Let $ABC$ be a right-angled triangle with $\angle B = 90^\circ$ and let $BD$ be the altitude from $B$ on to $AC$. Draw $DE \perp AB$ and $DF \perp BC$. Let $P, Q, R$ and $S$ be respectively the incentres of triangle $DF C, DBF, DEB$ and $DAE$. Suppose $S, R, Q$ are collinear. Prove that $P, Q, R, D$ lie on a circle.
Problem
Source: RMO 2015 Mumbai Region
Tags: geometry
Vfire
03.09.2018 22:31
Notice that $\angle RDF = \angle PDF + \angle BDF + \angle RDB = \tfrac{1}{2} \angle A + \angle C + \tfrac{1}{2} \angle A = 90^\circ$ so it suffices to show that $\angle SQP = 90^\circ$.
We have $\triangle QFP \sim \triangle BDC \sim \triangle ABC$ and $\triangle SER \sim \triangle ADB \sim \triangle ABC$. Since $ER \parallel QF$, we have $\angle SQP = 180^\circ - \angle PQF - \angle SRE = 180^\circ - \angle A - \angle C = 90^\circ$ as desired.
DynamoBlaze
06.09.2018 21:05
Vfire wrote:
Notice that $\angle RDF = \angle PDF + \angle BDF + \angle RDB = \tfrac{1}{2} \angle A + \angle C + \tfrac{1}{2} \angle A = 90^\circ$ so it suffices to show that $\angle SQP = 90^\circ$.
We have $\triangle QFP \sim \triangle BDC \sim \triangle ABC$ and $\triangle SER \sim \triangle ADB \sim \triangle ABC$. Since $ER \parallel QF$, we have $\angle SQP = 180^\circ - \angle PQF - \angle SRE = 180^\circ - \angle A - \angle C = 90^\circ$ as desired.
Do you mean $\angle RDP = \angle PDF + \angle BDF + \angle RDB$?
sanyalarnab
20.04.2022 07:12
Solution using spiral similarity: It's obvious that $CDB \sim BDA$. So we define the spiral similarity $\Phi$ centered at $D$ taking triangle $CDB$ to $BDA$. Thus $\Phi(F)=E$ and thus the corresponding incenters will also be mapped to each other that is $\Phi(P)=R,\Phi(Q)=S$ Thus $\Phi$ maps line $PQ$ to $RS$. Hence $DPQ \sim DRS \implies \angle DPQ=\angle DRS$. As $S,R,Q$ are collinear we get $\angle DPQ+\angle DRQ=180^o$ Hence $R,Q,P,D$ are concyclic