Is there a Pythagorean triple whose angles are 90, 45, and 45 degrees?

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.

• no, not integer lengths. Your triangle has lengths $(1,1,\sqrt 2)$ Jan 16 '18 at 1:22
• In the Wikipedia article Hippasus is the story of the 5th century BCE discovery of the irrationality of square root of two and other irrationals. Jan 16 '18 at 1:44
• No, but there are an infinite number that are close, eg (20, 21, 29) is a Pythagorean triple with the perpendicular legs almost equal, and the related (29, 29, 41) has an apex angle that's slightly less than 90°. You can find such relations from the continued fraction approximations to $\sqrt2$ Jan 16 '18 at 8:06
• @PM2Ring A slight expansion and that's a worthy answer. Legos click together with a tolerance, and once you're within that tolerance ... Jan 16 '18 at 8:13
• It was a big blow to the world view of the Pythagoreans when they discovered that this very constructible number (the diagonal of a square) that thus, to them, clearly existed, wasn't in a whole number ratio to the sides. Jan 16 '18 at 14:48

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.

• Is that sqrt(2*a) or sqrt(2)*a? It's hard to tell due to the font. Jan 16 '18 at 1:37
• Sorry, it's $\sqrt{2} \times a$, or more conventionally, $a\sqrt{2}$. Jan 16 '18 at 1:37
• And critically, sqrt(2) is irrational, and cannot be represented by a fraction. Jan 16 '18 at 20:14
• For completeness, $a\sqrt 2$ could be rational - for example, if $a$ were $\sqrt 2$. Of course, in that case, the sides would be irrational. It would be more correct to say the "either the hypotenuse or the sides must be of irrational length" Jan 16 '18 at 21:13

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.

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.

• Probably a typo: I guess it's a + b, a + b, 2a + b (12=5+7,17=2*5+7)? Jan 17 '18 at 7:56
• @Christoph: Thanks. I had messed up an edit putting it in nicer form Jan 17 '18 at 14:45