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. | ||
Revision as of 03:16, 2 March 2020
Contents
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?
![[asy] pair SS,R,T,X,Y,Z; SS = (2,2*sqrt(3)); R = (0,0); T = (4,0); X = (2,0); Y = (1,sqrt(3)); Z = (3,sqrt(3)); dot(SS); dot(R); dot(T); dot(X); dot(Y); dot(Z); label("$S$",SS,N); label("$R$",R,SW); label("$T$",T,SE); label("$X$",X,S); label("$Y$",Y,NW); label("$Z$",Z,NE); [/asy]](http://latex.artofproblemsolving.com/8/5/a/85a81732e12f5eaa1d40ce7d0ba5d9185110000f.png)
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. 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
