# Category Archives: Geometry

${\bf Proposition.}$ Suppose $T$ is a right triangle with area $A$ and perimeter $P$. Then the two legs of the triangle are the solutions of the quadratic polynomial $-2P x^2 + (P^2 + 4A) x - 4AP = 0.$
${\it Proof.}$ Let $x,y,z$ be the side lengths of $T$, where $z$ is the hypotenuse. Then $xy = 2A , x + y + z = P , x^2 + y^2 = z^2.$ The proposition follows from a modest computation. $\spadesuit$
For fun, one can check the solutions quadratic polynomial in the above proposition are invariant under the map $x \mapsto 2A / x$.