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HMMT 2001 G10

  • Geometry
  • Answer

AA is the center of a semicircle, with radius ADA D lying on the base. BB lies on the base between AA and DD, and EE is on the circular portion of the semicircle such that EBAE B A is a right angle. Extend EAE A through AA to CC, and put FF on line CDC D such that EBFE B F is a line. Now EA=1,AC=2,BF=2−24,CF=25+104E A=1, A C=\sqrt{2}, B F=\frac{2-\sqrt{2}}{4}, C F=\frac{2 \sqrt{5}+\sqrt{10}}{4}, and DF=25−104D F=\frac{2 \sqrt{5}-\sqrt{10}}{4}. Find DED E.

Dataset text NuminaMath olympiads. Quoted with source.

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\sqrt{2-\sqrt{2}}

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