2007 AMC 12B Problems/Problem 23

Problem 23

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution

Let $a$ and $b$ be the two legs of the triangle.

We have $\frac{1}{2}ab = 3(a+b+c)$ .

Then $ab=6\cdot (a+b+\sqrt {a^2 + b^2})$ .

We can complete the square under the root, and we get, $ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})$ .

Let $ab=p$ and $a+b=s$ , we have $p=6\cdot (s+ \sqrt {s^2 - 2p})$ .

After rearranging, squaring both sides, and simplifying, we have $p=12s-72$ .

Putting back $a$ and $b$ , and after factoring using $SFFT$ , we've got $(a-12)\cdot (b-12)=72$ .

Factoring 72, we get 6 pairs of $a$ and $b$

$(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$

And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$ .

Solution #2

We will proceed by using the fact that $[ABC] = r\cdot s$ , where $r$ is the radius of the incircle and $s$ is the semiperimeter ( $s = \frac{p}{2}$ ).

We are given $[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$ .

\begin{triangle}[thick] \draw(0,0) -- (90:2cm) node[midway,left]{ $opposite leg$ } -- (0:4cm) node[midway,above right]{ $hypotenuse$ } -- (0,0) node[midway,below]{ $adjacent leg$ }; \draw[fill=lightgray, thick] (0,0) -- (0:0.8cm) arc (0:90:0.8cm) node at (45:0.5cm) { $\gamma$ } -- cycle; \end{triangle}

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 12 Problems and Solutions

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2007 AMC 12B Problems/Problem 23

Contents

Problem 23

Solution

Solution #2

See Also