Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

A convex polyhedron has nn faces that are all congruent triangles with angles 36,7236^{\circ}, 72^{\circ}, and 7272^{\circ}. Determine, with proof, the maximum possible value of nn.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Answer: 36 Solution: Consider such a polyhedron with VV vertices, EE edges, and F=nF=n faces. By Euler's formula we have V+F=E+2V+F=E+2. Next, note that the number of pairs of incident faces and edges is both 2E2E and 3F3F, so 2E=3F2E=3F. Now, since our polyhedron is convex, the sum of the degree measures at each vertex is strictly less than 360=3610360=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 369=32436 \cdot 9=324. On the other hand, the total degree measure at all vertices is the total degree measure over all faces, which is 180F180F. Thus we have 180F324V180F \leq 324V, or 10F18V10F \leq 18V. Putting our three conditions together, we have 10F18V=18(E+2F)=9(2E)+3618F=9(3F)+3618F=9F+36 10F \leq 18V=18(E+2-F)=9(2E)+36-18F=9(3F)+36-18F=9F+36 Thus F36F \leq 36. F=36F=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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