2018 AIME II Problem 7

Below is the professionally curated solution for Problem 7 of the 2018 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AIME II solutions, or check the answer key.

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Concepts:similarityarea ratioperfect square

Difficulty rating: 2650

7.

Triangle ABCABC has side lengths AB=9,AB = 9, BC=53,BC = 5\sqrt{3}, and AC=12.AC = 12. Points A=P0,P1,P2,,P2450=BA = P_0, P_1, P_2, \ldots, P_{2450} = B are on segment AB\overline{AB} with PkP_k between Pk1P_{k-1} and Pk+1P_{k+1} for k=1,2,,2449,k = 1, 2, \ldots, 2449, and points A=Q0,Q1,Q2,,Q2450=CA = Q_0, Q_1, Q_2, \ldots, Q_{2450} = C are on segment AC\overline{AC} with QkQ_k between Qk1Q_{k-1} and Qk+1Q_{k+1} for k=1,2,,2449.k = 1, 2, \ldots, 2449. Furthermore, each segment PkQk,\overline{P_kQ_k}, k=1,2,,2449,k = 1, 2, \ldots, 2449, is parallel to BC.\overline{BC}. The segments cut the triangle into 24502450 regions, consisting of 24492449 trapezoids and 11 triangle. Each of the 24502450 regions has the same area. Find the number of segments PkQk,\overline{P_kQ_k}, k=1,2,,2450,k = 1, 2, \ldots, 2450, that have rational length.

Solution:

Since the 24502450 regions have equal areas, triangle APkQkAP_kQ_k (the union of the first kk regions) has area k2450\frac{k}{2450} of triangle ABC.ABC. Each triangle APkQkAP_kQ_k is similar to ABC,ABC, and lengths scale as the square root of areas, so PkQk=53k2450=53k352=6k14.P_kQ_k = 5\sqrt{3}\,\sqrt{\frac{k}{2450}} = 5\sqrt{3} \cdot \frac{\sqrt{k}}{35\sqrt{2}} = \frac{\sqrt{6k}}{14}.

This is rational exactly when 6k6k is a perfect square, which happens exactly when k=6j2k = 6j^2 for a positive integer j.j. The condition 6j224506j^2 \le 2450 gives j2408,j^2 \le 408, so j=1,2,,20.j = 1, 2, \ldots, 20. There are 2020 such segments.

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