2013 AMC 10B Problem 23

P21-25~8 minPrint

In triangle $ABC$ , $AB = 13$ , $BC = 14$ , and $CA = 15$ . Distinct points $D$ , $E$ , and $F$ lie on segments $\overline{BC}$ , $\overline{CA}$ , and $\overline{DE}$ , respectively, such that $\overline{AD} \perp \overline{BC}$ , $\overline{DE} \perp \overline{AC}$ , and $\overline{AF} \perp \overline{BF}$ . The length of segment $\overline{DF}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?

Show solution without answering

Key insight

Right angles at D and F both face AB, so A, F, D, B lie on one circle. Then angle AFE equals angle B, giving EF = 4.

Solution

Step A: the lengths we can get directly.

1. The $13$ - $14$ - $15$ triangle is famous: the height from $A$ down to $BC$ is $AD = 12$ , and it splits $BC$ into $BD = 5$ and $DC = 9$ . Check: $5 + 9 = 14$ , $5^2 + 12^2 = 13^2$ , and $9^2 + 12^2 = 15^2$ .
2. Triangle $ADC$ has a right angle at $D$ , and $DE$ is its height onto the long side $AC = 15$ . Find its area two ways:
$\frac{1}{2} \cdot 12 \cdot 9 = 54 = \frac{1}{2} \cdot 15 \cdot DE, \quad\text{so}\quad DE = \frac{108}{15} = \frac{36}{5}.$
3. Triangle $AED$ has a right angle at $E$ and hypotenuse $AD = 12$ . By the Pythagorean theorem,
$AE^2 = 12^2 - \left(\frac{36}{5}\right)^2 = \frac{3600 - 1296}{25} = \frac{2304}{25}, \quad\text{so}\quad AE = \frac{48}{5}.$

Step B: where is $F$ ?

We know $F$ is on $DE$ with a right angle at $F$ in triangle $AFB$ . There is also a right angle at $D$ in triangle $ADB$ . When a segment is seen at a right angle from a point, that point lies on the circle that has the segment as its diameter. So both $D$ and $F$ lie on the circle with diameter $AB$ . Now $A$ , $F$ , $D$ , $B$ are four points on one circle.
In a four-sided shape whose corners lie on a circle, opposite angles add up to $180^\circ$ . So $\angle AFD + \angle ABD = 180^\circ$ .
Since $D$ , $F$ , $E$ lie on one line, $\angle AFE = 180^\circ - \angle AFD$ . Comparing with step 5, $\angle AFE = \angle ABD$ , the angle at $B$ .

Step C: finish with similar triangles.

7. Triangle $AEF$ has a right angle at $E$ and an angle at $F$ equal to angle $B$ . Triangle $ADB$ has a right angle at $D$ and the angle $B$ . So the two triangles have the same shape.
8. In triangle $ADB$ , the side across from angle $B$ is $AD = 12$ and the side next to it is $BD = 5$ . In triangle $AEF$ , the side across from angle $F$ is $AE$ and the side next to it is $EF$ . So $AE : EF = 12 : 5$ , and
$EF = \frac{5}{12} \cdot AE = \frac{5}{12} \cdot \frac{48}{5} = 4.$
9. Finally, $DF = DE - EF = \frac{36}{5} - 4 = \frac{16}{5}$ . The fraction is in lowest terms, so $m + n = 16 + 5 = 21$ .

The answer is $\boxed{\textbf{(B)}\ 21}$ .

Why this works

Two right angles that both look at the same segment are a signal to draw the circle with that segment as diameter. The circle then carries a known angle (angle $B$ ) over to the point you care about. After that, the problem is right-triangle work inside the $13$ - $14$ - $15$ triangle, which splits into $5$ - $12$ - $13$ and $9$ - $12$ - $15$ pieces.

Alternative approach

Use coordinates and slopes. Put $D = (0, 0)$ , $B = (-5, 0)$ , $C = (9, 0)$ , $A = (0, 12)$ .

1. Line $AC$ has slope $\frac{0 - 12}{9 - 0} = -\frac{4}{3}$ . Line $DE$ is perpendicular to it, so its slope is $\frac{3}{4}$ , and it passes through $D$ . Every point on it has the form $(4t, 3t)$ , at distance $5t$ from $D$ .
2. Let $F = (4t, 3t)$ . Lines $AF$ and $BF$ are perpendicular, so their slopes multiply to $-1$ :
$\frac{3t - 12}{4t} \cdot \frac{3t}{4t + 5} = -1.$
3. Clear the fractions: $3t(3t - 12) = -4t(4t + 5)$ , so $9t^2 - 36t = -16t^2 - 20t$ , which gives $25t^2 = 16t$ . Since $F$ is not $D$ , $t \ne 0$ , so $t = \frac{16}{25}$ .
4. Then $DF = 5t = \frac{16}{5}$ , and $m + n = 21$ .

Concise solution

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

The $13$ - $14$ - $15$ triangle has altitude $AD = 12$ with $BD = 5$ and $DC = 9$ (check: $5^2 + 12^2 = 13^2$ , $9^2 + 12^2 = 15^2$ ).

In right triangle $ADC$ , $DE$ is the altitude to the hypotenuse $AC = 15$ , so
$DE = \frac{AD \cdot DC}{AC} = \frac{12 \cdot 9}{15} = \frac{36}{5}, \qquad AE = \frac{AD^2}{AC} = \frac{144}{15} = \frac{48}{5}.$

Now locate $F$ . Both $\angle ADB$ and $\angle AFB$ are right angles, so $D$ and $F$ lie on the circle with diameter $AB$ . In the cyclic quadrilateral $AFDB$ the angles at $F$ and $B$ are supplementary, so the exterior angle $\angle AFE$ (with $E$ on line $DF$ beyond $F$ ) equals $\angle ABD$ .

Triangle $AEF$ is right-angled at $E$ with $\angle AFE = \angle B$ , so it is similar to triangle $ADB$ , in which $\tan B = \frac{12}{5}$ . Hence
$EF = \frac{AE}{\tan B} = \frac{48}{5} \cdot \frac{5}{12} = 4.$
Finally $DF = DE - EF = \frac{36}{5} - 4 = \frac{16}{5}$ , so $m + n = 21$ .

The answer is $\boxed{\textbf{(B)}\ 21}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 3: Congruent Triangles
Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)