Difference between revisions of "2001 AMC 8 Problems/Problem 23"
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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20</math> | <math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20</math> | ||
− | ==Solution | + | ==Solution 1== |
There are <math> 6 </math> points in the figure, and <math> 3 </math> of them are needed to form a triangle, so there are <math> {6\choose{3}} =20 </math> possible triples of <math> 3 </math> of the <math> 6 </math> points. However, some of these created congruent triangles, and some don't even make triangles at all. | There are <math> 6 </math> points in the figure, and <math> 3 </math> of them are needed to form a triangle, so there are <math> {6\choose{3}} =20 </math> possible triples of <math> 3 </math> of the <math> 6 </math> points. However, some of these created congruent triangles, and some don't even make triangles at all. | ||
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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 | + | ==Solution 2== |
We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is <math>3+1 = \boxed{4}</math>, which is <math>\boxed{\text{D}}</math> | We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is <math>3+1 = \boxed{4}</math>, which is <math>\boxed{\text{D}}</math> | ||
-FIREDRAGONMATH16 | -FIREDRAGONMATH16 | ||
+ | |||
+ | ==Solution 3 (Multiple-choice abuse)== | ||
+ | |||
+ | Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: <math>\triangle RYX</math>, <math>\triangle RYT</math>, <math>\triangle RYZ</math>, <math>\triangle RST</math>. So the answer is <math>\boxed{4}</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 09:58, 15 December 2020
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 1
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
We can do casework in this problem like the solution above, but that would take too much time. Instead, we see that we can split this equilateral dot triangle to two halves by dropping an altitude from the top vertex of the big triangle. Using the smaller triangle we won't have to worry about extra unneeded cases. We can see that there are three distinct triangles in the half, and combining this with the larger equilateral triangle our answer is , which is
-FIREDRAGONMATH16
Solution 3 (Multiple-choice abuse)
Notice that 20 is obviously too high (There are only 20 triangles in total!) and you can count 4 distinct triangles quickly: , , , . So the answer is
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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