2024 AMC 12A Problems/Problem 24

Problem

A $\textit{disphenoid}$ is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?

$\textbf{(A) }\sqrt{3}\qquad\textbf{(B) }3\sqrt{15}\qquad\textbf{(C) }15\qquad\textbf{(D) }15\sqrt{7}\qquad\textbf{(E) }24\sqrt{6}$

Solution 1 (Definition of disphenoid)

Notice that any scalene $\textit{acute}$ triangle can be the faces of a $\textit{disphenoid}$ . (See proof in Solution 2.)

As a result, we simply have to find the smallest area a scalene acute triangle with integer side lengths can take on. This occurs with a $4,5,6$ triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles, degenerate, or non-acute). For this triangle, the semiperimeter is $\frac{15}{2}$ , so by Heron’s Formula:

\begin{align*} A&=\sqrt{\frac{15}{2}\cdot\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}}\\ &=\sqrt{\frac{15^2\cdot7}{16}}\\ &=\frac{15}{4}\sqrt{7} \end{align*}

The surface area is simply four times the area of one of the triangles, or $\boxed{\textbf{(D) }15\sqrt{7}}$ .

~eevee9406

Solution 2 (Disphenoid in Box)

Let the side lengths of one face of the disphenoid be $a, b, c$ . By the definition of a disphenoid with scalene faces, opposite sides must be the same length. Then the disphenoid can be constructed in a rectangular box with dimensions $p, q, r$ such that $a, b, c$ are the $3$ different diagonal lengths of the faces of the box and no two sides are parallel (someone feel free to insert a diagram). Then we have the system

$p^2 + q^2 = a^2$ $p^2 + r^2 = b^2$ $q^2 + r^2 = c^2$

for positive integers $a, b, c$ and positive $p, q, r$ .

Solving for $p, q, r$ , we have

$p^2 = \frac{a^2 + b^2 - c^2}{2}$ $q^2 = \frac{a^2 - b^2 + c^2}{2}$ $r^2 = \frac{-a^2 + b^2 + c^2}{2}$

(Notice that, by law of cosines (or by Pythagorean Inequality), these (parallelepiped box side lengths squared) $p^2, q^2, r^2$ each have the same sign as the cosine of the angle at a vertex of a triangular face of the disphenoid. Since they are squares and so non-negative, every angle is non-obtuse. Further, since they are squares of positive side lengths, every angle is acute. So $a, b, c$ are the side lengths of an $\textit{acute}$ triangle.)

WLOG, let $a < b < c$ . For $a<4$ , $a^2$ is less than or equal to the gap between squares greater than $a^2$ , so all such triangles are non-acute and fail.

The next smallest case works: $4^2 + 5^2 > 6^2$ so $a, b, c = 4, 5, 6$ .

Using Heron's Formula, the minimum total surface area of the disphenoid is $\boxed{\textbf{(D) }15\sqrt{7}}$ .

Bonus: by Heron’s Formula, the area of an $x-1, x, x+1$ triangle is $x \sqrt{3 (x-2)(x+2)}/4 = x \sqrt{3 (x^2-2)}/4$ .

~babyhamster

(Acute triangle observations and bonus formula by oinava)

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2024 AMC 12A Problems/Problem 24

Contents

Problem

Solution 1 (Definition of disphenoid)

Solution 2 (Disphenoid in Box)

See also