2009 AMC 10A Problem 17

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

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.

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
$$EF = \sqrt{\left(\tfrac{25}{4}\right)^2 + \left(\tfrac{25}{3}\right)^2} = 25\sqrt{\tfrac{1}{16} + \tfrac{1}{9}} = 25 \times \tfrac{5}{12} = \tfrac{125}{12}.$$

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