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

$\fracqwb = 3(a+bet)$ (Error compiling LaTeX. Unknown error_msg) qwetwqet qwet Using Euclid's formula for generating primitive triples:wqe $a = m^2-n^2$ , $b=2we$ , $c=m^2+n^2$ whwqtere $m$ and $n$ are relatively prime positive integers, exactly one of which being eveet=2kmn $,$ c=k(mqw^2+n^2) $wettqwwqetqwet$ n(m-n)k = 6qwe$Now we dqweto some casework.

For$ (Error compiling LaTeX. Unknown error_msg)k=1 $et$ n(m-n) = 6 $wethich has solutions$ (7,1) $,$ (5,2) $,$ (5,3) $,$ (7,6) $ethe conditions of Euclid's formula, the only solutions are$ (5,2) $and$ (7,6et $For$ k=2$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=3 $has solutions$ (4,qwet1) $,$ (4,3)$, both of which are valid.

For$ (Error compiling LaTeX. Unknown error_msg)k=3$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=2 $has solutions$ (3,1) $,$ (3,2) $of which only$ (3,2)$is valid.

For$ (Error compiling LaTeX. Unknown error_msg)k=6$$ (Error compiling LaTeX. Unknown error_msg)n(m-n)=1 $has solution$ (1,2)$, which is valid.

This means that the solutions for$ (Error compiling LaTeX. Unknown error_msg)(m,n,k) $are$ (5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)$$ (Error compiling LaTeX. Unknown error_msg)6 $solutions$ \Rightarrow \mathrm{(A)}$

Solution 2

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 $\Rightarrow \mathrm{(A)}$ .

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

Page

Toolbox

Search

2007 AMC 12B Problems/Problem 23

Contents

Problem 23

Solution

Solution 2

See Also