2018 AIME I Problems/Problem 7

Revision as of 01:14, 31 March 2018 by Unionpacificboy (talk | contribs) (Solution 3)


A right hexagonal prism has height $2$. The bases are regular hexagons with side length $1$. Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).

Solution 1

We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases.

Case 1: vertices are on one base. Then we can call one of the vertices $A$ for distinction. Either the triangle can have sides $1, 1, \sqrt{3}$ with 6 cases or $2, 2, 2$ with 2 cases. This can be repeated on the other base for $16$ cases.

Case 2: The vertices span two bases. WLOG call the only vertex on one of the bases $X$. Call the closest vertex on the other base $B$, and label clockwise $C, D, E, F, G$. We will multiply the following scenarios by $12$, because the top vertex can have $6$ positions and the top vertex can be on the other base. We can have $XCG, XDF$, but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is $2$ and the height is $2$, so $XBE$ is also correct! Those are the only three cases, so there are $12*3=36$ cases for this case.

In total there's $\boxed{052}$ cases.

Solution 2

If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \cdot 2\cdot 6 = 28$ ways. Then, if one length was $\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\sqrt{3}$ side, and two ways to put the other point. $2 \cdot 12 = 24$ for four ways to put the third point. Adding the number up, the final answer is $24+28 = \boxed{052}.$


Solution 3

To start, let's find the distances from any vertex (call it A, doesn't matter since the prism is symmetrical) to all the other 11 vertices. Using Pythagorean relations, we find that the distances are $1$, $\sqrt{3}$, $2$, $\sqrt{3}$, $1$, $\sqrt{7}$, $\sqrt{5}$, $2$, $\sqrt{5}$, $\sqrt{7}$, and $\sqrt{8}$.

We can clearly form an isosceles triangle using any vertex and any two neighboring edges that are equal. There are 12 total vertices, all of which are symmetrical to the vertex A. Thus, for each vertex, we can form 5 isosceles triangles, so have so far $12 \times 5 = 60$ isosceles triangles.

To check for overcounting, note that isosceles triangles can be split into two major categories: equilateral isosceles, and non-equilateral isosceles. We only counted non-equilateral isosceles triangles once since there is only one vertex whose two neighboring edges are equal. But equilateral triangles were counted three times (namely once for each vertex).

By observation, there are only 4 equilateral triangles (1-1-1 side lengths), two on each hexagonal face. Since we counted each two more times than we should of (three instead of once), we will subtract ($4$ eq. triangles) $\times$ ($2$ overcounts per eq. triangle) = $8$ overcounts.

Finally, we have $60 - 8$ overcounts = $\boxed{52}$ isoceles triangles.

See also

2018 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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