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HMMT Spring Team Round, 2021

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:
A convex polyhedron has $n$ faces that are all congruent triangles with angles $36^{\circ}$ , $72^{\circ}$ , and $72^{\circ}$ . Determine, with proof, the maximum possible value of $n$ .

Solution

Solution:
Consider such a polyhedron with $V$ vertices, $E$ edges, and $F = n$ faces. By Euler's formula we have $V + F = E + 2$ .

Next, note that the number of pairs of incident faces and edges is both $2E$ and $3F$ , so $2E = 3F$ .

Now, since our polyhedron is convex, the sum of the degree measures at each vertex is strictly less than $360 = 36 \cdot 10$ . As all angle measures of the faces of our polyhedron are divisible by $36$ , the maximum degree measure at a given vertex is $36 \cdot 9 = 324$ . On the other hand, the total degree measure at all vertices is the total degree measure over all faces, which is $180F$ . Thus we have $180F \leq 324V$ , or $10F \leq 18V$ .

Putting our three conditions together, we have
$10F \leq 18V = 18(E + 2 - F) = 9(2E) + 36 - 18F = 9(3F) + 36 - 18F = 9F + 36$
Thus $F \leq 36$ .

$F = 36$ is attainable by taking a $9$ -gon antiprism with a $9$ -gon pyramid attached on the top and the bottom. Thus the answer is $36$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.