IOQM Previous Papers jan 17 2020-21

Take $$CD = 1$$, $$AB = 3$$ and let $$h$$ be the distance between the parallel sides, so $$[ABCD] = \frac{1}{2}(3+1)h = 2h$$. Since $$E$$ is the midpoint of $$BD$$ and $$B$$ is at height $$h$$ while $$D$$ lies on line $$CD$$, the point $$E$$ is at height $$\frac{h}{2}$$ above $$CD$$. Hence $$[CDE] = \frac{1}{2} \times 1 \times \frac{h}{2} = \frac{h}{4}$$, and $$n = \frac{2h}{h/4} = 8$$.

The two representations give $$5b^2 + 3 = 3(b+2)^2 + 5$$, which simplifies to $$2b^2 - 12b - 14 = 0$$, that is $$b^2 - 6b - 7 = 0$$. The positive root is $$b = 7$$, so $$N = 5 \times 49 + 3 = 248$$. The product of its digits is $$2 \times 4 \times 8 = 64$$.

Since $$(k^2+k)^2 = k^2(k+1)^2$$ and $$2k+1 = (k+1)^2 - k^2$$, each term telescopes as $$\frac{1}{k^2} - \frac{1}{(k+1)^2}$$. The sum therefore equals $$1 - \frac{1}{(N+1)^2}$$, so $$\frac{1}{(N+1)^2} = 0.0001$$. This gives $$(N+1)^2 = 10000$$ and $$N = 99$$.

Put $$AB = CD = x$$ and $$BC = y$$, so $$2x + y = 20$$. From the right triangle $$ABE$$ we get $$x^2 + \left(\frac{y}{2}\right)^2 = 81$$, and substituting $$y = 20 - 2x$$ gives $$x^2 + (10-x)^2 = 81$$, that is $$2x^2 - 20x + 19 = 0$$. The area is $$xy = 20x - 2x^2$$, which equals 19 by the same equation.

The inequality means $$-5 < |x| - 2020 < 5$$, so $$2015 < |x| < 2025$$. The integer values of $$|x|$$ are $$2016, 2017, \ldots, 2024$$, which is 9 values. Each gives two integers $$x$$ of opposite signs, so there are $$9 \times 2 = 18$$ solutions.

Writing everything in prime powers, $$4^3 = 2^6$$ and $$6^7 = 2^7 \cdot 3^7$$, so the number is $$2^{5+6+7} \cdot 3^{6+7} \cdot 5^3 = 2^{18} \cdot 3^{13} \cdot 5^3$$. A perfect square needs every exponent even, and here the exponents of 3 and 5 are odd. Multiplying by $$3 \times 5 = 15$$ fixes both, so the least such integer is 15.

Let $$M$$ be the midpoint of $$BC$$, so $$AM$$ is perpendicular to $$BC$$. Here $$BC = \frac{2929}{61} + 61 = \frac{6650}{61}$$, so $$BM = \frac{3325}{61}$$ and $$DM = BM - BD = \frac{3325 - 2929}{61} = \frac{396}{61}$$. The right triangles $$ADM$$ and $$CDE$$ share the angle at $$D$$, hence they are similar and $$\frac{DM}{DA} = \frac{DE}{DC}$$, giving $$AD = \frac{396}{61} \times \frac{61}{11} = 36$$. Therefore $$AE = AD - DE = 36 - 11 = 25$$.

All five entries must be single digits, so $$3k \le 9$$ forces $$k \le 3$$, giving the candidates 12334, 23465 and 34596. Only the last is a perfect square, since $$186^2 = 34596$$. Hence $$m = 186$$ and the sum of its digits is $$1+8+6 = 15$$.

The bisector gives $$\frac{AD}{DB} = \frac{CA}{CB} = \frac{4}{6}$$, so $$AD = 2$$ and $$DB = 3$$, and $$DMCN$$ is a parallelogram whose diagonal $$CD$$ bisects angle $$C$$, making it a rhombus of side $$DM = 4 \times \frac{3}{5} = \frac{12}{5}$$. The bisector length satisfies $$CD^2 = CA \cdot CB - AD \cdot DB = 24 - 6 = 18$$, and for a rhombus $$\left(\frac{CD}{2}\right)^2 + \left(\frac{MN}{2}\right)^2 = \left(\frac{12}{5}\right)^2$$. This gives $$MN^2 = \frac{126}{25}$$, so $$|p-q| = 101$$ and its digit sum is 2.

With five scores the median is the third one. Taking the scores $$0, 0, 100, 100, 100$$ gives median 100 and mean 60, a difference of 40, while $$0, 0, 0, 100, 100$$ gives median 0 and mean 40, again a difference of 40. Pushing the two extreme scores to the ends is optimal, so the largest possible difference is 40.

If $$r$$ is a common zero, subtracting $$r(r^2+ar+b)$$ from $$r^3+br+a$$ gives $$a(1-r^2) = 0$$, so either $$a = 0$$ or $$r = \pm 1$$. When $$a = 0$$ a common zero exists exactly for $$b \le 0$$, giving 6 pairs; $$r = 1$$ needs $$a+b = -1$$ and $$r = -1$$ needs $$b = a-1$$, giving 10 pairs each. The only pair common to more than one family is $$(0, -1)$$, so by inclusion and exclusion the count is $$6+10+10-1-1-1+1 = 24$$.

Each chord is tangent to the inner circle, so at every vertex the direction of travel turns through the exterior angle $$180^\circ - 75^\circ = 105^\circ$$. Returning to the start after $$n$$ chords requires the total turning $$105n$$ to be a whole number of full turns, so $$105n = 360k$$ and $$n = \frac{24k}{7}$$. The smallest positive integer solution takes $$k = 7$$, giving $$n = 24$$.

Modulo 3 the expression is $$2(-1)^n - 2$$, which is 2 for odd $$n$$, and 2 is not a quadratic residue modulo 3, so $$n$$ must be even. Writing $$n = 2m$$ with $$m \ge 4$$ we have $$5^{2m} < 5^{2m} + 2^{2m} - 65 < (5^m+1)^2$$, so no square arises there. Checking $$n = 2, 4, 6$$ gives $$36 = 6^2$$, $$576 = 24^2$$ and 15624 which is not a square, so the sum is $$2 + 4 = 6$$.

Here $$55 \times 60 \times 65 = 214500 = 2^2 \cdot 3 \cdot 5^3 \cdot 11 \cdot 13$$. If every factor were at most 19, the primes 11 and 13 would force two of the factors to be exactly 11 and 13, leaving $$\frac{214500}{143} = 1500$$ as a product of three distinct integers at most 19, which is impossible. The largest factor is therefore at least 20, and $$5 \times 11 \times 13 \times 15 \times 20 = 214500$$ shows 20 is attainable.

There are $$6! = 720$$ seatings in all, and the forbidden events are the three couples occupying a horizontally adjacent pair or a vertical pair. Counting the seatings that avoid all of these, by inclusion and exclusion over the three couples, leaves 96. Hence 96 arrangements are possible.

Rearranging gives $$x - y = 2\Delta \cdot \frac{x-y}{xy}$$, and since the triangle is scalene $$x \neq y$$, so $$2\Delta = xy = 3780$$. But the area equals $$\frac{1}{2}xy \sin C$$ where $$C$$ is the angle between these two sides, so $$\sin C = 1$$ and $$C = 90^\circ$$. The largest side is the hypotenuse $$\sqrt{60^2 + 63^2} = \sqrt{7569} = 87$$.

A number has exactly four divisors when it is either $$p^3$$ or $$pq$$ for distinct primes $$p$$ and $$q$$. The two digit cubes of primes are $$27$$ and there are 29 two digit numbers that are a product of two distinct primes. Together this gives $$1 + 29 = 30$$ such numbers.

The expression under the radical is a perfect square, since $$1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} = \left(1 + \frac{1}{k} - \frac{1}{k+1}\right)^2$$. So each term equals $$1 + \frac{1}{k(k+1)}$$ and the sum telescopes to $$40 + \left(1 - \frac{1}{41}\right) = 40 + \frac{40}{41}$$. Thus $$a = 40$$, $$b = 40$$ and $$a+b = 80$$.

Ratios of areas are unchanged by an affine map, so take the unit square with $$A = (0,0)$$, $$B = (1,0)$$, $$C = (1,1)$$, $$D = (0,1)$$, giving $$E = \left(\frac{1}{2}, 0\right)$$ and $$F = \left(1, \frac{1}{2}\right)$$. Solving $$y = 2x-1$$ with $$y = 1 - \frac{x}{2}$$ gives $$P = (0.8, 0.6)$$, and the four triangles then have areas $$0.3, 0.1, 0.2, 0.4$$ of the whole. The largest is $$0.4 \times 100 = 40$$ square units.

Suppose there are $$n$$ women, the first works $$T$$ hours and the last works $$\frac{T}{5}$$ hours, the individual working times forming an arithmetic progression. The total labour is $$n \times \frac{T + \frac{T}{5}}{2} = \frac{3nT}{5}$$ woman hours, while the job needs $$45n$$ woman hours. Equating gives $$\frac{3nT}{5} = 45n$$, so $$T = 75$$ hours.

Let the first year ages be $$a, b, c$$. Since $$A$$ received half the total, $$a = b+c$$, so the total age is $$2a$$ and six years later it is $$2a+15$$. The condition on $$A$$ gives $$\frac{a+5}{2a+15} = \frac{3}{7}$$, so $$a = 10$$ and $$b+c = 10$$, and the condition on $$C$$ gives $$\frac{c+5}{35} = \frac{c}{10}$$, so $$c = 2$$ and $$b = 8$$. Finally $$B$$ gives $$\frac{13N}{35} = \frac{2N}{5} - 1$$, which yields $$N = 35$$.

Reflecting $$A$$ in each bisector lands on line $$BC$$, so $$P, Q, R, S$$ are the midpoints of $$A$$ with the points at signed positions $$-c$$, $$a-b$$, $$c$$ and $$a+b$$ on line $$BC$$ measured from $$B$$, and all four lie on the midline. Halving the distances gives $$a-b+c = 14$$, $$|b+c-a| = 12$$ and $$a+b-c = 16$$, whose solution is $$a = 15$$, $$b = 14$$, $$c = 13$$. By Heron's formula the area is $$\sqrt{21 \times 8 \times 7 \times 6} = 84$$.

The small circle in the corner $$A$$ is the image of $$\Gamma$$ under a homothety at $$A$$, which gives $$r_A = r \cdot \frac{1 - \sin\frac{A}{2}}{1 + \sin\frac{A}{2}}$$, and similarly for $$r_B$$ and $$r_C$$. Combining these with $$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$$ yields the identity $$r = \sqrt{r_A r_B} + \sqrt{r_B r_C} + \sqrt{r_C r_A}$$. Substituting the given values gives $$r = 20 + 30 + 24 = 74$$.

A line $$y = 16 + mx$$ through the source touches the disc when $$\frac{|6m+6|}{\sqrt{m^2+1}} = 2$$, which reduces to $$4m^2 + 9m + 4 = 0$$. The two slopes satisfy $$m_1m_2 = 1$$ and $$m_1 - m_2 = \frac{\sqrt{17}}{4}$$, and each tangent meets the $$X$$ axis at $$x = -\frac{16}{m}$$. The shadow length is $$16 \cdot \frac{|m_1-m_2|}{|m_1m_2|} = 4\sqrt{17}$$, so $$m+n = 4+17 = 21$$.

Here $$\langle 91 \rangle = 100$$, $$\langle 120 \rangle = 121$$, $$\langle 143 \rangle = 144$$ and $$\langle 180 \rangle = 169$$, so the equation reduces to $$\frac{\langle N \rangle}{N} = \frac{91 \cdot 120 \cdot 143 \cdot 180}{100 \cdot 121 \cdot 144 \cdot 169} = \frac{21}{22}$$. Writing $$N = 22t$$ forces $$21t$$ to be a perfect square, so $$t = 21m^2$$ and $$N = 462m^2$$ with $$\langle N \rangle = (21m)^2$$, which is indeed the closest square since $$21m^2 < 22m^2$$. The smallest value is $$N = 462$$, and $$4^2 + 6^2 + 2^2 = 56$$.

The six disks form a triangular array whose symmetry group has 6 elements, namely the identity, two rotations and three reflections. Without symmetry there are $$\frac{6!}{2!2!2!} = 90$$ colorings; the rotations split the disks into two 3-cycles and fix none of these colorings, while each reflection fixes 6 of them, since its two 2-cycles must take two different colors and the two fixed disks then take the third. By Burnside's lemma the answer is $$\frac{90 + 0 + 0 + 6 + 6 + 6}{6} = 18$$.

A lattice point $$P$$ lies on such a path exactly when the taxicab distances satisfy $$d(A,P) + d(P,B) \le 14$$, and this total always has the same parity as $$d(A,B) = 10$$. Writing the condition as $$f(x) + g(y) \le 14$$ where $$f(x) = |x+3| + |x-3|$$ and $$g(y) = |y-2| + |y+2|$$, one has $$f(x) = 6$$ for $$|x| \le 3$$ and $$2|x|$$ otherwise, with a similar description for $$g$$. Counting the lattice points satisfying this gives 87.

A number is a sum of at least two consecutive positive integers exactly when it has an odd divisor greater than 1, that is, when it is not a power of 2. The powers of 2 in the given set are $$1, 2, 4, 8, 16, 32, 64$$, which is 7 numbers. Hence the number of good numbers is $$100 - 7 = 93$$.

We need $$a > b$$ and $$(a-b) \mid ab$$. Taking $$a = 27$$ and $$b = 18$$ gives $$a - b = 9$$ and $$c = \frac{27 \times 18}{9} = 54$$, so $$a+b+c = 27+18+54 = 99$$. Since the bound 99 is attained, the largest possible value is 99.

Put $$d = a+1$$, so $$a^2+a+2 = d^2-d+2$$ and the conditions say $$b \equiv 1 \pmod d$$ with $$b \mid d^2-d+2$$. Writing $$d^2-d+2 = bk$$ and reducing modulo $$d$$ gives $$k \equiv 2 \pmod d$$, and any $$k \ge d+2$$ makes $$b$$ too small, so $$k = 2$$ and $$b = \frac{d^2-d+2}{2}$$, which satisfies $$b \equiv 1 \pmod d$$ only for odd $$d$$. Requiring $$100 \le b \le 999$$ gives the odd values $$d = 15, 17, \ldots, 45$$, so there are 16 pairs.