$AS$ bisects $RQ$ if and only if $\dfrac{\triangle{RSA}}{\triangle{QSA}}=1$. Since $\angle{RSA}+\angle{QSA}=180^o$, so we have: \[ \dfrac{\triangle{RSA}}{\triangle{QSA}}=\dfrac{RS \cdot RA}{QS \cdot QA}=\dfrac{RS}{QS}\cdot \dfrac{RA}{QA} \] Because $\dfrac{RS}{QS}=\dfrac{BQ}{BR}=\dfrac{PB}{PR}$, $\dfrac{RA}{QA}=\dfrac{PR}{PA}$ Hence: $\dfrac{RS}{QS}\cdot \dfrac{RA}{QA}=\dfrac{PB}{PR}\cdot \dfrac{PR}{PA}=1$ Therefore: $\triangle{RSA}=\triangle{QSA}$, thus $AS$ bisects $RQ$.

Nice solution ! Second Solution HINT If $ M$ is the common point of $ AS,QR$, then $ PBMA$ is inscribed(produce PB to PBB', so $ \angle MPB= \angle SBB' = \angle MAB$). Now playing a little with angles , we find that $ \triangle MSB$ is isosceles ect. Babis

Third solution. We have $ AB$ is $ A$ - symmedian line of triangle $ AQR.$ On the other hand, $ \widehat{QAB}=\widehat{RAS},$ so $ AS$ is median of triangle $ AQR.$ Hence $ SA$ bisects $ QR$ April (Style like Babis') P.S. Thank you Mr. Babis for your remark.

April wrote: Third solution. We have $ AB$ is $ A$ - symmedian line of triangle $ AQR.$ Hallo April ! I see that '' $ RP$ is $ R$ - symmedian line of triangle $ ABR.$ ? Can you explain a little more your arguments ? Probably I do not see something. Thank you - Babis

Hi, Mr. Babis! We have $ AQBR$ is a harmonic quadrilateral, so $ A,\,B$ and $ T$ the intersection of tangent lines at $ Q$ and $ R$ of the given circle are collinear. And you known that $ AT$ is $ A$ - symmedian line of triangle $ AQR,$ i.e. $ AB$ is $ A$ - symmedian line of triangle $ AQR.$ Good luck. April.

April wrote: Hi, Mr. Babis! We have $ AQBR$ is a harmonic quadrilateral, so $ A,\,B$ and $ T$ the intersection of tangent lines at $ Q$ and $ R$ of the given circle are collinear. And you known that $ AT$ is $ A$ - symmedian line of triangle $ AQR,$ i.e. $ AB$ is $ A$ - symmedian line of triangle $ AQR.$ Good luck. April. Thank you for your nice explanation. Of course , I would not ask you this , but in the first message I did not read carefully the image. In our language letter $ P$ is our $ R$ , and by reading $ P$ as $ R$, I did not see the right triangle. A!( a typo): I think that at the end of the second sentence to the first message you wanted to write ''.....median of triangle $ AQR$ ''instead of ''triangle $ APQ$''. Now , I have read carefully your solution and I see that it is really very nice , short and genious ! Sorry , and again many thanks. babis

denote $ M$ as the intersection of $ SA,QR$. $ PA,PB$ are tangent to the circle so we get that $ P(AB,QR)=-1$ hence $ S(AB,QR)=-1$ now if we take the intersections with line $ QR$ we get that: $ (MB',QR)=-1$ where $ B'$ is the intersection of $ SB,QR$ but we had $ BS \parallel QR$ so $ B' \rightarrow \infty$ so we get that: $ (M\infty,QR)=-1$ so $ M$ is the midpoint of $ QR$ as wanted...

shobber wrote: Since $\angle{RSA}+\angle{QSA}=180^o$, so we have: \[ \dfrac{\triangle{RSA}}{\triangle{QSA}}=\dfrac{RS \cdot RA}{QS \cdot QA}=\dfrac{RS}{QS}\cdot \dfrac{RA}{QA} \] Hi shobber i have some question, can you explain me : - How we have from ? - How we get from ?

shobber meant $\triangle RSA$ to be Area of $\triangle RSA$ Now,FOR THE FIRST PART $\frac{\triangle RSA}{\triangle QSA}=\frac{\frac{1}{2}RS\times RA \times \sin(\angle {ARS}}{\frac{1}{2}QS \times QA \times \sin(\angle{SQA}}$ FOR the second part, first observe, $PR \times PQ=(PA)^2=PB \times PB$ WHICH GIVES, $\frac{PQ}{PB}=\frac{PB}{PR}$......(2) then observe that $\triangle RSB$ IS SIMILAR TO $\triangle PQB$ WHICH GIVES, $\frac{RS}{PQ}=\frac{BR}{PB}=\frac{QS}{PB}$ (SINCE, BR=QS ) so $\frac{RS}{QS}=\frac{PQ}{PB}=\frac{PB}{PR}$ (USING (2)) simillarly, $\frac{RA}{QA}=\frac{PR}{PA}=\frac{PR}{PB}$

my solution (using angle chasing) let SA intersects QR at X <AXP = <ASB = <ABP so, A,B,P,X cyclic. also, denote O is the circumcentre of the circle(gamma). <OAP = <OBP = 90 <OAP + <OBP = 180 so, A,B,O,P cyclic. so, A,O,X,P cyclic. <OXP + <OAP = 180 so, <OXQ = <OXP = 90 thus QX = RX (proved) any nicer solution

USAJMO 2011 P5

Rename some points and use phantom points. We will solve the following equivalent problem: Quote: In $\triangle ABC$ with circumcircle $\Omega$, let $D$ be a point on $\Omega$ such that $AD$ is the $A$-symmedian. Let $T$ be the midpoint of $AD$ and let $BT$ meet the line through $C$ parallel to $AD$ at $E$. Show that $E$ lies on $\Omega$. We now employ Barycentric Coordinates. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Since $D$ is on the $A$-symmedian, it can be parameterized by $D=(t:b^2:c^2)$ but it is also on $\Omega$ so $\text{Pow}_\Omega (D)=0$. Of course, $\Omega$ has equation $a^2yz+b^2zx+c^2xy=0$, so plugging in $D$ gives $a^2b^2c^2+2tb^2c^2=0 \Rightarrow t=-\frac{a^2}2$. So$$D=\left(-\frac{a^2}2,b^2,c^2\right)=(-a^2:2b^2:2c^2).$$Thus, the point at infinity along the line $A$-symmedian has coordinates $$P_\infty=(b^2+c^2:-b^2:-c^2).$$ Normalize $D$ by multiplying each component by $\frac{1}{-a^2+2b^2+2c^2}$. The midpoint formula on normalized $D$ and $A$ gives $T=(-a^2+b^2+c^2:b^2:c^2)$. Note that $E$ is the intersection of cevians $BT$ and $CP_\infty$. Points on $BT$ can be parameterized by $(-a^2+b^2+c^2:s:c^2)$ while points on $CP_\infty$ can be parameterized by $(b^2+c^2:-b^2:t)$. Hence, their intersection must have coordinates $$E=[(b^2+c^2)(-a^2+b^2+c^2):-b^2(-a^2+b^2+c^2):c^2(b^2+c^2)].$$It remains to check that $\text{Pow}_\Omega (E)=0$. Plugging $E$ into the circumcircle equation yields $$a^2[-b^2(-a^2+b^2+c^2)c^2(b^2+c^2)]+b^2[(b^2+c^2)(-a^2+b^2+c^2)c^2(b^2+c^2)]+c^2[(b^2+c^2)(-a^2+b^2+c^2)(-b^2)(-a^2+b^2+c^2)]$$which is clearly true after factoring.

clean and conceptual $\angle BED= \angle AEC$ because it's already known for us that $AB \parallel CD$. we already know that $PA$ is a tangent of the circle $O$ we know that $\angle ADC = \angle PAC $. from this information, we can show that $\triangle ACP \sim \triangle DAP$ which means that $$\frac{|AC|}{|AD|}=\frac{|PC|}{|PA|}=\frac{|EC|}{|ED|} \implies |AC|\times|ED|=|EC|\times|AD|=\frac{|AE|\times|CD|}{2}$$ $\textbf{Claim:} \triangle EAC \sim \triangle DME$ it's easy to show that claim is true since $\angle BED=\angle AEC$ and $\angle EDC=\angle EAC$ $$\frac{|AC|}{|AE|}=\frac{|DM|}{|ED|} \implies |AE|\times|ED|=|AE|\times|DM|=\frac{|AE|\times|CD|}{2}\implies |DM|=\frac{|CD|}{2} $$so $M$ is the midpoint of $CD$