Geometry: Determining the length of a side of a triangle

Question

Triangle $ABC$ has all sides of integral length. Vertex $A$ is at $(0,0)$, $B$ lies on the line joining $(0,0)$ and $(3,6)$ and $C$ lies on the line joining $(0,0)$ and $(2,-1)$. Two of the three sides of the triangle are primes, whose difference is $50$. Find the minimum possible value of the third side.

Answer 1

First, notice that you are talking about a right angle triangle: just shift the two axis and consider $B$ and $C$ belonging to $x$ and $y$ axis respectively.

Let's call $z$ the length of $BC$. It is the hypothenuse of the triangle.

$z^2=x^2+y^2$

Since two out of the three number are prime, it falls that they are all coprimes (otherwise a common divisor of two numbers would divide also the third, which is forbidden by definition).

$x$ and $y$ are coprime thus at least one is odd.

If they were both odd, then $x^2+y^2$ would be even and $z$ as well (square have same parity as number), thus $4$ would divide $z^2$; but $z^2=(2t+1)^2+(2u+1)^2=4(t^2+u^2+t+u)+2$, which is impossible.

Thus $x$ is odd and $y$ even (or vice-versa) and also $z$ is odd by construction.

The two numbers with same parity are the one separated by $50$ and are the two prime numbers.

$z=x+50$

The third number is $y=\sqrt{2500+100x}$

Brute force give you

$x=3$ then $z=53$ then $y=52.9...$ NOK!

$x=5$ then $z=55=5\times 11$ NOK!

$x=7$ then $z=57=3\times 19$ NOK!

$x=11$ then $z=61$ then $y=60$ OK!