Here is my solution : Angles BAC=DAC=45. so AC is the bisector of angle A of triangle APQ. It is easy to see that PQ is the bisector of angle DPQ. But we know that the bisectors of angels QAP, QPD and PQB are concurrent. so QC is the bisector of PQB. let CQP=CQB=x. then AQP is 180-2x. but we have BCQ is 90-x. so we are done ...

An easy solution: just draw the segment CH where H lies on PQ and ∠CHP=90. but we know that : ∠BCN=∠QPN and ∠NCP=∠NPC so we can conclude that ∠QCP = ∠BCP = ∠CPD (note that AD and BC are parallel) then it's obvious that Triangles PHC and PDC are congruent. so CH is equal to CD. so CH=CD=BC so triangles HCQ and QCB are congruent so ∠HCB=2∠QCB but HCBQ is cyclic so ∠HCB=∠PQA=2∠QCB and the proof is completed.

MNik wrote: Here is my solution : Angles BAC=DAC=45. so AC is the bisector of angle A of triangle APQ. It is easy to see that PQ is the bisector of angle DPQ. But we know that the bisectors of angels QAP, QPD and PQB are concurrent. so QC is the bisector of PQB. let CQP=CQB=x. then AQP is 180-2x. but we have BCQ is 90-x. so we are done ... $ PC $ is the besictor of angle $ \angle NPD $

No , PC is the bisector of QPD. QPC = QPN + NPC = NCB + NCP = BCP = DPC so QPC is equal to DPC and PC is the bisector of QPD.

by the conditions $\angle QPC=\angle BCP=\angle DPC$ also $\angle CAP=\angle CAQ=45$ so $C$ is the $A$-excenter of $APQ$ now $\angle AQP=180-2\angle BQC=180-2(90-\angle BCQ)=2\angle BCQ$

Let $PQ \cap BC = R$ . Now, $NP=NC \implies \angle NPC = \angle NCP \implies \angle QPN + \angle NPC = \angle NCB + \angle NCP \implies \angle RPC = \angle RCP \therefore RP = RC$. Little bit of angle chasing shows that $\angle BCQ = \frac{1}{2} AQP \iff QC$ is the internal bisector of $\angle BQP \iff QC$ is the external bisector of $\angle RQB \iff \frac{QR}{QB} = \frac{CR}{CB}$. This follows from $\frac{CR}{RQ} = \frac{RP}{RQ} = \frac{AB}{QB} = \frac{CB}{QB}$.

Iranian Olympiad, second round 2015

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second day

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Dear Mathlinkers, have also a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=587739 Sincerely Jean-Louis

Dear Mathlinkers, see also http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20II.pdf p. 18-19. Sincerely Jean-Louis

Easy. Note that we only need to prove ∠PQC = ∠BQC or in fact we need to prove C is A-excenter of APQ. we know AC is angle bisector of A so we need to prove ∠DPC = ∠CPQ. ∠CPQ = ∠CPN + ∠NPQ = ∠NCP + ∠BCN = ∠BCP = 90 - ∠PCD = ∠DPC. we're Done.

Let $K = CN \cap PQ$ and $L = PN \cap CB$, then $KLCP$ is cyclic. Since $NP = NC$, we have that $KLCP$ is a isosceles trapezium with $KL \parallel PC$. Let $R$ and $S$ be the foot of height from $C$ and $N$ to $PQ$, respectively. Notice that $\angle RPC = \angle KPC = \angle LCP = \angle CPD$ so $\triangle DPC \cong \triangle RPC$ $\Rightarrow$ $CB = CD = CR$ and as $QBCR$ is cyclic, thus $\angle RCQ = \angle QCB$. Finally, let $T = SN \cap BC$ then $ST \parallel RC$ and $SBTQ$ is cyclic, therefore $\angle BCQ = \frac{1}{2}\angle BCR = \frac{1}{2}\angle BTS = \frac{1}{2}\angle BQS = \frac{1}{2}\angle AQP$ $\blacksquare$

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