Let $ABC$ be an acute-angled triangle, and $M$ be the midpoint of the minor arc $BC$ of its circumcircle. A circle $\omega$ touches the side $AB, AC$ at points $P, Q$ respectively and passes through $M$. Prove that $BP + CQ = PQ$.
Problem
Source: Sharygin Correspondence Round 2024 P3
Tags: geometry
InterLoop
06.03.2024 14:06
neat
Let $X$ be the point such on $PQ$ such that $BP = PX$. Let $N = AM \cap PQ$.
Claim 1. $IP = IQ$ and furthermore $\angle MPB = \angle MPQ = \angle MQP = \angle MQC$.
$\textsf{\textbf{Proof of Claim.}}$ Note that $AP = AQ$, which implies $\triangle PAQ$ is isosceles, and thus the angle bisector of $\angle PAQ$ is the same as the altitude from $A$ to $PQ$ and the perpendicular bisector of $PQ$. Since the angle bisector of $\angle PAQ$ is $AM$, we thus know that $AM \cap PQ = N$ is the midpoint of $PQ$, and $\angle PNA = \angle PNI = 90^{\circ}$. Since $N$ is the midpoint this clearly means that $\triangle PIQ$ is isosceles with $IP = IQ$.
For the next part of the proof, simply note using the alternate segment theorem of tangents that
$$\angle MPB = \angle MQP = \angle MPQ = \angle MQC$$so the claim is proven.
$\blacksquare$
Claim 2. $\triangle BPM \cong \triangle XPM$, and thus $BM = MX$.
$\textsf{\textbf{Proof of Claim.}}$ We know $BP = XP$ (by definition), $PM = PM$ and also $\angle MPB = \angle MPX$ by $\textbf{\textsf{Claim 1}}$. Thus using SAS criterion of congruence, the claim is proven.
$\blacksquare$
Claim 3. $\triangle CQM \cong \triangle XQM$, and thus $CQ = QX$.
$\textsf{\textbf{Proof of Claim.}}$ We know that $MB = MC$ since $M$ is the midpoint of arc, thus we have $MC = MX$ since $MX = MB$. Next, we know that $QM = QM$ and $\angle MQX = \angle MQC$. We also get $\angle MXQ = \angle MCQ$ since
$$\angle MCQ = \angle MCA = 180^{\circ} - \angle MBA = 180^{\circ} - \angle MBP = 180^{\circ} - \angle MXP = \angle MXQ$$Thus since these two pairs of angles are equal, we must have $\angle QMX = \angle QMC$. However now we can use SAS criterion of congruence so we are done, and we get $CQ = QX$.
$\blacksquare$
Hence we have
$$PQ = PX + XQ = BP + CQ$$so we are done.
Tintarn
06.03.2024 14:21
Define $X$ on $PQ$ with $XB=XP$. We want to show $XQ=QC$.
Define $C'$ on $QC$ such that $XQ=QC'$. It suffices to show that $C'$ is on $(ABM)$ so that $C=C'$ and the claim follows.
But an easy angle chase shows that $M$ lies on both angular bisectors of $\angle BPQ$ and $\angle PQC$ so that $BMXP$ and $C'MXQ$ are kites and we are now done by another easy angle chase.
idkk
06.03.2024 15:05
Easily done by Trig Claim: $AM$bisects $\angle{A}$ and $O$ lies on $AM$. Proof: As $MC=MB$ as its midpoint. So $\angle{MAB}=\angle{MCB}=\angle{MBC}=\angle{MAC}$ So $AM$ bisects $\angle{A}$. and $OP=OQ$ and $AP=AQ$[tangent from same point] and $AO=AO$[common] $\Delta AOP$ and $\Delta AOQ$ are congruent. [SSS] so $\angle{PAO}=\angle{OAQ}$ hence it also lies on bisector hence on $AM$ . Now $\angle{OAP}=\frac{\angle{A}}{2}$ Let $AO=x$ and $AM=d$. Then $OM=d-x$ and $OP=\sin(\frac{\angle{A}}{2})x$ Both of them are equal so $x=\frac{d}{\sin(\frac{\angle{A}}{2})+1}$ So $AP=AQ=\cos(\frac{\angle{A}}{2})\frac{d}{\sin(\frac{\angle{A}}{2})+1}$ Let $AO$ intersect $PQ$ at $D$ Then $D$ is midpoint of $PQ$ And $DP=\frac{\sin(\frac{\angle{A}}{2})\cos(\frac{\angle{A}}{2})}{\sin(\frac{\angle{A}}{2})+1}d$ so $QP=\frac{2\sin(\frac{\angle{A}}{2})\cos(\frac{\angle{A}}{2})}{\sin(\frac{\angle{A}}{2})+1}d$ By angle chasing $\angle{AMB}=\angle{ACB}$ and $\angle{CBM}=\frac{\angle{A}}{2}$ and hence $\angle{ABM}=\frac{\angle{A}}{2}+\angle{B}=90+\frac{\angle{B}}{2}-\frac{\angle{C}}{2}$ So sine rule gives. $AB=\frac{\sin(\angle C)}{\cos(\frac{\angle B-\angle C}{2})}d$ Similar computation for $AC$ Gives $AC=\frac{\sin(\angle B)}{\cos(\frac{\angle B-\angle C}{2})}d$ as $QC+PB=AB+AC-AO-AP=d(\frac{\sin(B)+\sin(C)}{\cos(\frac{B-C}{2})})-2\frac{\cos(\frac{A}{2})}{\sin(\frac{A}{2})+1}d$ Now observe that $\frac{\sin(B)+\sin(C)}{\cos(\frac{B-C}{2})}=\frac{\sin(\frac{B+C}{2}+\frac{B-C}{2})+\sin(\frac{B+C}{2}-\frac{B-C}{2})}{\cos(\frac{B-C}{2})}$ By using $sin(x+y)$ and $\sin(x-y)$ formula It simplfies to $\frac{2\cos(\frac{B-C}{2})\sin(\frac{B+C}{2})}{cos(\frac{B-C}{2})}=2cos(\frac{A}{2})$ so $QC+PB=d(2\cos(\frac{A}{2}))-2\frac{\cos(\frac{\angle{A}}{2})}{\sin(\frac{\angle{A}}{2})+1})$ $=\frac{2\cos(\frac{A}{2})\sin(\frac{A}{2})}{\sin(\frac{\angle{A}}{2})+1}d=PQ$
sami1618
09.06.2024 00:57
Claim: $MP=MQ$ Notice $M$ lies on the angle bisector and $AP=AQ$. Claim: $PM$ bisects $\angle BPQ$ $$\angle BPM=\angle PQM=\angle QPM$$Claim: $QM$ bisects $\angle CQP$ $$\angle CQM=\angle QPM=\angle PQM$$Claim: $\angle PBM+\angle QCM=180^{\circ}$ $$\angle PBM+\angle QCM=\angle ABM+\angle ACM=180^{\circ}$$ To finish notice we can 'fold' over triangle $PBM$ and $QCM$ onto $PQM$, implying $BP+CQ=PQ$.
Attachments:
ehuseyinyigit
06.08.2024 02:26
An easier solution here: Let $\angle APQ=\angle AQP=\angle PMQ=\alpha$ and $\angle BMP=\theta$. Thus, $\angle A=180-2\alpha\Longleftrightarrow \angle BMC=2\alpha$ which leads to $\angle QMC=\alpha-\theta$. By Incenter Lemma $BM=MC=IM$. Then, using $BM=MC$ we can rotate the $\triangle BPM$ with the angle $2\alpha-2\theta$ such that point $B$ coincides with point $Q$ and the rotation of $P$ be $P'$. You get that $PM=P'M$, $BP+CQ=QP'$ and $\angle PMQ=\alpha=\alpha-\theta+\theta=\angle QMC+\angle CMP'=\angle QMP'$ which shows that $PMP'Q$ is a kite, implying $QP'=BP+CQ=PQ$.
khina
06.08.2024 03:04
By angle chasing $MP$ bisects $\angle{BPQ}$ and $MQ$ bisects $\angle{CQP}$. Now let $D, E, F$ be the projections of $M$ onto $PQ, AB, AC$. Then evidently $PQ = PE + QF$, but $BE = CF$ as $MBE$ and $MCF$ are congruent (in fact rotations of each other), so we are done.
L13832
28.10.2024 19:30
Let the center of $\odot(MPQ)$ be $O$ and $I$ be the incenter of $\triangle APQ$, note that $O\in AM$ because $\triangle APO\cong \triangle AQO$. We have $\angle BPM=\angle PQM=\angle QPM$ and $\angle CPM=\angle QPM=\angle PQM$, so $M$ is the excenter of $\triangle APQ$ and by Fact 5 on $\triangle APQ$, $IO=PO=QO$ and similarly $IM=BM=SM$ by Fact 5 on $\triangle ABC$. Now we drop perpendiculars from $M$ onto line $AB, AC$ to intersect them at points $R,S$ respectively and we luse the fact that $AM$ is perpendicular bisector of $PQ$, so we get $$ BP+CQ=RP-BR+CS+SQ=RP+SQ=PQ $$[asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.6) + fontsize(8.9); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.11429365529535, xmax = 7.687343165059565, ymin = -7.286417200118921, ymax = 7.165139940037366; /* image dimensions */ /* draw figures */ draw(circle((-0.5,0.5), 5.70087712549569), linewidth(0.6)); draw(circle((-1.568349272152271,-2.0871343831428515), 3.2919240624387522), linewidth(0.6)); draw((-4.834883024886092,-1.679064199088736)--(0.7594084180977125,0.24059158190228747), linewidth(0.6)); draw((-4.834883024886092,-1.679064199088736)--(-0.5,-5.20087712549569), linewidth(0.6)); draw((0.7594084180977125,0.24059158190228747)--(-0.5,-5.20087712549569), linewidth(0.6)); draw((-4,5)--(-0.5,-5.20087712549569), linewidth(0.6)); draw(circle((-2.784395088473757,1.456685697461149), 3.7466556008105587), linewidth(0.6)); draw((-4,5)--(-5.201646415445623,-4.613171323564987), linewidth(0.6)); draw((-4,5)--(4,-3), linewidth(0.6)); draw((-5.201646415445623,-4.613171323564987)--(-0.5,-5.20087712549569), linewidth(0.6)); draw((-0.5,-5.20087712549569)--(2.850438562747845,-1.8504385627478448), linewidth(0.6)); draw((-5,-3)--(4,-3), linewidth(0.6)); /* dots and labels */ dot((-4,5),linewidth(3pt) + dotstyle); label("$A$", (-4.273221889095961,5.149528023120832), NE * labelscalefactor); dot((-5,-3),linewidth(3pt) + dotstyle); label("$B$", (-5.41413429489776,-3.293223779812578), NE * labelscalefactor); dot((4,-3),linewidth(3pt) + dotstyle); label("$C$", (4.188545120600715,-3.2361781595224874), NE * labelscalefactor); dot((-0.5,-5.20087712549569),linewidth(3pt) + dotstyle); label("$M$", (-0.47018053642329777,-5.632094211706293), NE * labelscalefactor); dot((-4.834883024886092,-1.679064199088736),linewidth(3pt) + dotstyle); label("$P$", (-5.223982227264127,-1.90511368608704), NE * labelscalefactor); dot((0.7594084180977125,0.24059158190228747),linewidth(3pt) + dotstyle); label("$Q$", (0.8418687302487711,0.3576959187532207), NE * labelscalefactor); dot((-1.568349272152271,-2.0871343831428515),linewidth(3pt) + dotstyle); label("$O$", (-1.382910461064737,-2.361478648407765), NE * labelscalefactor); dot((-2.6366714874189947,1.026529531850388),linewidth(3pt) + dotstyle); label("$I$", (-2.5618532803932625,1.1373193960511256), NE * labelscalefactor); dot((-5.201646415445623,-4.613171323564987),linewidth(3pt) + dotstyle); label("$R$", (-5.6042863625313935,-4.605273046484662), NE * labelscalefactor); dot((2.850438562747845,-1.8504385627478448),linewidth(3pt) + dotstyle); label("$S$", (2.933541474218736,-1.7339768252167682), NE * labelscalefactor); dot((-2.0376911334847834,-0.7192204655978417),linewidth(3pt) + dotstyle); label("$J$", (-2.0104122842557266,-0.593064419414956), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */
The motivation to drop perpendiculars was fairly obvious because then we'd get congruent triangles and a new way to express the length conditions which is always worth exploring.
TestX01
29.10.2024 01:50
Note that by angle chase, as $APQO$ is clearly cyclic, $O$ is centre of the circle, we have \[\measuredangle APQ=\measuredangle AOQ=\measuredangle MOQ=2\measuredangle MPQ\]Which tells us $PM$ bisects $\angle BPQ$. Similar result holds for $Q$, now reflect $B$ over $PM$ to get $B'$, reflect $C$ over $QM$ to get $C'$. By the bisector property these lie on $PQ$. Note that \[\measuredangle MCA=\measuredangle QC'M=\measuredangle MBP=\measuredangle PB'M=\angle QB'M\]Which means $B'=C'$ and the conclusion is trivial as $BP+CQ=B'P+QC'=B'P+B'Q=PQ$.