High School Pythagorean triples that sum to 60

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Pythagorean triples that sum to 60 can be derived from known triples like 3-4-5 and 5-12-13 by scaling. To ensure all possible triples are found, the equation x^2 + y^2 = z^2, along with the constraint x + y + z = 60, can be utilized. By substituting z into the first equation, a new equation emerges: 60x + 60y - xy = 1800. This can be rearranged to (60 - x)(60 - y) = 1800, allowing for the identification of positive factor pairs of 1800 that are both less than 60. This method guarantees the discovery of all Pythagorean triples summing to 60.
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I'm trying to find pythagorean triples that sum to 60. Just from memory, I kow that 3-4-5 and 5-12-13, scaled to some factor, will give triples that sum to 60. These seem to be the only ones that sum to 60, but how can I be sure that there aren't more triples that sum to 60?
 
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What do you mean by "sum up"? I don't see the ##60##.
You can find all by ##x = u^2 - v^2 \; , \; y = 2uv \; , \; z = u^2 + v^2##.
 
Let x,y,z be the side lengths of a triangle you describe, so that x^2+y^2=z^2 and x+y+z=60. You can solve for z in the last equation and plug it into the first, getting x^2+y^2=(60-(x+y))^2=3600-120(x+y)+x^2+2xy+y^2. This gives the equation 60x+60y-xy=1800, which you can rearrange into (60-x)(60-y)=1800. Now you just need to look for positive factor pairs of 1800 with both factors smaller than 60 (since 0<x,y<60).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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