Difference between revisions of "2003 AMC 12B Problems/Problem 10"

m (vertex not vertice)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 +
==Problem==
 
Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
 
Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
 
<center><asy>
 
<center><asy>
Line 25: Line 26:
 
==Solution==
 
==Solution==
 
Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of <math>\boxed{\text{(B) }2}</math>
 
Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of <math>\boxed{\text{(B) }2}</math>
 +
===Solution 2===
 +
Take <math>{5 \choose 2}</math> to realize there are 10 ways to choose 2 different triangles. Then divide by 5 for each vertex of a pentagon to get <math>\boxed{\text{(B) }2}</math>
 +
 +
==See Also==
 +
{{AMC12 box|year=2003|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:06, 3 February 2020

Problem

Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?

[asy] size(200); defaultpen(0.9); real r = 5/dir(54).x, h = 5 tan(54*pi/180); pair A = (5,0), B = A+10*dir(72), C = (0,r+h), E = (-5,0), D = E+10*dir(108); draw(A--B--C--D--E--cycle); label("\(A\)",A+(0,-0.5),SSE); label("\(B\)",B+(0.5,0),ENE); label("\(C\)",C+(0,0.5),N); label("\(D\)",D+(-0.5,0),WNW); label("\(E\)",E+(0,-0.5),SW); // real l = 5*sqrt(3); pair ab = (h+l)*dir(72), bc = (h+l)*dir(54); pair AB = (ab.y, h-ab.x), BC = (bc.x,h+bc.y), CD = (-bc.x,h+bc.y), DE = (-ab.y, h-ab.x), EA = (0,-l); draw(A--AB--B^^B--BC--C^^C--CD--D^^D--DE--E^^E--EA--A, dashed); // dot(A); dot(B); dot(C); dot(D); dot(E); dot(AB); dot(BC); dot(CD); dot(DE); dot(EA); [/asy]

$\text {(A) } 1 \qquad \text {(B) } 2 \qquad \text {(C) } 3 \qquad \text {(D) } 4 \qquad \text {(E) } 5$

Solution

Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\boxed{\text{(B) }2}$

Solution 2

Take ${5 \choose 2}$ to realize there are 10 ways to choose 2 different triangles. Then divide by 5 for each vertex of a pentagon to get $\boxed{\text{(B) }2}$

See Also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png