Difference between revisions of "2009 AMC 8 Problems/Problem 20"
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<math> \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9</math> | <math> \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9</math> | ||
+ | |||
+ | ==Solution== | ||
+ | Assume the base of the triangle is on the bottom four points, because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length <math>1</math>, there are <math>3</math> triangles. For a triangle with a base of length <math>2</math>, there are <math>3</math> triangles. For length <math>3</math>, there are <math>2</math>. In total, the number of non-congruent triangles is <math>3+3+2=\boxed{\textbf{(D)}\ 8}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=19|num-a=21}} | {{AMC8 box|year=2009|num-b=19|num-a=21}} |
Revision as of 17:33, 25 December 2012
Problem
How many non-congruent triangles have vertices at three of the eight points in the array shown below?
Solution
Assume the base of the triangle is on the bottom four points, because a congruent triangle can be made by reflecting the base on the top four points. For a triangle with a base of length , there are
triangles. For a triangle with a base of length
, there are
triangles. For length
, there are
. In total, the number of non-congruent triangles is
.
See Also
2009 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |