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

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

$ab = 6(a+b+c)$

Using Euclid's formula for generating primitive triples: $a = m^2-n^2$ , $b=2mn$ , $c=m^2+n^2$ where $m$ and $n$ are relatively prime positive integers, exactly one of which being even.

Since we do not want to restrict ourselves to only primitives, we will add a factor of k. $a = k(m^2-n^2)$ , $b=2kmn$ , $c=k(m^2+n^2)$

$(m^2-n^2)\cdot 2mn \cdot k^2 = 6(2m^2 + 2mn)k$

$mn(m-n)(m+n)k = 6m(m+n)$

$n(m-n)k = 6$

Now we do some casework.

For $k=1$

$n(m-n) = 6$ which has solutions $(7,1)$ , $(5,2)$ , $(5,3)$ , $(7,6)$

Removing the solutions that do not satisfy the conditions of Euclid's formula, the only solutions are $(5,2)$ and $(7,6)$

For $k=2$

$n(m-n)=3$ has solutions $(4,1)$ , $(4,3)$ , both of which are valid.

For $k=3$

$n(m-n)=2$ has solutions $(3,1)$ , $(3,2)$ of which only $(3,2)$ is valid.

For $k=6$

$n(m-n)=1$ has solution $(1,2)$ , which is valid.

This means that the solutions for $(m,n,k)$ are

$(5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)$

$6$ solutions $\Rightarrow \mathrm{(A)}$

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

Problem 23

Solution

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