Difference between revisions of "2001 AMC 8 Problems/Problem 23"
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However, if we add these up, we accounted for only <math> 1+4+6+6=17 </math> of the <math> 20 </math> possible triplets. We see that the remaining triplets don't even form triangles; they are <math> SYR, RXT, </math> and <math> TZS </math>. Adding these <math> 3 </math> into the total yields for all of the possible triplets, so we see that there are only <math> 4 </math> possible non-congruent, non-degenerate triangles, <math> \boxed{\text{D}} </math> | However, if we add these up, we accounted for only <math> 1+4+6+6=17 </math> of the <math> 20 </math> possible triplets. We see that the remaining triplets don't even form triangles; they are <math> SYR, RXT, </math> and <math> TZS </math>. Adding these <math> 3 </math> into the total yields for all of the possible triplets, so we see that there are only <math> 4 </math> possible non-congruent, non-degenerate triangles, <math> \boxed{\text{D}} </math> | ||
+ | ==Solution 2== | ||
+ | Note that there are 4 cases in Solution 1. Each of these cases corresponds to a unique triangle. Thus there are <math> 4 </math> non congruent <math> \boxed{\text{D}} </math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=22|num-a=24}} | {{AMC8 box|year=2001|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:37, 22 December 2017
Contents
[hide]Problem
Points , and are vertices of an equilateral triangle, and points , and are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?
Solution
There are points in the figure, and of them are needed to form a triangle, so there are possible triples of of the points. However, some of these created congruent triangles, and some don't even make triangles at all.
Case 1: Triangles congruent to There is obviously only of these: itself.
Case 2: Triangles congruent to There are of these: and .
Case 3: Triangles congruent to There are of these: and .
Case 4: Triangles congruent to There are again of these: and .
However, if we add these up, we accounted for only of the possible triplets. We see that the remaining triplets don't even form triangles; they are and . Adding these into the total yields for all of the possible triplets, so we see that there are only possible non-congruent, non-degenerate triangles,
Solution 2
Note that there are 4 cases in Solution 1. Each of these cases corresponds to a unique triangle. Thus there are non congruent
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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