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HMMT 2023 G8

  • Geometry
  • Answer

Triangle ABCA B C with ∠BAC>90∘\angle B A C>90^{\circ} has AB=5A B=5 and AC=7A C=7. Points DD and EE lie on segment BCB C such that BD=DE=ECB D=D E=E C. If ∠BAC+∠DAE=180∘\angle B A C+\angle D A E=180^{\circ}, compute BCB C.

Dataset text NuminaMath olympiads. Quoted with source.

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\sqrt{111}

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