8/25/2023

 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 $\frac{PQ}{EF}=\frac{P{Q}_x}{E{F}_x}$, where the subscript $x$ refers to the $x$-distance between the two mentioned points. We know that $$E{F}_x=EZ=FC-ZC=FC-EB=(5-2)-(1)=2$$

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 $$EF⟶y=2x-4$$

$$AC⟶y=\frac{-4}{5}x+4$$

$$AG⟶y=\frac{-3}{5}x+4$$

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 $$P{Q}_x=\frac{40}{13}-\frac{20}{7}=\frac{20}{91}$$

So, we have $$\frac{P{Q}_x}{E{F}_x}=\frac{\frac{20}{91}}{2}=\boxed{{\displaystyle \frac{10}{91}}}$$

I'll upload a synthetic solution later 😉