Library / Slovenia — National Math Olympiad /101 of 122

Slovenia — National Math Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Slovenia

Let $K_1$ and $K_2$ be the circles centered at $O_1$ and $O_2$ , respectively, meeting at the points $A$ and $B$ . Let $p$ be the line through the point $A$ meeting the circles $K_1$ and $K_2$ again at $C_1$ and $C_2$ . Assume that $A$ lies between $C_1$ and $C_2$ . Denote the intersection of the lines $C_1O_1$ and $C_2O_2$ by $D$ . Prove that the points $C_1, C_2, B$ and $D$ lie on the same circle.

Solution

Write $\angle ABC_1 = \alpha$ and $\angle C_2BA = \beta$ . Then $\angle C_2BC_1 = \alpha + \beta$ . The points $C_1, C_2, B$ and $D$ are concyclic if and only if $\angle C_2BC_1 = \angle C_2DC_1$ . Let us show that $\angle C_2DC_1 = \alpha + \beta$ .

The central angle equals twice the inscribed angle, so $\angle AO_1C_1 = 2\angle ABC_1 = 2\alpha$ and $\angle C_2O_2A = 2\angle C_2BA = 2\beta$ . The triangle $AO_1C_1$ is isosceles with the apex at $O_1$ , so $\angle O_1C_1A = \frac{\pi}{2} - \alpha$ . The triangle $C_2O_2A$ is isosceles with the apex at $O_2$ , so $\angle AC_2O_2 = \frac{\pi}{2} - \beta$ . We get
$\begin{aligned} \angle C_2DC_1 &= \pi - \angle C_1C_2D - \angle DC_1C_2 \\ &= \pi - \left(\frac{\pi}{2} - \alpha\right) - \left(\frac{\pi}{2} - \beta\right) = \alpha + \beta, \end{aligned}$
so $C_1, C_2, B$ and $D$ lie on the same circle.

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.