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HMMT 2021 G5

  • Geometry
  • Answer

Let AEFA E F be a triangle with EF=20E F=20 and AE=AF=21A E=A F=21. Let BB and DD be points chosen on segments AEA E and AFA F, respectively, such that BDB D is parallel to EFE F. Point CC is chosen in the interior of triangle AEFA E F such that ABCDA B C D is cyclic. If BC=3B C=3 and CD=4C D=4, then the ratio of areas [ABCD][AEF]\frac{[A B C D]}{[A E F]} can be written as ab\frac{a}{b} for relatively prime positive integers a,ba, b. Compute 100a+b100 a+b.

Dataset text NuminaMath olympiads. Quoted with source.

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