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Difference between revisions of "2008 Mock ARML 1 Problems/Problem 3"

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== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 17:33, 29 May 2008

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

[asy] pointpen = black; pathpen = black + linewidth(0.62); pair v(int n){ return dir(n * 60); } D(MP("A",v(0))--MP("B",v(1),N)--MP("C",v(2),N)--MP("D",v(3),SW)--MP("E",v(4))--MP("F",v(5))--cycle); D(v(0)--v(3));D(v(1)--v(5));D(v(1)--v(3));D(v(2)--v(4)); D(D(MP("G",IP(v(0)--v(3),v(1)--v(5)),NE))--D(MP("H",IP(v(1)--v(3),v(2)--v(4)),NW)),dashed); D(MP("H'",IP(v(2)--v(4),v(0)--v(3)),SW)); [/asy]

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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