HMMT November

There exist a triangle with side lengths $4^{2}, 5^{2}, 6^{2}$, which has perimeter 77. If the sides have lengths $a^{2}, b^{2}, c^{2}$ with $0<a<b<c$, then $a^{2}+b^{2}>c^{2}$ by the triangle inequality. Therefore $(b-1)^{2}+b^{2} \geq a^{2}+b^{2}>c^{2} \geq(b+1)^{2}$. Solving this inequality gives $b>4$. If $b \geq 6$, then $a^{2}+b^{2}+c^{2} \geq 6^{2}+7^{2}>77$. If $b=5$, then $c \geq 7$ is impossible, while $c=6$ forces $a=4$, which gives a perimeter of 77.