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2008 Mock ARML 1 Problems/Problem 3

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Problem

In regular hexagon ABCDEF with side length 1, AD intersects BF at G, and BD intersects EC at H. Compute the length of GH.

Solution

pointpen = black; pathpen = black + linewidth(0.62);pair v(int n){ return dir(n * 60); }D(MP("A",v(0))--MP("B&...

Let H' be the foot of the perpendicular from H to \overline{AD}. Since \angle CDA is an inscribed angle with measure \frac{120}{2} = 60^{\circ}, it follows that \triangle CDH' is a 30-60-90 \triangle, and DH' = \frac{1}{2} and CH' = BG = \frac{\sqrt{3}}{2}. Also, H'G = CB = 1. Note that \triangle DH'H \sim \triangle DGB by ratio 1/3. Thus HH' = BG/3 = \frac{\sqrt{3}}{6}.

By the Pythagorean Theorem, HG^2 = H'H^2 + H'H^2 = \left(\frac{\sqrt{3}}{6}\right)^2 + 1 = \frac{13}{12}. Thus HG = \boxed{\frac{\sqrt{39}}{6}}.

See also

2008 Mock ARML 1 (Problems, Source)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8
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Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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