Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.5 AIME, harder Prove it Italy

Let $ABC$ be a right triangle, with hypotenuse $AC$ , and let $H$ be the foot of the altitude drawn from $B$ to $AC$ . Knowing that the lengths $AB$ , $BC$ and $BH$ form the sides of a new right triangle, determine the possible values of $\frac{AH}{CH}$ .

Solution

Since triangles $ABH$ and $BCH$ are right triangles, we have $BH < AB$ and $BH < HC$ . Hence the larger of $AB$ and $BC$ must be the hypotenuse of the new right triangle.

Suppose first that $BC > AB$ . Then we have $AB^2 + BH^2 = BC^2$ and, analyzing the right triangle $BCH$ , $CH^2 + BH^2 = BC^2$ . It follows that $AB = CH$ .

From the similarity between triangles $AHB$ and $ABC$ we have $AB : AH = (AH + CH) : AB$ and therefore $CH : AH = (AH + CH) : CH$ . Letting $x$ be the desired ratio, we find $1/x = x + 1$ , from which $x = \frac{\sqrt{5} - 1}{2}$ (the negative value of the solution of the algebraic equation must be discarded).

If $AB > BC$ , the same reasoning proves that $CH / AH = \frac{\sqrt{5} - 1}{2}$ , that is $AH / CH = \frac{\sqrt{5} + 1}{2}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.