Question

In rectangle $ABCD,$ $AB=6$ and $BC=3.$ Point $E$ between $B$ and $C,$ and point $F$ between $E$ and $C$ are such that $BE=EF=FC.$ Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q,$ respectively.

The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t?$

$7$

$9$

$12$

$15$

$20$

Solution:

Since $BC=3$ and $BE=EF=FC$ , we have $BE=1$ and $BF=2$ . Also $AD=3$ .

From $\triangle APD\sim\triangle EPB$ , $\frac{BP}{BD}=\frac{BE}{AD+BE}=\frac14.$ From $\triangle AQD\sim\triangle FQB$ , $\frac{BQ}{BD}=\frac{BF}{AD+BF}=\frac25.$

Therefore $BP=\frac14BD$ , $PQ=\left(\frac25-\frac14\right)BD=\frac3{20}BD$ , and $QD=\frac35BD$ . Thus $BP:PQ:QD=\frac14:\frac3{20}:\frac35=5:3:12.$ The sum is $5+3+12=20$ .

Thus, the correct answer is E.

2016 AMC 10A Problem 19

Solution:

Problem 19 in Other Years