olymmath

AHSME 1952 Problem 22

    On hypotenuse ABAB of a right triangle ABCABC a second right triangle ABDABD is constructed with hypotenuse ABAB. If BC‾=1\overline{BC}=1, AC‾=b\overline{AC}=b, and AD‾=2\overline{AD}=2, then BD‾\overline{BD} equals: (A)  b2+1(B)  b2−3(C)  b2+1+2\textbf{(A) \ }\sqrt{b^2+1} \textbf{(B) \ }\sqrt{b^2-3} \textbf{(C) \ }\sqrt{b^2+1}+2 (D)  b2+5(E)  b2+3\textbf{(D) \ }b^2+5 \textbf{(E) \ }\sqrt{b^2+3}

    Dataset text NuminaMath-1.5. Quoted with source.

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