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2008 AMC 12B Problems/Problem 15

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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 1

[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 $\boxed{C}$.

Solution 2

[asy] real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((a+b+a,a+b+a)--(a+b+a,-b)--(-b,-b)--(-b,a+a+b)--cycle,dashed); 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 area of the largest square is $(1+\sqrt{3})^2=4+2\sqrt{3}$. The area of region $R$ is $1+12\frac{1}{2}\frac{\sqrt{3}}{2}=1+3\sqrt{3}$. The total area of four small 45-45-90 triangles at corner is $4*\frac{1}{2}(\frac{\sqrt{3}+1-2}{2})^2=2-\sqrt{3}$. $S=4+2\sqrt{3}-(2-\sqrt{3})=2+3\sqrt{3}$, $S-R=1$.


~Bran Qin

See Also

Region with Squares and Triangles

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

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