2016 AMC 10A Problem 19

P16-20~5 minPrint

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 $r+s+t$ ?

Show solution without answering

Key insight

Because BC is parallel to AD, lines from A cut BD in the ratios 1 to 3 and 2 to 3. Use twentieths.

Solution

1. Lengths on side $BC$ . $BC = 3$ is split into three equal parts, so $BE = 1$ and $BF = 2$ . Also $AD = BC = 3$ .
2. Why similar triangles appear. Sides $BC$ and $AD$ are parallel. The segment $AE$ crosses the diagonal $BD$ at $P$ , making two triangles, $PBE$ and $PDA$ , that form an hourglass. The angles at $P$ are vertical angles, and the angles at $B$ and $D$ are equal because $BC \parallel AD$ . So the triangles are similar.
3. First crossing point. Matching sides in the hourglass:
$\frac{BP}{PD} = \frac{BE}{DA} = \frac{1}{3}.$
So $BD$ is cut into $1 + 3 = 4$ equal parts, and $BP$ is $\tfrac14$ of $BD$ .
4. Second crossing point. In the same way, triangles $QBF$ and $QDA$ are similar, so
$\frac{BQ}{QD} = \frac{BF}{DA} = \frac{2}{3}.$
So $BD$ is cut into $2 + 3 = 5$ equal parts, and $BQ$ is $\tfrac25$ of $BD$ .
5. Put both on one scale. Fourths and fifths both fit into twentieths. Measure $BD$ as $20$ small units:
- $BP = \tfrac14 \cdot 20 = 5$ units.
- $BQ = \tfrac25 \cdot 20 = 8$ units, so $PQ = 8 - 5 = 3$ units.
- $QD = 20 - 8 = 12$ units.
6. So $BP : PQ : QD = 5 : 3 : 12$ . These have no common factor, so $r + s + t = 5 + 3 + 12 = 20$ .

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

Why this works

Whenever two sides are parallel, a line joining them makes an hourglass of similar triangles, and it cuts a diagonal in the same ratio as the two parallel pieces. With two such lines on one diagonal, rewrite both fractions with a common denominator so the pieces can be read off. The length $AB = 6$ was never used, a sign that the problem is only about ratios.

Alternative approach

Coordinates. Put $A = (0,0)$ , $B = (6,0)$ , $C = (6,3)$ , $D = (0,3)$ , so $E = (6,1)$ and $F = (6,2)$ .

Line $AE$ is $y = \tfrac{x}{6}$ and line $AF$ is $y = \tfrac{x}{3}$ .
Line $BD$ is $x + 2y = 6$ .
For $P$ : $x + \tfrac{x}{3} = 6$ , so $x = 4.5$ . For $Q$ : $x + \tfrac{2x}{3} = 6$ , so $x = 3.6$ .

Along $BD$ the $x$ -coordinate drops evenly from $6$ to $0$ , so the pieces are proportional to $6 - 4.5 = 1.5$ , then $4.5 - 3.6 = 0.9$ , then $3.6$ . Multiplying by $\tfrac{10}{3}$ gives $5 : 3 : 12$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(E)}\ 20}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 5: Similar Triangles