Is there a Pythagorean triple (a.k.a. an integer triangle) whose angles are 90, 45 and 45 degrees? I am trying to connect LEGO roads at angles other than the standard 90 degrees.
Is there a Pythagorean triple whose angles are 90, 45, and 45 degrees?
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No, since if the perpendicular sides are $a$ in length, the hypotenuse would be $a\sqrt2$. But $\sqrt2$ is irrational, so $a\sqrt2$ is not an integer.
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In your context you might be interested in isosceles triangles that are almost right. As others have said, a right isosceles triangle has sides that are $a,a,a\sqrt 2$ and as $\sqrt 2$ is not rational we cannot have an integer sided one. However, if we find a rational number that is close to $\sqrt 2$ we can find isosceles triangles that are close to right. We have $\sqrt 2 \approx 1.414213$, while $\frac 75 = 1.4$ is not so far away, so a $5,5,7$ triangle is close to right. In fact the angle is $\arccos \left(\frac 1{50}\right)\approx 88.85^\circ$. You might have enough give to tolerate that. If not, given one triangle in the list is $a,a,b$ the next is $a+b,a+b,2a+b$, so the next is $12,12,17$, then $29,29,41$, and so on. The get closer and closer to right as you progress. If you are interested in where this comes from, you could look up Pell's equation.
You cannot have an integer Pythagorean Triple whose angles are $45°, 45°$ and $90°$.
Assume on the triangle we have sides $a$. Then by Pythagoras' Theorem,
$$a^2+a^2=2a^2=(a\sqrt{2})^2$$
This means the hypotenuse is no longer an integer length, because now it measures $a\sqrt2$. This means no such Pythagoren Triple exists.