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Difference between revisions of "2001 AMC 8 Problems/Problem 23"
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==Solution 2==
 
==Solution 2==
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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>
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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>
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==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 00:38, 23 December 2017

Problem

Points $R$, $S$ and $T$ are vertices of an equilateral triangle, and points $X$, $Y$ and $Z$ 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]

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 20$

Solution

There are $6$ points in the figure, and $3$ of them are needed to form a triangle, so there are ${6\choose{3}} =20$ possible triples of $3$ of the $6$ points. However, some of these created congruent triangles, and some don't even make triangles at all.

Case 1: Triangles congruent to $\triangle RST$ There is obviously only $1$ of these: $\triangle RST$ itself.

Case 2: Triangles congruent to $\triangle SYZ$ There are $4$ of these: $\triangle SYZ, \triangle RXY, \triangle TXZ,$ and $\triangle XYZ$.

Case 3: Triangles congruent to $\triangle RSX$ There are $6$ of these: $\triangle RSX, \triangle TSX, \triangle STY, \triangle RTY, \triangle RSZ,$ and $\triangle RTZ$.

Case 4: Triangles congruent to $\triangle SYX$ There are again $6$ of these: $\triangle SYX, \triangle SZX, \triangle TYZ, \triangle TYX, \triangle RXZ,$ and $\triangle RYZ$.

However, if we add these up, we accounted for only $1+4+6+6=17$ of the $20$ possible triplets. We see that the remaining triplets don't even form triangles; they are $SYR, RXT,$ and $TZS$. Adding these $3$ into the total yields for all of the possible triplets, so we see that there are only $4$ possible non-congruent, non-degenerate triangles, $\boxed{\text{D}}$

Solution 2

Note that there are 4 cases in Solution 1. Each of these cases corresponds to a unique triangle. Thus there are $4$ non congruent $triangles \boxed{\text{D}}$

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. AMC Logo.png

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