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Problem 445

AMC 12 late, AIME early

Geometry Difficulty 4.8 Find the answer

10. A. As shown in Figure 3, in the right triangle $\triangle ABC$ , the hypotenuse $AB$ is 35 units long, and the square $CDEF$ is inscribed in $\triangle ABC$ with a side length of 12. Then the perimeter of $\triangle ABC$ is $\qquad$

Your answer

Official solution

10. A. 84.

Let $BC = a, AC = b$ . Then,
$a^{2} + b^{2} = 35^{2} = 1225.$

Since Rt $\triangle AFE \sim \text{Rt} \triangle ACB$ , we have,
$\frac{FE}{CB} = \frac{AF}{AC} \Rightarrow \frac{12}{a} = \frac{b-12}{b}.$

Thus, $12(a + b) = ab$ .
From equations (1) and (2), we get
$\begin{array}{l} (a + b)^{2} = a^{2} + b^{2} + 2ab \\ = 1225 + 24(a + b). \end{array}$

Solving this, we get $a + b = -25$ (discard), 49.
Therefore, $a + b + c = 49 + 35 = 84$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.