2001 AMC 8 Problems/Problem 23: Difference between revisions
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==Solution 2== | ==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> | 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> triangles \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 23:38, 22 December 2017
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
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
