2018 AIME II

Geometry Level 4

Triangle A B C ABC has sides A B = 9 , B C = 5 3 AB=9, BC=5\sqrt3 and A C = 12. AC=12. Points A = P 0 , P 1 , P 2 , , P 2450 = B A=P_0, P_1, P_2, \ldots , P_{2450}=B are on the segment A B \overline{AB} with P k P_k between P k 1 P_{k-1} and P k + 1 P_{k+1} for k = 1 , 2 , , 2449 , k=1,2,\ldots , 2449, and points A = Q 0 , Q 1 , Q 2 , , Q 2450 = C A=Q_0, Q_1, Q_2, \ldots , Q_{2450}=C for k = 1 , 2 , , 2449 k=1,2,\ldots , 2449 . Furthermore, each segment P k Q k , k = 1 , 2 , , 2449 , \overline{P_kQ_k}, k=1,2, \ldots , 2449, is parallel to B C . \overline{BC}. The segments cut the triangle into 2450 2450 regions, consisting of 2449 2449 trapezoids and 1 1 triangle. Each of the 2450 2450 regions have the same area. Find the number of segments P k Q k , k = 1 , 2 , , 2450 , \overline{P_kQ_k}, k=1,2, \ldots , 2450, that have rational length.


The answer is 20.

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1 solution

Patrick Corn
Jan 15, 2021

Each segment is the base of a triangle T k = A P k Q k T_k = AP_kQ_k that is similar to A B C . ABC. The area of T k T_k equals k 2450 \frac{k}{2450} times the area of A B C , ABC, so the length P k Q k P_kQ_k equals k 2450 \sqrt{\frac{k}{2450}} times B C . BC. This is k 2450 5 3 = 1 7 3 k 2 . \sqrt{\frac{k}{2450}} \cdot 5 \sqrt{3} = \frac17 \sqrt{\frac{3k}{2}}. This is rational if and only if 3 k / 2 3k/2 is a square, which happens if and only if k = 6 n 2 k = 6n^2 for some n , n, where 1 k 2450. 1 \le k \le 2450. This inequality is satisfied for n = 1 , 2 , , 20 , n = 1, 2, \ldots, 20, so the answer is 20 . \fbox{20}.

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