Difference between revisions of "2001 AMC 8 Problems/Problem 23"
Harsha12345 (talk | contribs) (→Solution) |
Harsha12345 (talk | contribs) (→easy solution) |
||
| Line 30: | Line 30: | ||
Note that it has to be less than 6C3 because some groups of three points make lines. | Note that it has to be less than 6C3 because some groups of three points make lines. | ||
One possible triangle is the one connecting all the midpoints, another is the original equilateral triangle, a third one is a right triangle and the fourth one is an obtuse triangle by connecting two midpoints and a vertex that is not the vertex between them. Because there are no more answer choices from 5-19, our answer is 4. | One possible triangle is the one connecting all the midpoints, another is the original equilateral triangle, a third one is a right triangle and the fourth one is an obtuse triangle by connecting two midpoints and a vertex that is not the vertex between them. Because there are no more answer choices from 5-19, our answer is 4. | ||
| + | -harsha12345 | ||
==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 08:22, 2 February 2019
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
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, 
easy solution
Note that it has to be less than 6C3 because some groups of three points make lines. One possible triangle is the one connecting all the midpoints, another is the original equilateral triangle, a third one is a right triangle and the fourth one is an obtuse triangle by connecting two midpoints and a vertex that is not the vertex between them. Because there are no more answer choices from 5-19, our answer is 4. -harsha12345
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. 
