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Problem 163

AMC 10/12, early questions

Geometry Difficulty 3.6 Find the answer Slovenija 2016 · Slovenia · 2016

Take an equilateral triangle $ABC$ with the side of length $a$ . Draw the squares $BADE$ and $CBFG$ on the sides $AB$ and $BC$ , respectively (see figure). How long is the line segment $DG$ ?

(A) $(\sqrt{2} + 1)a$
(B) $(2\sqrt{2} + 1)a$
(C) $(\sqrt{3} + 1)a$
(D) $\sqrt{3}a$
(E) 1

Your answer

Official solution

Denote the intersections of the segment $DG$ with the lines $AB$ and $BC$ by $K$ and $L$ respectively. Due to symmetry the segment $DG$ is parallel to the segment $AC$ , so $KBL$ is an equilateral triangle. The triangles $DKA$ and $LGC$ are one half of an equilateral triangle with altitudes of length $a$ and sides $|DK| = |GL| = \frac{2a}{\sqrt{3}}$ . The side of the equilateral triangle $KBL$ therefore measures
$|KL| = a - |AK| = a - \frac{|DK|}{2} = a - \frac{a}{\sqrt{3}}.$
The length of the segment $DG$ is

|DG| = 2|DK| + |KL| = \frac{4a}{\sqrt{3}} + \left(a - \frac{a}{\sqrt{3}}\right) = a + \frac{3a}{\sqrt{3}} = (1 + \sqrt{3})a.

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