# 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 $(1,1,\sqrt 2)$ – Will Jagy 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. – Somos 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 $\sqrt2$ – PM 2Ring 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 ... – Neil_UK 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. – Arthur 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. – posfan12 Jan 16 '18 at 1:37
• Sorry, it's $\sqrt{2} \times a$, or more conventionally, $a\sqrt{2}$. – Landuros Jan 16 '18 at 1:37
• And critically, sqrt(2) is irrational, and cannot be represented by a fraction. – Arcanist Lupus 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" – AFischbein 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)? – Christoph Jan 17 '18 at 7:56
• @Christoph: Thanks. I had messed up an edit putting it in nicer form – Ross Millikan Jan 17 '18 at 14:45