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 1: Let the triangle be $\triangle A B C$, with $A C$ as the hypotenuse, and let $D, E, F, G$ be on sides $A B, B C, A C$, $A C$, respectively, such that they all lie on the circle. We have $A G=1, G F=24$, and $F C=3$. By power of a point, we have \[ \begin{aligned} & A D=\sqrt{A G \cdot A F}=\sqrt{1(1+24)}=5 \\ & C E=\sqrt{C F \cdot C G}=\sqrt{3(3+24)}=9 \end{aligned} \] Now, let $B D=B E=x$. 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 $\triangle A B C$ is $\frac{1}{2}(x+5)(x+9)=\frac{1}{2} \cdot 384=192$. $\fbox{192}$.