2018 AIME I Problems/Problem 7

Revision as of 17:48, 7 March 2018 by Expilncalc (talk | contribs) (Added solution)

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.

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