14th European Girls' Mathematical Olympiad (EGMO 2025)

Problem 4 Let ABC be an acute triangle with incentre I and AB≠AC. Let lines BI and CI intersect the circumcircle of ABC at P≠B and Q≠C, respectively. Consider points R and S such that AQRB and ACSP are parallelograms (with AQ ∥ RB, AB ∥ QR, AC ∥ SP , and AP ∥ CS). Let T be the point of intersection of lines RB and SC. Prove that points R, S, T , and I are concyclic.

SOLUTION 1

WLOG, assume that $AB < AC$ .

$\textbf{Claim:}$ $I$ is the center of spiral similarity which maps $BR$ to $CS$ .

$\textbf{Proof:}$ Note that
$\angle IBR = 360^\circ - \angle ABR - \angle ABI = 360^\circ - (180^\circ - \angle QAB) - \frac{\angle ABC}{2} = 180^\circ + \frac{\angle ACB}{2} - \frac{\angle ABC}{2}$
and
$\angle ICS = \angle ICA + \angle ACS = \frac{\angle ACB}{2} + 180^\circ - \angle CAP = 180^\circ + \frac{\angle ACB}{2} - \frac{\angle ABC}{2}$
$\Rightarrow \angle IBR = \angle ICS$
Furthermore,
$\frac{IB}{BR} = \frac{IB}{AQ} = \frac{IB}{IQ} = \frac{IC}{IP} = \frac{IC}{AP} = \frac{IC}{CS}$
$\Rightarrow \triangle IBR \sim \triangle ICS \quad \blacksquare$
By the claim and the fact that $T = RB \cap SC$ ,
we can conclude that $T, I, R, S$ are concyclic. $\blacksquare$

SOLUTION 2

The main idea is to show that $\angle RTS=\angle RIS$ .
We may assume WLOG that $AB<AC$ .
$\boldsymbol{Claim:}$ $\triangle IBR \sim \triangle ICS$ .
$\boldsymbol{Proof:}$ By Incentre-Excentre Lemma, $BQ=QI$ and $CP=PI$ .
Since $\angle BAC=\angle BQI=\angle CPI$ , then $\triangle BQI \sim \triangle CPI$ , so $\frac{IB}{IC}=\frac{BQ}{CP}$ .
$QA=QB=BR$ and $PA=PC=CS$ , giving $\frac{IB}{IC}=\frac{BR}{CS}$ .
Now it remains to prove that $\angle IBR=\angle ICS.$ It follows from angle chase:
$\angle IBR=2\pi-\angle ABR-\angle ABI = 2\pi - (\pi- \angle QAB)-\frac{\angle B}{2}= \pi +\frac{\angle C}{2} - \frac{\angle B}{2}$ and $\angle ICS= \angle ICA +\angle ACS = \frac{\angle C}{2}+ \pi - \angle PAC=\pi + \frac{\angle C}{2}-\frac{\angle B}{2}$ .
Therefore $\angle IBR=\angle ICS$ and the claim follows.

We have already proved that $\angle IBR=\angle ICS$ , giving $\angle TBI=\angle TCI$ , so $T, B, C, I$ lie on a circle. Hence $\angle BIC=\angle BTC$ .
From the claim, we get that $\angle BIR=\angle CIS$ , which yields $\angle BIC =\angle RIS$ .
Therefore $\angle RTS = \angle RIS$ , so points $R, S, T, I$ are concyclic.