JEE JEE 2D Geometry Questions

We have $$\triangle PQR$$ with $$P(5, 4)$$, $$Q(-2, 4)$$, $$R(a, b)$$, area = 35 sq. units, and orthocentre $$O\left(2, \dfrac{14}{5}\right)$$.

Since $$P$$ and $$Q$$ have the same y-coordinate (y = 4), the base $$PQ$$ is horizontal with length $$|5 - (-2)| = 7$$.

Area = $$\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 7 \times |b - 4| = 35$$.

The orthocentre has y-coordinate $$\dfrac{14}{5}$$, which is less than 4. Since PQ is horizontal (slope = 0), the altitude from R to PQ is vertical: $$x = a$$. The orthocentre lies on this altitude, so $$a = 2$$.

Slope of PR: $$\dfrac{b - 4}{a - 5} = \dfrac{b - 4}{2 - 5} = \dfrac{b - 4}{-3}$$.

The altitude from Q(-2, 4) is perpendicular to PR, so its slope = $$\dfrac{3}{b - 4}$$.

Altitude from Q(-2, 4) with slope $$\dfrac{3}{b-4}$$:

This passes through the orthocentre $$(2, 14/5)$$:

This is consistent with our area calculation ($$b = -6$$ gives $$|b - 4| = 10$$).

So $$R = (2, -6)$$.

The answer is $$\boxed{3}$$, which corresponds to Option 4.

We must find the area of the set

$$S=\Bigl\{(x,y)\in\mathbb{R}\times\mathbb{R}\;:\;x\gt 0,\;y\gt\frac1x,\;5x-4y-1\gt 0,\;4x+4y-17\lt 0\Bigr\}.$$

Rewrite the straight-line inequalities in the usual $$y=m x+c$$ form:

$$5x-4y-1\gt0\;\Longrightarrow\;4y\lt5x-1\;\Longrightarrow\;y&lt\frac{5x-1}{4},$$
$$4x+4y-17&lt0;;\Longrightarrow\;4y&lt17-4x\;\Longrightarrow\;y&lt\frac{17-4x}{4}=\,\frac{17}{4}-x.$$

Therefore every point of $$S$$ must satisfy

$$y&gt\frac1x \quad\text{and}\quad y&lt\min\!\Bigl\{\frac{5x-1}{4},\,\frac{17}{4}-x\Bigr\}.$$

Call the two lines

$$\ell_1:\;y=\frac{5x-1}{4}\qquad \ell_2:\;y=\frac{17}{4}-x.$$

Intersection of the two lines
Set $$\frac{5x-1}{4}=\frac{17}{4}-x$$ to obtain

$$5x-1=17-4x\;\Longrightarrow\;9x=18\;\Longrightarrow\;x=2$$
and hence $$y=\frac{5(2)-1}{4}=\frac94.$$
Thus $$\ell_1\cap\ell_2=(2,\tfrac94).$$

Which line is lower?
Solve $$\frac{5x-1}{4}\lt\frac{17}{4}-x\quad\Longrightarrow\quad x&lt2.$$
Hence
for $$0&lt x&lt2$$ the lower (controlling) upper-boundary is $$\ell_1,$$
for $$x&gt2$$ the lower upper-boundary is $$\ell_2.$$

Feasibility with the hyperbola
The region exists only where the hyperbola is below the chosen line:

1. For $$y&lt\ell_1$$ (i.e. $$0&lt x&lt2$$) we need $$\frac1x&lt\frac{5x-1}{4}\;.$$ Multiply by $$4x\;(x&gt0):\quad4&lt5x^{2}-x\;,$$
$$5x^{2}-x-4&gt0;;\Longrightarrow\;x&gt1\quad(\text{rejected root }x&lt0).$$

2. For $$y&lt\ell_2$$ (i.e. $$x&gt2$$) we need $$\frac1x&lt\frac{17}{4}-x.$$ Multiply by $$4x\;(x&gt0):\quad17x-4x^{2}-4&gt0,$$
$$-4x^{2}+17x-4&gt0;;\Longrightarrow\;4x^{2}-17x+4&lt0.$$
This quadratic has roots $$x=\frac{17\pm15}{8}\;=\;0.25,\;4,$$ so the inequality holds for $$0.25&lt x&lt4.$$ Intersecting with $$x&gt2$$ leaves the range $$2&lt x&lt4.$

Thus the region splits into two x-intervals

• $$1&lt x&lt2:\quad$$\frac$$1x&lt y&lt$$\frac{5x-1}{4}$$,$$
• $$2&lt x&lt4:\quad$$\frac$$1x&lt y&lt$$\frac{17}{4}$$-x.$$

Area computation

Area $$A=$$\int_{1}^{2}\Bigl(\frac{5x-1}{4}-\frac1x\Bigr$$)\,dx+$$\int_{2}^{4}\Bigl(\frac{17}{4}-x-\frac1x\Bigr$$)\,dx.$$

First integral:

$$$$\int_{1}^{2}\!\Bigl(\frac{5x}{4}-\frac14-\frac1x\Bigr$$)dx =\Bigl[$$\frac{5x^{2}$$}{8}\Bigr]_{1}^{2}-\Bigl[$$\frac{x}{4}$$\Bigr]_{1}^{2}-\bigl[$$\ln$$ x\bigr]_{1}^{2}$$ $$=$$\frac{5}{8}$$(4-1)-$$\frac$$14(2-1)-$$\ln$$2 =$$\frac{15}{8}-\frac$$14-$$\ln$$2 =$$\frac{13}{8}-\ln$$2.$$

Second integral:

$$$$\int_{2}^{4}\!\Bigl(\frac{17}{4}-x-\frac1x\Bigr$$)dx =\Bigl[$$\frac{17x}{4}$$\Bigr]_{2}^{4}-\Bigl[$$\frac{x^{2}$$}{2}\Bigr]_{2}^{4}-\bigl[$$\ln$$ x\bigr]_{2}^{4}$$ $$=$$\frac{17}{4}$$(4-2)-$$\frac$$12(16-4)-($$\ln$$4-$$\ln$$2)$$ $$=$$\frac{17}{2}$$-6-$$\ln$$2 =$$\frac{5}{2}-\ln$$2.$$

Adding:

$$A=\Bigl($$\frac{13}{8}-\ln$$2\Bigr)+\Bigl($$\frac{5}{2}-\ln$$2\Bigr) =$$\frac{13}{8}+\frac{20}{8}$$-2$$\ln$$2 =$$\frac{33}{8}-\ln$$4.$$

Final result

Area $$=\displaystyle$$\frac{33}{8}-\log_e4$$.$$ Hence the correct option is

Option B which is: $$\dfrac{33}{8}-$$\log_e4$$$$.

Let the two concurrent lines be$$4x-3y=12\alpha \qquad -(1)$$and$$4\alpha x+3\alpha y=12 \qquad -(2)$$with $$\alpha\neq 0$$.

Coordinates of the intersection
From $$(1)$$ we get $$4x-3y=12\alpha$$.
Divide $$(2)$$ by $$\alpha$$ to obtain $$4x+3y=\dfrac{12}{\alpha}\qquad -(3)$$.

Add $$(1)$$ and $$(3)$$:
$$8x=12\alpha+\dfrac{12}{\alpha}\;\Longrightarrow\;x=\dfrac{3}{2}\left(\alpha+\dfrac{1}{\alpha}\right).$$

Subtract $$(1)$$ from $$(3)$$:
$$6y=\dfrac{12}{\alpha}-12\alpha\;\Longrightarrow\;y=2\left(\dfrac{1}{\alpha}-\alpha\right).$$

Eliminating $$\alpha$$ to get the locus $$S$$
Let $$t=\alpha+\dfrac{1}{\alpha}=\dfrac{2}{3}x$$ and$$s=\alpha-\dfrac{1}{\alpha}=-\dfrac{y}{2}.$$ For any real $$\alpha\neq 0$$ we have $$(\alpha+\tfrac{1}{\alpha})^{2}-(\alpha-\tfrac{1}{\alpha})^{2}=4.$$ Substituting $$t$$ and $$s$$, $$\left(\dfrac{2}{3}x\right)^{2}-\left(-\dfrac{y}{2}\right)^{2}=4.$$ This simplifies to $$\dfrac{4}{9}x^{2}-\dfrac{y^{2}}{4}=4 \;\Longrightarrow\; \dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1.$$ Hence $$S$$ is the hyperbola $$\boxed{\dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1}.$

Equation of the required tangent $$T$$
The tangent is parallel to the line $$4x-\dfrac{3}{$$\sqrt$$2}y=0$$ whose slope is $$m=\dfrac{4$$\sqrt$$2}{3}.$$
Let the tangent cut the axes at $$(p,0)$$ and $$(0,q)\;(q\gt 0).$$ In intercept form it is $$\dfrac{x}{p}+\dfrac{y}{q}=1,$$ whose slope is $$-\dfrac{q}{p}.$$ Parallelism gives $$-\dfrac{q}{p}=m=\dfrac{4$$\sqrt$$2}{3}\;\Longrightarrow\;q=-mp.$$

Tangency condition
For the hyperbola $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ with $$a^{2}=9,\;b^{2}=16,$$ the line $$y=mx+c$$ is a tangent iff $$c^{2}=a^{2}m^{2}-b^{2}.$$ Here $$y=mx+c$$ with $$c=-mp,$$ so $$c^{2}=(-mp)^{2}=m^{2}p^{2}.$$ Using the tangency criterion, $$m^{2}p^{2}=a^{2}m^{2}-b^{2} \;\Longrightarrow\; p^{2}=a^{2}-\dfrac{b^{2}}{m^{2}}.$$

Compute each term:
$$a^{2}m^{2}=9$$\left$$(\dfrac{4$$\sqrt$$2}{3}$$\right$$)^{2}=9$$\cdot$$\dfrac{32}{9}=32,$$ and therefore $$c^{2}=a^{2}m^{2}-b^{2}=32-16=16\quad$$\Rightarrow$$\quad c=4 \;(c\gt 0).$$

Since $$c=-mp,$$ we have $$4=-mp\;\Longrightarrow\;p=-\dfrac{4}{m} =-\dfrac{4}{\dfrac{4$$\sqrt$$2}{3}}=-\dfrac{3}{$$\sqrt$$2}.$$ Also $$q=-mp=-$$\left$$(\dfrac{4$$\sqrt$$2}{3}$$\right$$)\!$$\left$$(-\dfrac{3}{$$\sqrt$$2}$$\right$$)=4.$$

Required product
$$pq=$$\left$$(-\dfrac{3}{$$\sqrt$$2}$$\right$$)(4)=-\dfrac{12}{$$\sqrt$$2}=-6$$\sqrt$$2.$$

Thus the correct value is $$\boxed{-6$$\sqrt$$2}$$, which corresponds to
Option A.

Let the position vectors of the vertices be $$\mathbf{p},\;\mathbf{q},\;\mathbf{r}$$ for the points $$P,\;Q,\;R$$ respectively, and let $$\mathbf{s}$$ be the position vector of the point $$S$$.

The given vector condition is
$$\overrightarrow{SP}+5\,\overrightarrow{SQ}+6\,\overrightarrow{SR}=\vec{0}.$$

Using $$\overrightarrow{SP}=\mathbf{p}-\mathbf{s},\;\overrightarrow{SQ}=\mathbf{q}-\mathbf{s},\;\overrightarrow{SR}=\mathbf{r}-\mathbf{s},$$ we get

$$\bigl(\mathbf{p}-\mathbf{s}\bigr)+5\bigl(\mathbf{q}-\mathbf{s}\bigr)+6\bigl(\mathbf{r}-\mathbf{s}\bigr)=\vec{0}.$$

Simplifying,
$$\mathbf{p}+5\mathbf{q}+6\mathbf{r}-12\mathbf{s}= \vec{0}\;\Longrightarrow\; \mathbf{s}= \dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}.$$

Next, let $$E$$ and $$F$$ be the mid-points of $$PR$$ and $$QR$$.

Mid-point position vectors:
$$\mathbf{e}= \dfrac{\mathbf{p}+\mathbf{r}}{2}, \qquad \mathbf{f}= \dfrac{\mathbf{q}+\mathbf{r}}{2}.$$

Length of $$EF$$
$$\overrightarrow{EF}= \mathbf{f}-\mathbf{e}= \dfrac{\mathbf{q}+\mathbf{r}}{2}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{q}-\mathbf{p}}{2}.$$
Hence $$|EF|=\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|.$$

Length of $$ES$$
$$\overrightarrow{ES}= \mathbf{s}-\mathbf{e} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}-6\mathbf{p}-6\mathbf{r}}{12} =\dfrac{-5\mathbf{p}+5\mathbf{q}}{12} =\dfrac{5}{12}\,(\mathbf{q}-\mathbf{p}).$$
Therefore $$|ES|=\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|.$$

Required ratio
$$\dfrac{|EF|}{|ES|}= \dfrac{\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|}{\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|} =\dfrac{1}{2}\times\dfrac{12}{5} =\dfrac{6}{5}=1.2.$$

Hence, $$\dfrac{\text{length of }EF}{\text{length of }ES}=1.2,$$ which lies in the range 1.15-1.25.

Line through (4, -9) with slope m > 0: $$y + 9 = m(x - 4)$$.

x-intercept A: $$y = 0 \Rightarrow x = 4 + 9/m$$. OA = 4 + 9/m.

y-intercept B: $$x = 0 \Rightarrow y = -9 - 4m$$. OB = |−9 − 4m| = 9 + 4m (since m > 0).

Sum = OA + OB = 4 + 9/m + 9 + 4m = 13 + 9/m + 4m.

By AM-GM: $$9/m + 4m \geq 2\sqrt{36} = 12$$. Equality when 9/m = 4m, m = 3/2.

Minimum sum = 13 + 12 = 25.

The correct answer is Option (2): 25.

Let the initial side length of the equilateral triangle be $$a$$.

• Perimeter of 1st triangle = $$3a$$

• Perimeter of 2nd triangle (sides are halved) = $$\frac{3a}{2}$$

• Perimeter of 3rd triangle = $$\frac{3a}{4}$$

This creates an infinite geometric series with first term $$A_1 = 3a$$ and common ratio $$r = \frac{1}{2}$$:

$$P = \frac{A_1}{1 - r} = \frac{3a}{1 - \frac{1}{2}} = 6a \implies a = \frac{P}{6}$$

• Area of 1st triangle = $$\frac{\sqrt{3}}{4}a^2$$

• Area of 2nd triangle (sides halved means area is quartered) = $$\frac{1}{4}\left(\frac{\sqrt{3}}{4}a^2\right)$$

This creates an infinite geometric series with first term $$A_1 = \frac{\sqrt{3}}{4}a^2$$ and common ratio $$r = \frac{1}{4}$$:

$$Q = \frac{A_1}{1 - r} = \frac{\frac{\sqrt{3}}{4}a^2}{1 - \frac{1}{4}} = \frac{\frac{\sqrt{3}}{4}a^2}{\frac{3}{4}} = \frac{\sqrt{3}}{3}a^2 = \frac{a^2}{\sqrt{3}}$$

$$Q = \frac{\left(\frac{P}{6}\right)^2}{\sqrt{3}} = \frac{P^2}{36\sqrt{3}}$$

$$P^2 = 36\sqrt{3}Q$$

Two lines from origin O intersect $$3x + 4y = 12$$ at P and Q such that $$\triangle OPQ$$ is isosceles with $$\angle POQ = 90°$$ and $$OP = OQ$$.

The distance from O to the line $$3x + 4y = 12$$ is $$d = \frac{12}{\sqrt{9+16}} = \frac{12}{5}$$.

Since $$\angle POQ = 90°$$ and $$OP = OQ$$, the perpendicular from O to PQ bisects both $$PQ$$ and $$\angle POQ$$. Let M be the foot of the perpendicular from O to line PQ, so that $$OM = d = 12/5$$.

Because $$\angle POM = \angle QOM = 45°$$, we have
$$OP = \frac{OM}{\cos 45°} = \frac{12/5}{1/\sqrt{2}} = \frac{12\sqrt{2}}{5},$$
$$PM = OP \sin 45° = \frac{12\sqrt{2}}{5} \cdot \frac{1}{\sqrt{2}} = \frac{12}{5},$$
and hence
$$PQ = 2 \cdot PM = \frac{24}{5}.$$

Next, we calculate
$$l = OP^2 + PQ^2 + QO^2 = 2 \cdot OP^2 + PQ^2,$$
where
$$OP^2 = QO^2 = \left(\frac{12\sqrt{2}}{5}\right)^2 = \frac{288}{25},$$
$$PQ^2 = \left(\frac{24}{5}\right)^2 = \frac{576}{25}.$$
It follows that
$$l = \frac{576}{25} + \frac{576}{25} = \frac{1152}{25} = 46.08.$$

The greatest integer less than or equal to $$l$$ is $$\lfloor 46.08 \rfloor = 46$$.

The correct answer is Option (2): 46.

Since $$A(a, -2)$$ and $$B(a, 6)$$ have the same x-coordinate, side $$AB$$ is vertical (along $$x = a$$). The perpendicular bisector of $$AB$$ is horizontal, passing through the midpoint $$(a, 2)$$, so it is the line $$y = 2$$.

The circumcenter lies on this perpendicular bisector, so:

$$\frac{a}{4} = 2 \implies a = 8$$

$$A(8, -2)$$, $$B(8, 6)$$, $$C(2, -2)$$. Circumcenter: $$(5, 2)$$.

$$AB = |6 - (-2)| = 8$$

$$AC = |8 - 2| = 6$$

$$BC = \sqrt{(8-2)^2 + (6-(-2))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

 $$6^2 + 8^2 = 100 = 10^2$$, so this is a right triangle with the right angle at $$A$$.

For a right triangle, the circumradius equals half the hypotenuse:

$$\alpha = \frac{BC}{2} = \frac{10}{2} = 5$$

$$\beta = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 8 \times 6 = 24$$

$$\gamma = AB + AC + BC = 8 + 6 + 10 = 24$$

$$\alpha + \beta + \gamma = 5 + 24 + 24 = 53$$

The correct answer is Option (2): 53.

Orthocenter property:

$$(AP\perp BC),(BP\perp AC)$$

Find (C):

$$Slope(AP=-1\Rightarrow BC)slope(=1\Rightarrow y=x+5)$$

$$Slope(BP=-\frac{2}{3}\Rightarrow AC)slope(=\frac{3}{2}\Rightarrow y=\frac{3}{2}x-\frac{11}{2})$$

Intersection ⇒ (C(21,26))
$$\alpha+\beta=47$$

Circumcenter of (PAB):

Solve perpendicular bisectors ⇒ (h + k = -21)

Final:
$$(\alpha+\beta)+2(h+k)=47+2(-21)=5$$

The sides are $$AB: x-y+4=0$$, $$BC: 3x+5y-4=0$$, and $$AC: x+y-2=0$$.

Shifting a line $$ax+by+c=0$$ inward by $$d$$ units: $$\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm d$$.

For $$AC (x+y-2=0)$$, the distance from origin is $$\frac{|-2|}{\sqrt{2}} = \sqrt{2}$$. Shifting it 1 unit "inward" (towards the triangle's centroid) results in $$x+y-(2-\sqrt{2})=0$$.

This line is closest to the origin because the original distance was $$\sqrt{2}$$ and we moved 1 unit closer.

Answer: C ($$x+y-(2-\sqrt{2})=0$$)

The line segment joining $$(5, 2)$$ and $$(2, a)$$ subtends an angle $$\frac{\pi}{4}$$ at the origin.

Let $$O = (0,0)$$, $$P = (5, 2)$$, $$Q = (2, a)$$.

$$\cos\frac{\pi}{4} = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}||\vec{OQ}|}$$

$$\frac{1}{\sqrt{2}} = \frac{5(2) + 2(a)}{\sqrt{5^2 + 2^2} \cdot \sqrt{2^2 + a^2}} = \frac{10 + 2a}{\sqrt{29} \cdot \sqrt{4 + a^2}}$$

$$\frac{1}{2} = \frac{(10 + 2a)^2}{29(4 + a^2)}$$

$$29(4 + a^2) = 2(10 + 2a)^2$$

$$116 + 29a^2 = 2(100 + 40a + 4a^2) = 200 + 80a + 8a^2$$

$$21a^2 - 80a - 84 = 0$$

By Vieta's formulas for $$21a^2 - 80a - 84 = 0$$:

Sum of roots = $$\frac{80}{21}$$

Product of roots = $$\frac{-84}{21} = -4$$

Discriminant = $$80^2 + 4(21)(84) = 6400 + 7056 = 13456 = 116^2$$

$$a_1 = \frac{80 + 116}{42} = \frac{196}{42} = \frac{14}{3}, \quad a_2 = \frac{80 - 116}{42} = \frac{-36}{42} = -\frac{6}{7}$$

For both values: $$10 + 2a > 0$$, confirming the angle is in the correct range.

The product of all possible values of $$a$$ is $$\frac{14}{3} \times \left(-\frac{6}{7}\right) = -4$$.

Therefore, the answer is Option D: $$\mathbf{-4}$$.

A line through $$(3, 5)$$ with intercepts on positive axes at A$$(a, 0)$$ and B$$(0, b)$$.

Line: $$\frac{x}{a} + \frac{y}{b} = 1$$. Since it passes through $$(3, 5)$$: $$\frac{3}{a} + \frac{5}{b} = 1$$.

Area of triangle OAB = $$\frac{1}{2}ab$$.

We need to minimize $$ab$$ subject to $$\frac{3}{a} + \frac{5}{b} = 1$$ with $$a > 3, b > 5$$ (positive intercepts with point (3,5) between them).

By AM-GM: $$1 = \frac{3}{a} + \frac{5}{b} \geq 2\sqrt{\frac{15}{ab}}$$.

$$\frac{1}{4} \geq \frac{15}{ab} \implies ab \geq 60$$.

Minimum area = $$\frac{1}{2} \times 60 = 30$$.

Equality when $$\frac{3}{a} = \frac{5}{b}$$, i.e., $$a = 6, b = 10$$.

Let inner rectangle sides be $$4$$ and $$2$$, inclined at an angle $$\theta$$ to the outer rectangle sides.

Outer dimensions: $$a = 4\cos\theta + 2\sin\theta$$ and $$b = 4\sin\theta + 2\cos\theta$$.

The outer area is maximized when it forms a symmetric configuration around the inner block ($$\theta = 45^\circ$$).

Substituting $$\theta = 45^\circ$$: $$a = b = \frac{4}{\sqrt{2}} + \frac{2}{\sqrt{2}} = 3\sqrt{2}$$.

 $$(a + b)^2 = (3\sqrt{2} + 3\sqrt{2})^2 = (6\sqrt{2})^2 = 72$$.

Given points $$P(3, 2, 3)$$, $$Q(4, 6, 2)$$, and $$R(7, 3, 2)$$. 

Find the direction vectors from P: $$\vec{PQ} = Q - P = (4-3, 6-2, 2-3) = (1, 4, -1)$$

$$\vec{PR} = R - P = (7-3, 3-2, 2-3) = (4, 1, -1)$$

Compute the dot product and magnitudes: $$\vec{PQ} \cdot \vec{PR} = (1)(4) + (4)(1) + (-1)(-1) = 4 + 4 + 1 = 9$$

$$|\vec{PQ}| = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$$

$$|\vec{PR}| = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}$$

Apply the angle formula: $$ \cos(\angle QPR) = \frac{\vec{PQ} \cdot \vec{PR}}{|\vec{PQ}||\vec{PR}|} = \frac{9}{3\sqrt{2} \cdot 3\sqrt{2}} = \frac{9}{18} = \frac{1}{2} $$

$$ \angle QPR = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} $$

The correct answer is Option (4): $$\frac{\pi}{3}$$.

For $$x$$ to be real, the discriminant $$D \ge 0$$:

$$[2b(a+c)]^2 - 4(a^2 + b^2)(b^2 + c^2) \ge 0$$

Expanding and simplifying:

$$b^2(a^2 + c^2 + 2ac) - (a^2b^2 + a^2c^2 + b^4 + b^2c^2) \ge 0$$

$$2ab^2c - a^2c^2 - b^4 \ge 0 \implies -(b^2 - ac)^2 \ge 0$$

This is only possible if $$b^2 - ac = 0$$, meaning $$b^2 = ac$$ (Geometric Progression).

If $$D=0$$, $$x = \frac{2b(a+c)}{2(a^2+b^2)} = \frac{b(a+c)}{a^2+ac} = \frac{b(a+c)}{a(a+c)} = \frac{b}{a}$$.

Since $$b^2=ac$$, we can also write $$x = \frac{\sqrt{ac}}{a} = \sqrt{\frac{c}{a}}$$.

For a triangle: $$a+b > c$$, $$b+c > a$$, and $$a+c > b$$.

Using $$b = \sqrt{ac}$$, the set of $$x$$ (where $$x = b/a = \sqrt{c/a}$$) must satisfy:

$$1 + x > x^2$$, $$x + x^2 > 1$$, and $$1 + x^2 > x$$.

Solving these gives $$x \in (\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2})$$.

So $$\alpha = \frac{\sqrt{5}-1}{2}$$ and $$\beta = \frac{\sqrt{5}+1}{2}$$.

$$\alpha^2 + \beta^2 = (\frac{6-2\sqrt{5}}{4}) + (\frac{6+2\sqrt{5}}{4}) = \frac{12}{4} = 3$$.

Value = $$12(3) = \mathbf{36}$$

We have an isosceles triangle ABC with $$A(-1, 0)$$, $$\angle A = \frac{2\pi}{3}$$, $$AB = AC$$, and B on the positive x-axis. Also $$BC = 4\sqrt{3}$$.

Since $$\angle A = 120°$$ and $$AB = AC$$, the base angles are each $$\frac{180° - 120°}{2} = 30°$$.

Using the sine rule: $$\frac{BC}{\sin A} = \frac{AB}{\sin C}$$

$$\frac{4\sqrt{3}}{\sin 120°} = \frac{AB}{\sin 30°}$$

$$AB = 4$$

Since B is on the positive x-axis and $$AB = 4$$, we have $$B = (-1 + 4, 0) = (3, 0)$$.

Now, since the triangle is isosceles with $$AB = AC = 4$$ and $$\angle A = 120°$$, the direction from A to C makes an angle of $$120°$$ with AB.

Direction of AB is along the positive x-axis (angle $$0°$$). So AC is at angle $$120°$$ from AB.

$$C = A + 4(\cos 120°, \sin 120°) = (-1 + 4(-\frac{1}{2}), 0 + 4 \cdot \frac{\sqrt{3}}{2}) = (-3, 2\sqrt{3})$$

Or at angle $$-120°$$: $$C = (-3, -2\sqrt{3})$$.

Let's check $$BC$$: $$BC = \sqrt{(3-(-3))^2 + (0-2\sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3}$$. Correct!

Line BC passes through $$B(3, 0)$$ and $$C(-3, 2\sqrt{3})$$:

Equation: $$y = -\frac{\sqrt{3}}{3}(x - 3) = -\frac{\sqrt{3}}{3}x + \sqrt{3}$$

Intersection with $$y = x + 3$$: $$x + 3 = -\frac{\sqrt{3}}{3}x + \sqrt{3}$$

$$x = \frac{3(\sqrt{3} - 3)}{3 + \sqrt{3}} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{9 - 3} = \frac{3(\sqrt{3} - 3)(3 - \sqrt{3})}{6}$$

$$(\sqrt{3} - 3)(3 - \sqrt{3}) = -(3 - \sqrt{3})(3 - \sqrt{3}) = -(3 - \sqrt{3})^2 = -(9 - 6\sqrt{3} + 3) = -(12 - 6\sqrt{3})$$

$$x = \frac{3 \cdot (-(12 - 6\sqrt{3}))}{6} = \frac{-3(12 - 6\sqrt{3})}{6} = \frac{-36 + 18\sqrt{3}}{6} = -6 + 3\sqrt{3}$$

$$\beta = x + 3 = -6 + 3\sqrt{3} + 3 = -3 + 3\sqrt{3} = 3(\sqrt{3} - 1)$$

$$\alpha = -6 + 3\sqrt{3} = 3(\sqrt{3} - 2)$$

For the other case $$C = (-3, -2\sqrt{3})$$, slope = $$\frac{\sqrt{3}}{3}$$, equation: $$y = \frac{\sqrt{3}}{3}(x-3) = \frac{\sqrt{3}}{3}x - \sqrt{3}$$.

Intersection: $$x + 3 = \frac{\sqrt{3}}{3}x - \sqrt{3}$$, so $$x(1 - \frac{\sqrt{3}}{3}) = -\sqrt{3} - 3$$.

$$x = \frac{-3(\sqrt{3}+3)}{3-\sqrt{3}} \cdot \frac{3+\sqrt{3}}{3+\sqrt{3}} = \frac{-3(\sqrt{3}+3)(3+\sqrt{3})}{6}$$

$$= \frac{-3(3\sqrt{3}+3+9+3\sqrt{3})}{6} = \frac{-3(12+6\sqrt{3})}{6} = -(6+3\sqrt{3})$$

$$\alpha = -6-3\sqrt{3}$$, $$\beta = -3-3\sqrt{3}$$.

Now compute $$\frac{\beta^4}{\alpha^2}$$. Using the first case:

$$\alpha = 3(\sqrt{3}-2)$$, $$\beta = 3(\sqrt{3}-1)$$

$$\frac{\beta^4}{\alpha^2} = \frac{81(\sqrt{3}-1)^4}{9(\sqrt{3}-2)^2} = \frac{9(\sqrt{3}-1)^4}{(\sqrt{3}-2)^2}$$

$$\frac{9(28-16\sqrt{3})}{7-4\sqrt{3}} = \frac{9 \cdot 4(7-4\sqrt{3})}{7-4\sqrt{3}} = 36$$

Let the centre of the circle be $$O$$ and take the complex plane with $$O$$ at the origin.
Because the radius is $$2$$, every vertex of the regular octagon can be written as $$A_k :\; 2\,\zeta^{\,k}, \qquad k = 0,1,\dots ,7$$ where $$\zeta = e^{\,i\pi/4}$$ is a primitive eighth root of unity.

Let the variable point on the circle be $$P :\; 2\,e^{\,i\theta}, \qquad 0\le \theta \lt 2\pi.$$ The distance between $$P$$ and $$A_k$$ is therefore

$$PA_k \;=\; \bigl|2e^{\,i\theta} - 2\zeta^{\,k}\bigr| = 2\,\bigl|e^{\,i\theta} - \zeta^{\,k}\bigr|.$$ Hence the required product is

$$\begin{aligned} PA_1\cdot PA_2\cdots PA_8 &= \prod_{k=0}^{7} 2\,\bigl|e^{\,i\theta}-\zeta^{\,k}\bigr| \\ &= 2^{8}\;\prod_{k=0}^{7} \bigl|e^{\,i\theta}-\zeta^{\,k}\bigr|. \end{aligned}$$

To evaluate the remaining product, recall the factorisation identity

$$\prod_{k=0}^{7}\bigl(z-\zeta^{\,k}\bigr) = z^{8}-1.$$

Taking modulus on the unit circle (i.e. $$|z|=1$$, put $$z=e^{\,i\theta}$$) gives

$$\prod_{k=0}^{7}\bigl|e^{\,i\theta}-\zeta^{\,k}\bigr| = \bigl|e^{\,i8\theta}-1\bigr|.$$

Using $$e^{\,ix}-1 = 2i\sin\dfrac{x}{2}\,e^{\,ix/2},$$ we have

$$\bigl|e^{\,i8\theta}-1\bigr| = 2\bigl|\sin 4\theta\bigr|.$$ Therefore

$$PA_1\cdot PA_2\cdots PA_8 = 2^{8}\;\cdot 2\bigl|\sin 4\theta\bigr| = 2^{9}\bigl|\sin 4\theta\bigr|.$$

The maximum value of $$|\sin 4\theta|$$ is $$1$$, so the maximum value of the whole product is

$$2^{9}\times 1 = 512.$$

Hence the required maximum value is 512.

Let the lamp post be at point $$O$$. The person is at point $$P$$, a distance $$y$$ (in cm) from $$O$$. The tip of the shadow is at point $$T$$, so $$PT = x$$ and $$OT = y + x$$.

Heights (in cm): lamp post = $$400$$, person = $$160$$.
Because $$\triangle OTP$$ and $$\triangle PT(\text{top of person})$$ are similar,

$$\frac{400}{\,y + x\,} = \frac{160}{\,x\,}$$.

Cross-multiplying:
$$400x = 160(y + x)$$
$$400x = 160y + 160x$$
$$240x = 160y$$
$$\Rightarrow\; x = \frac{2}{3}\,y$$ $$-(1)$$

Differentiating $$(1)$$ with respect to time $$t$$,

$$\frac{dx}{dt} = \frac{2}{3}\,\frac{dy}{dt}$$ $$-(2)$$

The person walks away at $$\frac{dy}{dt} = 60\text{ cm s}^{-1}$$. Substituting in $$(2)$$:

$$\frac{dx}{dt} = \frac{2}{3}\times 60 = 40\text{ cm s}^{-1}$$.

Speed of the tip of the shadow relative to the ground:
$$\frac{d}{dt}(y + x) = \frac{dy}{dt} + \frac{dx}{dt} = 60 + 40 = 100\text{ cm s}^{-1}$$.

Required speed (tip with respect to the person) = ground speed of tip − ground speed of person:

$$100 - 60 = 40\text{ cm s}^{-1}$$.

Hence, the speed of the tip of the shadow relative to the person is 40 cm s$$^{-1}$$.

Let the angles of the triangle be $$A \le B \le C$$ and the sides opposite to them be $$a,\,b,\,c$$ respectively.
Because the triangle is obtuse, $$C$$ is the obtuse angle. We are told that

$$C-A=\frac{\pi}{2} \qquad -(1)$$

All three vertices lie on a circle of radius $$R=1$$, so by the sine rule

$$a=2\sin A,\; b=2\sin B,\; c=2\sin C \qquad -(2)$$

The sides are in arithmetic progression, hence

$$2b=a+c \;\Longrightarrow\; 2\sin B=\sin A+\sin C \qquad -(3)$$

Using $$C=A+\frac{\pi}{2}$$ from $$(1)$$, we get $$\sin C=\sin\!\left(A+\frac{\pi}{2}\right)=\cos A$$. Also, since $$A+B+C=\pi$$,

$$B=\pi-A-C=\pi-A-\left(A+\frac{\pi}{2}\right)=\frac{\pi}{2}-2A \qquad -(4)$$

Now write every term in $$(3)$$ in terms of $$A$$:

$$\sin B=\sin\!\left(\frac{\pi}{2}-2A\right)=\cos 2A$$ $$\sin C=\cos A$$

Substituting in $$(3)$$,

$$2\cos 2A=\sin A+\cos A \qquad -(5)$$

Square both sides of $$(5)$$:

$$4\cos^2 2A=\sin^2 A+\cos^2 A+2\sin A\cos A=1+\sin 2A \qquad -(6)$$

Rewrite $$\cos^2 2A=\dfrac{1+\cos 4A}{2}$$ and put this in $$(6)$$:

$$2\bigl(1+\cos 4A\bigr)=1+\sin 2A\;\;\Longrightarrow\;\;1+2\cos 4A=\sin 2A \qquad -(7)$$

Set $$t=2A$$ (note $$0\lt t\lt \dfrac{\pi}{2}$$ because $$A\lt \dfrac{\pi}{4}$$). Equation $$(7)$$ becomes

$$1+2\cos 2t=\sin t \qquad -(8)$$

Using $$\cos 2t=1-2\sin^2 t$$, convert $$(8)$$ entirely to $$\sin t$$:

$$1+2\bigl(1-2\sin^2 t\bigr)=\sin t \;\;\Longrightarrow\;\;3-4\sin^2 t-\sin t=0$$

This is a quadratic in $$\sin t$$:

$$4\sin^2 t+\sin t-3=0 \qquad -(9)$$

Solving $$(9)$$,

$$\sin t=\frac{-1\pm\sqrt{1+48}}{8}=\frac{-1\pm7}{8}$$

The negative root is extraneous, hence

$$\sin t=\frac{3}{4} \;\;\Longrightarrow\;\; \sin 2A=\frac{3}{4} \qquad -(10)$$

Because $$0\lt 2A\lt \dfrac{\pi}{2}$$, $$\cos 2A=\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt7}{4} \qquad -(11)$$

Half-angle formulae give

$$\sin A=\sqrt{\frac{1-\cos 2A}{2}}=\sqrt{\frac{1-\dfrac{\sqrt7}{4}}{2}} \;=\;\sqrt{\frac{4-\sqrt7}{8}}$$ $$\cos A=\sqrt{\frac{1+\cos 2A}{2}}=\sqrt{\frac{1+\dfrac{\sqrt7}{4}}{2}} \;=\;\sqrt{\frac{4+\sqrt7}{8}} \qquad -(12)$$

From $$(11)$$, $$\sin B=\cos 2A=\dfrac{\sqrt7}{4}$$ and from $$(12)$$, $$\sin C=\cos A$$.

Area of a triangle with circum-radius $$R$$ is $$\Delta=\dfrac{abc}{4R}$$. Using $$R=1$$ and $$(2)$$,

$$\Delta=2\sin A\sin B\sin C=2(\sin A)(\cos 2A)(\cos A) \qquad -(13)$$

Compute the product $$\sin A\cos A$$ from $$(12)$$:

$$\sin A\cos A=\sqrt{\frac{4-\sqrt7}{8}}\, \sqrt{\frac{4+\sqrt7}{8}} =\frac{\sqrt{(4-\sqrt7)(4+\sqrt7)}}{8} =\frac{\sqrt{16-7}}{8}=\frac{3}{8} \qquad -(14)$$

Put $$(11)$$ and $$(14)$$ in $$(13)$$:

$$\Delta=2\left(\frac{3}{8}\right)\left(\frac{\sqrt7}{4}\right) =\frac{3\sqrt7}{16} \qquad -(15)$$

Finally,

$$64\Delta = 64\left(\frac{3\sqrt7}{16}\right)=12\sqrt7$$ $$\bigl(64\Delta\bigr)^2=(12\sqrt7)^2=144\times7=1008$$

Therefore, the required value is 1008.

Place the triangle on the coordinate plane so that the right angle is at the origin.
Let $$A(0,0)$$, $$B(1,0)$$ and $$C(0,3)$$. Thus $$AB=1$$, $$AC=3$$ and $$\angle BAC=\frac{\pi}{2}$$.

Step 1 : Circumcircle of $$\triangle ABC$$
For a right-angled triangle the hypotenuse is the diameter of the circumcircle.
Hence the circumcircle has
center $$O$$ = midpoint of $$BC = \left(\tfrac{1}{2},\tfrac{3}{2}\right)$$ and
radius $$R = \dfrac{BC}{2} = \dfrac{\sqrt{1^{2}+3^{2}}}{2} = \dfrac{\sqrt{10}}{2} = \sqrt{2.5}$$.

Step 2 : Coordinates of the required smaller circle
The required circle touches the positive $$x$$-axis (side $$AB$$) and the positive $$y$$-axis (side $$AC$$).
If its radius is $$r$$, its centre must therefore be $$P(r,r)$$, because the distance from $$P$$ to each axis must be $$r$$.

Step 3 : Condition for internal tangency with the circumcircle
For two circles with centres $$P$$ and $$O$$ and radii $$r$$ and $$R$$, internal tangency means
$$OP + r = R$$ $$-(1)$$

First compute $$OP$$:
$$OP = \sqrt{(r-0.5)^{2} + (r-1.5)^{2}}$$.

Substitute into $$(1)$$:
$$\sqrt{(r-0.5)^{2} + (r-1.5)^{2}} + r = \sqrt{2.5}$$.

Step 4 : Solve for $$r$$

Rearrange:
$$\sqrt{(r-0.5)^{2} + (r-1.5)^{2}} = \sqrt{2.5} - r.$$ Square both sides:
$$(r-0.5)^{2} + (r-1.5)^{2} = (\sqrt{2.5} - r)^{2}.$$ Compute each side:
Left side: $$(r^{2}-r+0.25) + (r^{2}-3r+2.25) = 2r^{2} - 4r + 2.5.$$ Right side: $$2.5 - 2r\sqrt{2.5} + r^{2}.$$ Set them equal:
$$2r^{2} - 4r + 2.5 = 2.5 - 2r\sqrt{2.5} + r^{2}.$$ Simplify:
$$r^{2} - 4r + 2r\sqrt{2.5} = 0.$$ Factor $$r$$:
$$r\bigl(r - 4 + 2\sqrt{2.5}\bigr) = 0.$$ Since $$r>0$$, we obtain
$$r = 4 - 2\sqrt{2.5}.$$

Step 5 : Numerical value
$$\sqrt{2.5} \approx 1.58114 \;\;\Longrightarrow\;\; r \approx 4 - 2(1.58114) \approx 4 - 3.16228 \approx 0.83772.$$ Rounded to two decimal places, $$r \approx 0.84$$. Accepting truncation gives $$0.83$$.

Hence the admissible values are 0.83 or 0.84.

Given $$A = \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$$, $$B$$ is the image of $$A$$ in the y-axis, and $$C$$ is the image of $$B$$ in the x-axis.

Find coordinates of B and C.

$$B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$$ (reflection of A in y-axis)

$$C = \left(-\frac{3}{\sqrt{a}}, -\sqrt{a}\right)$$ (reflection of B in x-axis)

Set up the area of triangle ACD.

$$D = (3\cos\theta, a\sin\theta)$$ is in the fourth quadrant ($$\cos\theta > 0, \sin\theta < 0$$).

Area $$= \frac{1}{2}|x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)|$$

$$= \frac{1}{2}\left|\frac{3}{\sqrt{a}}(-\sqrt{a} - a\sin\theta) + \left(-\frac{3}{\sqrt{a}}\right)(a\sin\theta - \sqrt{a}) + 3\cos\theta(\sqrt{a} + \sqrt{a})\right|$$

$$= \frac{1}{2}\left|(-3 - 3\sqrt{a}\sin\theta) + (-3\sqrt{a}\sin\theta + 3) + 6\sqrt{a}\cos\theta\right|$$

$$= \frac{1}{2}\left|-6\sqrt{a}\sin\theta + 6\sqrt{a}\cos\theta\right|$$

$$= 3\sqrt{a}|\cos\theta - \sin\theta|$$

Maximize the area.

The maximum of $$|\cos\theta - \sin\theta| = \sqrt{2}$$.

Maximum area $$= 3\sqrt{a} \cdot \sqrt{2} = 12$$

$$\sqrt{a} = \frac{12}{3\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

$$a = 8$$

Answer: 8

Let the original point be denoted by $$P(a,\,b)\,. $$ We shall follow the three prescribed transformations one after another and track the co-ordinates at every stage.

First, we perform a reflection about the line $$y=x.$$ For any point $$\bigl(x,\,y\bigr)$$, reflection in the line $$y=x$$ interchanges its co-ordinates. Hence

$$P(a, b)\;\longrightarrow\;P_1\bigl(b$$, $$\$$, $$a\bigr).$$

Next, we translate the point by 2 units along the positive $$x$$-axis. Translation along the positive $$x$$-direction by 2 units simply adds 2 to the $$x$$-coordinate while leaving the $$y$$-coordinate unchanged. Therefore

$$P_1\bigl(b,\,a\bigr)\;\longrightarrow\;P_2\bigl(b+2,\,a\bigr).$$

Now we rotate the point anticlockwise about the origin through an angle $$\dfrac{\pi}{4}\,.$$ For rotation of a point $$(x,\,y)$$ by an angle $$\theta$$ anticlockwise about the origin, the standard formula is

$$\bigl(x,\,y\bigr)\;\longrightarrow\;\Bigl(x\cos\theta-y\sin\theta,\;x\sin\theta+y\cos\theta\Bigr).$$

Here $$\theta=\dfrac{\pi}{4}$$, so $$\cos\theta=\dfrac{1}{\sqrt2}$$ and $$\sin\theta=\dfrac{1}{\sqrt2}\,.$$ Applying the formula to $$P_2\bigl(b+2,\,a\bigr)$$ we obtain the final point $$P_3(x_f,\,y_f):$$

$$\begin{aligned} x_f &= (b+2)\cos\frac{\pi}{4} - a\sin\frac{\pi}{4} = (b+2)\cdot\frac1{\sqrt2} - a\cdot\frac1{\sqrt2} = \frac{(b+2)-a}{\sqrt2},\\[4pt] y_f &= (b+2)\sin\frac{\pi}{4} + a\cos\frac{\pi}{4} = (b+2)\cdot\frac1{\sqrt2} + a\cdot\frac1{\sqrt2} = \frac{(b+2)+a}{\sqrt2}. \end{aligned}$$

We are told that the final co-ordinates are $$\left(-\dfrac1{\sqrt2},\;\dfrac7{\sqrt2}\right).$$ Hence

$$\frac{(b+2)-a}{\sqrt2}=-\frac1{\sqrt2}\quad\text{and}\quad \frac{(b+2)+a}{\sqrt2}=\frac7{\sqrt2}.$$

Multiplying both equations by $$\sqrt2$$ eliminates the denominator:

$$\begin{aligned} (b+2)-a &= -1,\\ (b+2)+a &= 7. \end{aligned}$$

Simplifying each equation gives

$$\begin{aligned} -b + a &= 3 \quad\Longrightarrow\quad a - b = 3,\\ a + b &= 5. \end{aligned}$$

Now we solve these simultaneous linear equations. Adding them yields

$$2a = 8 \;\;\Longrightarrow\;\; a = 4.$$

Substituting $$a=4$$ in $$a+b=5$$ gives

$$4 + b = 5 \;\;\Longrightarrow\;\; b = 1.$$

Finally, we compute the required expression $$2a + b$$:

$$2a + b = 2(4) + 1 = 8 + 1 = 9.$$

Hence, the correct answer is Option B.

Let the square have vertices $$O, A, B, C$$ in that order, with $$O$$ at the origin. The length of every side is given to be $$2$$.

The side $$OA$$ passes through the origin and makes an angle of $$30^{\circ}$$ with the positive $$x$$-axis. Hence the direction ratios of $$OA$$ are $$\bigl(\cos 30^{\circ},\; \sin 30^{\circ}\bigr).$$ Using the length $$2,$$ the exact coordinates of $$A$$ are obtained by simple trigonometry: $$\begin{aligned} x_A &= 2\cos 30^{\circ}=2\left(\frac{\sqrt3}{2}\right)=\sqrt3,\\[2pt] y_A &= 2\sin 30^{\circ}=2\left(\frac12\right)=1. \end{aligned}$$ Thus $$A(\sqrt3,\;1).$$

Write the vector $$\overrightarrow{OA}$$ in component form: $$\overrightarrow{OA}=(\sqrt3,\,1).$$ To obtain the adjacent side of the square we need a vector of the same length that is perpendicular to $$\overrightarrow{OA}.$$ Rotating a vector $$(x,\,y)$$ through $$90^{\circ}$$ counter-clockwise produces $$(-y,\,x).$$ Therefore $$\overrightarrow{w}=(-1,\;\sqrt3)$$ is perpendicular to $$\overrightarrow{OA}$$ and has the same magnitude because $$\bigl|(-1,\sqrt3)\bigr|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.$$

If instead we had rotated clockwise, we would get $$(1,-\sqrt3),$$ whose $$y$$-component is negative; that would force two vertices below the $$x$$-axis, contradicting the condition that the whole square lies above it. Hence we retain $$\overrightarrow{w}=(-1,\;\sqrt3).$$

The remaining two vertices are obtained by adding this vector once to $$A$$ and once to $$O$$:

For $$B:$$ $$\begin{aligned} B &= A+\overrightarrow{w}\\ &=\bigl(\sqrt3,\,1\bigr)+\bigl(-1,\,\sqrt3\bigr)\\ &=\bigl(\sqrt3-1,\,1+\sqrt3\bigr). \end{aligned}$$

For $$C:$$ $$\begin{aligned} C &= O+\overrightarrow{w}\\ &=\bigl(0,\,0\bigr)+\bigl(-1,\,\sqrt3\bigr)\\ &=\bigl(-1,\,\sqrt3\bigr). \end{aligned}$$

Collecting all four vertices with their $$x$$-coordinates:

$$\begin{aligned} O &: (0,\,0) &\Rightarrow&\ x_O=0,\\ A &: (\sqrt3,\,1) &\Rightarrow&\ x_A=\sqrt3,\\ B &: (\sqrt3-1,\,1+\sqrt3) &\Rightarrow&\ x_B=\sqrt3-1,\\ C &: (-1,\,\sqrt3) &\Rightarrow&\ x_C=-1. \end{aligned}$$

Now add these four $$x$$-coordinates explicitly:

$$\begin{aligned} x_O+x_A+x_B+x_C&=0+\sqrt3+(\sqrt3-1)+(-1)\\ &=\sqrt3+\sqrt3-1-1\\ &=2\sqrt3-2. \end{aligned}$$

Thus the required sum of the $$x$$-coordinates is $$2\sqrt3-2.$

Hence, the correct answer is Option A.