## Thursday, February 22, 2007

### Pythagorean x3. Topic: Geometry/NT. Level: AMC.

Problem: (2007 AMC12B - #23) How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

Solution: Well, basically, you should know the Pythagorean triple generating formula, i.e. $a = k(u^2-v^2)$, $b = 2kuv$, $c = k(u^2+v^2)$. Substitute accordingly and we have to solve the diophantine equation

$\frac{1}{2} \cdot k(u^2-v^2) \cdot 2kuv = 3 \cdot (k(u^2-v^2)+2kuv+k(u^2+v^2))$

which conveniently simplifies to

$kv(u-v) = 6$.

Obviously then $k = 1, 2, 3, 6$ so look at these cases:

$k = 1$: We can take $v = 1, 2, 3, 6$ to get the triples $(14, 48, 50); (20, 21, 29); (16, 30, 34); (13, 84, 85)$.

$k = 2$: We can take $v = 1, 3$ to get the triples $(16, 30, 34); (14, 48, 50)$ both of which are already counted.

$k = 3$: We can take $v = 1, 2$ to get the triples $(18, 24, 30); (15, 24, 39)$.

$k = 6$: We can take $v = 1$ to get the triple $(18, 24, 30)$, which is already counted.

So we have $4+2 = 6$ triangles. QED.

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Comment: Not too hard if you knew the generating formula for Pythagorean triples. It was a little annoying having to check for repeated triples, but at least there weren't that many.

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Practice Problem: (2007 AMC 12B - #24) How many pairs of positive integers $(a, b)$ are there such that $\text{gcd}(a, b) = 1$ and

$\frac{a}{b}+\frac{14b}{9a}$

is an integer?