Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 8 Problems/Problem 23"

(→‎easy solution)
m (→‎Solution 3 (Elimination))
 
(19 intermediate revisions by 10 users not shown)
Line 15: Line 15:
 
<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 (Complementary Counting)==
−
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 triplets of the <math> 6 </math> points. However, some of these created congruent triangles, and some don't even make triangles at all.
  
 
'''Case 1: Triangles congruent to <math> \triangle RST </math>''' There is obviously only <math> 1 </math> of these: <math> \triangle RST </math> itself.
 
'''Case 1: Triangles congruent to <math> \triangle RST </math>''' There is obviously only <math> 1 </math> of these: <math> \triangle RST </math> itself.
Line 27: Line 27:
  
 
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.  
+
==Solution 2 (Brute Force)==
−
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
+
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 by dropping an altitude from the top vertex of the big triangle. 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 <math>3+1 = \boxed{4}</math>, which is <math>\boxed{\text{D}}</math>
 +
 
 +
-FIREDRAGONMATH16
 +
 
 +
==Solution 3 (Elimination)==
 +
 
 +
Notice that 20 is obviously too high (There are only 20 ways to choose 3 of the points to form a triangle or a line in total!) and you can count 4 distinct triangles quickly: <math>\triangle RYX</math>, <math>\triangle RYT</math>, <math>\triangle RYZ</math>, <math>\triangle RST</math>. So the answer is <math>\boxed{\text{(D) } 4}</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}}
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 15:35, 23 July 2023

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 1 (Complementary Counting)

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 triplets 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 (Brute Force)

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 by dropping an altitude from the top vertex of the big triangle. 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 $3+1 = \boxed{4}$, which is $\boxed{\text{D}}$

-FIREDRAGONMATH16

Solution 3 (Elimination)

Notice that 20 is obviously too high (There are only 20 ways to choose 3 of the points to form a triangle or a line in total!) and you can count 4 distinct triangles quickly: $\triangle RYX$, $\triangle RYT$, $\triangle RYZ$, $\triangle RST$. So the answer is $\boxed{\text{(D) } 4}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_23&oldid=196004"