# Three positive numbers a, b, c satisfy $a^2 + b^2 = c^2$; is it necessarily true that there exists a right triangle with side lengths a,b and c?

If so, how could you go about constructing it? If not, why not?

I am new to proofs and I was reading a book where they posed this question. I understand that if we are given any right triangle, the relation between sides a, b, c is given by pythagorus' theorem; although, given $a^2 + b^2 = c^2$, does it necessarily imply that there exists a triangle with sides a, b, c?

## 6 Answers

Suppose you draw a right triangle such that the two legs have lengths $a$ and $b$ respectively. This is clearly possible, since one can construct line segments of any integer length and can construct right angles. Now, the hypotenuse has to have some length $c$, and this $c$ must satisfy $$a^2+b^2=c^2$$ by the Pythagorean theorem. However, there is exactly one $c$ that satisfies this for any given $a$ and $b$.

You're right to be suspicious of the converse, but since the equation uniquely defines one parameter given the others, it turns out not to be a problem.

Yes, the triangle , $\{ (0, 0), (0, b), (a, 0) \}$.

• Maybe elaborate a bit and explain why this triangle has the given side lengths. – Anon Jun 18 '16 at 20:06

Yes, it does. You can construct a pair of perpendicular lines, then measure $a$ units from the intersection down one and $b$ units down the other. Draw the line between those two points and you have your triangle.

Take a right angle triangle with base length $a$ and height $b$. Then its hypotenuse, which we temporarily denote by $d$, should satisfy $d^2= a^2 + b^2$. Therefore, $d^2 = c^2$, which gives $d =c$ because they are both positive numbers.

If $a^2 + b^2 = c^2$, then:

• $(a+b)^2-c^2 = 2ab \ge 0$ and so $c \le a+b$.

• $(b-a)^2 -c^2 = -2ab \le 0$ and so $b-a \le c$, that is, $b \le a+c$.

• $(a-b)^2 -c^2 = -2ab \le 0$ and so $a-b \le c$, that is, $a \le b+c$.

This proves that there exists a triangle with sides $a,b,c$.

By the law of cosines, we have $c^{2}=a^{2}+b^{2}-2ab\cos \hat c$, and so $\cos \hat c=0$, that is, $\hat c = 90^\circ$. Therefore, the triangle is a right triangle.

Yes indeed the converse for Pythagoras is also true. When such a construction is made (SSS) the relation is satisfied. The vertices lie on a semicircle, and by Thale's thm a right angle is enclosed.