Can you correct Calvin's mistake?

By Ceva's Theorem the segments $AP$ , $BQ$ , $CR$ meet at a point if and only if:

$\dfrac{AR}{RB} \cdot \dfrac{BP}{PC} \cdot \dfrac{CQ}{QA} = 1$

Since $AP$ is a median, $BP = PC$ and we get $\dfrac{AR}{RB}\cdot\dfrac{CQ}{QA} = 1$ .

Now let the unknown lengths $BC, CA$ be $a, b$ , respectively. Since $CR$ is the angle bisector from $C$ , we have $\dfrac{AR}{RB} = \dfrac{AC}{CB} = \dfrac{b}a$ , and thus $\dfrac{b}a \cdot \dfrac{CQ}{QA} = 1$ .

On the other hand, $CQ+QA = b$ and combining the last two equations soon leads to $CQ = \dfrac{ab}{a+b}, QA = \dfrac{b^2}{a+b}$ . Now we use that $BQ$ is an altitude, so that by Pythagoras'Theorem, $BC^2 - CQ^2 = AB^2 - QA^2$ which leads to:

$(a-13)(a+13)(a+b) = b^2(a-b)$

Now, we need to find a solution to this equation in positive integers $a,b$ so that there exists a triangle with sides $13, a, b$ , and $13, a, b$ are all distinct. Note that any prime factor, $p$ , of $a + b$ must be a factor of $b^2(a-b)$ and hence such a prime must divide either $b$ or $a-b$ . This suggests that $a$ and $b$ might have common factors. On the other hand, if $p$ is a common factor of $a$ and $b$ , then $p^3$ divides $b^2(a-b) = (a-13)(a+13)(a+b)$ and unless $p = 13$ (which is easily seen to be impossible) $p$ cannot divide $a-13$ or $a + 13$ , so that $p^3$ divides $a +b$ .

This suggests that the common factors of $a$ and $b$ cannot be too large!

Specifically, if we try $p = 3$ , we need $27$ to divide $a + b$ . Let us try $a+b =27, a = 3x, b = 27-3x$ . Substituting this into our equation yields:

$(3x-13)(3x+13) = (9-x)^2(2x-9)$

which is correct for $x=4$ . Thus, $a=12, b=15$ is a solution and its unique.

Therefore $a+b=12+15=\boxed{27}$ .