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> | ||
+ | == 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. | ||
==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 09:22, 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?
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.
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.