Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Brazil

Given a circle and its center OO, a point AA inside the circle and a distance hh, construct a triangle BACBAC with A=90\angle A = 90^\circ, BB and CC on the circle and the altitude from AA with length hh.

Solution

Let HH on BCBC such that AH=hAH = h. Since BACBAC is a right angle, BHCH=h2BH \cdot CH = h^2.

But the power of HH with respect to the given circle is HBHC=R2OH2HB \cdot HC = R^2 - OH^2.

So OH=R2h2OH = \sqrt{R^2 - h^2} is determined and HH is the intersection of the circle with center OO and radius R2h2\sqrt{R^2 - h^2} and the circle with center AA and radius hh. So HH is determined. To determine BB and CC, it is enough to trace a perpendicular to AHAH passing through HH.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.