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Difference between revisions of "2001 AMC 8 Problems/Problem 24"

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<math>\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9</math>
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<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9</math>
  
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==Solution==
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Each half has <math> 3 </math> red triangles, <math> 5 </math> blue triangles, and <math> 8 </math> white triangles. There are also <math> 2 </math> pairs of red triangles, so <math> 2 </math> red triangles on each side are used, leaving <math> 1 </math> red triangle, <math> 5 </math> blue triangles, and <math> 8 </math> white triangles remaining on each half. Also, there are <math> 3 </math> pairs of blue triangles, using <math> 3 </math> blue triangles on each side, so there is <math> 1 </math> red triangle, <math> 2 </math> blue triangles, and <math> 8 </math> white triangles remaining on each half. Also, we have <math> 2 </math> red-white pairs. This obviously can't use <math> 2 </math> red triangles on one side, since there is only <math> 1 </math> on each side, so we must use <math> 1 </math> red triangle and <math> 1 </math> white triangle per side, leaving <math> 2 </math> blue triangles and <math> 7 </math> white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than <math> 3 </math> blue pairs, so the remaining blue triangles must be paired with white triangles, yielding <math> 4 </math> blue-white pairs, one for each of the remaining blue triangles. This uses <math> 2 </math> blue triangles and <math> 2 </math> white triangles on each side, leaving <math> 5 </math> white triangles apiece, which must be paired with each other, so there are <math> 5 </math> white-white pairs, <math> \boxed{\text{B}} </math>.
  
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==Solution==
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==Video Solutions==
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Each half has <math> 3 </math> red triangles, <math> 5 </math> blue triangles, and <math> 8 </math> white triangles. There are also <math> 2 </math> pairs of red triangles, so <math> 2 </math> red triangles on each side are used, leaving <math> 1 </math> red triangle, <math> 5 </math> blue triangles, and <math> 8 </math> white triangles remaining on each half. Also, there are <math> 3 </math> pairs of blue triangles, using <math> 3 </math> blue triangles on each side, so there is <math> 1 </math> red triangle, <math> 2 </math> blue triangles, and <math> 8 </math> white triangles remaining on each half. Also, we have <math> 2 </math> red-white pairs. This obviously can't use <math> 2 </math> red triangles on one side, since there is only <math> 1 </math> on each side, so we must use <math> 1 </math> red triangle and <math> 1 </math> white triangle per side, leaving <math> 2 </math> blue triangles and <math> 7 </math> white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than <math> 3 </math> blue pairs, so the remaining blue triangles must be paired with white triangles, yielding <math> 4 </math> blue-white pairs, one for each of the remaining blue triangles. This uses <math> 2 </math> blue triangles and <math> 2 </math> white triangles on each side, leaving <math> 5 </math> white triangles apiece, which must be paired with each other, so there are <math> 5 </math> white-white pairs, <math> \boxed{\text{B}} </math>.
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https://www.youtube.com/watch?v=D5iOE6w2LMk
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-jchoi1267
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2001|num-b=23|num-a=25}}
 
{{AMC8 box|year=2001|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 11:56, 1 October 2023

Problem

Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?

[asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy]

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$

Solution

Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue triangles on each side, so there is $1$ red triangle, $2$ blue triangles, and $8$ white triangles remaining on each half. Also, we have $2$ red-white pairs. This obviously can't use $2$ red triangles on one side, since there is only $1$ on each side, so we must use $1$ red triangle and $1$ white triangle per side, leaving $2$ blue triangles and $7$ white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than $3$ blue pairs, so the remaining blue triangles must be paired with white triangles, yielding $4$ blue-white pairs, one for each of the remaining blue triangles. This uses $2$ blue triangles and $2$ white triangles on each side, leaving $5$ white triangles apiece, which must be paired with each other, so there are $5$ white-white pairs, $\boxed{\text{B}}$.

Video Solutions

https://www.youtube.com/watch?v=D5iOE6w2LMk -jchoi1267

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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