Today I finished the 2016 AMC 10B.
My favorite problem was #19, which I thought was really, really cool! :)
Problem:
Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$. and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?
Solution 1:
Drop perpendiculars from $P,Q,E$, to the line $DC$, and let the foot of each altitude be $X, Y, Z$, respectively. By similar triangles, note that $\frac{PQ}{EF}=\frac{PQ_x}{EF_x}$, where the subscript $x$ refers to the $x$-distance between the two mentioned points. We know that $$EF_x = EZ = FC - ZC = FC - EB = (5-2) - (1) = 2$$
Now, using coordinate geometry, let $D=(0,0)$, $C=(5,0)$, $B=(5,4)$, and $A=(0,4)$. Forming the equations for the lines, we see that $$EF \longrightarrow y = 2x - 4$$
$$AC \longrightarrow y=\frac{-4}{5}x+4$$
$$AG \longrightarrow y=\frac{-3}{5}x+4$$
Setting the linear equation of $EF$ equal to that of $AC$ and $AG$, we see that the $x$-coordinates of $P,Q$ are $\frac{20}{7}$ and $\frac{40}{13}$. Thus $$PQ_x=\frac{40}{13}-\frac{20}{7}=\frac{20}{91}$$
So, we have $$\frac{PQ_x}{EF_x}=\frac{\frac{20}{91}}{2}=\boxed{\frac{10}{91}}$$
I'll upload a synthetic solution later 😉
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