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Difference between revisions of "2008 AMC 12B Problems/Problem 15"

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filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
 
filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
 
</asy>
 
</asy>
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The equilateral triangles form trapezoids with side lengths <math> 1, 1, 1, 2</math> (half a unit hexagon) on each face of the unit square.  The four triangles "in between" these trapezoids are isosceles triangles with two side lengths <math> 1</math> and an angle of <math> 30^{\circ}</math> in between them, so the total area of these triangles (which is the area of <math> S - R</math>) is, by the [[Law of Sines]], <math> 4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1</math> which makes the answer <math> \boxed{C}</math>.
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The equilateral triangles form trapezoids with side lengths <math> 1, 1, 1, 2</math> (half a unit hexagon) on each face of the unit square.  The four triangles "in between" these trapezoids are isosceles triangles with two side lengths <math> 1</math> and an angle of <math> 30^{\circ}</math> in between them, so the total area of these triangles (which is the area of <math> S - R</math>) is, <math> 4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1</math> which makes the answer <math> \boxed{C}</math>.
  
 
==See Also==  
 
==See Also==  

Revision as of 22:23, 31 December 2017

Problem

On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and $S$ be the smallest convex polygon that contains $R$. What is the area of the region that is inside $S$ but outside $R$?

$\textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$

Solution

[asy] real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0)); draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0)); filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9)); filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9)); filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9)); filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9)); [/asy] The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is the area of $S - R$) is, $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$ which makes the answer $\boxed{C}$.

See Also

Region with Squares and Triangles

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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

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