2018 AIME II Problem 7

Attempt Problem 7 of the 2018 AIME II below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

Triangle ABCABC has side lengths AB=9,AB = 9, BC=53,BC = 5\sqrt{3}, and AC=12.AC = 12. Points A=P0,A = P_0, P1,P_1, P2,P_2, …,\ldots, P2450=BP_{2450} = B are on segment AB‾\overline{AB} with PkP_k between Pk−1P_{k-1} and Pk+1P_{k+1} for k=1,k = 1, 2,2, …,\ldots, 2449,2449, and points A=Q0,A = Q_0, Q1,Q_1, Q2,Q_2, …,\ldots, Q2450=CQ_{2450} = C are on segment AC‾\overline{AC} with QkQ_k between Qk−1Q_{k-1} and Qk+1Q_{k+1} for k=1,k = 1, 2,2, …,\ldots, 2449.2449. Furthermore, each segment PkQk‾,\overline{P_kQ_k}, k=1,k = 1, 2,2, …,\ldots, 2449,2449, 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,k = 1, 2,2, …,\ldots, 2450,2450, that have rational length.
Answer: 20
Concepts:similarityarea ratioperfect square
Small hint:
The regions have equal areas, so triangle APkQkAP_kQ_k has area k2450\frac{k}{2450} of the whole, and lengths scale as the square root of the area ratio
Big hint:
PkQk=53 k2450=6k14,P_kQ_k = 5\sqrt{3}\,\sqrt{\frac{k}{2450}} = \frac{\sqrt{6k}}{14}, which is rational exactly when 6k6k is a perfect square
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=53 k2450=53⋅k352=6k14. \begin{aligned} P_kQ_k &= 5\sqrt{3}\,\sqrt{\frac{k}{2450}} \\ &= 5\sqrt{3} \cdot \frac{\sqrt{k}}{35\sqrt{2}} \\ &= \frac{\sqrt{6k}}{14}. \end{aligned}
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 6j2≤24506j^2 \le 2450 gives j2≤408,j^2 \le 408, so j=1,2,…,20.j = 1, 2, \ldots, 20. There are 2020 such segments.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.