Let $AD$ be the diameter of a circle $\omega$ and $BC$ is a chord of $\omega$ which is perpendicular to $AD$. Let $M,N,P$ be points on the segments $AB,AC,BC$ respectively, such that $MP\parallel AC$ and $PN\parallel AB$. The line $MN$ cuts the line $PD$ in the point $Q$ and the angle bisector of $\angle MPN$ in the point $R$. Prove that the points $B,R,Q,C$ are concyclic.
Problem
Source: Peru EGMO TST 2020 #5
Tags: geometry, angle bisector
Kagebaka
17.03.2021 22:00
This is a really cool Miquel Point configuration, and pretty tricky to unpack in my opinion. We first introduce the following general lemma: Lemma: In triangle $\triangle ABC,$ let $M,N,P$ be points on $AB,AC,BC,$ respectively, such that $MP\parallel AC$ and $PN\parallel AB.$ Then $(AMN)$ passes through the $A$-Dumpty Point (the midpoint of the $A$-symmedian) of $\triangle ABC.$ Proof: Let $D$ be the $A$-Dumpty Point. It's well-known that $D$ is the center of the spiral similarity sending $BA$ to $AC,$ so it suffices to show that the same spiral similarity also sends $M$ to $N.$ However, this just follows from the fact that $\triangle BMP\sim\triangle PNC,$ since we have \[\frac{BM}{AM}=\frac{BM}{PN}=\frac{PM}{CN}=\frac{AN}{CN},\]so we're done by the Gliding Principle. $\Box$ Back to the main problem: Let the center of $(ABC)$ be $O,$ $(AMN)$ meet $\omega$ at $X$ (the Miquel Point of $BMNC$), and $MN\cap BC=Y.$ By our lemma, $O$ lies on $(AMNX);$ since we know that $O$ lies on the angle bisector of $\angle NAM,$ this implies that $O$ is on the perpendicular bisector of $MN,$ but it is also on the perpendicular bisector of $AX$ by the definition of a circle, so in fact $AX\parallel MN.$ We also note that $O$ being the midpoint of arc $MN$ must furthermore imply that $O\to D$ under the spiral similarity sending $MN\to BC.$ On the other hand, we know that $\triangle PNM\sim\triangle CAB\sim\triangle PMC,$ so \[\frac{RM}{RN}\stackrel{\text{Angle Bisector Theorem}}{=}\frac{PM}{PN}=\frac{PC}{PB}\]implies that the spiral similarity sending $MN\to BC$ also sends $R\to P$ by the Gliding Principle. Finally, because $\triangle CAB$ is isosceles, $\triangle PNM$ and $\triangle PMC$ are as well, meaning that \[\measuredangle RPB=\measuredangle RPN+\measuredangle NPB=\measuredangle MPR+\measuredangle CPM=\measuredangle CPR\implies 90^\circ=\measuredangle CPR\implies RP\parallel OD;\]thus the spiral similarity sending $RP\to OD$ must actually be a homothety, hence $O,R,X$ and $D,P,Q,X$ are collinear. From here, noting that $DQ\perp AX\parallel MN,$ we find that $PRXY$ is a cyclic kite, so $XY$ is tangent to $(ABC)$ and $YQ\cdot YR=YX^2 = YB\cdot YC,$ which gives the desired result by Power of a Point. $\blacksquare$
A-Thought-Of-God
24.04.2021 05:03
My bash.
[asy][asy]
import math;
import olympiad;
size(5cm);
defaultpen(fontsize(11pt));
pair O = origin;
pair C = dir(330);
pair B = dir(210);
pair A = dir(90);
pair D = dir(270);
pair P = 0.6B+0.4C;
pair M = 0.6B+0.4A;
pair N = 0.6A+0.4C;
pair Q = extension(P,D,M,N);
pair H = bisectorpoint(M,P,N);
pair R = extension(P,H,M,N);
pair F1 = foot(M,P,B);
pair F2 = foot(N,C,P);
draw(M--F1,dotted);
draw(N--F2,dotted);
draw(P--R);
draw(circumcircle(B,Q,C),linewidth(1.5)+green);
draw(D--Q,linewidth(0.3)+blue);
draw(M--P--N--cycle,linewidth(1)+blue+red);
draw(A--B--C--cycle,linewidth(0.2)+red);
draw(circle(O,1));
draw(A--D);
draw(anglemark(R,P,M,t=8),blue+red);
draw(anglemark(N,P,R,t=6),blue+red);
draw(rightanglemark(M,Q,P,s=2));
draw(B--D--C);
dot("$F_1$",F1,NW);
dot("$F_2$",F2,NE);
dot("$Q$",Q,dir(330));
dot("$M$",M,W);
dot("$N$",N,E);
dot("$P$",P,S);
dot("$D$",D,S);
dot("$A$",A,dir(90));
dot("$B$",B,W);
dot("$C$",C,E);
dot("$R$",R,dir(100));
[/asy][/asy]
We begin with the following claim.
Claim. We have $\overline{PD} \perp \overline{MN}$.
Proof. Toss on the complex plane with $\odot(ABC)$ being unit circle. WLOG, assume $a=i, d=-i$. Since $B,C$ are reflections over $\overline{AD}$ we have $bc=-1$. Alternatively, just note that $b=-\overline{c}$. Since $P \in \overline{BC},$ we have $p-bc\overline{p}=b+c$ or $\overline{p} = b+c-p$. Since $\triangle BPM\sim \triangle BCA$ and $\triangle CPN \sim \triangle CBA$ we can easily get (using well-known formuals) that $$m=\frac{p(b-a)-b(c-a)}{b-c}$$and $$n=\frac{p(c-a)-c(b-a)}{c-b}$$We need to prove that
\begin{align*}
\left(\frac{m-n}{d-p}\right) + \overline{\left(\frac{m-n}{d-p}\right)} &= 0\\
\iff \frac{p(b+c-2a)-2bc+ab+ac}{(b-c)(-i-p)} &= -\frac{(p-b-c)(\frac{1}{b}+\frac{1}{c} - \frac{2}{a})-\frac{2}{bc} + \frac{1}{ab} + \frac{1}{ac}}{\left(\frac{1}{b}-\frac{1}{c}\right)(i-(p-b-c))} \\
\iff \frac{p(b+c-2i)+2+ib+ic}{i+p} &= \frac{(p-b-c)(\frac{ic+ib+2}{-i})-\frac{2i-c-b}{-i}}{(i-p+b+c)} \\
\iff \frac{pib+pic+2p+2i+2ib+2ic}{i+p} &= \frac{pic+pib+2p+\cancel{i}-b^2i-2b-c^2i+\cancel{i}-2c-\cancel{2i}+c+b}{p-b-c-i} \\
\iff p^2ib+p^2ic+2p^2+2ip-bp-cp-pib^2+pi-2pb-2ib+b^2-1 &= (i+p)(pic+pib+2p-b^2i-c^2i-b-c) \\
+pi-pic^2-2pc-2ic-1+c^2+pb+pc-2ip+2+bi+ci \\
\iff p^2ib+p^2ic+2p^2+2ip-pib^2-pic^2 &=-pc-pb+2pi+b^2+c^2-ib-ic+p^2ic+p^2ib+2p^2-b^2ip-pc^2i-pb-pc \\
-2pb-2pc-ib-ic+b^2+c^2\\
\end{align*}
which is true, both expressions are just rearranged. This establishes the claim. $\square$
Next, note that $$\angle RPB=\angle RPM+\angle MPB=\frac{1}{2}\cdot \angle NPM+\angle C=\frac{1}{2} \cdot (180^\circ-2\angle B)+\angle B=90^\circ$$implying $\overline{RP} \perp\overline{BC}$. Let $P$ divide $BC$ in $r$ to $1$. Note that by similarity $$MF_1 = \left(\frac{r}{r+1}\right) \cdot \sqrt{c^2-\frac{a^2}{4}}$$and $$NF_2 = \left(\frac{1}{r+1}\right) \cdot \sqrt{c^2-\frac{a^2}{4}}$$Let the length of $RP$ be $\ell$. Dropping perpendicular feets from $R$ onto $\overline{NF_2}$ and $\overline{MF_1}$ and calling them $X,Y$ respectively. Applying pythagorean theorem in $\triangle RNX$ and $\triangle RYM$. We have $$r=\frac{\frac{r}{r+1} \cdot b}{\frac{1}{r+1} \cdot b} = \frac{MP}{PN} = \frac{MR}{NR} = \sqrt{\frac{\left(\ell-\frac{r}{r+1} \cdot \sqrt{c^2 - \frac{a^2}{4}}\right)^2+(\frac{ra}{2(r+1)})^2}{\left(\frac{1}{r+1} \cdot \sqrt{c^2 - \frac{a^2}{4}}-\ell \right)^2 + (\frac{a}{2(r+1)})^2}}$$Solving this equation yields $$\ell=\frac{2r}{(r+1)^2} \cdot \sqrt{c^2-\frac{a^2}{4}}$$Also note that $$DC^2=(AD)^2-(AC)^2=4R^2-c^2=4\left(\frac{c^2}{2\sqrt{c^2-\frac{a^2}{4}}}\right)^2-c^2=\frac{c^4}{c^2-\frac{a^2}{4}}-c^2=\frac{\frac{c^2a^2}{4}}{c^2-\frac{a^2}{4}}$$Now we present our next claim.
Claim. We have $\triangle BPR \sim \triangle DCN$.
Proof.Note that since $\angle BPR=\angle DCN=90^\circ$. It suffices to show that $$\frac{BP}{PR}=\frac{DC}{CN}$$or
\begin{align*}
\frac{\frac{r}{r+1} \cdot a}{\frac{2r}{(r+1)^2} \cdot \sqrt{c^2-\frac{a^2}{4}}}&=\frac{\sqrt{\frac{\frac{c^2a^2}{4}}{c^2-\frac{a^2}{4}}}}{\frac{1}{r+1}\cdot c} \\
\iff ca&=2\sqrt{\frac{c^2a^2}{4}}=ca \\
\end{align*}which is true. This establishes our claim. $\square$
Note that $QNCD$ is cyclic. To finish, combing previous facts together, we have $$\angle RBC=\angle RBP=\angle NDC=\angle NQC=\angle RQC$$which shows $QRCB$ is cyclic, finishing the problem. $\blacksquare$
Jalil_Huseynov
15.04.2022 18:13
Let $\angle MPR =\angle NPR=\alpha$. Since $AMPN$ is parallelogram and $AB=AC$ we get $\angle BAC=2\alpha $ and $\angle DAC=\alpha$. Since $MP||AC$ and $\angle DAC=\angle MPR$ we get $RP||AD$. Hence, $PR\perp BC$. Let $BC\cap MN=T$. $AD\cap BC$ be origin of coordinate plane and let $A=(0,a), B=(b,0), P=(-p,0).$ Then we can easly get that $D=(0,\frac{-b^2}{a}),C=(-b,0),M=(\frac{b-p}{2},\frac{a(b+p)}{2b}),N=(\frac{-p-b}{2},\frac{a(b-p)}{2b}), T=(\frac{-(p^2+b^2)}{2p},0)$. So we get $\text{(Slope of MN)}\cdot \text{(Slope of PD)}=-1$, hence $PD\perp MN \implies TQ\cdot TR=TP^2$. Also $TP^2=(\frac{b^2-p^2}{2p})^2=\frac{(b-p)^2}{2p}\cdot \frac{(b+p)^2}{2p}=TC\cdot TB$. So we get $TQ\cdot TR=TP^2=TB\cdot TC$. Hence, $BCQR$ is cyclic. So we are done.