2001 AMC 8 Problems/Problem 24
Contents
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?
Solution
Each half has red triangles, blue triangles, and white triangles. There are also pairs of red triangles, so red triangles on each side are used, leaving red triangle, blue triangles, and white triangles remaining on each half. Also, there are pairs of blue triangles, using blue triangles on each side, so there is red triangle, blue triangles, and white triangles remaining on each half. Also, we have red-white pairs. This obviously can't use red triangles on one side, since there is only on each side, so we must use red triangle and white triangle per side, leaving blue triangles and white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than blue pairs, so the remaining blue triangles must be paired with white triangles, yielding blue-white pairs, one for each of the remaining blue triangles. This uses blue triangles and white triangles on each side, leaving white triangles apiece, which must be paired with each other, so there are white-white pairs, .
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 | |
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All AJHSME/AMC 8 Problems and Solutions |
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