2013 AMC 10B Problem 23

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 $\boxed{\textbf{(B)}\ 21}$.

Two right angles on the same segment are an invitation to draw the circle with that segment as diameter; the cyclic quadrilateral then transports a known angle ($\angle B$) to the point we care about. After that, everything is right-triangle bookkeeping in the famous $13$-$14$-$15$ triangle, whose altitude splits it into $5$-$12$-$13$ and $9$-$12$-$15$ pieces.

Coordinates: $D = (0,0)$, $B = (-5, 0)$, $C = (9, 0)$, $A = (0, 12)$. Line $AC$ is $4x + 3y = 36$, so $E = \left(\frac{144}{25}, \frac{108}{25}\right)$. Write $F = tE$ and impose $(F - A)\cdot(F - B) = 0$: this gives $\frac{1296}{25}t^2 - \frac{576}{25}t = 0$, so $t = \frac{4}{9}$ and $DF = \frac{4}{9} \cdot \frac{36}{5} = \frac{16}{5}$.

• Assuming F is the foot of the perpendicular from B to DE, or mislabeling the 13-14-15 altitude segments (BD = 5, DC = 9, AD = 12).
• Reporting $EF = 4$ or $DE = \frac{36}{5}$ (giving $m + n = 41$) instead of $DF$.