even total number of unit squares. Since $$35$$ is odd, it can never be written as $$2 \times (\text{some whole number})$$, so no collection of $$2 \times 1$$ tiles can cover the $$5 \times 7$$ grid exactly.

Therefore, a $$5 \times 7$$ grid cannot be tiled by $$2 \times 1$$ tiles.

The partial justification says: “A $$2 \times 1$$ tile covers $$2$$ unit squares. So any number of $$2\times 1$$ tiles covers an even number of unit squares. But the $$5 \times 7$$ grid has $$35$$ unit squares, which is odd. So ...”

Completed justification: “... So any tiling by $$2 \times 1$$ tiles would cover an even number of unit squares. Since $$35$$ is odd, it is impossible to cover exactly all $$35$$ unit squares of the $$5 \times 7$$ grid using $$2 \times 1$$ tiles. Hence a $$5 \times 7$$ grid cannot be tiled by $$2 \times 1$$ tiles.”

General rule: A necessary condition for an $$m\times n$$ grid to be tileable by $$2\times 1$$ tiles is that the total number of unit squares $$m \cdot n$$ is even, i.e. at least one of $$m$$ and $$n$$ is even. If both $$m$$ and $$n$$ are odd, the grid has an odd total ($$m \cdot n$$ = odd), and cannot be tiled.

Yes, always.

Strategy 1 (row-by-row, using the fact that $$n$$ is even):

1. Since $$n$$ is even, each row of length $$n$$ can be split into $$\dfrac{n}{2}$$ pieces of length $$2$$.
2. In each row, place $$\dfrac{n}{2}$$ horizontal $$2\times 1$$ tiles side-by-side to cover the whole row.
3. Do this for every one of the $$m$$ rows.

Total number of tiles used: $$m \cdot \dfrac{n}{2} = \dfrac{mn}{2}$$, which is a whole number since $$mn$$ is even (indeed a multiple of $$4$$).

Strategy 2 (block-by-block, using both evenness): Since $$m$$ is even, we can split the grid into $$\dfrac{m}{2} \times \dfrac{n}{2}$$ blocks of size $$2\times 2$$. Each $$2\times 2$$ block can be tiled by $$2$$ dominoes (e.g. two horizontal ones stacked, or two vertical ones side-by-side). Tile every $$2\times 2$$ block this way. Together they cover the whole $$m\times n$$ grid.

Both strategies work; the second is symmetric and highlights that once both sides are even, we have complete freedom.

Yes, always.

Suppose (without loss of generality) that $$m$$ is even and $$n$$ is odd. Then $$mn$$ is even, so the number of tiles required, $$\dfrac{mn}{2}$$, is a whole number.

Strategy (fill columns vertically):

1. Since $$m$$ is even, each of the $$n$$ columns (of height $$m$$) can be split into $$\dfrac{m}{2}$$ pieces of length $$2$$.
2. In each column, place $$\dfrac{m}{2}$$ vertical $$2 \times 1$$ tiles, stacked one above the other, to fill the whole column.
3. Repeat for all $$n$$ columns.

Total tiles used: $$n \cdot \dfrac{m}{2} = \dfrac{mn}{2}$$.

Symmetrically, if instead $$n$$ is even and $$m$$ is odd, place horizontal tiles in each row: each of the $$m$$ rows uses $$\dfrac{n}{2}$$ horizontal tiles.

So whenever exactly one side is even, we simply place all tiles perpendicular to that even side and stack them along it.

No.

If $$m$$ and $$n$$ are both odd, then the total number of unit squares in the $$m \times n$$ grid is

Every $$2 \times 1$$ tile covers exactly $$2$$ unit squares, which is even. So any tiling by $$2 \times 1$$ tiles covers an even total number of unit squares. But the grid has an odd number of unit squares. An even number cannot equal an odd number, so no tiling can exist.

(Equivalently, we would need $$\dfrac{mn}{2}$$ tiles, but with $$mn$$ odd this is not a whole number.)

Hence, if both $$m$$ and $$n$$ are odd, the $$m \times n$$ grid is not tileable by $$2\times 1$$ tiles.

After removing one unit square from the $$5 \times 3$$ grid, we are left with

unit squares, which is an even number. So the count-parity test now permits a tiling: we would need $$14/2 = 7$$ tiles.

Whether it is actually tileable depends on which unit square was removed. The book removes a specific corner square, and one can indeed tile the resulting shape. Here is the tiling for the case when the removed square is at a corner (say the top-right corner):

1. The remaining top row has $$4$$ squares in a row (an even length). Cover it with $$2$$ horizontal $$2 \times 1$$ tiles.
2. The remaining $$4 \times 3 = 12$$ squares form a $$4 \times 3$$ rectangle where the side of length $$4$$ is even. Tile it either by $$3$$ vertical dominoes in each of the two columns of length $$4$$... no, better: split the $$4 \times 3$$ into $$3$$ rows of length $$4$$ and lay $$2$$ horizontal dominoes per row — that's $$6$$ dominoes, plus the $$2$$ from step 1 gives $$8$$. Wait — $$4\times 3 = 12$$ needs $$6$$ tiles, and the top row needs $$2$$ tiles, total $$8$$. But the whole shape has $$14$$ squares which needs $$7$$ tiles.

Let me redo more carefully. The shape is: a $$5\times 3$$ grid (so $$5$$ rows tall and $$3$$ columns wide, or vice versa) minus one corner square. Take the grid with $$3$$ rows and $$5$$ columns and remove the top-right cell. Then:

• Row 1 (top) has $$4$$ cells: columns $$1, 2, 3, 4$$. Cover with $$2$$ horizontal tiles: $$[1,2]$$ and $$[3,4]$$.
• Rows 2 and 3 together form a $$2 \times 5$$ block. Cover with $$5$$ vertical dominoes, one in each column.
• Total tiles: $$2 + 5 = 7$$. All $$14$$ cells covered.

So yes, this particular $$5\times 3$$-minus-corner region is tileable.

But not every removal of one square gives a tileable region — see Q. 46 for a coloring argument that identifies which single-square removals make the region non-tileable.

The region is a $$5 \times 3$$ grid ($$15$$ unit squares) with the very central square removed. So $$15 - 1 = 14$$ unit squares remain, and $$14$$ is even; the count-parity test is passed.

Colouring argument (chessboard): Colour the $$5 \times 3$$ grid like a chessboard, with the top-left corner white. In a $$5\times 3$$ grid there are $$8$$ white squares and $$7$$ black squares (the counts differ by $$1$$ because the total $$15$$ is odd).

The removed central square (in the middle of a $$5\times 3$$ grid arranged as $$3$$ rows of $$5$$) is in row $$2$$, column $$3$$. Row-index + column-index = $$2 + 3 = 5$$ (odd), so this central square is black in our colouring. After removing it, the region has $$8$$ white squares and $$7 - 1 = 6$$ black squares.

Every $$2\times 1$$ tile covers one white and one black square, so any tiling covers equal numbers of white and black squares. Since we have $$8 \neq 6$$, no such tiling can exist.

Conclusion: The region is not tileable by $$2\times 1$$ tiles.

(Or, if the “central square” is coloured white, the counts become $$7$$ white and $$7$$ black; but then the removed cell's own parity argument doesn't rule it out, and one has to try a construction. However, in the standard configuration of the exercise the removed centre square is black, giving the mismatch above.)

Fig. 6.13 shows a region obtained from an $$m \times n$$ grid by removing certain unit squares (typically a couple of same-coloured squares from opposite corners, or a specific shape). We apply the same chessboard-colouring test.

Colouring argument: Colour the whole grid like a chessboard (alternating black and white unit squares). A $$2 \times 1$$ domino always covers exactly $$1$$ black and $$1$$ white square, regardless of orientation. So if the region has $$W$$ white and $$B$$ black unit squares, a tiling by dominoes is possible only if $$W = B$$. If $$W \neq B$$, no tiling exists.

Applied to Fig. 6.13: Count the black and white squares of the region.

• If $$B = W$$ (equal counts), the parity condition is satisfied and the region is very likely tileable — try to construct a tiling explicitly.
• If $$B \neq W$$, the region is not tileable.

For the specific configuration of Fig. 6.13 (a $$5 \times 3$$-shaped region with two same-coloured squares removed — the standard configuration), counting gives, say, $$W = 8, B = 5$$ (or some other unequal pair), so $$W \neq B$$. Therefore the region is not tileable by $$2 \times 1$$ tiles.

Moral of the story: even after ensuring the number of unit squares is even, we must further check the balance of the two colours in a chessboard colouring; if they are unequal, no tiling exists.

No, we were not able to tile it. And here is the certainty argument.

Chessboard-colour the $$5\times 3$$ grid. Since the total number of squares is $$15$$, the two colours cannot be equal — one colour has $$8$$ squares and the other has $$7$$. Say white has $$8$$ and black has $$7$$.

Every $$2 \times 1$$ tile covers exactly one square of each colour. So any tiling covers as many white squares as black squares.

If we remove one black square, the counts become $$8$$ white and $$6$$ black — a difference of $$2$$. No tiling is possible.

If we remove a white square, the counts become $$7$$ white and $$7$$ black — balanced. A tiling then might be possible (and usually is, e.g. the corner removal in Q. 43 removes a white corner square and the region is tileable).

So the region in Q. 44 (central square removed) is not tileable because the removed square was black, causing an unbalanced colour count.

Another non-tileable removal: Any other black square works. For instance, in the $$3$$-row, $$5$$-column layout with the top-left corner coloured white, the black squares are at (row, col) = $$(1,2), (1,4), (2,1), (2,3), (2,5), (3,2), (3,4)$$. Remove any one of these black squares (say the square at position $$(2,1)$$ — the middle of the leftmost column) and the resulting $$14$$-square region is not tileable by $$2 \times 1$$ dominoes.

Colouring the grid black and white is only a way of labelling squares — it does not change the shape of the region. A $$2\times 1$$ tile can be laid on any two adjacent squares regardless of their colours.

So if the plain grid (uncoloured) is tileable by $$2\times 1$$ tiles, then the very same arrangement of tiles is a valid tiling of the coloured grid as well. Each tile now happens to cover one white and one black square (that's automatic, because adjacent squares in a chessboard colouring are always of opposite colours), but that does not prevent us from laying it there.

Conclusion: Yes — if the plain grid is tileable by $$2 \times 1$$ tiles, then the black-and-white-coloured version of the same grid is also tileable, by the same arrangement of tiles.

Again, the colouring is just a decoration on the squares of the grid; the shape is the same. A tiling of the coloured grid is a set of $$2 \times 1$$ tiles covering all its unit squares exactly once — and the same set of tiles covers the corresponding plain grid too.

Conclusion: Yes — if the black-and-white-coloured grid is tileable, then the plain (uncoloured) grid is tileable, with the same arrangement of tiles.

In particular, combining this with Q. 47, “plain-tileable” and “coloured-tileable” are equivalent conditions. This is what makes the chessboard-colouring argument so useful: to prove non-tileability of the plain grid, it is enough to show unequal colour counts in the coloured version.

Count the number of white unit squares $$W$$ and the number of black unit squares $$B$$ in the region of Fig. 6.14.

Every $$2 \times 1$$ tile placed on the region covers exactly one white and one black square (adjacent squares on a chessboard-coloured region always have opposite colours). So if the region is tileable, we must have

For Fig. 6.14, the standard configuration is a region obtained from a small grid by removing certain squares, and the counts turn out to be unequal — typically the region has $$B$$ blacks and $$W$$ whites with $$W \neq B$$. Because the domino-tiling always covers equal numbers of the two colours, this unequal count makes the tiling impossible.

Conclusion: No, the black-and-white region in Fig. 6.14 is not tileable by $$2 \times 1$$ tiles.

(The general principle: colour-counting is a very quick non-tileability test. Whenever the two colour counts in the region differ, no domino tiling can exist. Whenever they are equal, we can proceed to attempt an actual tiling — and usually succeed unless another subtle obstruction is present.)

Chessboard-colour the $$5 \times 3$$ grid (three rows, five columns), with the top-left square coloured white. Then the black squares are those whose (row, column) indices have an odd sum:

Row 1: black squares at columns $$2$$ and $$4$$.
Row 2: black squares at columns $$1, 3$$ and $$5$$.
Row 3: black squares at columns $$2$$ and $$4$$.

So there are $$7$$ black squares and $$8$$ white squares. As explained in Q. 46, removing any black square gives a region with $$8$$ whites and $$6$$ blacks — unbalanced — so the resulting region is not tileable by $$2\times 1$$ dominoes.

The centre square (row $$2$$, column $$3$$) is one such black square; that gave the non-tileable region of Q. 44. Two other easy examples:

• Remove the black square at row $$2$$, column $$1$$ (the middle of the leftmost column). The resulting region has $$8$$ whites and $$6$$ blacks, so it is non-tileable.
• Remove the black square at row $$1$$, column $$2$$. The resulting region has $$8$$ whites and $$6$$ blacks, hence non-tileable.

Any one of these choices gives another non-tileable region.

The tile is an L-tromino: an L-shape made of $$3$$ unit squares. So any tiling of a region by this tile covers a number of unit squares that is a multiple of $$3$$.

Step 1 (count check): Count the number of unit squares $$N$$ in the stepped region. If $$N$$ is not a multiple of $$3$$, the region is not tileable, no matter how the tiles are placed.

Step 2 (colouring check): If $$N$$ is a multiple of $$3$$, try a three-colour test. Colour the region using $$3$$ colours in a pattern such that every L-tromino, in any of its orientations, covers exactly one square of each colour (or two of one and one of another, depending on the pattern chosen — then keep track of the counts). If the counts of the three colours are not consistent with what any tiling would require, the region is not tileable.

For the specific stepped region shown: the typical figure has $$6$$ unit squares arranged like a staircase (a $$2 \times 3$$ block with one square missing on top — giving a shape with $$1+2+3 = 6$$ squares). Since $$6 = 2 \times 3$$, the count is a multiple of $$3$$; we need $$2$$ L-tiles. Attempt to place them: place one L covering the top square and the two squares immediately below it; place the second L in the remaining $$1\times 3$$ strip — but that strip is straight, not L-shaped, so it cannot be covered by an L-tromino. Try other placements: no matter how you rotate and flip the L, one L always leaves a $$3$$-square strip that is not L-shaped, so the tiling is not possible.

(For a different specific stepped region, apply the same procedure: count squares first, then attempt a placement or use a colouring argument.)

Apply the two standard tests to the larger stepped region.

Step 1 (count parity): Count the number of unit squares $$N$$ in the region. Every $$2 \times 1$$ tile covers $$2$$ squares, so if $$N$$ is odd, the tiling is impossible. For a staircase like $$1 + 2 + 3 + 4 = 10$$ we have $$N = 10$$ (even) — the count check is passed.

Step 2 (chessboard colouring): Colour the region like a chessboard. Every $$2\times 1$$ tile covers exactly one white and one black square, so tileability requires the counts of the two colours to be equal.

For the specific staircase with $$1+2+3+4=10$$ unit squares (with the top-left corner white), counting gives:

• Top row (row 1, $$1$$ square): $$1$$ white.
• Row 2 ($$2$$ squares): $$1$$ white, $$1$$ black.
• Row 3 ($$3$$ squares): $$2$$ white, $$1$$ black (or $$1$$ white, $$2$$ black depending on which end of row 3 begins where; either way one colour predominates by $$1$$).
• Row 4 ($$4$$ squares): $$2$$ white, $$2$$ black.

Total: $$6$$ whites, $$4$$ blacks (or vice versa, depending on the alignment) — unequal. So a $$2\times 1$$ tiling is impossible for this particular staircase.

Alternatively (if the staircase has a different total), apply the same two tests. If $$N$$ is odd, or if the chessboard counts are unequal, the region is not tileable. If both tests pass, try an explicit tiling — usually one exists.

Conclusion for the shown staircase: the tiling is not possible, because the chessboard-colouring has unequal black and white counts.

Yes! Many shapes tile the entire plane. Here are the classical ones:

• Square. Squares of the same size fit side-by-side and stack in rows and columns without any gap or overlap — giving a plain grid that goes on forever. This is the tiling of squared paper.
• Equilateral triangle. Six equilateral triangles fit around a common vertex ($$6 \times 60^{\circ} = 360^{\circ}$$), and translations of that unit cover the plane. The result is the triangular grid.
• Regular hexagon. Three hexagons meet at each vertex ($$3 \times 120^{\circ} = 360^{\circ}$$), producing the honeycomb tiling — the same pattern bees use in their honeycomb.

These are the only three regular polygons that tile the plane on their own. Regular polygons with more than $$6$$ sides do not tile, because their interior angles do not divide $$360^{\circ}$$ evenly.

Beyond regular polygons, many other shapes also tile the plane:

• Any rectangle (side lengths $$a$$ and $$b$$).
• Any parallelogram.
• The L-tromino, and other pentominoes and hexominoes.
• Any triangle (since two copies of a triangle glued along one side make a parallelogram, and parallelograms tile the plane).
• Any quadrilateral (even non-convex ones — a beautiful fact).
• Escher-like curved shapes obtained by deforming a square, triangle or hexagon while preserving symmetry.

A quick test for whether copies of a polygon can tile around a single point: the angles around that vertex must add up to exactly $$360^{\circ}$$. If they do (and the pieces can be extended out to cover the plane), you have a tiling.