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2003 AIME I Problems/Problem 6

Revision as of 18:35, 27 December 2021 by Fuzimiao2013 (talk | contribs) (Fixed the cubes to not just be three black hexagons (which draw(unitcube) inconveniently does))
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Problem

The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers. Find $m + n + p.$

Solution

[asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,1,1)--(1,0,0)--(1,1,0)--(1,1,1), blue); [/asy]

[asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((0,1,1)--(1,0,0)--(1,1,0)--(0,1,1), blue); [/asy]

[asy] size(120); import three; unitsize(1cm); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1)); draw((0,1,1)--(0,1,0)); draw((1,1,1)--(1,1,0)); draw((1,0,1)--(1,0,0)); draw((1,0,0)--(0,1,0)--(1,1,1)--(1,0,0), blue); [/asy]

Since there are $8$ vertices of a cube, there are ${8 \choose 3} = 56$ total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal), those which lie in a plane perpendicular to one face of the cube (whose sides are one edge, one face diagonal and one space diagonal of the cube) and those which lie in a plane oblique to the edges of the cube, whose sides are three face diagonals of the cube.

Each face of the cube contains ${4\choose 3} = 4$ triangles of the first type, and there are $6$ faces, so there are $24$ triangles of the first type. Each of these is a right triangle with legs of length $1$, so each triangle of the first type has area $\frac 12$.

Each edge of the cube is a side of exactly $2$ of the triangles of the second type, and there are $12$ edges, so there are $24$ triangles of the second type. Each of these is a right triangle with legs of length $1$ and $\sqrt 2$, so each triangle of the second type has area $\frac{\sqrt{2}}{2}$.

Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are $8$ vertices of the cube, so there are $8$ triangles of the third type. Each of the these is an equilateral triangle with sides of length $\sqrt 2$, so each triangle of the third type has area $\frac{\sqrt 3}2$.

Thus the total area of all these triangles is $24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}$ and the answer is $12 + 288 + 48 = \boxed{348}$.

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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