# 2008 iTest Problems/Problem 52

## Problem

A triangle has sides of length $48, 55$, and $73$. A square is inscribed in the triangle such that one side of the square lies on the longest side of the triangle, and the two vertices not on that side of the square touch the other two sides of the triangle. If $c$ and $d$ are relatively prime positive integers such that $c/d$ is the length of a side of the square, ﬁnd the value of $c + d$.

## Solution $[asy] draw((0,0)--(73,0)--(2304/73,2640/73)--(0,0)); draw((21.1,0)--(21.1,192720/7969)--(45.1,192720/7969)--(45.1,0)); draw((2304/73,2640/73)--(2304/73,0),dotted); [/asy]$

Let $a$ be the altitude to the longest side, and let $x$ be the length of the square. Using similar triangles to write a proportion, $$\frac{a-x}{x} = \frac{a}{73}$$ $$73a-73x=ax$$ $$x = \frac{73a}{a+73}$$

Note that $48^2 + 55^2 = 73^2$, so the sides are part of a right triangle. That means the altitude to the hypotenuse is $\tfrac{48 \cdot 55}{73}$. Substitute that to get $$\frac{73 \cdot \frac{48 \cdot 55}{73}}{73 + \frac{48 \cdot 55}{73}}$$ $$\frac{48 \cdot 55}{\frac{73^2 + 48 \cdot 55}{73}}$$ $$\frac{48 \cdot 55 \cdot 73}{73^2 + 48 \cdot 55}$$ $$\frac{192720}{7969}$$ The prime factorization of $7969$ is $13 \cdot 613$. Since both of the numbers are not a factor of $192720$, we can confirm that $192720$ and $7969$ are relatively prime, so $c+d = \boxed{200689}$.

## See Also

 2008 iTest (Problems) Preceded by:Problem 51 Followed by:Problem 53 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
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