Library / Irish Mathematical Olympiad /217 of 236

Irish Mathematical Olympiad

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it Ireland

Let $ABC$ be a triangle. Let $AD$ be the angle bisector of the angle $\angle BAC$ and let $BE$ be the angle bisector of the angle $\angle ABC$ , with $D$ interior to the side $BC$ and $E$ to the side $AC$ . Let $M$ be a point interior to the side $BC$ such that $|CM| = |AE|$ , and let $N$ be a point interior to the side $AC$ such that $|CN| = |BD|$ . Show that the points $D, E, M$ and $N$ are all on a circle if and only if $|AB| = |AC|$ .

Solution

Let $a$ , $b$ and $c$ be the lengths of the sides of the triangle $ABC$ . Using that $AD$ is an angle bisector, that the two angles $\angle ADB$ and $\angle CDA$ are complementary (hence have equal sines) and the sine rule for the triangles $ABD$ and $ADC$ one obtains
$\frac{|BD|}{|DC|} = \frac{c}{b} \quad \text{and similarly} \quad \frac{|AE|}{|EC|} = \frac{c}{a}.$

The points $E$ , $D$ , $M$ and $N$ are all on a circle if and only if opposite angles of the quadrilateral $EDMN$ add up to $180^\circ$ . This is equivalent to $\angle CMN = \angle CED$ and $\angle CNM = \angle CDE$ which is equivalent to the triangles $CMN$ and $CED$ being similar. This, however, is equivalent to
$\frac{|CE|}{|CM|} = \frac{|CD|}{|CN|}.$
Using $a = c \cdot |CE|/|AE|$ and $b = c \cdot |CD|/|BD|$ , which was shown above, and $|AE| = |CM|$ , $|BD| = |CN|$ , it is now straightforward to see that $E$ , $D$ , $M$ , $N$ are concyclic if and only if $a = b$ .

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.