A triangle $ABC$ with positive integer side lengths has perimeter $35$, centroid $G$, and incenter $I$. If $\angle{GIC} = 90^{\circ}$, what is the length of $\overline{AB}$?
We use barycentric coordinates with reference triangle $\triangle{ABC}$. Letting $k=a+b+c$, we have the displacement vectors $\vec{GI} = \left(\frac{1}{3}-\frac{a}{k}, \frac{1}{3}-\frac{b}{k}, \frac{1}{3}-\frac{c}{k}\right)$ and $\vec{IC} = \left(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}-1\right)$. EFFT gives us$$a^2\left(\frac{4b+c}{3k}-\frac{1}{3}-\frac{2bc}{k^2}\right) + b^2\left(\frac{4a+c}{3k}-\frac{1}{3}-\frac{2ac}{k^2}\right) + c^2\left(\frac{a+b}{3k}-\frac{2ab}{k^2}\right) = 0$$. Multiplying by $3k^2$, writing $k=a+b+c$, and expanding out, the equation turns into$$a^2((a+b+c)(3b-a)-6bc)+b^2((a+b+c)(3a-b)-6ac)+c^2((a+b)(a+b+c)-6ab)=0.$$Expanding further and factoring an $a+b+c$ out, we have$$(a+b+c)((a+b)(4ab-a^2-b^2+c^2)-6abc) = 0.$$Notice that when $c=a+b$ the term $(a+b)(4ab-a^2-b^2+c^2)-6abc)$ vanishes, so we factor further to obtain$$(a+b+c)(c-a-b)(a^2+b^2+c(a+b)-4ab)=0\implies a^2+b^2+c(a+b)-4ab = 0\implies (a+b)^2+c(a+b) = 6ab\implies 35(35-c) = 6ab.$$From the triangle inequality, we have $c\leq 17$ and we also know $6|35-c$; it's now easy to find a triple $(a, b, c) = (10, 14, 11)$ satisfying the equation above, and the answer is $\boxed{11}$.
Any better solution than this?
This answer is based on expressions for $|CI|,|GI|,|CG|$ in terms of semiperimeter $\rho$ and radii of inscribed $r$ and circumscribed $R$ circle and the side length $|AB|=c$:
\begin{align} |CI|^2&=4rR\,\Big(\frac{\rho}c-1\Big) \tag{1}\label{1} ,\\ |GI|^2&= \tfrac19\,(\rho^2+5r^2-16rR) \tag{2}\label{2} ,\\ |CG|^2&= \tfrac19\,(4\,(\rho^2-\,r^2-4\,r\,R)-3\,c^2) \tag{3}\label{3} . \end{align}
From \eqref{2}
\begin{align} 4rR&= \tfrac14\,(\rho^2+5r^2-9\,|GI|^2) \tag{4}\label{4} , \end{align}
thus
\begin{align} |CG|^2&= \tfrac13\,\rho^2-r^2+|GI|^2-\tfrac13\,c^2 \tag{5}\label{5} , \end{align}
The cubic equation for the side length $a,b,c$ in terms of $\rho,r$ and $R$ is \begin{align} x^3-2\rho\,x^2+(\rho^2+r^2+4rR)\,x-4\rho r R&=0 \tag{6}\label{6} ,\\ \text{so}\quad c^3-2\rho\,c^2+(\rho^2+r^2+4rR)\,c-4\rho r R&=0 \tag{6a}\label{6a} , \end{align} substitution of \eqref{4} into \eqref{6a} gives
\begin{align} c^3-2\rho\,c^2+(\rho^2+r^2+\tfrac14\,(\rho^2+5r^2-9\,|GI|^2))\,c -\tfrac14\,\rho\,(\rho^2+5r^2-9\,|GI|^2)&=0 \tag{7}\label{7} , \end{align}
and we can express $r^2$ in terms of $\rho,|GI|$ and $c$:
\begin{align} r^2&= \frac{(\rho-c)(9\,|GI|^2-(\rho-2c)^2)}{5\,\rho-9\,c} \tag{8}\label{8} . \end{align}
Using the perpendicularity condition $|CG|^2=|CI|^2+|CG|^2$, after all substitutions and simplifications, we end up with cubic equation in $c$: \begin{align} (c-\rho)(6 c^2-7\rho\,c+27\,|GI|^2+2\,\rho^2)&=0 \tag{9}\label{9} . \end{align}
Solution $c=\rho$ means a degenerate triangle, so we need to consider just two expressions for $c$:
\begin{align} c_-&= \tfrac1{12}\Big(7\,\rho-\sqrt{\rho^2-648\,|GI|^2}\Big) \tag{11}\label{11} ,\\ c_+&= \tfrac1{12}\Big(7\,\rho+\sqrt{\rho^2-648\,|GI|^2}\Big) \tag{10}\label{10} . \end{align} Given $\rho=\tfrac{35}2$,
\begin{align} c_-&= \frac{245-\sqrt{1225-2592\,|GI|^2}}{24} \tag{12}\label{12} ,\\ c_+&= \frac{245+\sqrt{1225-2592\,|GI|^2}}{24} \tag{13}\label{13} . \end{align}
Equation \eqref{12} provides the range
\begin{align} c_-&\in (\tfrac{35}4,\tfrac{245}{24}) = (8.75,10.20833)=[9,10] \end{align}
and another one provides the range \begin{align} c_+&\in (\tfrac{245}{24},\tfrac{35}3) = (10.20833,11.67)=[11,11] , \end{align}
so we need to test just three integer values $c=9,10,11$.
And indeed, the only suitable value is $c=11$, for which
\begin{align} |GI|^2 &= \tfrac13 ,\quad r^2=\frac{39}4 ,\quad 4\,r\,R=88 , \end{align}
the cubic equation \eqref{6} for side lengths becomes \begin{align} x^3-35 x^2+404x-1540&=0 ,\\ \text{or}\quad (x-11)(x^2-24x+140)&=0 ,\\ \end{align}
and the other two sides are $10$ and $14$ units long.