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2008 Mock ARML 1 Problems/Problem 3

Revision as of 21:26, 5 December 2020 by Yofro (talk | contribs) (→‎Solution 2)

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 1

[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 $HG = \boxed{\frac{\sqrt{39}}{6}}$.

Solution 2

We use coordinates. Consider the diagram in Solution 1. Let $D=(0,0)$. By 30-60-90 triangle ratios, we can get that $C=(\frac12, \frac{\sqrt3}{2})$, and $H'=(\frac12, 0)$. In addition, $B=(\frac32, \frac{\sqrt3}{2})$ and $E=(\frac12, -\frac{\sqrt3}{2})$. The line $\overline{CE}$ is represented by $x=\frac12$. The line $\overline{BD}$ is represented by $y=\frac{\sqrt3}{3}x$. The intersection ($H$) is then $(\frac12, \frac{\sqrt3}{6})$. We additionaly can get that $G=(\frac32,0)$. Then, by 

the distance formula, the length of $GH$ is $\sqrt{1^2+\left(\frac{\sqrt3}{6}\right)^2}=\boxed{\frac{\sqrt{39}}{6}}}$ (Error compiling LaTeX. Unknown error_msg).

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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