2009 AMC 10A Problem 17

P16-20~6 minPrint

Rectangle $ABCD$ has $AB=4$ and $BC=3$ . Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$ , and $A$ and $C$ lie on $DE$ and $DF$ , respectively. What is $EF$ ?

Show solution without answering

Key insight

DB is the height of right triangle DEF to its long side EF; the triangles DBE and DBF are enlarged copies of the 3-4-5 triangles DAB and DCB.

Solution

Place the rectangle with $A$ bottom left, $B$ bottom right, $C$ top right and $D$ top left. Then $DA = 3$ , $AB = 4$ , and the diagonal $DB$ is $5$ (a $3$ - $4$ - $5$ right triangle).

1. $E$ is on line $DA$ past $A$ (below the rectangle), and $F$ is on line $DC$ past $C$ (to the right). $EF$ passes through $B$ and makes a right angle with $DB$ .
2. $EF$ is split by $B$ into two pieces, $BE$ and $BF$ . Find each one.
3. Piece $BE$ . Compare triangle $DAB$ (right angle at $A$ ) with triangle $DBE$ (right angle at $B$ ). They share the angle at $D$ , so they have the same shape. In triangle $DAB$ the side next to the angle at $D$ is $DA = 3$ ; in triangle $DBE$ it is $DB = 5$ . So triangle $DBE$ is triangle $DAB$ enlarged by $\frac{5}{3}$ , and
$BE = AB \times \frac{5}{3} = 4 \times \frac{5}{3} = \frac{20}{3}.$
4. Piece $BF$ . Compare triangle $DCB$ (right angle at $C$ ) with triangle $DBF$ (right angle at $B$ ). They also share the angle at $D$ . The side next to that angle goes from $DC = 4$ to $DB = 5$ , an enlargement by $\frac{5}{4}$ :
$BF = CB \times \frac{5}{4} = 3 \times \frac{5}{4} = \frac{15}{4}.$
5. Add the pieces:
$EF = \frac{20}{3} + \frac{15}{4} = \frac{80}{12} + \frac{45}{12} = \frac{125}{12}.$

The answer is $\boxed{\textbf{(C)}\ \frac{125}{12}}$ .

Why this works

A line drawn at a right angle to the diagonal makes new right triangles that share an angle with the rectangle's own right triangles, so all of them have the $3$ - $4$ - $5$ shape. Find the enlargement factor from one pair of matching sides, then multiply. When a length is cut into two pieces by a point, work out each piece on its own.

Alternative approach

Use coordinates: $A = (0,0)$ , $B = (4,0)$ , $C = (4,3)$ , $D = (0,3)$ .

1. Line $DB$ has slope $-\frac{3}{4}$ , so the line through $B$ at a right angle to it has slope $\frac{4}{3}$ : $y = \frac{4}{3}(x - 4)$ .
2. It meets the line through $D$ and $A$ ( $x = 0$ ) at $E = (0, -\frac{16}{3})$ .
3. It meets the line through $D$ and $C$ ( $y = 3$ ) where $\frac43(x - 4) = 3$ , so $x = \frac{25}{4}$ and $F = (\frac{25}{4}, 3)$ .
4. From $E$ to $F$ you go $\frac{25}{4}$ across and $3 + \frac{16}{3} = \frac{25}{3}$ up, so
$EF = \sqrt{\left(\tfrac{25}{4}\right)^2 + \left(\tfrac{25}{3}\right)^2} = 25\sqrt{\tfrac{1}{16} + \tfrac{1}{9}} = 25 \times \tfrac{5}{12} = \tfrac{125}{12}.$

Concise solution

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

Draw the rectangle with $D$ at the top left, $A$ below it, and $C$ to its right. The diagonal $DB$ has length $5$ . Since $A$ is on $DE$ , the point $E$ lies on line $DA$ extended beyond $A$ ; since $C$ is on $DF$ , the point $F$ lies on line $DC$ extended beyond $C$ . Thus $\angle EDF = \angle ADC = 90^\circ$ , and $DB$ is the altitude from $D$ to the hypotenuse $EF$ of right triangle $DEF$ .

Look at $\triangle DBE$ : it is right-angled at $B$ and shares $\angle BDA$ with $\triangle DAB$ , which is right-angled at $A$ . So $\triangle DAB \sim \triangle DBE$ with $DA \to DB$ , i.e. scale factor $\frac{DB}{DA} = \frac{5}{3}$ . Hence
$BE = AB \cdot \frac{5}{3} = \frac{20}{3}.$

Similarly $\triangle DCB \sim \triangle DBF$ (right angles at $C$ and $B$ , common angle at $D$ ) with scale factor $\frac{DB}{DC} = \frac{5}{4}$ , so
$BF = CB \cdot \frac{5}{4} = \frac{15}{4}.$

Therefore
$EF = BE + BF = \frac{20}{3} + \frac{15}{4} = \frac{80 + 45}{12} = \frac{125}{12}.$

The answer is $\boxed{\textbf{(C)}\ \frac{125}{12}}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 5: Similar Triangles
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)