Maths Olympiad Prep

Library / /642 of 740

, 2013

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let XYZ\triangle X Y Z be a right triangle with XYZ=90\angle X Y Z=90^{\circ}. Suppose there exists an infinite sequence of equilateral triangles X0Y0T0,X1Y1T1,X_{0} Y_{0} T_{0}, X_{1} Y_{1} T_{1}, \ldots such that X0=XX_{0}=X, Y0=YY_{0}=Y, XiX_{i} lies on the segment XZX Z for all i0i \geq 0, YiY_{i} lies on the segment YZY Z for all i0i \geq 0, XiYiX_{i} Y_{i} is perpendicular to YZY Z for all i0i \geq 0, TiT_{i} and YY are separated by line XZX Z for all i0i \geq 0, and XiX_{i} lies on segment Yi1Ti1Y_{i-1} T_{i-1} for i1i \geq 1.
Let P\mathcal{P} denote the union of the equilateral triangles. If the area of P\mathcal{P} is equal to the area of XYZX Y Z, find XYYZ\frac{X Y}{Y Z}.

Solution

Solution:

Let a=XYa = X Y, b=YZb = Y Z, ra=X1Y1r a = X_{1} Y_{1}. Then [P]=[XYT0](1+r2+r4+)[\mathcal{P}] = [X Y T_{0}](1 + r^{2} + r^{4} + \cdots), [XYZ]=[XYY1X1](1+r2+r4+)[X Y Z] = [X Y Y_{1} X_{1}](1 + r^{2} + r^{4} + \cdots), YY1=ra3Y Y_{1} = r a \sqrt{3}, and b=ra3(1+r+r2+)b = r a \sqrt{3}(1 + r + r^{2} + \cdots) (although we can also get this by similar triangles).

Hence a234=12(ra+a)(ra3)\frac{a^{2} \sqrt{3}}{4} = \frac{1}{2}(r a + a)(r a \sqrt{3}), or 2r(r+1)=1r=3122 r(r+1) = 1 \Longrightarrow r = \frac{\sqrt{3} - 1}{2}. Thus XYYZ=ab=1rr3=1\frac{X Y}{Y Z} = \frac{a}{b} = \frac{1 - r}{r \sqrt{3}} = 1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.