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==
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==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==
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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>
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~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

Problem

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 $24+28 = \boxed{052}.$

~kevinmathz~

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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