Difference between revisions of "2018 AIME I Problems/Problem 7"
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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). | 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== | + | ==Solution 1== |
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. | ||
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In total there's <math>\boxed{052}</math> cases. | In total there's <math>\boxed{052}</math> cases. | ||
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| + | ==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 <math>4+ 2 \cdot 2\cdot 6 = 28</math> ways. Then, if one length was <math>\sqrt{3}</math> but no side on the diameter, there would be twelve was to put the <math>\sqrt{3}</math> side, and two ways to put the other point. <math>2 \cdot 12 = 24</math> for four ways to put the third point. Adding the number up, the final answer is <math>24+28 = \boxed{052}.</math> | ||
| + | |||
| + | ~kevinmathz~ | ||
== See also == | == See also == | ||
{{AIME box|year=2018|n=I|num-b=6|num-a=8}} | {{AIME box|year=2018|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:30, 8 March 2018
Contents
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 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 
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.
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 
ways. Then, if one length was 
but no side on the diameter, there would be twelve was to put the side, and two ways to put the other point. 
for four ways to put the third point. Adding the number up, the final answer is 
~kevinmathz~
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 | ||
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