olymmath

HMMT 2013 G9

  • Geometry
  • Answer

Pentagon ABCDEA B C D E is given with the following conditions: (a) ∠CBD+∠DAE=∠BAD=45∘,∠BCD+∠DEA=300∘\angle C B D+\angle D A E=\angle B A D=45^{\circ}, \angle B C D+\angle D E A=300^{\circ} (b) BADA=223,CD=753\frac{B A}{D A}=\frac{2 \sqrt{2}}{3}, C D=\frac{7 \sqrt{5}}{3}, and DE=1524D E=\frac{15 \sqrt{2}}{4} (c) AD2⋅BC=AB⋅AE⋅BDA D^{2} \cdot B C=A B \cdot A E \cdot B D

Compute BDB D.

Dataset text NuminaMath olympiads. Quoted with source.

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\sqrt{39}

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