Library / Slovenian Mathematical Olympiad /41 of 68

Slovenian Mathematical Olympiad, 2008

Geometry Difficulty 5.7 AIME, harder Prove it Slovenia

In the triangle $ABC$ the lengths of the sides are given: $|AB| = 15$ cm, $|BC| = 14$ cm and $|CA| = 13$ cm. Let $D$ be the foot of the altitude from $A$ and let $E$ be a point on this altitude such that $\angle BAD = \angle DEC$ . Denote the intersection of lines $AB$ and $CE$ by $F$ . Find $|EF|$ .

Solution

First, let us find the lengths of the segments $AD$ and $CD$ . Write $|AD| = v$ and $|CD| = x$ . By Pythagoras's theorem $v^2 = |AC|^2 - x^2 = |AB|^2 - (|BC| - x)^2$ , so $13^2 - x^2 = 15^2 - (14 - x)^2$ or, equivalently, $13^2 = 15^2 - 14^2 + 2 \cdot 14 \cdot x$ . We see that $x = 5$ and $v = 12$ .

Triangles $EDC$ and $ADB$ are similar because $\angle BAD = \angle DEC$ . So, $\frac{|EC|}{|CD|} = \frac{|AB|}{|AD|}$ and $\angle DBA = \angle ECD$ . This implies $|EC| = \frac{15}{9} \cdot 5 = \frac{25}{3}$ and $\angle FCB = \angle CBF$ . The triangle $BFC$ is isosceles with the apex at $B$ , so $|CF| = |FB|$ .

Let $G$ be the midpoint of $BC$ . Then $FG$ is perpendicular to $BC$ . The triangle $FBG$ is similar to the triangle $ABD$ , so $\frac{|FB|}{|BG|} = \frac{|AB|}{|BD|}$ and $|FB| = \frac{|AB| \cdot |BG|}{|BD|} = \frac{15 \cdot 7}{9} = \frac{35}{3}$ .

The length of the segment $EF$ is $|EF| = |CF| - |CE| = |FB| - |CE| = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.