olymmath

AIME 2018b Problem 7

  • Paper
  • Answer

Triangle ABCABC has side lengths AB=9AB = 9, BC=BC = 535\sqrt{3}, and AC=12AC = 12. Points A=P0,P1,P2,...,P2450=BA = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B are on segment AB‾\overline{AB} with PkP_{k} between Pk−1P_{k-1} and Pk+1P_{k+1} for k=1,2,...,2449k = 1, 2, ..., 2449, and points A=Q0,Q1,Q2,...,Q2450=CA = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C are on segment AC‾\overline{AC} with QkQ_{k} between Qk−1Q_{k-1} and Qk+1Q_{k+1} for k=1,2,...,2449k = 1, 2, ..., 2449. Furthermore, each segment PkQk‾\overline{P_{k}Q_{k}}, k=1,2,...,2449k = 1, 2, ..., 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_{k}Q_{k}}, k=1,2,...,2450k = 1, 2, ..., 2450, that have rational length.

Dataset text AIME 1983 to 2026. Quoted with source.

Show the answer

20

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Locator aime/2018b/p7 · Tag Q7MWPA657F · Edit this page on GitHub · Keys: j next, k previous