2016 AMC 10A Problem 19

Since $BC = 3$ is split into three equal parts, $BE = 1$ and $BF = 2$, while $AD = BC = 3$.

Because $BC \parallel AD$, triangles $PBE$ and $PDA$ are similar (vertical angles at $P$, alternate interior angles at $B$ and $D$). Hence
$$\frac{BP}{PD} = \frac{BE}{AD} = \frac13, \qquad\text{so}\qquad BP = \frac14 BD.$$

Likewise triangles $QBF$ and $QDA$ are similar, giving
$$\frac{BQ}{QD} = \frac{BF}{AD} = \frac23, \qquad\text{so}\qquad BQ = \frac25 BD.$$

Measure everything in twentieths of $BD$: $BP = \frac{5}{20}BD$, $BQ = \frac{8}{20}BD$, so $PQ = \frac{3}{20}BD$ and $QD = \frac{12}{20}BD$. Thus
$$BP : PQ : QD = 5 : 3 : 12,$$
already in lowest terms, and $r + s + t = 20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.

A line from a vertex of a rectangle to a point on the opposite side crosses the diagonal in a ratio equal to the ratio of the two parallel segments it connects. The two cevians from $A$ give two such ratios on the same diagonal; converting both to a common denominator ($20$) lets you read off the three pieces. The actual length $AB = 6$ never matters, a sign that the problem is about ratios, not lengths.

Coordinates: $A(0,0)$, $B(6,0)$, $C(6,3)$, $D(0,3)$, $E(6,1)$, $F(6,2)$. Parametrize $BD$ as $(6-6t, 3t)$. Meeting $AE$ ($y = x/6$) gives $t = \frac14$; meeting $AF$ ($y = x/3$) gives $t = \frac25$. The pieces are $\frac14, \frac{3}{20}, \frac35$ of $BD$, i.e. $5:3:12$.

• Dividing BD into the ratios 1 : 3 and 2 : 3 and then combining them without a common denominator, or reporting BP : BQ : BD instead of BP : PQ : QD.
• Using $BE : BC$ or $BE : AB$ instead of $BE : AD$ as the similarity ratio.