Let $ABC$ be a triangle with $AB < BC$. The bisector of angle $C$ meets the line parallel to $AC$ and passing through $B$, at point $P$. The tangent at $B$ to the circumcircle of $ABC$ meets this bisector at point $R$. Let $R'$ be the reflection of $R$ with respect to $AB$. Prove that $\angle R'P B = \angle RPA$.
Problem
Source: Sharygin 2018
Tags: geometry
Fumiko
04.04.2018 22:52
Let $T$ be the isogonal conjugate of $R$ wrt $\Delta PAB$ ,now $P,R'$ are isogonal conjugates wrt $\Delta ATB$ now proving $P,T,R'$ are collinear is equivalent to $\angle ATP=\angle BTP$ which is equivalent to $\angle ABP+\angle PRB=\angle PAB+\angle ARP$ which is just simple angle chasing noting that $XA$ is tangent to $\odot (ARP)$ where $X$ is the point where $C-$ bisector meets $\odot (ABC)$ $\blacksquare$
anantmudgal09
04.04.2018 22:57
TheDarkPrince wrote: Let $ABC$ be a triangle with $AB < BC$. The bisector of angle $C$ meets the line parallel to $AC$ and passing through $B$, at point $P$. The tangent at $B$ to the circumcircle of $ABC$ meets this bisector at point $R$. Let $R'$ be the reflection of $R$ with respect to $AB$. Prove that $\angle R'P B = \angle RPA$. Observe that $P$ and $R$ are isogonal conjugates in $\triangle ABC$. Let $I_C$ be the $C$-excenter; then $$\frac{PA}{RA}=\frac{PI_C}{RI_C}=\frac{PB}{RB}.$$Now apply the lemma from IZHO 2015/2 in triangle $APB$ to conclude.
mastermind.hk16
05.04.2018 02:30
Let $M=CP \cap (ABC)$. Then $MB=MA$. $\gamma = \angle MBR= \angle MCB= \angle MCA= \angle CPB$ so $\triangle RMB \sim \triangle BMP$. Then $\frac{R'B}{BP}=\frac{RB}{BP}=\frac{MB}{MP}=\frac{MA}{MP}$. Moreover $\angle R'BP= \angle R'BA + \angle ABP = \angle RBA + \angle ABP =2 \gamma +\angle BAC $ and $\angle AMP = 180- \angle AMC = 180- \angle ABC=\angle BCA +\angle CAB=2 \gamma +\angle BAC$. Hence $\triangle AMP \sim \triangle R'BP$ by SAS. $\angle R'PB = \angle MPA= \angle RPA$ and we are done.
TheDarkPrince
05.04.2018 09:52
[asy][asy]
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[/asy][/asy]
Let $\angle APR=\angle APC =\alpha$ and $\angle R'PB = \beta$.
From law of sines in triangle $PBA$, we have \[\frac{\sin A}{\sin \left(\alpha + \frac{C}{2}\right)}=\frac{\sin \angle ABP}{\sin \angle APB}=\frac{AP}{AB}.\]So, $AP=\frac{\sin A\cdot AB}{\sin\left(\frac{C}{2}+\alpha\right)}$.
Also from law of sines in triangle $APC$, we have \[\frac{\sin \frac{C}{2}}{\sin \alpha}=\frac{AP}{AC}=\frac{\sin A\cdot AB}{AC\sin\left(\frac{C}{2}+\alpha\right)}.\]
So, $\cos \frac{C}{2}+\cot \alpha \sin\frac{C}{2}=\frac{\sin\left(\frac{C}{2}+\alpha\right)}{\sin \alpha}=\frac{\sin A\cdot AB}{AC\cdot \sin \frac{C}{2}}$. So, \[\cot \alpha = \frac{\frac{\sin A\cdot AB}{AC\cdot \sin \frac{C}{2}}-\cos \frac{C}{2}}{\sin \frac{C}{2}}=\frac{\frac{2\sin A\cdot \cos \frac{C}{2}}{\sin B}-\cos \frac{C}{2}}{\sin \frac{C}{2}}=\frac{2\sin A\cos \frac{C}{2}-\sin B\cos \frac{C}{2}}{\sin B\sin \frac{C}{2}}.\]
We know from angle chasing, $PB=BC$. So, from law of sines in triangles $PBR'$ and $BRC$ we have \[\sin B\cot \beta - \cos B=\frac{\sin (B-\beta)}{\sin \beta}=\frac{PB}{BR'}=\frac{BC}{BR}=\frac{\sin\left(A-\frac{C}{2}\right)}{\sin \frac{C}{2}}.\]
So, $\cot \beta=\frac{\sin\left(A-\frac{C}{2}\right)+\cos B\sin\frac{C}{2}}{\sin B\sin \frac{C}{2}}$.
As both $\alpha,\beta$ are less than $180^{\circ}$, showing $\cot\alpha= \cot \beta$ we would get $\alpha=\beta$ and we would be done.
So we need \[\sin\left(A+\frac{C}{2}\right)+\sin\left(A-\frac{C}{2}\right)-\sin B\cos \frac{C}{2}=2\sin A\cos \frac{C}{2}-\sin B\cos \frac{C}{2}=\sin\left(A-\frac{C}{2}\right)+\cos B\sin \frac{C}{2}\]or \[\sin\left(A+\frac{C}{2}\right)=\sin B\cos \frac{C}{2}+\cos B\sin \frac{C}{2}=\sin\left(B+\frac{C}{2}\right)=\sin \left(180^{\circ}-B-\frac{C}{2}\right)=\sin\left(A+\frac{C}{2}\right).\]
So, we are done.
USER378398
06.04.2018 09:45
simple angle trace and a final use of sine rule to prove similarity the trig bash works well too
USER378398
06.04.2018 10:11
what would be the expected cut-off score for the olympiad??
BobaFett101
16.01.2019 08:22
The problem is asking us to show that lines $PR, PR'$ are isogonal in $\angle APB$. We show that $C, R'$ are isogonal conjugates in $\triangle APB$, which finishes. Claim: $P$ and $R$ are isogonal conjugates in $\triangle ABC$. Proof: Both points lie on the angle bisector of $\angle C$. In addition, it is well known that the tangent to $(ABC)$ at $A$ and the line through $A$ parallel to $BC$ are isogonal lines in $\angle A$ (prove with angle chase). Claim: $C$ and $R'$ are isogonal conjugates in $\triangle APB$. Proof: From the above claim, we know lines $BR, CP$ are isogonal in $\angle ABC$. Thus, $\angle RBA = 180 - \angle PBC$. By reflection, $\angle R'BA = \angle RBA$, so lines $BR'$ and $BC$ are isogonal in $\angle ABP$. Similarly, $AR$ and $AP$ are isogonal in $\angle BAC$, so by the equal angle obtained upon reflection, $\angle R'AB = 180 - \angle CAP$, so $AR'$ and $AC$ are isogonal in $\angle PAB$. This proves the claim and finishes the problem.