very easy WE WANA SHOW THAT: $ RQ*QL*LC*QC^{-1}=RP*PK*KC*PC^{-1}$ so thats enough to show$ {RQ}*{AC}={BC}*{RP}$

Very easy, too many ways to solve.

one more, easy geometry...

Here is my solution by metriccal Lemma:${ CR=\frac{a+b}{2}\cos\frac{C}{2}}$ We have $ S_{RPK}=\frac{1}{2}PK\cdot d(R,PK)$ but $ PK=\frac{a}{4}\tan\frac{C}{2}$ and $ d(R,PK)=RP\sin\angle KPC=(CR-CP)\cos\frac{C}{2}=(\frac{a+b}{2\cos\frac{C}{2}}-\frac{a}{2\cos\frac{C}{2}})\cos\frac{C}{2}=\frac{b}{2}$ Now $ S_{RPK}=\frac{ab}{8}\tan\frac{C}{2}$ similarly we get $ d(R,LQ)=\frac{a}{2}$ thus $ S_{RQL}=\frac{ba}{8}\tan\frac{C}{2}$ we are done.

I could remind the level of problem1 of last year imo;;;

Very easy problem,I solved it using trigonometry but there are simpler ways also

Dont just say its easy, come with a solution too, please.

color wrote: Dont just say its easy, come with a solution too, please. I will post it later,now I dont have enough time

My idea is showing $ S_{RKC}-S_{PKC}=S_{RLC}-S_{QLC}$, it can easily simplified to showing $ RC=\frac{BC+CA}{2cos\frac{C}{2}}$, which is trivial. (Here $ S_{RKC}=\frac{1}{4}RC\cdot BC sin \frac{C}{2}$ and $ S_{PKC}=\frac{1}{8}\cdot CA^{2}tan\frac{C}{2}$, etc)

Since $ \angle PQO=\angle QPO=90-\frac{C}{2}$ we have $ OP=OQ$ then the midpoint of $ PQ$ is the same of $ CR$ so $ CQ=PR$ therefore $ (RQL)=(CPL)$ and $ (RPK)=(CQK)$. Now let $ T$ be the projection of $ Q$ over $ BC$ so $ (CQL)=(CQT)$, so we only need to prove $ (PQL)=(QKT)$ but $ QT$ is parallel to $ PK$ then $ KT=QU=PV$ where $ U$ is the projection of $ Q$ over $ OP$ and $ V$ the projection of $ P$ over $ OQ$, also $ QL=QT$ then $ QL\cdot PV=QT\cdot KT$ and the result follows.

Dear MLs, M.T.

Proof. (Started from Jose Gabriel's idea) We have $ \widehat{OQP}=\widehat{OPQ}=90^\circ-\frac{C}{2}\implies OP=OQ.$ On the other hand, $ OC=OR,$ so $ CQ=PR$ and $ CP=QR,$ also. Since $ \triangle CQL\sim\triangle CPK\Rightarrow\frac{CQ}{CP}=\frac{QL}{PK}\Rightarrow\frac{PR}{QR}=\frac{QL}{PK}\Rightarrow PR\cdot PK=QR\cdot QL$ Note that $ \widehat{RPK}=\widehat{RQL},$ so $ S_{RPK}=S_{RQL}$

There must be a better solution, but I cannot resist any solution that uses potential. Denote the center of the circumcircle by O, and let the angle at C be gamma. Angle KPC = Angle LQC = 90-gamma/2. Therefore OQR is isosceles and OQ=OR, which then implies that the potentials of Q and P on the circle are equal. Thus PR.PC = QR.QC. Since the triangles CQL and CPK are similar we have PC.QL = PK.QC. Dividing these two proportions leads to PR.PK = QR.QL. Further, the above remark on the angles also shows that Angle LQR = Angle KPR = 90+gamma/2. Thus the areas of the triangles in question are equal (equal angle and product of the corresponding sides).

I hope this is right. Has anyone posted this solution? Well, when I drew the diagram I found two cases are possible. It is either $ P$ or $ Q$ that is closer to $ R$. I am just going to describe if $ Q$ is closer to $ R$ but the other case is exactly the same. Let: $ O$ be the circumcentre of $ \triangle ABC$. Since $ \angle RCB = \angle RCA \Rightarrow \angle CPK = \angle LQC$ and thus $ \angle RQL = \angle RPK$. So essentially we need to prove that $ RP \times PK = RQ \times QL$. Cosider $ \triangle OPQ$. Since $ \angle OQP = \angle OPQ = 90^{\circ}-\frac{C}{2}\Rightarrow OQ=OP$ and $ \angle QOP = C$. We know $ RQ = RP+PQ = RP+2OQ \sin{\frac{C}{2}}$ and we also know that $ \triangle PKC$ and $ \triangle LQC$ are similar. Therefore $ \frac{PK}{QL}= \frac{PC}{QC}$. $ \frac{PC}{QC}= \frac{QC+PQ}{QC}= 1+\frac{2 OQ \sin{\frac{C}{2}}}{QC}$. Now it is easily deduced that $ RP = QC$. So $ \frac{RQ}{RP}= \frac{PC}{QC}= \frac{PK}{QL}$ and hence the result. Edit: Oops, now I see that its the same as shunka's

Very nice and easy problem ! Quote: In the triangle $ ABC$ the bisector of the angle $ \widehat{BCA}$ intersects the circumcircle $ w=\mathcal C(O,R)$ again at the point $ R$, the perpendicular bisector of the side $ [BC]$ at the point $ P$, and the perpendicular bisector of the side $ [AC]$ at the point $ Q$. The midpoint of the side $ [BC]$ is the point $ K$ and the midpoint of the side $ [AC]$ is the point $ L$. Prove that the triangles $ RPK$ and $ RQL$ have the same area, i.e. $ [RPK]=[RQL]$. Proof. Observe that $ O\in PK\cap QL$ and prove easily that $ m(\widehat{OQP})=m(\widehat{OPQ})=90^{\circ}-\frac{C}{2}$ $ \implies$ $ \widehat{OQP}\equiv\widehat{OPQ}$ $ \implies$ $ OP=OQ=d$ $ \implies$ the points $ P$ and $ Q$ have the same power w.r.t. the circumcircle $ w$, i.e. $ p_{w}(P)=p_{w}(Q)=R^{2}-d^{2}$ $ \implies$ $ \boxed{PR\cdot PC=QR\cdot QC}$. Observe that $ \frac{PK}{PC}=\frac{QL}{QC}=\sin\frac{C}{2}$. Therefore, $ [RPK]=[RQL]$ because $ (\ \Longleftarrow\ )$ $ \left\|\begin{array}{ccc}[RPK] & = & \frac{1}{2}\cdot PR\cdot PK\cdot \sin\widehat{RPK}=\frac{1}{2}\cdot PR\cdot PC\sin \frac{C}{2}\cdot \sin\left(90^{\circ}+\frac{C}{2}\right)=\frac{1}{4}\cdot PR\cdot PC\cdot \sin C\\\ [RQL] & = & \frac{1}{2}\cdot QR\cdot QL\cdot \sin\widehat{RQL}=\frac{1}{2}\cdot QR\cdot QC\sin \frac{C}{2}\cdot\sin\left(90^{\circ}+\frac{C}{2}\right)=\frac{1}{4}\cdot QR\cdot QC\cdot \sin C\end{array}\right\|$ and $ PR\cdot PC=QR\cdot QC$.

$ \triangle{RQA}\sim \triangle{BCA}\Longrightarrow \frac{RQ}{RA}= \frac{BC}{BA}$ $ \triangle{RPB}\sim \triangle{ACB}\Longrightarrow \frac{RP}{RB}= \frac{AC}{AB}$ Since we have $ RA=RB$, so we get $ \frac{RQ}{RP}= \frac{BC}{AC}$. But because $ \angle{LQR}= \angle{KPR}$, hence $ \frac{\triangle{LQR}}{\triangle{KPR}}= \frac{QL}{PK}\cdot \frac{QR}{PR}= \frac{CA}{CB}\cdot \frac{CB}{CA}= 1$. Done.

Dear MLs, M.T.

April wrote: Proof. (Started from Jose Gabriel's idea) We have $ \widehat{OQP}=\widehat{OPQ}=90^\circ-\frac{C}{2}\implies OP=OQ.$ On the other hand, $ OC=OR,$ so $ CQ=PR$ and $ CP=QR,$ also. Since $ \triangle CQL\sim\triangle CPK\Rightarrow\frac{CQ}{CP}=\frac{QL}{PK}\Rightarrow\frac{PR}{QR}=\frac{QL}{PK}\Rightarrow PR\cdot PK=QR\cdot QL$ Note that $ \widehat{RPK}=\widehat{RQL},$ so $ S_{RPK}=S_{RQL}$ April , your nice remarks give perhaps the best and more beautifull solution to this problem .As I always use to say , many olympiad problems are based on elementary statements and very clear relations. (Of course , the solution after yours is very similar to this , done by you). Babis

$ \frac{h_{1}}{h_{2}}= \frac{KP}{QL}= \frac{a}{b}$ Call O the circumcenter, we have: $ \angle COQ = \beta$ and $ \angle POR = 180-\angle PRO-\angle OPR = 180-\left ( \frac{\gamma}{2}+\alpha-90 \right )-\left ( \frac{\gamma}{2}+90 \right ) =$$ 180-\alpha-\gamma = \beta$ then $ \angle COQ = \angle POR$ Then $ \triangle CQO \cong \triangle POR$ (ALA) and so $ CQ = PR \ \ \ \Longleftrightarrow \ \ \ \frac{PR}{QR}= \frac{b}{a}$

Refer to diagram. Notice RPK=180-KPC=90+C/2=180-LQC=RQL, so it suffices to prove RP*PK=RQ*QL by sine area formula. Notice similar triangles KPC and LQC yields $KP*QC=PC*LQ$. Then $$KPC=LQC=PQO\implies OP=OQ\implies RP*CP=CQ*QR\implies RP=CQ\implies RP*PK=CQ*PK=LQ*PC=LQ*QR,$$as desired. $\blacksquare$ Note: RP*CP=CQ*QR implies RP=CQ if the triangle is not isosceles (otherwise, both sides are 0). Fortunately, this is easily patched as if it's isosceles, the angle bisector is a perp. bisector and they concur at circumcenter, implying RP=RQ=RO and QL=OL=OK=PK. I think some people missed this case.

Drop altitudes from $R$ to $QL$ and $PK$ at $D_1$ and $D_2$, respectively. Claim: We have the similarities $(\triangle D_1RQ \sim \triangle D_2RP) \sim (\triangle KCP \sim \triangle QCL)$. Proof: Via angle chase, we have $$\angle KPC = \angle LQC = \angle OPQ.$$Therefore we have the desired similar triangles. $\square$ It suffices to show that $RD_1 \cdot QL = RD_2 \cdot PK \iff \frac{RD_1}{RD_2} = \frac{PK}{LQ}$. This is equivalent to showing that \begin{align*} \frac{RD_1}{RD_2} &= \frac{PK}{LQ} \\ \iff \frac{RQ}{RP} &= \frac{PK}{LQ} \\ \iff \frac{RQ}{RP} & = \frac{CP}{CQ}. \end{align*} However, observe that as $\triangle OQP$ is isosceles and $OR = OC$, we have that $RQ = CP$ and $RP = CQ$. Therefore we have the desired final equivalence. $\blacksquare$

First of all, note that $\triangle CQL$ ~ $\triangle CPK$. So $\angle RQL=\angle RPK$. To get the areas equal, we must show that $RP \cdot PK= RQ \cdot QL$. By the aforementioned similarity, we deduce that $PK/QL=CK/CL=BC/AC=sin BAC/sinCBA$.Note that $\angle RBA=\angle PBC$, therefore $\angle RBP=\angle ABC$. Similarly $\angle RAQ=\angle CAB$. ($*$) Also, it is easy to observe that $\angle RPB=\angle RQA$. Now we will apply Law of Sines to the triangles $RPB$ and $RQA$. By also using ($*$), we get that $RP/RQ=sinABC/sinBAC$. So, $PK/QL=RQ/RP$, as desired.

My overcomplicated solution:

Claim. $RQ=CP$ Proof. Let $O$ be the circumcenter of $\Delta ABC$ and $M$ be the midpoint of $RC$. Clearly, $M$ is also the projection of $O$ on $CR$. We also have,\begin{align*} \angle OQP &= \angle CQL \\ &= 90^{\circ} - \angle LCR \\ &= 90^{\circ} - \angle KCR \\ &= \angle CPK \end{align*}Thus, $OP=OQ$. This implies that $M$ is the midpoint of $PQ$, which gives $RQ = RM + MQ = CM + MP = CP$, as desired. $\blacksquare$ Equivalent to the claim above is $PR=CQ$, so letting $X$ be the projection of $Q$ onto $BC$, it follows that\begin{align*} \dfrac{PR}{QR} &= \dfrac{CQ}{CP} \\ &= \dfrac{QX}{PK} \\ &= \dfrac{QL}{PK} \end{align*}which implies that $PK \cdot PR = QR \cdot QL$. However,\begin{align*} \angle LQR &= 90^{\circ} + \angle LCR \\ &= 90^{\circ} + \angle RCB \\ &= \angle KPR \end{align*}so\begin{align*} [LQR] &= \dfrac 12 \cdot LQ \cdot QR \cdot \sin{LQR} \\ &= \dfrac 12 \cdot PK \cdot PR \cdot \sin{KPR} \\ &= [KPR] \end{align*}as desired. $\blacksquare$

Denote $\frac{m\angle C}{2} = \theta$ Also let $CK = a$ and $CL = b$ Claim: $\triangle AQR \sim \triangle ACB$ Proof: $m \angle AQR = \theta$. Also, $m\angle ARQ = m\angle B$. By $AA$ similarity we are done. Claim: $CP = QR$ Proof: By similarity ratios in the triangles proved before we have that $AQ = CB \cdot \frac{AQ}{AC} = \frac{CB}{2} \cdot \frac{AQ}{AL} = CP$. The last step follows from $\triangle CKP \sim \triangle ALQ$ Claim: $m \angle RPK = m\angle RQL$. Proof: Note that both are $90 + \theta$ by basic angle chasing. Now, with $CP = QR$ we can finish the problem. $$CP = QR$$$$CR = CP + CQ$$$$CR = \frac{a+b}{\cos{\theta}}$$$$(a-b) \cdot CR = \frac{(a-b)(a+b)}{\cos{\theta}}$$$$a \cdot \tan{\theta} \cdot \left( CR - \frac{a}{\cos{\theta}} \right) = b \cdot \tan{\theta} \cdot \left( CR - \frac{b}{cos{\theta}} \right)$$$$KP \cdot PR = RQ \cdot QL$$$$\frac12 KP \cdot PR \cdot \sin{\angle RPK} = \frac12 RQ \cdot QL \cdot \sin{\angle RQL}$$$$[RPK] = [RQL]$$as desired and we are done.

As $CR$ is the angle bisector, then $\angle KPC = \angle CQL$. Let $O$ be the circumcenter of $\triangle ABC$. Since $K$, $P$, $O$ are collinear and $L$, $Q$, $O$ are collinear, then $\angle KPC = \angle OPQ$ and $\angle LQC = \angle OQP$. Thus, $\angle OPQ = \angle OQP$ which implies $\triangle OQP$ is isosceles. We may conclude $CP = QR$ and $CQ = PR$. Now, \begin{align*} [RPK] &= \tfrac{1}{2} \cdot KP \cdot PR \cdot \sin \angle KPR \\ [RQL] &= \tfrac{1}{2} \cdot LQ \cdot QR \cdot \sin \angle LQR \end{align*}But, $\angle KPR = 180^{\circ} - \angle KPC$ and $\angle LQR = 180^{\circ} - \angle LQC$. Thus, $\angle KPR = \angle LQR$. Therefore, it suffices to show $KP \cdot PR = LQ \cdot QR$. As $\triangle CPK \sim \triangle CQL$, thus \begin{align*} \dfrac{CP}{PK} = \dfrac{CQ}{QL} \quad \longrightarrow \quad CP \cdot LQ = CQ \cdot KP \quad \longrightarrow \quad QR \cdot LQ = PR \cdot KP. \end{align*}Thus, we have succesfully prove $KP \cdot PR = LQ \cdot QR$ which implies $[RPK] = [RQL]$, as desired.

Let the circumcenter of $\triangle ABC$ be $O={KP}\cap{QL}$. Note that $\angle OPQ=90-\frac12\angle ACB=\angle CQL=\angle OQP$ and $\angle CPK=\angle CQL=\angle PQO\implies \triangle POQ$ is equilateral. Now for \begin{align*}[RPK]&=[RKL]\\ \iff \frac{RQ\cdot QL \sin({180-\angle CQL})}{2}&=\frac{RP\cdot KP \sin({180-\angle CPK})}{2}\end{align*}Since $180-\angle CQL=180-\angle CPK$ we just have to prove $RQ\cdot QL=RP\cdot KP$. Because of all the angle chasing we also have $\triangle CKP \sim \triangle CLQ$. So, \begin{align*}&\implies \frac{CP}{PK}=\frac{RQ}{PK}=\frac{CQ}{LQ}=\frac{RP}{LQ}\\ &\implies RP\cdot KP=RQ\cdot QL \end{align*}

Compute $RP = RC - PC = 2R \sin {B + \frac C2} - R \sin A \frac{1}{\cos \frac C2}, PK = R \sin A \tan \frac C2$, so the area of $RPK$ is $\frac 12 (\sin 90 + \frac C2) R^2 (2\sin B + \frac C2 - \sin A \frac{1}{\cos \frac C2}) (\sin A \tan \frac C2)$, and we can find the area of $RQL$ symmetrically. We can cancel out terms that are obviously symmetric in $A,B$. It remains to show $(2 \cos \frac C2 \sin B + \frac C2 - \sin A)(\sin A)$ is symmetric in $A,B$. By product to sum, we have $2 \cos \frac C2 \sin B + \frac C2 = \sin A + \sin B$, so plugging this back in we have $\sin B \sin A$, so we are done.

If $AC = BC$ the result is clear, so assume otherwise. Let $\alpha = \angle ACR$. Claim: $RC = \frac{a+b}{2\cos\alpha}$. Proof. From Law of Sines on $\triangle RAB$ we get $$\frac{RA}{AB} = \frac{1}{2\cos \alpha}.$$From Ptolemy, $$RC = \frac{RA(a+b)}{AB} = \frac{a+b}{2\cos\alpha}. \quad \square$$ We have $$[RPK] = [RCK] - [PCK] = \frac{a}{4}RC\sin\alpha - \frac{a^2}{8}\tan\alpha,$$and similarly $$[RQL] = [RCL] - [QCL] = \frac{b}{4}RC\sin\alpha - \frac{b^2}{8}\tan\alpha.$$Then rearranging gives that $RC = \frac{a+b}{2\cos\alpha}$ is equivalent to $[RPK] = [RQL]$, as desired.

Very easy indeed Claim:$\color{blue}{[\Delta RPK]=[\Delta RQL]}$ Proof: Call $(ABC)$ as $\mho$. Clearly, $O=PK \cap QL$ is the circumcenter of triangle $ABC$. Also, $\angle{ACR}=\angle{BCR}=\frac{\angle C}{2}$ thus, $\angle{OQP}=\angle{OPQ}=90^{\circ}-\frac{\angle C}{2} \implies \boxed{OP=OQ}$. Hence, $R^{2}-OP^{2}=R^{2}-OQ^{2}$ implying power of $P$ and $Q$ with respect to $\mho$ are equal. Thus we have $RP.PC=RQ.QC$ but note that, $\sin \left(\frac{C}{2}\right)=\frac{PK}{PC}=\frac{QL}{QC} \implies \boxed{RP.PK=RQ.QL}$. We also have, $\angle{RPK}=\angle{RQL}=90^{\circ}+\frac{\angle{C}}{2}$ thus, $\frac{1}{2}RP.PK.\sin (\angle{RPK})=\frac{1}{2}RQ.QL.\sin(\angle{RQL}) \implies [\Delta RPK]=[\Delta RQL]$ as desired. $\blacksquare$ ($\mathcal{QED}$)

Note that $$[\triangle RPK] = [\triangle RQL] \iff PR \cdot PK \cdot \sin \angle RPK = QR \cdot QL \cdot \sin \angle RQL.$$However, $\angle RPK = 90^\circ + \angle RCK = 90^\circ + \angle RCL = \angle RQL,$ so it suffices to show that $PR \cdot PK = QR \cdot QL.$ However, because $\frac{PK}{PC} = \sin \angle PCK = \sin \angle QCL = \frac{QL}{QC},$ it is enough to show that $PR \cdot PC = QR \cdot QC,$ or that $P, Q$ are symmetric with respect to the midpoint of $RC.$ To show this, let $O$ be the circumcircle of the triangle, then $\angle OPQ = 90^\circ - \angle PCK = 90^\circ - \angle QCL = \angle PQO,$ so $OP=OQ.$ If $P=Q,$ it follows that $O=P=Q$ and the bisector is also a perpendicular bisector, and it is easy to show that $O$ is the midpoint of $RC,$ so this case is finished. If $P \neq Q,$ we would be done immediately. QED

why this is a very skididi rizz Ohio mogging fanum tax tiktok rizz party cameraman blue shirt guy gyatt gigachad problem

idk what to do im just dum