Difference between revisions of "2003 AMC 12B Problems/Problem 10"
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| + | ==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> | ||
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==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> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2003|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:38, 4 January 2014
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]](http://latex.artofproblemsolving.com/a/3/b/a3b1113e62996ab9b1c6c83bc4836bbb32070801.png)
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 
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| 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. 
