Let $ABC$ be a isosceles triangle with $AB=AC$ an $D$ be the foot of perpendicular of $A$. $P$ be an interior point of triangle $ADC$ such that $m(APB)>90$ and $m(PBD)+m(PAD)=m(PCB)$. $CP$ and $AD$ intersects at $Q$, $BP$ and $AD$ intersects at $R$. Let $T$ be a point on $[AB]$ and $S$ be a point on $[AP$ and not belongs to $[AP]$ satisfying $m(TRB)=m(DQC)$ and $m(PSR)=2m(PAR)$. Show that $RS=RT$
Problem
Source: Turkey NMO 2012
Tags: trigonometry, geometry, circumcircle, projective geometry, geometry proposed
NgoNgang
30.01.2013 20:05
Any solution?
xeroxia
16.03.2013 13:14
Let $\angle PAR = \angle QBR= \alpha$, $\angle PBC = \beta$, $\angle BAD = \theta$. Apply, trigonometric form of Ceva Theorem for point $P$ inside $\triangle ABC$. \[\frac{\sin (\theta + \alpha)}{\sin (\theta - \alpha)} \cdot \frac{\cos (\alpha + \beta + \theta)}{\sin (\alpha - \beta)} \cdot \frac{\sin \beta}{\cos (\beta + \alpha)} = 1 \quad (..1..)\] $2\cos(\alpha + \beta + \theta)\sin \beta = \sin (\alpha + 2\beta + \theta) - \sin (\alpha + \theta)$. $2\sin(\alpha - \beta )\cos (\alpha + \beta) = \sin (\alpha + 2\beta + \theta) - \sin (\alpha - \theta)$. Rewriting $(..1..)$, \[\frac{\sin (\theta + \alpha)}{\sin (\theta - \alpha)} \cdot \frac{\sin (\alpha + 2\beta + \theta) - \sin (\alpha + \theta)}{\sin (\alpha + 2\beta + \theta) - \sin (\alpha - \theta)} = 1 \\ \\ \Rightarrow (\sin (\theta + \alpha) - \sin(\theta - \alpha))\sin(\alpha + 2 \beta + \theta) = \sin^2(\theta + \alpha) - \sin^2(\theta - \alpha) \\ \\ \Rightarrow \sin (\alpha + 2\beta + \theta ) = \sin (\theta + \alpha) + \sin(\theta - \alpha) = 2\sin\theta \cos \alpha \quad (..2..)\] Apply trigonometric form of Ceva for point $R$ inside $\triangle ATS$, \[\frac{\sin (\alpha + 2\beta + \theta)}{\sin \theta} \cdot \frac{\sin \alpha}{ \sin (2\alpha)}\cdot \frac{\sin \angle RST}{\sin \angle RTS} = 1 \quad (..3..)\] Since $\sin (2\alpha) = 2\cos \alpha \sin \alpha$ and $\sin (\alpha + 2\beta + \theta) = 2\sin\theta \cos \alpha$ (by $(..2..)$), rewriting of $(..3..)$ will be \[\frac{2\sin\theta \cos \alpha}{\sin \theta} \cdot \frac {1}{2\cos \alpha} \cdot \frac{\sin \angle RST}{\sin \angle RTS} = 1 \Rightarrow \sin \angle RST = \sin \angle RTS \Rightarrow RT = RS\].
polya78
07.06.2013 20:15
Actually, $P$ can be any point in $\triangle ADC$, as long as $S$ is chosen so that $\angle ARS =\angle PCR$, and $T$ such that $\angle TRB =\angle DQC$. (It's easy to see that this condition is satisfied in the actual problem.) Let $M=QC \cap RS$, $w$ the circumcircle of $\triangle RMC$, and $N, B'$ the intersections of $AC,BR$ with $w$. Applying Pascal's theorem to inscribed hexagon $RRB'NCM$ yields that $S,N,B'$ are collinear. Since $\angle MRQ = \angle MCR$, $AD$ is tangent to $w$ at $R$. So $\angle RB'C =\angle CRD =\angle BRD$, which means that $R$ is the midpoint of the hypotenuse of right triangle $\triangle BCB'$. Also $\angle TRB =\angle DQC =\angle MCB' =\angle MRB'$ and $\angle RB'S =\angle RCN =\angle RBT$, so $\triangle RB'S$ and $RBT$ are congruent.
Attachments:
turkey1.pdf (428kb)