A7. On the sides of an equilateral triangle, three squares are drawn. The sides of the squares that are parallel to the sides of the triangle are extended until they intersect. The three intersection points form another equilateral triangle. The length of the side of the original triangle is 1. What is the length of the side of the large equilateral triangle?
(A) $1+2 \sqrt{2}$
(B) $5-\frac{1}{2} \sqrt{3}$
(C) $3 \sqrt{2}$
(D) $1+2 \sqrt{3}$
(E) $2 \sqrt{6}$
A7. On the sides of an equilateral triangle, three squares are drawn. The sides of the squares that are parallel to the sides of the triangle are extended until they intersect. The three intersection points form another equilateral triangle. The length of the side of the original triangle is 1. What is the length of the side of the large equilateral triangle? (D) $1+2 \sqrt{3}$

Solution In $\triangle A B C$, $\angle A$ is half of $60^{\circ}$, so $30^{\circ}$. Furthermore, $\angle C$ is a right angle, so $\triangle A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $|B C|=1$. It is thus half of an equilateral triangle with side lengths of 2: $|A B|=2$. We now calculate the length of $A C$ using Pythagoras: $|A C|=\sqrt{2^{2}-1^{2}}=\sqrt{3}$. Therefore, the required length is $\sqrt{3}+1+\sqrt{3}$.

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