2024 AMC 12A Problem 24

A disphenoid exists (as the tetrahedron formed by the face-plane midpoints of a box) exactly when the common face triangle is acute, and its total surface area is $4$ times one face’s area. Write the integer side lengths as $u\lt v\lt w.$ If $u\le3,$ then $w\ge v+1$ and $u^2+v^2\le(v+1)^2$ (with equality only for $(u,v,w)=(3,4,5)$), so the triangle is not acute. Thus $u\ge4.$ The triangle $(4,5,6)$ is acute because $4^2+5^2\gt6^2.$ It also has the least possible area. The largest angle of any acute triangle is at least $60^\circ,$ so a candidate with $(u,v)\ne(4,5)$ has area at least $\tfrac12uv\sin60^\circ$ $\ge\tfrac12(4)(6)\tfrac{\sqrt3}{2}=6\sqrt3,$ which is greater than $\tfrac{15\sqrt7}{4},$ the area of $(4,5,6).$ By Heron’s formula with $s=\tfrac{15}2,$ that area is $\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}.$ The total surface area is $4\cdot\tfrac{15\sqrt7}{4}=15\sqrt7.$ Thus, the correct answer is D.