GeometryDifficulty 5.5AIME, harderProve itUnited States
Problem:
Let △XYZ be a right triangle with ∠XYZ=90∘. Suppose there exists an infinite sequence of equilateral triangles X0Y0T0,X1Y1T1,… such that X0=X, Y0=Y, Xi lies on the segment XZ for all i≥0, Yi lies on the segment YZ for all i≥0, XiYi is perpendicular to YZ for all i≥0, Ti and Y are separated by line XZ for all i≥0, and Xi lies on segment Yi−1Ti−1 for i≥1. Let P denote the union of the equilateral triangles. If the area of P is equal to the area of XYZ, find YZXY.
Solution
Solution:
Let a=XY, b=YZ, ra=X1Y1. Then [P]=[XYT0](1+r2+r4+⋯), [XYZ]=[XYY1X1](1+r2+r4+⋯), YY1=ra3, and b=ra3(1+r+r2+⋯) (although we can also get this by similar triangles).
Hence 4a23=21(ra+a)(ra3), or 2r(r+1)=1⟹r=23−1. Thus YZXY=ba=r31−r=1.
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Source: MathNet,
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