SOLUTION: if x and y are positive numbers with x>y, show that a triangle with sides of lengths 2xy, x^2-y^2, and x^2+y^2 is always a right triangle

Question 430318: if x and y are positive numbers with x>y, show that a triangle with sides of lengths 2xy, x^2-y^2, and x^2+y^2 is always a right triangle
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
We know that must be the longest side of the triangle (it cannot be , I will demonstrate why afterwards). By the Pythagorean theorem,

. Therefore this is a right triangle.

The only "assumption" we had to make was that , but this is easy to prove. By the AM-GM inequality, . Multiplying both sides by 2, . Equality occurs only when , but this cannot be true, so is strictly greater than .

RELATED QUESTIONS
if x and y are positive numbers with x>y, show that a triangle with sides of lengths... (answered by richard1234)

1.The sum of two positive real numbers is 7. The sum of their squares is 26.1/2. Find the (answered by ankor@dixie-net.com)

show that ((x^2y)/2(x+y))+ ((x+y)/2xy)-(x)^1/2 >= 0 for positive real numbers x and... (answered by josmiceli)

If X and Y are integers, what is always even? A) x+y /2 B) 2(x+y/2) C)2(x+y) D) x^2... (answered by Theo,greenestamps)

Pythagorean triples are given by the formulas x^2 - y^2, 2xy, and x^2 + y^2. Use the... (answered by Fombitz)

Show that if x and y are positive rational numbers with x < y then there is a rational... (answered by ikleyn)

show that for any integers x and y :2xy<= x^2 +y^2 (answered by ikleyn)

Let x and y be real numbers. Find the maximum value of (x + y)^2, if x and y satisfy x^2 (answered by CPhill,ikleyn,Edwin McCravy)

if x and y are positive real numbers, use the fact that t+(1/t)>=2 for positive numbers... (answered by venugopalramana)