Library / /460 of 740

, 2023

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:
A right triangle and a circle are drawn such that the circle is tangent to the legs of the right triangle. The circle cuts the hypotenuse into three segments of lengths $1$ , $24$ , and $3$ , and the segment of length $24$ is a chord of the circle. Compute the area of the triangle.

Solution

Solution:

Let the triangle be $\triangle ABC$ , with $AC$ as the hypotenuse, and let $D, E, F, G$ be on sides $AB, BC, AC, AC$ , respectively, such that they all lie on the circle. We have $AG = 1$ , $GF = 24$ , and $FC = 3$ .
By power of a point, we have
$\begin{aligned} & AD = \sqrt{AG \cdot AF} = \sqrt{1(1+24)} = 5 \\ & CE = \sqrt{CF \cdot CG} = \sqrt{3(3+24)} = 9 \end{aligned}$
Now, let $BD = BE = x$ . By the Pythagorean Theorem, we get that
$\begin{aligned} (x+5)^2 + (x+9)^2 & = 28^2 \\ (x+5)^2 + (x+9)^2 - ((x+9)-(x+5))^2 & = 28^2 - 4^2 \\ 2(x+5)(x+9) & = 768 \\ (x+5)(x+9) & = 384. \end{aligned}$
The area of $\triangle ABC$ is $\frac{1}{2}(x+5)(x+9) = \frac{1}{2} \cdot 384 = 192$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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