Problem
NT-B2-M07-P014 A Pythagorean Triangle With Fixed Perimeter
#14
★★★★☆ Level 4 of 5
Find all right triangles with integer sides and perimeter \(60\).
Let the legs be \(x,y\); then the hypotenuse is \(60-x-y\). Substitute into the Pythagorean theorem.
We have \(x^2+y^2=(60-x-y)^2\). Expanding and simplifying gives \(xy-60x-60y+1800=0\), so \((x-60)(y-60)=1800\). Since \(x,y<60\), put \(a=60-x\), \(b=60-y\). Then \(ab=1800\), and the hypotenuse is \(a+b-60\), so we need \(a<60\), \(b<60\), \(a+b>60\). The suitable divisor pairs are \((36,50)\) and \((40,45)\), with swaps. We get the side triples \(24,10,26\) and \(20,15,25\). The answer is \((10,24,26)\) and \((15,20,25)\).
No diagram is needed: despite the geometric wording, the problem is arithmetic.