Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it United States

Problem:

All subscripts in this problem are to be considered modulo 66, that means for example that ω7\omega_{7} is the same as ω1\omega_{1}. Let ω1,,ω6\omega_{1}, \ldots, \omega_{6} be circles of radius rr, whose centers lie on a regular hexagon of side length 11. Let PiP_{i} be the intersection of ωi\omega_{i} and ωi+1\omega_{i+1} that lies further from the center of the hexagon, for i=1,,6i=1, \ldots, 6. Let QiQ_{i}, i=16i=1 \ldots 6, lie on ωi\omega_{i} such that Qi,Pi,Qi+1Q_{i}, P_{i}, Q_{i+1} are colinear. Find the number of possible values of rr.

Solution

Solution:

Consider two consecutive circles ωi\omega_{i} and ωi+1\omega_{i+1}. Let Qi,QiQ_{i}, Q_{i}' be two points on ωi\omega_{i} and Qi+1,Qi+1Q_{i+1}, Q_{i+1}' on ωi+1\omega_{i+1} such that Qi,PiQ_{i}, P_{i} and Qi+1Q_{i+1} are colinear and also Qi,PiQ_{i}', P_{i} and Qi+1Q_{i+1}'. Then QiQi=2QiPiQi=2Qi+1PiQi+1=Qi+1Qi+1Q_{i} Q_{i}' = 2 \angle Q_{i} P_{i} Q_{i}' = 2 \angle Q_{i+1} P_{i} Q_{i+1}' = \angle Q_{i+1} Q_{i+1}'. Refer to the center of ωi\omega_{i} as OiO_{i}. The previous result shows that the lines OiQiO_{i} Q_{i} and Oi+1Qi+1O_{i+1} Q_{i+1} meet at the same angle as the lines OiQiO_{i} Q_{i}' and Oi+1Qi+1O_{i+1} Q_{i+1}', call this angle ψi\psi_{i}. ψi\psi_{i} is a function solely of the circles ωi\omega_{i} and ωi+1\omega_{i+1} and the distance between them (we have just showed that any two points QiQ_{i} and QiQ_{i}' on ωi\omega_{i} give the same value of ψi\psi_{i}, so ψi\psi_{i} can't depend on this.) Now, the geometry of ωi\omega_{i} and ωi+1\omega_{i+1} is the same for every ii, so ψi\psi_{i} is simply a constant ψ\psi which depends only on rr. We know 6ψ=0mod2π6 \psi = 0 \bmod 2\pi because Q7=Q1Q_{7} = Q_{1}.

We now compute ψ\psi. It suffices to do the computation for some specific choice of QiQ_{i}. Take QiQ_{i} to be the intersection of OiOi+1O_{i} O_{i+1} and ωi\omega_{i} which is further from Oi+1O_{i+1}. We are to compute the angle between OiQiO_{i} Q_{i} and Oi+1Qi+1O_{i+1} Q_{i+1} which is the same as OiOi+1Qi+1\angle O_{i} O_{i+1} Q_{i+1}. Note the triangle OiPiOi+1\triangle O_{i} P_{i} O_{i+1} is isosceles, call the base angle ξ\xi. We have

OiOi+1Qi+1=OiOi+1Pi+PiOi+1Qi+1=ξ+(π2Oi+1PiQi+1)=ξ+(π2(πQiOi+1PiPiQiOi+1))=ξπ+2(ξ+(1/2)PiOiOi+1)=ξπ+2(ξ+(1/2)ξ)=4ξπ\angle O_{i} O_{i+1} Q_{i+1} = \angle O_{i} O_{i+1} P_{i} + \angle P_{i} O_{i+1} Q_{i+1} = \xi + \left(\pi - 2 \angle O_{i+1} P_{i} Q_{i+1}\right) = \xi + \left(\pi - 2\left(\pi - \angle Q_{i} O_{i+1} P_{i} - \angle P_{i} Q_{i} O_{i+1}\right)\right) = \xi - \pi + 2\left(\xi + (1/2) \angle P_{i} O_{i} O_{i+1}\right) = \xi - \pi + 2(\xi + (1/2) \xi) = 4\xi - \pi.

So we get 6(4ξπ)=0mod2π6(4\xi - \pi) = 0 \bmod 2\pi. Noting that ξ\xi must be acute, ξ=π/12,π/6,π/4,π/3\xi = \pi/12, \pi/6, \pi/4, \pi/3 or 5π/125\pi/12. rr is uniquely determined as (1/2)secξ(1/2) \sec \xi so there are 55 possible values of rr.

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