Track / Stage 7 / 290 of 300 #1690 of 2000

Problem 1690

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.9 Prove it

Let $O$ be the center of the equilateral triangle $ABC$ . Pick two points $P_1$ and $P_2$ other than $B$ , $O$ , $C$ on the circle $\odot(BOC)$ so that on this circle $B$ , $P_1$ , $P_2$ , $O$ , $C$ are placed in this order. Extensions of $BP_1$ and $CP_1$ intersects respectively with side $CA$ and $AB$ at points $R$ and $S$ . Line $AP_1$ and $RS$ intersects at point $Q_1$ . Analogously point $Q_2$ is defined. Let $\odot(OP_1Q_1)$ and $\odot(OP_2Q_2)$ meet again at point $U$ other than $O$ .

Prove that $2\,\angle Q_2UQ_1 + \angle Q_2OQ_1 = 360^\circ$ .

Remark. $\odot(XYZ)$ denotes the circumcircle of triangle $XYZ$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Restate the problem in terms of angles:
We need to prove that $2 \angle Q_2UQ_1 + \angle Q_2OQ_1 = 360^\circ$ . By angle chasing, this is equivalent to proving $2 \angle AP_1O + \angle AOQ_1 = 2 \angle AP_2O + \angle AOQ_2$ . We will show that $2 \angle AP_1O + \angle AOQ_1 = 180^\circ$ , which will imply $2 \angle AP_2O + \angle AOQ_2 = 180^\circ$ analogously.

2. **Claim 1: Points $A, S, P_1, O, R$ lie on a circle $\omega_1$ :
- Proof:** Note that $\angle SAO = 30^\circ = \angle OBC = \angle OP_1C = 180^\circ - \angle OP_1S$ , so $S \in \odot(AOP_1)$ . Similarly, $R \in \odot(AOP_1)$ . This proves our claim. $\square$

3. **Claim 2: $AR = BS$ :
- Proof:** Using the cross-ratio, we have $\frac{BP_1}{CP_1} = (B, C; P_1, O) \stackrel{B}{=} (A, C; R, \text{midpoint}(\overline{AC})) = \frac{AR}{CR}$ . Similarly, $\frac{CP_1}{BP_1} = \frac{AS}{BS}$ . So, $\frac{AR}{CR} = \frac{BS}{AS}$ , which implies $AR = BS$ since $AB = AC$ . $\square$

4. **Claim 3: $L = \overline{AO} \cap \overline{RS}$ and $Q_1$ are isotomic conjugates with respect to segment $RS$ :
- Proof:** Let $k = \frac{AR}{CR} = \frac{BS}{CR}$ . Since $\overline{AL}$ is the internal angle bisector of $\angle SAR$ , we have $\frac{RL}{SL} = k$ . It suffices to show $\frac{SQ_1}{RQ_1} = k$ . Using Ceva's Theorem in $\triangle ARS$ with points $Q_1, B, C$ , we obtain:
$\frac{SQ_1}{RQ_1} = \frac{BS}{BA} \cdot \frac{CA}{CR} = \frac{BS}{CR} = k$
This proves our claim. $\square$

5. Finish the proof:
- Since $\overline{AO}$ is the internal angle bisector of $\angle SAR$ , $O$ is the midpoint of arc $\widehat{RS}$ of $\omega_1$ not containing $A$ . Let $T = \overline{OQ_1} \cap \omega_1 \ne O$ . By Claim 3, $T$ is the reflection of $A$ across the perpendicular bisector of segment $\overline{BC}$ . Hence,
$2 \angle AP_1O + \angle AOQ_1 = 2(\angle AP_1R + \angle RP_1O) + \angle AOT = 2 \angle AP_1R + 2 \angle RP_1O + (\angle AOS - \angle TOS)$
Simplifying further,
$= 2 \angle ASR + 2 \angle RAO + (\angle ARS - \angle ASR) = \angle ASR + \angle SAR + \angle ARS = 180^\circ$
This completes the proof of the problem. $\blacksquare$

The final answer is $\boxed{ 2 \angle Q_2UQ_1 + \angle Q_2OQ_1 = 360^\circ }$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.