Difference between revisions of "2018 AIME I Problems/Problem 7"
Expilncalc (talk | contribs) (Added solution) |
Treetor10145 (talk | contribs) (Added Problem) |
||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | A right hexagonal prism has height <math>2</math>. The bases are regular hexagons with side length <math>1</math>. Any <math>3</math> of the <math>12</math> vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | ||
+ | |||
+ | ==Solution== | ||
We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases. | We can consider two cases: when the three vertices are on one base, and when the vertices are on two bases. | ||
Line 6: | Line 10: | ||
In total there's <math>\boxed{052}</math> cases. | In total there's <math>\boxed{052}</math> cases. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2018|n=I|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Revision as of 03:09, 8 March 2018
Problem
A right hexagonal prism has height . The bases are regular hexagons with side length . Any of the vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles).
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 for distinction. Either the triangle can have sides with 6 cases or with 2 cases. This can be repeated on the other base for cases.
Case 2: The vertices span two bases. WLOG call the only vertex on one of the bases . Call the closest vertex on the other base , and label clockwise . We will multiply the following scenarios by , because the top vertex can have positions and the top vertex can be on the other base. We can have , but we are not done! Don't forget that the problem statement implies that the longest diagonal in a base is and the height is , so is also correct! Those are the only three cases, so there are cases for this case.
In total there's cases.
See also
2018 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.