Library / Saudi Arabian Mathematical Competitions /181 of 395

Saudi Arabian Mathematical Competitions, 2012

Geometry Difficulty 5.5 AIME, harder Prove it Saudi Arabia

Let $ABCD$ be a convex quadrilateral such that $AB = AC = BD$ . The lines $AC$ and $BD$ meet at point $O$ , the circles $ABC$ and $ADO$ meet again at point $P$ , and the lines $AP$ and $BC$ meet at point $Q$ . Show that $\overline{COQ} = \overline{DOQ}$ .

Solution

We shall prove that the circles $ADO$ and $BCO$ meet again at the incenter $I$ of the triangle $ABO$ , so the line $IO$ is the radical axis of the circles $ADO$ and $BCO$ . Noticing further that the lines $AP$ and $BC$ are the radical axes of the pairs of circles $(ABC, ADO)$ and $(ABC, BCO)$ , respectively, it follows that the lines $AP$ , $BC$ and $IO$ are concurrent (at point $Q$ ), and the conclusion follows.

To show that the point $I$ lies on the circle $ADO$ , notice that
$\begin{aligned} \overline{AIO} &= 90^\circ + \frac{1}{2} \overline{ABO} = 90^\circ + \frac{1}{2} \overline{ABD} = 90^\circ + \frac{1}{2} (180^\circ - 2\overline{ADB}) \\ &= 180^\circ - \overline{ADB} = 180^\circ - \overline{ADO}. \end{aligned}$
Similarly, the point $I$ lies on the circle $BCO$ , for
$\begin{aligned} \overline{BIO} &= 90^\circ + \frac{1}{2} \overline{BAO} = 90^\circ + \frac{1}{2} \overline{BAC} = 90^\circ + \frac{1}{2} (180^\circ - 2\overline{ACB}) \\ &= 180^\circ - \overline{ACB} = 180^\circ - \overline{BCO}. \end{aligned}$

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.