Help my diagram has too many points

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Help my diagram has too many points

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MarkBcc168

1595 posts

MarkBcc168

#1 h Jul 17, 2024, 12:01 PM • 5 Y

Y by OronSH, peace09, Rounak_iitr, ehuseyinyigit, Funcshun840

Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ . A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$ . Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$ . Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$ .

Prove that $\angle BXP = \angle CXQ$ .

Kian Moshiri, United Kingdom

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math90

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math90

#2 h Jul 17, 2024, 12:02 PM • 2 Y

Y by Rounak_iitr, ehuseyinyigit

Solution

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popop614

271 posts

popop614

#3 h Jul 17, 2024, 12:03 PM • 3 Y

Y by peace09, OronSH, Rounak_iitr

Liked this one :3

sol

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bin_sherlo

707 posts

bin_sherlo

#4 h Jul 17, 2024, 12:05 PM • 2 Y

Y by Rounak_iitr, ehuseyinyigit

Claim $1$ : $AD$ is the angle bisector of $\angle A$ .
Proof $1$ : Invert from $A$ . $B^*C^*\parallel \overline{P^*D^*Q^*}$ and $\overline{P^*D^*Q^*}$ is tangent to $(AB^*C^*D^*)$ hence $D^*$ is the midpoint of the arc $B^*C^*$ not containing $A$ which gives that $AD^*$ is the angle bisector. $\square$

Claim $2$ : $PQ\parallel BC$ .
Proof $2$ : Let $O$ be the circumcenter of $(ADPQ)$ . Both $O$ and $D$ lye on the perpendicular bisector of $PQ$ and $PQ\perp OD\perp BC$ which gives the result. $\square$

Claim $3$ : $A,K,D$ are collinear.
Proof $3$ : By trigonometric ceva, we have
$1=\frac{BD}{DC}.\frac{AC}{AB}=\frac{BM}{BP}.\frac{CQ}{CN}=\frac{\sin BPM}{\sin PMB}.\frac{\sin CNQ}{\sin NQC}=\frac{\sin APK}{\sin KPQ}.\frac{\sin PQK}{\sin KQA}=\frac{\sin KAP}{\sin QAK}$ Thus, $\angle KAP=\angle QAK\iff AK$ is the angle bisector of $\angle A\iff A,K,D$ are collinear. $\square$

Claim $4$ : $X\in (IDM),(JDN).$
Proof $4$ :
$\angle IMD=\angle IPQ=\angle IJQ=\angle IJK=\angle IXK=\angle IXD$ $\angle DNJ=\angle PQJ=\angle PIJ=\angle KIJ=\angle KXJ=\angle DXJ$ These give that $X,I,D,M$ and $X,J,D,N$ are cyclic. $\square$

Claim $5$ : $XM\parallel AB$ and $XN\parallel AC$ .
Proof $5$ : We have $KA.KD=KI.KP$ and $KX.KD=KI.KM$ By dividing these we get $\frac{KA}{KX}=\frac{KP}{KM}\iff AP\parallel XM$ Similarily we have $KA.KD=KJ.KQ$ and $KX.KD=KJ.KN$ By dividing these we get $\frac{KA}{KX}=\frac{KQ}{KN}\iff AQ\parallel XN$ . which gives the desired result. $\square$

Claim $6$ : $X\in (PDC),(QDB)$ .
Proof $6$ : $1=\frac{DB}{BM}=\frac{DA}{AX}\implies AX=AD$ .
Let $PC\cap (APDQ)=K$ and $PC\cap AD=T$ . Let's show that $AK\parallel XC$ which proves the result since $\frac{TA}{TK}=\frac{TP}{TD}\overset{?}{=}\frac{TX}{TC}$
$\frac{TA}{AD}=\frac{TA}{AX}\overset{?}{=}\frac{TK}{KC}\implies \frac{DK}{AP}=\frac{TK}{TA}\overset{?}{=} \frac{KC}{AD}$ $CKD\sim CDP$ and $ADC\sim APD$ hence
$\frac{DP}{DC}=\frac{DK}{KC}\overset{?}{=}\frac{AP}{AD}$ which is true. Thus, $X,P,D,C$ are cyclic and similarily we get that $X,Q,D,B$ are cyclic. $\square$

Claim $7$ : $\angle PXB=\angle CXQ$ .
Proof $7$ :
$\angle PXB=\angle DXB-\angle DXP=\angle DQB-\angle DCP$ $\angle CXQ=\angle CXD-\angle QXD=\angle CPD-\angle QBD$ $\angle DQB-\angle DCP=\angle CPD-\angle QBD\iff \frac{\angle A}{2}=\angle QDC=\angle DQB+\angle QBD=\angle DCP+\angle CPD=\angle BDP=\frac{\angle A}{2}$ As desired. $\blacksquare$

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MarkBcc168

1595 posts

MarkBcc168

#5 h Jul 17, 2024, 12:06 PM • 4 Y

Y by peace09, OronSH, levifb, Rounak_iitr

We spam lots of midpoints. Let $P_1$ , $Q_1$ , $B_1$ , $C_1$ , and $A_1$ be the midpoints of $DP$ , $DQ$ , $DB$ , $DC$ , and $DA$ , respectively. Also, it's well-known that $AD$ bisects $\angle BAC$ .

Part 1. $\boldsymbol K$ lies on $\boldsymbol{AD}$ .

Notice that $MP\parallel BP_1$ and $\triangle BDP\cup P_1\sim\triangle BDA\cup A_1$ . These two give
$\measuredangle APK = \measuredangle BPM = \measuredangle PBP_1 = \measuredangle ABA_1,$ and similarly, $\measuredangle AQK = \measuredangle ACA_1$ . Thus, point $K$ in $\triangle APQ$ and point $A_1$ in $\triangle ABC$ are isogonal conjugate, implying that $K\in AD$ .

Part 2. $\boldsymbol{XA=AD}$

First, notice that $\measuredangle IXK = \measuredangle IJK = \measuredangle IPQ = \measuredangle KMD$ , so $XIMD$ is cyclic. Thus, $\measuredangle MXD = \measuredangle KID = \measuredangle PID = \measuredangle PAD$ , so $MX\parallel AB$ . Similarly, $NX\parallel AC$ . Thus, $\triangle XMN$ and $\triangle ABC$ are homothetic at $D$ with ratio $1:2$ , so we are done.

Part 3. Finish

Notice that $\measuredangle BXP = \measuredangle B_1AP_1$ , and similarly $\triangle CXQ = \measuredangle C_1AQ_1$ . However, observe
$\begin{align*} \triangle APD\cup P_1\sim\triangle ADC\cup C_1 &\implies \measuredangle P_1AD = \measuredangle C_1AC \\ \triangle ABD\cup B_1\sim\triangle AQC\cup Q_1 &\implies \measuredangle B_1AD = \measuredangle Q_1AC \\ \end{align*}$ Subtracting these two equations gives $\measuredangle B_1AP_1 = \measuredangle Q_1AC_1$ .

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squarc_rs3v2m

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squarc_rs3v2m

#6 h Jul 17, 2024, 12:38 PM

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Let $L$ be the midpoint of minor arc $BC$ ; $D$ is $AL \cap BC$ by shooting lemma. Note that if $PQ$ meets $AD$ at $T$ , dilating at $A$ gives $APDTQ \stackrel{+}{\sim} ABLDC$ and so $PT/TQ = BD/DC = MD/DN$ means $MP$ , $DT$ , $QN$ intersect at a dilation centre $K$ . Now we do a long length bash: $\frac{AK}{KT} = \frac{AP}{PT} \frac{\sin \angle KPA}{\sin \angle TPK} = \frac{AB}{BD} \frac{\sin \angle MPB}{\sin \angle BMP} - \frac{AB}{BP},$ but $BP \cdot BA = BD^2$ and $\frac{AB}{AC} = \frac{BD}{DC}$ so $\frac{AB \cdot AC}{BD \cdot DC} = \frac{AK}{KT}$ . $BD \cdot DC = AD \cdot DL$ so $AB \cdot AC \cdot KT = AK \cdot AD \cdot DL$ ; $\frac{AD}{DT} = \frac{AL}{LD}$ by dilation at $A$ so $x \cdot KT = AK \cdot DT$ where $x = \frac{AB \cdot AC}{AL}$ . But, inverting at $K$ to fix $(PQIJ)$ , $X$ swaps with $T$ , so $XK \cdot KT = AK \cdot KD$ and this rearranges to $XA \cdot KT = AK \cdot DT$ and so we get $\frac{AX}{AQ} = \frac{AC}{AQ} \cdot \frac{AB}{AL} = \frac{AL}{AD} \cdot \frac{AB}{AL}$ by dilation at $A$ , so by easy angle chase since $\triangle BPD \sim \triangle BDA$ we get $\triangle QDB \sim \triangle QAX$ which is the key idea. This gives us Miquel diagram so that $XBDQ$ and similarly $XCDP$ is cyclic and now it's an easy angle chase.

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TestX01

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TestX01

#7 h Jul 17, 2024, 12:46 PM • 2 Y

Y by GeoKing, ehuseyinyigit

how do people writeup so quickly

Claim: $K$ lies on $AD$

Intersect $MP$ with $AD$ at $K_1$ , and $NQ$ with $AD$ at $K_2$ . First of all, $PQ\parallel BC$ by homothety. Now, by Menelaus,
$\frac{AK_1}{K_1D}=2\frac{BP}{PA}=2\frac{CQ}{QA}=\frac{AK_2}{K_2D}$ by addendo on parallel lines.

Claim: $XPDC$ , $XQDB$ cyclic

There are clearly synth ways to do this, however, the fast solution I found is by inversion. Invert at $D$ swapping $B,C$ . $A$ is sent to the midpoint of arc $BC$ , and $P,Q$ to points such $DA'P'C$ , $DA'Q'B$ are isosceles trapeziums. In fact, as $\measuredangle QDP=\measuredangle QAP=\angle B+\angle C$ , $BQ'$ is parallel to $CP'$ hence we actually have the two trapeziums congruent.

Now, $N'$ is the midpoint of $BD$ , $M'$ midpoint of $DC$ . In addition, $K'=(DN'Q')\cap (DM'P')$ , and $J'=(DM'P')\cap P'Q'$ , $I'=(DN'Q')\cap P'Q'$ . Note that reconstructing $I',J'$ as midpoints of $Q'A', P'A'$ , for example $Q'N'=J'C'=DJ'$ from parallels and trapezium, hence they work.

Let $Y$ be the reflection of $K'$ over $A'$ , by our previous lemma $D,A',K',Y$ collinear. By Reim from parallel lines & homothety $DP'YQ'$ is cyclic. Dilating factor half at $A'$ gives $X'$ as the midpoint of $A'D$ , note inversion means $(K'I'X'J')$ cyclic. Now the collinearities are trivial, say by $180^\circ$ rotation at $X'$ .

Now, in our initial diagram, we want $XP,XQ$ isogonal, or $\measuredangle PXC=\measuredangle PDC$ , $\measuredangle BXQ=\measuredangle CDQ$ . The two angles are obviously equal by Fact 5 and alternate segment theorem.

Attachments:

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Reason: diagram

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Aiden-1089

278 posts

Aiden-1089

#8 h Jul 17, 2024, 12:53 PM

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TestX01 wrote:

how do people writeup so quickly

They probably type their solutions up beforehand (since they got the problems from TSTs/training), then release them as the problems go online.

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ohhh

46 posts

ohhh

#9 h Jul 17, 2024, 2:27 PM

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MarkBcc168 wrote:

Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ . A circle $\Gamma$ is internally tangent to $\omega$ and also tangent to $BC$ at $D$ . Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$ . Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$ .

Prove that $\angle BXP = \angle CXQ$ .

I think the problem is incomplete, since apparently everyone assumed that the point of tangency is A

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ihatemath123

3445 posts

ihatemath123

#10 h Jul 17, 2024, 2:37 PM • 1 Y

Y by GrantStar

It's well known that $\overline{AD}$ bisects $\angle A$ and that $\overline{PQ} \parallel \overline{BC}$ .

Claim: $A$ , $K$ and $D$ are collinear.
Proof: $\overline{BC}$ is sent to $\overline{PQ}$ via a homothety centered at $A$ while $\overline{PQ}$ is sent to $\overline{MN}$ via a homothety centered at $K$ . The composite of these two homotheties is one homothety centered at $D$ , so with an argument similar to that of Monge's theorem, the claim follows.

Clam: $A$ is the midpoint of $\overline{XD}$ .
Proof: By Reim's theorem, the tangent to $(IJKX)$ at $K$ is parallel to $\overline{BC}$ ; if $XJKK$ is cyclic, by Reim's theorem again, $XJDN$ is cyclic; since both quadrilaterals $AJDQ$ and $XJDN$ are cyclic, it follows from Reim's theorem again that $\overline{XN} \parallel \overline{AC}$ , which implies the claim.

Now, just take a $\sqrt{mn}$ -inversion centered at $X$ which swaps points $M$ and $N$ , which finishes.

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#12 h Jul 18, 2024, 2:11 PM • 2 Y

Y by ehuseyinyigit, Rounak_iitr

Note that since $\Gamma$ is tangent to $\omega$ , we have that $PQ \parallel BC$ .
Claim $1: AD$ is the angle bisector of $BAC$ .
Proof $: \angle BDA = \angle ADP + \angle PDB = \angle AQP + \angle PAD = \angle C + \angle PAD$ so $\angle PAD = \angle DAC$ .
Claim $2: K$ lies on $AD$ .
Proof $: \frac{\sin{PAK}}{\sin{QAK}} = \frac{PK}{KQ}.\frac{\sin{APK}}{\sin{AQK}}$ . Note that $MB^2 = BD^2 = BP.BA$ so $PAM$ is tangent to $MN$ . similarly $QAN$ is tangent to $MN$ . Note that $\frac{PK}{KQ}.\frac{\sin{APK}}{\sin{AQK}} = \frac{QC}{CN}.\frac{MB}{BP} = \frac{QC}{PB}.\frac{MB}{NC} = \frac{AC}{AB}.\frac{BD}{CD} = 1$ so $K$ lies on angle bisector of $BAC$ which is $AD$ .
Claim $3: XIDM$ and $XJDN$ are cyclic.
Proof $:$ Note that $\angle DMI = \angle QPI = \angle QJI = \angle KJI = \angle KXI = \angle DXI$ . similarly we can prove $XJDN$ is cyclic.
Claim $4: XM \parallel AB$ and $XN \parallel AC$ .
Proof $:$ Note that $\angle MXD = \angle MID = \angle PID = \angle PAD$ so $XM \parallel AB$ . similarly we can prove $XN \parallel AC$ .
Let $PQ$ meet $XM$ and $XN$ at $T_M,T_N$ .
Claim $5 : XT_MDQ$ and $XT_NDP$ are cyclic.
Proof $:$ Note that $\angle T_MXD = \angle PAD = \angle PQD = \angle T_MQD$ . similarly we can prove $XT_NDP$ is cyclic.
Claim $6 : T_MBDQ$ and $T_NCDP$ are cyclic.
Proof $:$ Note that $T_MP \parallel BC$ and $T_MX \parallel AB$ so $T_MPBM$ is parallelogram and since $MB=BD$ , $T_MPDB$ is also parallelogram so $\angle BT_MQ = \angle BT_MP = \angle PDB = \frac{A}{2} = \angle QDC$ so $T_MBDQ$ is cyclic. we can similarly prove that $T_NQCN$ and $T_NQDC$ are parallelogram and so $T_NCDP$ is cyclic.

Note that since $T_MBDQ$ and $XT_MDQ$ are cyclic, we have that $XBDQ$ is cyclic as well. similarly we can prove $CDPX$ is cyclic. Now note that $\angle BXQ = \angle QDC = \frac{A}{2} = \angle PDB = \angle PXC \implies \angle BXP = \angle QXC$ .

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pi_quadrat_sechstel

588 posts

pi_quadrat_sechstel

#13 h Jul 18, 2024, 9:16 PM

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MarkBcc168 wrote:

Pascal and inversion forever!

Let $S$ be the $A$ -southpole of $ABC$ and let $\Omega$ be the circumcircle of $IJK$ . Let $E\neq J,G,L$ be the intersections of $DJ$ with $\Omega,AB,MP$ respectivly and let $F\neq I,H,R$ be the intections of $DI$ with $\Omega,AC,NQ$ respectivly.

There is a homothety $\Phi$ with center $A$ mapping $\Gamma$ to $\gamma$ . This map sends $P,Q,D$ to $B,C,S$ . Thus $PQ\parallel BC$ and $\frac{AP}{PB}=\frac{AQ}{QC}$ . We also get that $D$ lies on the $A$ -internal angle bisector of $ABC$ .

Let $K'$ be on $AD$ so that $\frac{AK'}{K'D}=\frac{1}{2}\cdot\frac{AP}{PB}=\frac{1}{2}\cdot\frac{AQ}{QC}$ . Then we have $\frac{AP}{PB}\cdot\frac{BM}{MD}\cdot\frac{DK'}{K'A}=\frac{AQ}{QC}\cdot\frac{QN}{ND}\cdot\frac{DK'}{K'A}=-1$ . So by Menelaus $K'$ lies on $MP$ and $NQ$ . Thus $K'=K$ and $K$ lies on $AD$ .

By Pascal's on $DDIPQA$ and $DDJQPA$ we get that $K,G$ and $H$ lie on a line parallel to $BC$ . By Pascal's on $DDIPQJ$ we get that $LP$ is parallel to $BC$ . Thus
$\frac{KL\cdot KI}{KR\cdot KJ}=\frac{KP\cdot KI}{KQ\cdot KJ}=1$ and $IJLR$ is a cyclic quadrilateral. So we have
$\frac{DE}{DL}:\frac{DF}{DR}=\frac{DE\cdot DJ}{DL\cdot DJ}:\frac{DF\cdot DI}{DR\cdot DI}=\frac{DE\cdot DJ}{DF\cdot DI}:\frac{DL\cdot DJ}{DR\cdot DI}=1$ and thus $EF\parallel BC$ . By Pascal's on $KKIFEJ$ we get that the tangent at $K$ to $\Omega$ is parallel to $BC$ . Thus $GH$ is tangent to $\Omega$ . By Pascal's on $KKXEFI$ and $KKXFEJ$ we get that $X$ lies on $ME$ and $NF$ .

By angle chasing $JGKA$ is cyclic and since $\frac{DG}{DE}:\frac{DA}{DX}=\frac{DG\cdot DJ}{DE\cdot DJ}:\frac{DA\cdot DK}{DX\cdot DK}=1$ we have $AB=AG\parallel EX=MX$ . Analogously $AC\parallel NX$ . So the homothety at $D$ with factor 2 maps $A,B,C$ to $X,M,N$ . In particular $A$ is the midpoint of $DX$ .

Inversion at $X$ with arbitrary radius. Denote the images of $A,B,C,D,P,Q$ with $A',B',C',D',P',Q'$ . Then $D'$ is the midpoint of $A'X$ and the circumcircles of $XB'D'C'$ and $A'Q'D'P'$ are externally tangent at $D'$ . So they are symmetric w.r.t. $D'$ . We have $\angle XAB=\angle CAX$ and thus $\angle A'B'X=\angle XC'A'$ . So $AX$ has the same inscribed angle w.r.t. the circumcircles of $XB'P'A'$ and $XA'Q'C'$ . Thus these two circles lie symmetric w.r.t. $D'$ . So $D'$ must be the midpoint of $B'Q'$ and $C'P'$ . Now $(XB'P'A')$ and $(XA'Q'C')$ are two circle with identical radii so that the arcs $B'P'$ and $Q'C'$ have the same length since they are symmetric w.r.t. $D'$ . Thus $\angle BXP=\angle B'XP'=\angle Q'XC'=\angle QXC$ .

This post has been edited 2 times. Last edited by pi_quadrat_sechstel, Jul 18, 2024, 9:24 PM

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Shreyasharma

679 posts

Shreyasharma

#14 h Jul 18, 2024, 10:01 PM • 2 Y

Y by math90, Rounak_iitr

Claim: $IJMN$ is cyclic.
Proof. Note that $PQ \parallel BC$ due to homothety so the result follows by Reim's on cyclic $IJPQ$ . $\square$

Claim: $K$ lies on $AD$ .
Proof. Note that,
$\begin{align*} \frac{\sin \angle KAP}{\sin \angle KAQ} &= \frac{KP \sin \angle APK}{KQ \sin \angle AQK}\\ &= \frac{KM \sin \angle BPM}{KN \sin \angle CQN}\\ &= \frac{BD \cdot CQ}{CD \cdot BP}\\ &= \frac{BD \cdot AC}{CD \cdot AB}\\ &= 1 \end{align*}$ whence the claim follows. $\square$

Claim: $XJDN$ and $XIDM$ are cyclic.
Proof. Note that,
$\begin{align*} \angle DXJ = \angle KIJ = \angle JND \end{align*}$ which suffices. $\square$

Claim: $A$ is the midpoint of $XD$ .
Proof. It suffices to show that $XM \parallel AB$ which is true as,
$\begin{align*} \angle NXD = \angle NJD = \angle QAD \end{align*}$ proving the claim. $\square$

Here I used a hint to use length ratios proven above to demonstrate a similarity.

Claim: $P$ and $Q$ are isogonal conjugates in $\triangle X BC$ .
Proof. Note that $\triangle DXP \sim \triangle NXC$ as,
$\begin{align*} \frac{DX}{DP} = \frac{2AD}{DP} = \frac{2AC}{CD} = \frac{XN}{NC} \end{align*}$ and we have that,
$\begin{align*} \angle XNC = \angle ACD = \angle AQP = \angle ADP \end{align*}$ as desired. In an identical manner we may prove $\triangle AXP \sim \triangle DCP$ so that,
$\begin{align*} \angle XCQ = \angle CXN = \angle DXP = \angle DCP \end{align*}$ as desired. Thus $P$ and $Q$ are isogonal in $\triangle XB C$ so the result follows. $\square$

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KST2003

173 posts

KST2003

#15 h Jul 19, 2024, 4:31 AM

Y by

Very nice problem

It does feel like two different problems combined into one though, because once you see that $A$ is the midpoint of $XD$ you can delete like half of the diagram.

Note that $PQ \parallel BC$ by homothety. Let $AD$ intersect $PQ$ at $D'$ . Then $PD':D'Q = BD:DC = MD:DN$ , so again by homothety, $K$ lies on $AD$ . Now note that $(A, D';K, D) \stackrel{P}{=} (B, P_\infty; M, D) = -1$ . Consider an inversion at $K$ with radius $\sqrt{KA \cdot KD}$ followed by a reflection around $K$ . This sends $A, D', K, D$ to $D, X, P_\infty, A$ , and since inversion preserves cross ratios, we see that $(D, X; P_\infty, A) = -1$ , and hence $A$ is the midpoint of $DX$ .

Now note that we have $\triangle ABD \sim \triangle ADQ$ by angle chasing, so $A$ is the $D$ -Dumpty point of $\triangle BDQ$ . It is well-known that this then gives $\triangle ABX \sim \triangle AXQ$ , and similarly we get $\triangle APX \sim \triangle AXC$ . Hence
$\angle BXP = \angle XPA - \angle XBA = \angle CXA - \angle QXA = \angle CXQ.$

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thepsserby

18 posts

thepsserby

#16 h Jul 20, 2024, 8:09 PM • 5 Y

Y by khina, kamatadu, Funcshun840, CyclicISLscelesTrapezoid, MS_asdfgzxcvb

Proposed by me

Will add some more details here later.

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VicKmath7

1388 posts

VicKmath7

#17 h Jul 21, 2024, 12:34 AM • 1 Y

Y by Rounak_iitr

Solution

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sami1618

896 posts

sami1618

#18 h Jul 22, 2024, 1:06 AM

Y by

The crux of the problem is that $A$ is the midpoint of $XD$ . By homothety $PQ\parallel BC$ and $AD$ bisects $\angle A$ . We invert about $D$ *. Let $X'$ be the midpoint of $AD$ . One notices that letting $P$ , $Q$ , $I$ , and $J$ to be the reflections of $C$ , $B$ , $D$ , and $M$ about $X'$ satisfies all the properties so the reflections must hold*. As the power of $X'$ in circle $(MDI)$ is $X'M\cdot X'I$ and the power of $X'$ in circle $(NDJ)$ is $X'N\cdot X'J$ , $K$ must lie on the angle bisector of $\angle BAC$ . Now $X'IJK$ is cyclic as $\angle X'KJ=\angle JQD=\angle ABC=\angle JIX'$ Thus $X=X'$ , as desired.

Now to finish, let $\mathcal{J}$ be an inversion about $A$ followed by a reflection about the $\angle A$ -bisector such that $\mathcal{J}\colon B\leftrightarrow Q$ and $\mathcal{J}\colon C\leftrightarrow P$ . As $\mathcal{J}\colon \Gamma \leftrightarrow BC$ , $\mathcal{J}$ preserves $D$ and $X$ . However after inversion $\angle BXP$ is mapped to $\angle CXQ$ , finishing the problem.

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La_Campanella

5 posts

La_Campanella

#19 h Jul 23, 2024, 5:15 AM • 1 Y

Y by MS_asdfgzxcvb

There's a good way to prove this elegant problem by using the inner product of vectors, as below:

Suppose that there're three points called $O, A, B$ in a plane. denote the vectors $\overrightarrow{OA}, \overrightarrow{OB}$ by $\vec{a}, \vec{b}$ respectively. make $\alpha=\angle AOB$ . Then
$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=2\cdot\frac{|OA||OB|(\sin\alpha)/2}{|\vec{a}||\vec{b}|\cos\alpha}=2\cdot\frac{S_{\triangle AOB}}{\vec{a}\cdot\vec{b}}.$ So necessarily and sufficiently we need to prove
$\tan\angle BXP=\tan\angle CXQ\quad\Leftrightarrow\quad\frac{S_{\triangle BXP}}{\overrightarrow{XB}\cdot\overrightarrow{XP}}=\frac{S_{\triangle CXQ}}{\overrightarrow{XC}\cdot\overrightarrow{XQ}}\quad\Leftrightarrow\quad\frac{S_{\triangle BXP}}{S_{\triangle CXQ}}=\frac{\overrightarrow{XB}\cdot\overrightarrow{XP}}{\overrightarrow{XC}\cdot\overrightarrow{XQ}}.$ Construct the projections of X w.r.t lines AB and AC and denote them with U, V, respectively. Then XU=XV because AX bisects $\angle UAV$ . Also, $PQ\parallel BC$ . Hence
$\frac{S_{\triangle BXP}}{S_{\triangle CXQ}}=\frac{BP\cdot XU}{CQ\cdot XV}=\frac{AB}{AC}.$ Thus, only the statement below need to be proved:
$\frac{\overrightarrow{XB}\cdot\overrightarrow{XP}}{\overrightarrow{XC}\cdot\overrightarrow{XQ}}=\frac{AB}{AC}.$ Similar to other solutions, it's proved that $XA=AD$ . Then $\overrightarrow{XA}=\overrightarrow{AD}$ . Set
$\frac{AP}{AB}=\frac{AD}{AS}=\frac{AQ}{AC}=\lambda, \angle BAS=\angle CAS=\alpha\quad\text{(see the figure below)}.$ Thus
$\begin{align*} \overrightarrow{XB}\cdot\overrightarrow{XP}&=(\overrightarrow{XA}+\overrightarrow{AB})(\overrightarrow{XA}+\overrightarrow{AP})=(\overrightarrow{AD}+\lambda\overrightarrow{AB})(\overrightarrow{AD}+\overrightarrow{AB})\\ &=AD^2+\lambda AB^2+(1+\lambda)AB\cdot AD\cdot\cos\alpha\\ &=\lambda AD\cdot AS+\lambda AB^2+\lambda(1+\lambda)AB\cdot AS\cos\alpha\\ &=\lambda AB\cdot AC+\lambda AB^2+\lambda(1+\lambda)AB\cdot\frac{AB+AC}{2}\\ &=\frac{\lambda(AB+AC)}{2}\cdot(2\lambda AB+(1+\lambda) AB)\\ &=\frac{\lambda(3+\lambda)(AB+AC)}{2}\cdot AB. \end{align*}$ The similar conclusion also holds for AC. So
$\frac{\overrightarrow{XB}\cdot\overrightarrow{XP}}{\overrightarrow{XC}\cdot\overrightarrow{XQ}}=\frac{\lambda(3+\lambda)(AB+AC)\cdot AB}{\lambda(3+\lambda)(AB+AC)\cdot AC}=\frac{AB}{AC}.$ That's what we desired.

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SHZhang

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SHZhang

#20 h Jul 23, 2024, 6:24 AM • 1 Y

Y by Rounak_iitr

By a homothety from $\omega$ to $\Gamma$ , $AP/AB = AQ/AC$ , so $PQ \parallel BC$ . Since $\angle CAD = \angle QPD = \angle PDB = \angle BAD$ , $AD$ bisects $\angle BAC$ . Since $\angle JIM = \angle JQP = \angle JNM$ , $(MNIJ)$ is cyclic. Since $\angle BAJ = \angle PQJ = \angle BNJ$ , $(ABNJ)$ is cyclic, and similarly $(ACMI)$ is cyclic. Let $\ell$ be the radical axis of $(ABNJ)$ and $(ACMI)$ . Since $DB \cdot DN = 2DB \cdot DC = DM \cdot DC$ , $D \in \ell$ , and since $KM \cdot KI = KJ \cdot KN$ from $(MNIJ)$ , $K \in \ell$ . Let $G$ be the second intersection of $(ABNJ)$ and $(ACMI)$ , so $G \in \ell$ .

Since $CD$ is tangent to $\Gamma$ , $CD^2 = CA \cdot CQ$ , so $CN^2 = CA \cdot CQ$ , and $CN$ is tangent to $(AQN)$ . This gives $\angle CAN = \angle QAN = \angle CNQ = \angle BNJ = \angle BAJ$ , so $\angle GNJ = \angle GAJ = \angle BAD + \angle BAJ = \angle CAD + \angle CAN = \angle GAN = \angle GJN$ . This gives $GN = GJ$ , and similarly $GM = GI$ . Also, $\angle GMN = \angle GAC = \angle GAB = \angle GNM$ , and $GM = GN$ . Therefore $G$ is the center of $(MNIJ)$ . Furthermore, since $\angle BJQ = \angle BJN = \angle BAN = \angle BAC + \angle CAN = \angle BAC + \angle BAJ = \angle JAC = \angle JAQ,$ $BJ$ , and similarly $CI$ , are tangent to $\Gamma$ .

Let $X'$ be the reflection of $D$ over $A$ . Since $\angle MX'D = \angle BAD = \angle PAD = \angle PID = \angle MID$ , $(X'IDM)$ is cyclic, and similarly $(X'JDN)$ is cyclic. Let $E$ be the second intersection of $CK$ with $(ACMI)$ and let $F$ be the second intersection of $BK$ with $(ABNJ)$ . Then $\angle APK = \angle API = 180^\circ - \angle AIC = \angle AEC = \angle AEK,$ so $(AEPK)$ , and similarly $(AFQK)$ , are cyclic. Since $KX' \cdot KD = KI \cdot KM = KA \cdot KG = KB \cdot KF$ , $(X'BDF)$ is cyclic, and since $\angle BFQ = \angle KFQ = \angle KAQ = \angle PAD = \angle PQD = \angle CDQ = 180^\circ - \angle BDQ,$ $(X'BDFQ)$ is cyclic, and similarly $(X'CDEP)$ is.

Since $\angle GMD = \angle GAC = \angle BAD = \angle MX'D$ , $GM$ is tangent to $(X'IDM)$ . Let $\Phi$ be the inversion about $(MNIJ)$ (so centered about $G$ ). Then $\Phi$ sends $(X'IDM)$ to itself, so $\Phi$ swaps $X'$ and $D$ . Also, since $\Phi$ swaps line $MI$ (containing $K$ ) with $(ACMIG)$ , $\Phi$ swaps $A$ and $K$ . Then the image of $\Gamma$ under $\Phi$ contains $I$ , $J$ , $K$ , and $X'$ , so $(IJKX')$ is cyclic, and since $X' \in KA$ , $X = X'$ .

To finish, it suffices to show that $P$ and $Q$ are isogonal conjugates with respect to $\triangle XBC$ . This is true since $\angle XBQ = \angle XDQ = \angle APQ = \angle PBC$ and similarly $\angle XCP = \angle QCB$ .

This post has been edited 1 time. Last edited by SHZhang, Jul 24, 2024, 5:20 AM
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Aiden-1089

278 posts

Aiden-1089

#21 h Aug 4, 2024, 3:05 PM • 1 Y

Y by Rounak_iitr

We claim that $P$ and $Q$ are isogonal conjugates w.r.t $\Delta XBC$ , which directly implies the result.

By taking a homothety from $\omega$ to $\Gamma$ , we see that $PQ // BC$ and $AD$ is the internal angle bisector of $\angle BAC$ .
Note that $(A,P;D,J) \stackrel{Q}{=} (C,\infty_{BC};D,N)=-1$ and $(A,Q;D,I) \stackrel{P}{=} (B,\infty_{BC};D,M)=-1$ .
Now $(A, PQ \cap AD; D, MP \cap AD) \stackrel{P}{=} (A,Q;D,I) = (A,P;D,J) \stackrel{Q}{=} (A, PQ \cap AD; D, NQ \cap AD)$ , so $MP \cap NQ = K$ lies on $AD$ .

$\measuredangle IXD = \measuredangle IJQ = \measuredangle IPQ = \measuredangle IMD \implies (XIDM)$ are concyclic.
By power of point, $KM \cdot KI = KD \cdot KX$ and $KP \cdot KI = KD \cdot KA$ , so $\frac{KM}{KP} = \frac{KX}{KA} \implies XM // AP$ .
Since $AB // XM$ , $\frac{DX}{DA} = \frac{DM}{DB} = 2 \implies A$ is the midpoint of $XD$ .

Now take an inversion at $D$ . Use $T'$ to denote the inverted image of any point $T$ .
Since $\angle BAD = \angle CAD$ , we have $\angle A'B'D = \angle A'C'D$ .
$X'$ is the midpoint of $A'$ and $D$ .
$\Gamma$ is the line passing through $A'$ parallel to $BC$ .
$Q'$ lies on $(A'C'D)$ and $A'Q' // BC$ .
Since $\angle A'B'D = \angle A'C'D = \angle A'Q'D$ and $B'D // A'Q'$ , we have that $DB'A'Q'$ is a parallelogram. Thus $B'Q'$ passes through $X$ .
Inverting back, we see that $(XBDQ)$ are conyclic.

Now $\measuredangle XBQ = \measuredangle XDQ = \measuredangle APQ = \measuredangle PBC$ , so $BP$ and $BQ$ are isogonal in $\angle XBC$ .
Similarly we may prove that $CP$ and $CQ$ are isogonal in $\angle XCB$ , so $P$ and $Q$ are isogonal conjugates w.r.t $\Delta XBC$ . It follows that $\angle BXP = \angle CXQ$ . $\square$

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gambi

82 posts

gambi

#22 h Aug 22, 2024, 12:33 AM

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Let $L$ be the midpoint of arc $BC$ in $\omega$ not containing $A$ , and let $D'$ be the reflection of $D$ over $L$ .

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.34, xmax = 22.76, ymin = -4.7, ymax = 6.82; /* image dimensions */ pen wwccqq = rgb(0.4,0.8,0.); pen ffzztt = rgb(1.,0.6,0.2); pen wwzzff = rgb(0.4,0.6,1.); /* draw figures */ draw((5.7,2.52)--(4.6,-1.04)); draw((4.6,-1.04)--(9.34,-1.)); draw((9.34,-1.)--(5.7,2.52)); draw(circle((6.959866907500713,0.18077146116550968), 2.656925814077833), red); draw((7.355218230266601,-3.929076102452998)--(2.590643382110303,-1.056956595931559)); draw((4.6,-1.04)--(6.982287424078149,-2.47605975326072)); draw((2.590643382110303,-1.056956595931559)--(4.7906433821103045,6.063043404068441)); draw((4.6,-1.04)--(2.590643382110303,-1.056956595931559)); draw((9.34,-1.)--(12.070643382110303,-0.9769565959315588)); draw((6.982287424078149,-2.47605975326072)--(9.34,-1.)); draw((12.070643382110303,-0.9769565959315588)--(7.355218230266601,-3.929076102452998)); draw((12.070643382110303,-0.9769565959315588)--(4.7906433821103045,6.063043404068441)); draw((4.91991570618593,-0.004636441798263189)--(8.28136984498474,0.02373025979497626)); draw(circle((6.593456715306735,0.8610950520123435), 1.8842055430829807), wwccqq); draw(circle((6.174954271208556,3.4927564006405407), 2.9193649853463612), ffzztt); draw((4.7906433821103045,6.063043404068441)--(7.355218230266601,-3.929076102452998)); draw((2.590643382110303,-1.056956595931559)--(8.349285160476356,1.54468630394426)); draw((4.712190187914333,0.9662938233814242)--(12.070643382110303,-0.9769565959315588)); draw((6.609356617889696,-1.0230434040684413)--(4.91991570618593,-0.004636441798263189)); draw(circle((5.613298865641586,-2.053057670597737), 1.43285045802674), wwzzff); draw((8.28136984498474,0.02373025979497626)--(4.2516440419678485,-2.4991031573291584)); /* dots and labels */ dot((5.7,2.52),linewidth(2.pt) + dotstyle); label("$A$", (5.32,2.5), NE * labelscalefactor); dot((4.6,-1.04),linewidth(2.pt) + dotstyle); label("$B$", (4.26,-0.98), NE * labelscalefactor); dot((9.34,-1.),linewidth(2.pt) + dotstyle); label("$C$", (9.42,-1.46), NE * labelscalefactor); dot((6.609356617889696,-1.0230434040684413),linewidth(2.pt) + dotstyle); label("$D$", (6.64,-0.82), NE * labelscalefactor); dot((6.982287424078149,-2.47605975326072),linewidth(2.pt) + dotstyle); label("$L$", (7.18,-2.9), NE * labelscalefactor); dot((7.355218230266601,-3.929076102452998),linewidth(2.pt) + dotstyle); label("$D'$", (7.56,-4.2), NE * labelscalefactor); dot((4.7906433821103045,6.063043404068441),linewidth(2.pt) + dotstyle); label("$X$", (4.64,6.16), NE * labelscalefactor); dot((2.590643382110303,-1.056956595931559),linewidth(2.pt) + dotstyle); label("$M$", (2.24,-1.24), NE * labelscalefactor); dot((12.070643382110303,-0.9769565959315588),linewidth(2.pt) + dotstyle); label("$N$", (12.16,-0.9), NE * labelscalefactor); dot((4.91991570618593,-0.004636441798263189),linewidth(2.pt) + dotstyle); label("$P$", (4.52,-0.02), NE * labelscalefactor); dot((8.28136984498474,0.02373025979497626),linewidth(2.pt) + dotstyle); label("$Q$", (8.36,0.1), NE * labelscalefactor); dot((6.199589385483384,0.5734953590735087),linewidth(2.pt) + dotstyle); label("$K$", (6.26,0.88), NE * labelscalefactor); dot((8.349285160476356,1.54468630394426),linewidth(2.pt) + dotstyle); label("$I$", (8.58,1.34), NE * labelscalefactor); dot((4.712190187914333,0.9662938233814242),linewidth(2.pt) + dotstyle); label("$J$", (4.8,1.04), NE * labelscalefactor); dot((6.795822020983922,-1.7495515786645806),linewidth(2.pt) + dotstyle); label("$Y$", (6.4,-1.84), NE * labelscalefactor); dot((4.2516440419678485,-2.4991031573291584),linewidth(2.pt) + dotstyle); label("$Z$", (3.94,-2.8), NE * labelscalefactor); dot((6.344886195537849,0.007388625622428354),linewidth(2.pt) + dotstyle); label("$R$", (6.42,0.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Claim 1. Points $A-K-D-L-D'$ are collinear.
Proof.
The homothety centered at $A$ mapping $\Gamma$ to $\omega$ maps $D$ to $L$ , $P$ to $B$ and $Q$ to $C$ , so $A-D-L$ collinear.
The homothety centered at $D$ of scale $2$ maps $B$ to $M$ , $C$ to $N$ and $L$ to $D'$ .
Thus
$DP\parallel LB\parallel D'M, \qquad DQ\parallel LC\parallel D'N, \qquad PQ\parallel MN,$ which means that $\triangle DPQ$ and $\triangle L'MN$ are homothetic, so $K-D-L-D'$ collinear. $\square$

Let $R=AD\cap PQ$ .
Claim 2. $XA=AD$ .
Proof.
The inversion centered at $K$ mapping $I$ to $P$ and $J$ to $Q$ must hence map $X$ to $R$ , so
$KI\cdot KM=KI\cdot KP\cdot \frac{KM}{KP}=KX\cdot KR\cdot \frac{KM}{KP}=KX\cdot KR\cdot \frac{KD}{KR} =KX\cdot KD,$ so $IDMX$ is cyclic, which implies
$\measuredangle MXD=\measuredangle MID=\measuredangle PID=\measuredangle PAD=\measuredangle BAD,$ so $AB\parallel MX$ , and so homothety centered at $D$ of scale $2$ maps $A$ to $X$ . $\square$

Let $Y$ be the midpoint of $DL$ and let $QD$ meet $(BDL)$ at $Z$ .

Claim 3. $XBDQ$ is cyclic.
Proof.
Consider the inversion centered around $D$ of mapping $B$ to $C$ .
It maps $A$ to $L$ , $X$ to $Y$ , $Q$ to $Z$ .
Recall $DQ\parallel CL$ . Also,
$\measuredangle LZD=\measuredangle LBD=\measuredangle LBC=\measuredangle BCL=\measuredangle PQD,$ so $LZ\parallel BC$ . Thus $LZDC$ is a parallelogram, and so $C-Y-Z$ are collinear.
Therefore the circle containing $B$ , $X$ , $Q$ also contains $D$ , the center of inversion. $\square$

By the symmetry of the heading $XCDP$ is also cyclic, so
$\measuredangle BXQ=\measuredangle CDQ=\measuredangle DAQ=\measuredangle PAD=\measuredangle PDB=\measuredangle PXC \, \blacksquare$

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kamatadu

478 posts

kamatadu

#23 h Oct 16, 2024, 10:31 AM • 3 Y

Y by DistortedDragon1o4, SilverBlaze_SY, GeoKing

Solved with SilverBlaze_SY and DistortedDragon1o4.

Troll problem due to the fact that there are millions of other seemingly important claims that you can prove (which we did) that end up not being used in the proof at all. :ninja:

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (5.76993,18.70486); pair B = (-3.75514,-11.06193); pair C = (33.82533,-11.40203); pair D = (12.46621,-11.20873); pair M = (-19.97650,-10.91513); pair N = (55.18445,-11.59532); pair P = (-1.18902,-3.04254); pair Q = (26.26700,-3.29101); pair I = (27.68259,9.05563); pair J = (-3.11803,5.14757); pair K = (9.62390,1.48843); pair X = (-0.92634,48.61845); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.6)); draw(B--C, linewidth(0.6)); draw(C--A, linewidth(0.6)); draw(circle((12.60864,4.52943), 15.73881), linewidth(0.6)); draw(A--D, linewidth(0.6)); draw(B--M, linewidth(0.6)); draw(C--N, linewidth(0.6)); draw(M--I, linewidth(0.6)); draw(N--J, linewidth(0.6)); draw(circle((9.84830,26.28453), 24.79711), linewidth(0.6)); draw(X--A, linewidth(0.6)); draw(circle((-3.50455,16.62872), 32.09345), linewidth(0.6) + linetype("4 4") + blue); draw(circle((23.45318,22.66332), 35.60940), linewidth(0.6) + linetype("4 4") + red); dot("$A$", A, NW); dot("$B$", B, S); dot("$C$", C, S); dot("$D$", D, S); dot("$M$", M, SW); dot("$N$", N, SE); dot("$P$", P, W); dot("$Q$", Q, 2*dir(10)); dot("$I$", I, dir(10)); dot("$J$", J, SW); dot("$K$", K, NE); dot("$X$", X, NW); [/asy]$

Claim: $PQ \parallel BC$ .
Proof. Note that the construction of $\odot(APQ)$ is done by taking a homothety that maps the midpoint of arc $\widehat{BC}$ of $\odot(ABC)$ not containing $A$ to the point $D$ . Note that from here it also follows that the line $AD$ is the angle bisector of $\angle BAC$ .

It follows that this homothety also maps $B\mapsto P$ and $C\mapsto Q$ . This gives us that $PQ \parallel BC$ . $\blacksquare$

Claim: $K$ lies on $AD$ .
Proof. Define $K'=MP\cap AD$ and $K''=NQ\cap AD$ . By Menelaus on $\triangle ABD$ with $MP$ as the transversal, we get that, $\frac{AP}{PB}\cdot \frac{BM}{MD}\cdot \frac{DK'}{K'A}=-1 \implies \frac{DK'}{K'A} = - \frac{PB}{AP}\cdot (-2) = 2\cdot \frac{PB}{AP} .$
Similarly we also get that, $\frac{DK''}{K''A} = 2\cdot \frac{QC}{AQ} .$
Now note that by Thales's theorem, as $PQ\parallel BC$ , we also get that $\frac{PB}{AP}=\frac{QC}{AQ}$ .

This finally gives us that $\frac{DK'}{K'A}= \frac{DK''}{K''A}$ , i.e., $K'\equiv K''\equiv K$ . $\blacksquare$

Claim: $IJMN$ is cyclic.
Proof. Firstly, by Thales's theorem, we have, $\frac{KP}{KM}=\frac{KQ}{KN}$ .

Now we have, $\begin{align*} \operatorname{Pow}_{\odot(APQ)}(K)&=KI\cdot KP =KJ\cdot KQ \\ &\implies KI\cdot KP\cdot \frac{KM}{KP}=KJ\cdot KQ \cdot \frac{KN}{KQ}\\ &\implies KI\cdot KM=KJ\cdot KN\\ &\implies IJMN \text{ is cyclic} .\blacksquare\end{align*}$
Claim: $IDMX$ is cyclic.
Proof. We have, $\measuredangle DXI=\measuredangle KXI=\measuredangle KJI =\measuredangle NJI=\measuredangle NMI=\measuredangle DMI .\blacksquare$
Claim: $AD=XA$ .
Proof. Note that, $\measuredangle MXD=\measuredangle MID=\measuredangle PID =\measuredangle PAD=\measuredangle BAD$ which gives us that $AB\parallel XM$ .

Now as $B$ is the midpoint of $MD$ , by Thales's theorem we also get that $A$ is the midpoint of $XD$ . $\blacksquare$

Claim: $\triangle CDP\stackrel{+}{\sim}\triangle CAX$ .
Proof. Firstly, we have, $\measuredangle CDP=\measuredangle BDP=\measuredangle DQP =\measuredangle DAP=\measuredangle QAD=\measuredangle QAX .$
Now we are going to prove that $\triangle CDA\stackrel{+}{=}\triangle DPA$ .

For this, we firstly have $\measuredangle DAC =\measuredangle PAD$ . Now we have $\measuredangle CDA =\measuredangle DPA$ due to the tangency. This gives us that $\triangle CDA\stackrel{+}{=}\triangle DPA$ .

Now to finish our original claim, we have, $\frac{CD}{CA}=\frac{DP}{DA}=\frac{DP}{AX}\implies \frac{CA}{AX}=\frac{CD}{DP} .$
So by our SAS criterion, we have that $\triangle CDP\stackrel{+}{\sim}\triangle CAX$ . $\blacksquare$

Claim: $XPDC$ is cyclic.
Proof. From $\triangle CDP\stackrel{+}{\sim}\triangle CAX$ , we get $\measuredangle AXC=\measuredangle DPC$ .

Thus to finish, we have, $\measuredangle DXC=\measuredangle AXC=\measuredangle DPC$ which gives us our desired claim. $\blacksquare$

Similarly, we also get that $XQDB$ is also cyclic.

Now to finish, we have, $\measuredangle BXP=\measuredangle BXD-\measuredangle PXD = \measuredangle BQD-\measuredangle PCD =\measuredangle BQD-\measuredangle CPQ .$
Similarly, we also get that $\measuredangle QXC =\measuredangle DPC-\measuredangle PQB$ .

Thus,
$\begin{align*} &\measuredangle BXP=\measuredangle QXC\\ &\iff\measuredangle BQD -\measuredangle CPQ =\measuredangle DPC-\measuredangle PQB\\ &\iff\measuredangle BQD+\measuredangle PQB =\measuredangle DPC +\measuredangle CPQ\\ &\iff\measuredangle PQD=\measuredangle DPQ\\ &\iff\measuredangle PAD=\measuredangle DAQ \end{align*}$ which is indeed true and we are done.

This post has been edited 5 times. Last edited by kamatadu, Oct 16, 2024, 10:34 AM

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SomeonesPenguin

125 posts

SomeonesPenguin

#24 h Nov 10, 2024, 9:36 PM

Y by

Really beautiful problem .

Clearly, $AD$ is the angle bisector of $\angle BAC$ and $PQ$ is parallel to $BC$ .

Claim 1: $K$ lies on $AD$

Proof: By trig Ceva, it suffices to prove that $\frac{\sin(\angle AQJ)}{\sin(\angle API)}\cdot\frac{\sin(\angle QPI)}{\sin(\angle API)}=1$ But this is equal to $\frac{\sin(\angle CQN)}{\sin(\angle CNQ)}\cdot\frac{\sin(\angle PMB)}{\sin(\angle MPB)}=\frac{CN}{CQ}\cdot\frac{BP}{BM}=\frac{DC}{CQ}\cdot\frac{BP}{BD}=\frac{AC}{CD}\cdot\frac{AB}{BD}=1$
The last ratio equality comes by PoP. Now let $IJ$ meet $BC$ at $S$ and let $T$ be the $D$ antipode in $\Gamma$ .

Claim 2: $S$ , $A$ and $T$ are collinear.

Proof: We have $-1=(M,D;B,\infty_{BC})\overset{P}{=}(I,D;A,Q)$ So $CI$ is tangent to $\Gamma$ , therefore $CD=CI=CN$ or $\angle DIN=90^\circ$ . This gives that $T$ , $I$ , $N$ are collinear and similarly $T$ , $J$ , $M$ are collinear. Now we also have $TI\cdot TN=TD^2=TJ\cdot TM$ So $MJIN$ is cyclic. Notice that by ratio lemma, it suffices to prove that $\frac{SJ}{SI}=\frac{AJ}{AI}\cdot\frac{TJ}{TI}$ Now we have $\frac{SJ}{SI}=\frac{JD^2}{ID^2}=\frac{TJ\cdot JM}{TI\cdot IN}$ and $\frac{AJ}{AI}=\frac{\sin(\angle ADJ)}{\sin(\angle ADI)}=\frac{\sin(\angle DJN)}{\sin(\angle NJI)}\cdot\frac{\sin(\angle JIM)}{\sin(\angle MID)}=\frac{\sin(\angle KNM)}{\sin(\angle KMN)}=\frac{KM}{KN}$ Now it remains to prove that $\frac{JM}{IN}=\frac{KM}{KN}$ which is true since $\triangle KMJ\sim\triangle KNI$ .

Claim 3: $X$ lies on the $D$ -Apollonian circle of $\triangle DJI$ .

Proof: This is equivalent to $\frac{JD}{ID}=\frac{XJ}{XI}=\frac{\sin(\angle XKJ)}{\sin(\angle XKI)}=\frac{\sin(\angle DKN)}{\sin(\angle DKM)}=\frac{DN}{DM}\cdot\frac{KM}{KN}=\frac{DN}{DM}\cdot\frac{JM}{IN}$ This is true since $\frac{JD}{DM}=\frac{TD}{DM}$ and $\frac{ID}{DN}=\frac{TD}{DN}$ .

Finally, note that this implies that $SX$ is tangent to $(XJI)$ so $SX^2=SJ\cdot SI=SD^2$ And since $\angle SAD=90^\circ$ we have that $A$ is the midpoint of $XD$ .

Claim 4: $P$ and $Q$ are isogonal conjugates in $\triangle XBC$ .

Proof: Note that $\angle BAX=\angle BDQ$ and $\frac{AX}{AB}=\frac{AD}{AB}=\frac{PD}{BD}=\frac{DQ}{BD}$ Therefore $\triangle BAX\sim\triangle BDQ$ so $\angle CBQ=\angle XBP$ and similarly $\angle BCP=\angle XCQ$ so the conclusion follows. $\blacksquare$

remarks
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iStud

268 posts

iStud

#25 h Jan 19, 2025, 9:42 AM

Y by

[ 200th post! ]

why is everyone using lengths (even inversion & cross ratios) bruh :rotfl:

funfact: I was stuck at this problem for about 30 minutes because I tried to prove that $J,I$ lies on the circle with diameter $DM,DN$ , respectively, which turns out later to be useless lol :blush:

Firstly, by the infamous Shooting Star lemma (also mentioned on a handout by Evan Chen), we have that $AD$ is the angle bisector of $\angle{A}$ . Let $K'=MP\cap AD$ . By Thales', we have $\frac{BP}{PA}=\frac{CQ}{QA}$ . Using Menelause's Theorem on $\triangle{ABD}$ with transversal $MK'$ , we have $\frac{AK'}{K'D}\cdot\frac{DM}{MB}\cdot\frac{CQ}{QA}=\frac{AK'}{K'D}\cdot\frac{DM}{MB}\cdot\frac{BP}{PA}=1$ , so $\overline{K,M,Q}$ . In other words, $K'=MP\cap NQ$ , so $K=K'$ . Hence, $K$ lies on $AD$ .

We have $MJIN$ is cyclic by Reim's on $PJIQ$ since that $PQ\parallel BC$ by the homothety at $A$ mapping $\Gamma$ to $\omega$ . Now $\angle{DMI}=\angle{NMI}=\angle{NJI}=\angle{KJI}=\angle{KXI}=\angle{DXI}$ , so $XIDM$ is cyclic. Similarly, we have $XJDN$ is also cyclic.

Next, observe that $\angle{MXD}=\angle{MID}=\angle{PID}=\angle{PAD}=\angle{BAD}$ , so $AB\parallel MX$ . Analogously, $AC\parallel NX$ . Since $B,C$ are the midpoints of $DM,DN$ , respectively, there exists a dilatation with center $D$ and scale $2$ that maps $\triangle{ABC}\mapsto\triangle{XMN}$ .

Let $PQ$ cuts $XM$ and $XN$ at $F,G$ , respectively. We have $FP=MB=BD$ and $GQ=NC=CD$ , so $BFPD$ and $CGQD$ are parallelograms. Thus $BF=DP=DQ=CG$ . Using the previous fact that $PQ\parallel BC$ , we have $BFQD$ and $CGPD$ are cyclic. After that, we can derive $\angle{FXD}=\angle{PAD}=\angle{PQD}=\angle{FQD}$ , so $X$ lies on $(BFQD)$ . With the same spirit, $X$ also lies on $(CGPD)$ .

After all, it's obvious that $\angle{BDQ}=\angle{CDP}$ , so $180^\circ-\angle{BDQ}=180^\circ-\angle{CDP}\Longleftrightarrow\angle{BXQ}=\angle{CXP}$ . This means that $XP$ and $XQ$ are isogonal w.r.t. $\angle{BXC}$ , therefore $\angle BXP = \angle CXQ$ , as desired. $\blacksquare$

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cosdealfa

27 posts

cosdealfa

#26 h Jan 29, 2025, 3:09 PM • 2 Y

Y by ehuseyinyigit, Kaus_sgr

My first G6

Although this felt a bit easy for G6
Solution(not the best writeup)

This post has been edited 6 times. Last edited by cosdealfa, Jan 29, 2025, 4:00 PM

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Ilikeminecraft

355 posts

Ilikeminecraft

#27 h Mar 12, 2025, 7:04 PM

Y by

yay gg?!!?!?! I didn't think I would ever be able to solve a g6
not purely synthetic, which is sad

Let $M_A$ be the arc midpoint of $BC,$ not containing $A.$
Claim: $A, D, M_A$ are collinear.
Proof: Well-known. Take a homothety about $A$ mapping curvilinear to the circumcircle. This finishes.
We also clearly have $PQ\parallel BC.$

Claim: $K$ lies on $AD.$
Proof: Note that $\frac{\sin\angle APJ}{\sin\angle \angle JPQ} \cdot \frac{\sin\angle IQP}{\sin\angle PQI} = \frac{\sin\angle MPB}{\angle PMB} \cdot\frac{\sin \angle QNC}{\angle CQN}= \frac{PB}{MB}\cdot\frac{CN}{CQ} = \frac{PB}{BD}\cdot\frac{CD}{CQ}$ However, $\frac{PB}{BD} = \frac{BD}{AB}, \frac{CD}{CQ} = \frac{AC}{CD},$ so this evaluates to 1, and so $AK$ bisects $\angle BAC.$

Claim: $A$ is midpoint of $XD.$
Proof: Observe that $\angle IMC = \angle IPQ = \angle IJK = \angle IXK,$ so $XIDM$ is cyclic. Thus, $\angle IMX = \angle IDX = \angle IPA,$ which implies $AP\parallel XM.$ Finally, observe that a $\times2$ homothety centered at $D$ finishes.

We can rephrase the problem as the following complex-able problem:

Quote:

Let $ABC$ be a triangle, and $D$ be the intersection of $A$ -angle bisector and $BC.$ Let $\omega$ be the circle passing through $A, D$ tangent to $BC.$ Let $X$ be the reflection of $D$ across $A.$ Let $\omega$ intersect $AB, AC$ at $P, Q.$ Show that $\angle PXB = \angle CXQ$ are equal.

Define $A = a, P = p, D = 1.$ Hence, $Q = \overline{p}.$ We also have $X = 2a-1.$
We get $C = \frac{2ap - a - p}{ap - 1}, B = \frac{2a\overline{p} - a-\overline{p}}{a\overline p - 1} = \frac{2a - ap - 1}{a - p}.$
Hence,
$\begin{align*} \angle QXB & = \arg \frac{Q - X}{B - X} \\ & = \arg\frac{\frac1p - 2a + 1}{\frac{2a-ap-1}{a-p}-2a+1} \\ & = \arg\frac{\left(\frac1p - 2a + 1\right)(a-p)}{2a-ap-1-2a^2+2ap+a-p} \\ & = \arg\frac{(a-p)(2ap - p - 1)}{(a- 1)p(2a-p-1)} \end{align*}$ while
$\begin{align*} \angle CXP & = \arg\frac{C - X}{P - X} \\ & = \arg \frac{\frac{2ap-a-p}{ap-1}-2a+1}{p-2a+1} \\ & = \arg\frac{(a-1)(2ap-p-1)}{(2a-p-1)(ap-1)} \end{align*}$ so
$\begin{align*} \angle CXP - \angle QXB & = \arg\frac{\frac{(a-p)(2ap - p - 1)}{(a- 1)p(2a-p-1)}}{\frac{(a-1)(2ap-p-1)}{(2a-p-1)(ap-1)}} \\ & = \arg \frac{(ap-1)(a-p)}{(a-1)^2p} \end{align*}$ while
$\overline{\frac{(ap-1)(a-p)}{(a-1)^2p}} = \frac{(ap-1)(a-p)}{(1-a)^2p} \implies \frac{(ap-1)(a-p)}{(1-a)^2p}\in\mathbb R$ which finishes

Here is another interesting thing one can get:
$CJ, BI$ are tangent to $(APQ).$
Note that $-1 = (MD;B\infty) \stackrel P= (JD;AQ),$ but $AQ$ passes through $C$ while $CD$ is tangent to $(APQ)$

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ihategeo_1969

205 posts

ihategeo_1969

#28 h Apr 8, 2025, 8:46 AM

Y by

We will define some new points (assume $AB \le AC$ ).

$\bullet$ Let $M_A$ be arc midpoint of minor arc $\widehat{BC}$ .
$\bullet$ Let $F=\overline{AD} \cap \overline{PQ}$ .
$\bullet$ Let $P'$ and $Q'$ be midpoints of $\overline{PD}$ and $\overline{QD}$ .
$\bullet$ Let $B'$ and $C'$ be midpoints of $\overline{BD}$ and $\overline{CD}$ .

Claim: $\overline{PQ} \parallel \overline{BC}$ .
Proof: By shooting lemma $D=\overline{AM_A} \cap \overline{BC}$ and hence if we $\sqrt{bc}$ invert at $A$ in $\triangle ABC$ then $\Gamma$ gets mapped to $\overline{M_AM_A}$ and $\overline{P^*Q^*}=\overline{M_AM_A}$ which is obviously parallel to $\overline{BC}$ . $\square$

Claim: $K \in \overline{AD}$ .
Proof: Take a $\frac12$ homothety at $D$ and we will prove $\overline{BP'} \cap \overline{CQ'} \cap \overline{AD}$ concur. See that $\angle APD=\angle ADC=\angle ABM_A$ and hence $\overline{PD} \parallel \overline{CM_A}$ and similarly $\overline{PQ} \parallel \overline{CM_A}$ . And so $-1=(P,D;P',\infty) \overset B= (A,D';\overline{BP'} \cap \overline{AD},M_A)$ And similarly $(A,D;\overline{CQ'} \cap \overline{AD})=-1$ and done. $\square$

Claim: $A$ is midpoint of $\overline{DX}$ .
Proof: We invert at $K$ fixing $\Gamma$ and note that cross ratio is preserved and hence $(X,D;A,\infty)=(F,A;D,K) \overset P= (B,\infty;M,D)=-1$ And done. $\square$

Claim: $(AB'DQ')$ is cyclic (and so is $(AC'DP')$ ).
Proof: Let $F'$ be midpoint of $\overline{PF}$ . Now by homothety at $A$ , $D$ is minor arc midpoint of $\widehat{PQ}$ and hence see that $\triangle APF \cup F' \sim \triangle ADQ \cup Q'$ and hence $\angle AQ'D=180^{\circ}-\angle AQ'Q=180 ^{\circ}-\angle AF'F=180 ^{\circ}-\angle AB'D$ And done. $\square$

So this means that $\angle XBQ=\angle AB'Q'=\angle ADQ'=\angle PBC$ and hence $\overline{BP}$ and $\overline{BQ}$ are isogonal in $\angle XBC$ and similarly $\overline{CP}$ and $\overline{CQ}$ are isogonal in $\angle XCB$ and hence $P$ and $Q$ are isogonal conjugates in $\triangle XBC$ as required.

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Om245

164 posts

Om245

#29 h Today at 7:55 AM • 1 Y

Y by GeoKing

Headsolve in 2hr (First G6 headsolve yay)

Claim : $D$ is feet of angle bisector
As $(APQ)$ tangent to $(ABC)$ we should have $PQ \parallel BC$ . Then as it tangent to $BC$ at $D$ , we get that $\angle DPQ = \angle PDB = \angle DQP$ Thus $\angle PAD = \angle DAQ$ which tell us $D$ is feet of angle bisector.

Diagram you may need :)

Claim : $\triangle AIJ \sim \triangle ANM$
Now as $PQ \parallel BC$ we have $\angle IMB = \angle IPQ = \angle IJQ = \angle IJN$ thus $I,J,M,N$ cyclic.
Notice that $P$ and $Q$ are humpty point of $\triangle AMD$ and $\triangle AND$ . Thus $(APM)$ tangent to $BC$ . $\angle AJI = \angle API = \angle MPB = \angle AMB$ and similarly $\angle AIJ = \angle ANM$ , which prove the claim.
Now time for cool inversion...
Notice that $A$ is miquel point of $\Box MNIJ$ , let $O$ be center of $(MNJI)$ . By miquel's theorem (and some well known stuff) we know that angle bisector of $\angle JAN$ and $\angle MAI$ is $\overline{A-K-O}$ only.
Do $\sqrt{AJ \cdot AN}$ inversion with reflection about angle bisector $AK$ . This will swap $J \leftrightarrow N$ , $I \leftrightarrow M$ , $K \leftrightarrow O$ . Let $X \leftrightarrow X'$ .
Also as $MN \leftrightarrow (AIJ)$ we get $B \leftrightarrow P$ and $C \leftrightarrow Q$ (As $B,P$ lie on line $AB$ and similarly other). Thus $\angle BXP = \angle PX'B$ and $CXQ = \angle QX'C$ .

Main Claim : $A$ is midpoint of $X'D$

Now after inversion we have $\angle KNM = \angle JIK = \angle JXA = \angle ANX'$ and similarly $\angle KMN = \angle AMX'$ . Thus $A$ and $K$ are isogonal conjugate in $\triangle X'MN$ . But as $X',A,K$ are collinear, we get $\angle MX'A = \angle AX'N$ which implies $\overline{X'-A-K-D}$ .
This implies $\angle AIX = \angle AJX$ .

Let $L = AX \cap (APQ)$ . Now know that $\angle AIX + \angle AXI = \angle AIX + \angle JIP = \angle IAL = \angle JIL = \angle PIL + \angle JIP$ which implies $\angle AIX = \angle PAL$ and similarly $\angle AJX = \angle QAL$ but as $\angle AJX = \angle AIX \Rightarrow \angle PAL = \angle QAL$ . Thus $L$ is midpoint of arc $PQ$ . Thus $2\angle PAL = \angle A = 2\angle AIX = 2\angle MX'A = \angle MX'N$

As $\angle BAD = \angle MX'D$ we get $AB \parallel X'M$ , as $B$ is midpoint of $MD$ we get $A$ is midpoint of $X'$ .

Diagram_2 for you :)

We need to prove $\angle BX'P = \angle CX'Q$ . Consider $Y = AD \cap (ABC)$ .
Now do second inversion as $-\sqrt{DB \cdot DC}$ . Note $B \leftrightarrow C$ , $A \leftrightarrow Y$ , As $(APQ)$ tangent to $BC$ and $(ABC)$ we should have $(APQ) \leftrightarrow LL$ where $LL$ is tangent to $(ABC)$ at $L$ (ya lack of notations.)

Thus image of $P$ and $Q$ are $P' = (LDC) \cap LL$ and $Q' = (LQB) \cap LL$ respectively. Notice if $H$ is midpoint of $DL$ then $X' \leftrightarrow H$ . By some angle chase $\angle BX'P = \angle BX'A - \angle PX'A = \angle DCH - \angle DP'H$ and $\angle CX'Q = \angle DBH - \angle DQ'H$ . Notice that as $LL \parallel BC$ and $LB = LC$ , we have $LP'CD \cong DBQ'L$
Time for some angle chase .....

Claim : $\angle DCH - \angle DP'H = \angle DBH - \angle DQ'H$

Diagram_3 for you :)

For now we only consider one quadrilateral $ABCD$ with $AB \parallel CD$ and $BC = AD$ . Now if $M$ and $N$ are midpoint of $BC$ and $AD$ , we need to prove that $\angle NCD - \angle NBD = \angle ABN - \angle ACN$ . Consider $U = BN \cap CD$ and $V = CN \cap AB$ . Now we will have $\angle NBD = \angle ABD - \angle ABN = \angle ABD - \angle NUD$ , Thus $\angle BND = \angle DUN + \angle NCD = \angle ABD - (\angle NBD - \angle NCD)$ But we also have $\angle ACN = \angle ACD - \angle NCD = \angle ACD - \angle NVA$ give implies $\angle BND = \angle NVA + \angle ABN = \angle ACD - (\angle ACN - \angle ABN)$ As $A,B,C,D$ cyclic we get $\angle ACD = \angle ABD$ , thus $\angle NBD - \angle NCD = \angle ACN - \angle ABN$ which is equivalent to $\angle DCH - \angle DP'H = \angle DBH - \angle DQ'H$ , thus $\angle BXP = \angle CXQ$ $\blacksquare$ .

This post has been edited 3 times. Last edited by Om245, Today at 7:59 AM
Reason: I may messed up with I and J, as I was headsolving... not my fault :)

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