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Source: 2021 China TST, Test 1, Day 2 P5

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JustPostChinaTST

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JustPostChinaTST

#1 h Mar 17, 2021, 1:28 PM • 4 Y

Y by samrocksnature, brickmaster8, Whitemountain, Rounak_iitr

Given a triangle $ABC$ , a circle $\Omega$ is tangent to $AB,AC$ at $B,C,$ respectively. Point $D$ is the midpoint of $AC$ , $O$ is the circumcenter of triangle $ABC$ . A circle $\Gamma$ passing through $A,C$ intersects the minor arc $BC$ on $\Omega$ at $P$ , and intersects $AB$ at $Q$ . It is known that the midpoint $R$ of minor arc $PQ$ satisfies that $CR \perp AB$ . Ray $PQ$ intersects line $AC$ at $L$ , $M$ is the midpoint of $AL$ , $N$ is the midpoint of $DR$ , and $X$ is the projection of $M$ onto $ON$ . Prove that the circumcircle of triangle $DNX$ passes through the center of $\Gamma$ .

This post has been edited 1 time. Last edited by JustPostChinaTST, Mar 17, 2021, 6:05 PM
Reason: clarification

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#2 h Mar 20, 2021, 11:48 AM • 14 Y

Y by amar_04, enzoP14, samrocksnature, Aritra12, R.e.d.a.c.t.e.d, Mathematicsislovely, brickmaster8, mijail, Pluto04, Whitemountain, PIartist, DaRKst0Rm, centslordm, R8kt

Let $O',Y$ be the circumcenter of $\Gamma,\Omega$ and $Z$ be the projection of $N$ on $MO$ and $C'$ lie on $\Gamma$ such that $CC' \parallel AB$ . Clearly, $\overline{D-O'-O},~\odot(ZMXN),~O$ is the midpoint of $AY$ , $O'$ is the midpoint of $RC'$ .Consider $\begin{align*} \odot(DNXO') &\iff OD \cdot OO' = ON \cdot OX \\& \iff OD \cdot OO' = OM \cdot OZ \\& \iff \odot(MZDO') \\& \iff O'Z \perp MO \iff O'N \perp OM \\& \iff O'N \perp LY \iff C'D \perp LY \end{align*}$ Let $Y'$ be the reflection of $C$ across $Y$ and $S$ is the intersection of $Y'P$ on $CR$ . Hence, $\triangle CC'A \cup \{D\} \sim \triangle CSY' \cup \{Y\}$ . Hence, $C'D \perp SY$ and it suffices to show that $S,Y,L$ are collinear.

Claim: $\overline{Y-P-R}$ .
Proof: Let $Y'P \cap \Gamma = A'$ and $P'$ lie on $\Gamma$ such that $PP' \parallel CY'$ . By angle chasing, $\square A'RCP'$ is a kite and $(A',C;R,P')=-1$ . Then, project it to $CY'$ throught $P$ .

Let $L' = PR \cap AC, B'' = AY \cap CR$ and $LY \cap CR = S*$ . $(A,C;Q,R)=(A,C;L,L') =(B'',C;S*,R)$ Hence, $AB'' \cap S*Q$ lies on $\Gamma$ . Let $B' = AY \cap \Gamma$ . It suffices to show that $AB' \cap SQ$ lies on $\Gamma$ . Equivalent to $A'P, QB'$ and $CR$ are concurrent.

Consider $\triangle CQB$ , $P$ is $B-$ Dumpty point. Let $T$ be the isogonal conjugate of $P$ which is $B-$ Humpty point and $BT \cap CQ = T'$ which is the midpoint of $CQ$ . Let $E$ be the foot of the altitude from $Q$ to $BC$ , $\angle BAY = \theta, \angle PQR = \alpha$ . By angle chaseing, $\angle CTT' = \angle BCT' = \angle CET$ . Therefore, $\odot(CT'TE)$ and $\angle QET = 90^{\circ} - 2 \theta$ .

Notice that $A'C$ is the diameter of $\Gamma$ . Let $G = B'Q \cap A'P$ , By angle chasing, $\square BETQ \sim CB'GA'$ which mean $\overline{G-C-R}$ . Therefore, $A'P, QB'$ and $CR$ are concurrent.

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#3 h Mar 21, 2021, 12:30 PM • 7 Y

Y by k12byda5h, enzoP14, samrocksnature, Mathematicsislovely, brickmaster8, mijail, TheHimMan

As above shows, it's equivalent to prove $C'D \perp LY$ . Here is another finishing (I'll just adopt above's diagram for laziness):

Since $R$ is the midpoint of arc $PQ$ , we have $\angle QAR= \angle RCP$ , hence from $CR \perp AB$ we get $90^{\circ}= \angle QAC+ \angle RCA = \left( \angle QAC- \angle QAR \right) + \left( \angle RCA + \angle RCP \right) =\angle RAC + \angle ACP$ . Because $YC \perp AC$ , we have ghen $\angle YPC = \angle YCP = 90^{\circ} - \angle ACP =\angle CAR$ , hence $Y,P,R$ are collinear.

Further, $YPR \perp C'P$ , so $YP$ is tangent to $(C',C'P)$ (which also passes through $Q$ by symmetry). Since $YP=YC$ , $Y$ lies on the radical axis of $(C',C'P)$ and $(AC)$ . But $LQ \cdot LP = LA \cdot LC$ , so $L$ is also on the radical axis, therefore $C'D \perp YL$ .

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#4 h Mar 30, 2021, 6:29 PM • 2 Y

Y by samrocksnature, brickmaster8

As in the figure above let $C'$ be the point diametrically opposite to $R$ in circle $\Gamma$ .It suffices to show that $O'N \perp OM$ (on a point $X_1$ ) for then we would have that $OX_1\cdot OM=OD\cdot OO'=ON\cdot OX$
Clearly this is equivalent to showing that $C'D \perp YL$ .
(1) $C'P$ is tangent to $\Omega$
Some angle chasing gives that angles $\angle PBC=\angle PCA=\angle AQP=\angle CRC'=\angle CPC'$
Trivially $CC'$ is parallel to $AB$
(2) $PQ$ , $CC'$ concur on a point $Z$ on circle $\Omega$
trivial angle chasing: $\angle QPC=\angle QAC=\angle ZCA$
(3)Let $CP$ meet $AB$ on $P'$ . $C',D,P'$ are collinear
$AP=AP'$ because the bisector $CR$ of $PCQ$ is perpendicular to $AB$ and $AP'P=AQC=P'PA$
also $CC'=AP$ and the fact that $CC'=AP'$ is just what we need.
Let $F$ be the pole of $PZ$ wrt to circle $\Omega$ then $L$ is the pole of $CZ$ and $CZ\perp LY$ .
Clearly $FZ$ is parallel to $QC'$ and then triangles $P'QC'$ and $CZF$ are homothetic wrt $P$ and $P'C'$ is parallel to $CZ$ which is perpendicular to $LY$ done.

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173 posts

#5 h Apr 13, 2021, 5:59 PM • 2 Y

Y by brickmaster8, Hoto_Mukai

Let $Z$ and $S$ be the centers of $\Omega$ and $\Gamma$ respectively. Let $F$ be the foot of perpendicular from $C$ and let $Q'$ be the reflection of $Q$ over $F$ . First, notice that $\triangle CQQ'$ is isosceles, and since $\triangle CQQ'\sim\triangle APQ'$ , we see that the latter is also isosceles. As $RA$ internally bisects $\angle PAQ$ , it follows that $AR\perp CQ'$ , and hence $R$ is the orthocenter of $\triangle AQ'C$ . As $O,D,S$ are collinear, to show that $S$ lies on $(DNX)$ , it suffices to show
$OD\cdot OS=ON\cdot OX.$ To eliminate $X$ , let $Y$ be the foot of perpendicular from $N$ to $OM$ . Then $OD\cdot OS=OY\cdot OM$ if and only if $D,S,Y,M$ are concyclic or equivalently, $\angle SYM=90^\circ$ . Thus all we have to show is that $Y,N,S$ are collinear, or equivalently, $NS\perp MO$ . Now let $T$ be the midpoint of $Q'R$ . Since $R$ is the orthocenter and $S$ is the center of $(ARC)$ , it follows that $RTDS$ is a parallelogram, so $NS$ passes through $T$ . Now homotheties centered at $A$ and $R$ with ratio 2 respectively show that $MO\parallel LZ$ , and $NS\parallel Q'D$ , we just have to show that $LZ\parallel Q'D$ . Let $K$ be the Miquel point of cyclic quadrilateral $AQPC$ . Clearly $K$ lies on $\overline{Q'L}$ . Then
$DZ^2-Q'Z^2=DC^2-CZ^2-(Q'B^2-BZ^2)=DC^2-Q'B^2=DA^2-Q'B^2$ and
$\begin{align*} (LD^2-DA^2)-(LQ'^2-Q'B^2)&=LA\cdot LC-LQ'^2+Q'B^2\\ &=LK\cdot KQ'+Q'P\cdot Q'C-LQ'^2\\ &=LK\cdot LQ'+Q'K\cdot Q'L-LQ'^2 &=LQ'^2-LQ'^2=0 \end{align*}$ and we're done.

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#6 h Apr 30, 2021, 5:07 PM

Y by

Is there a way do draw this diagram (using only ruler and compass)?

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OlyMan

#7 h May 8, 2021, 3:44 AM • 3 Y

Y by Mango247, Mango247, Mango247

hydo2332 wrote:

Is there a way do draw this diagram (using only ruler and compass)?

Why not, the students who would have written the actual test, must have drawn this
by ruler and compass....

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#8 h Nov 6, 2022, 4:42 AM

Y by

hydo2332 wrote:

Is there a way do draw this diagram (using only ruler and compass)?

I think that it is best to think about the problems with drawings by hand, but I also think that thinking about how the construction would be is a nice exercise (it has even helped me to solve some problems). yes it is possible to build it with ruler and compass, if you can't do it I recommend you look at the solutions above to help you, for example you would know that you only need the construction of $\text{B-Dumpty poin}$

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cliid

#9 h Oct 28, 2023, 10:36 PM

Y by

By the given conditions, note that $\triangle ABC$ is isosceles. Let $O_1$ be the center of $\Omega$ , and $O_2$ be the center of $\Gamma$ . Draw $F$ to be the intersection of the line passing through $C$ parallel to $AB$ with $\Gamma$ , so $AB \parallel CF$ and $F \in \Gamma$ . Now, I claim $\overline{R O_2 F}$ collinear. This is true since $\measuredangle BKC = \measuredangle FCK = \measuredangle FCR = 90^{\circ}$ , so $RF$ is a diameter of $\Gamma$ . The condition $(X,N,D,O_2)$ concyclic is equivalent to $ON \cdot OX = OD \cdot OO_2$ , but since $\measuredangle MXO = 90^{\circ}$ , $ON \cdot OX = OH \cdot OM$ where $H$ is the foot from $N$ to $MO$ . So, it suffices to show $OH \cdot OM = OD \cdot OO_2$ . This is just proving $(H,M,D,O_2)$ concyclic. Note that $\measuredangle O_2 D M = 90^{\circ}$ , so it is enough to prove $MH \perp HO_2$ . But since $MH \perp HN$ , it suffices to prove $\overline{HNO_2}$ collinear, or equivalently $MH \perp NO_2$ . Now, since $M$ and $O$ are the midpoints of $AL$ and $AO_1$ , respectively, it suffices to prove $LO_1 \perp NO_2$ . Note that $N$ and $O_2$ are the midpoints of $RD$ and $RF$ , respectively, Thus, $NO_2 \parallel DF$ , so it suffices to prove $LO_1 \parallel DF$ . Note that we can completely erase the point $O_2$ at this point.
Let $\Pi$ be the circle $(AKC)$ with diameter $AC$ , and $\Delta$ be the circle passing through $P$ and $Q$ , centered at $F$ . Then, $\mathrm{Pow}_\Pi L = LA \cdot LC = \mathrm{Pow}_\Gamma L = LQ \cdot LP = \mathrm{Pow}_\Delta L$ , so $L$ lies on the radical axis of $\Pi$ and $\Delta$ .
Moreover, I claim that $\overline{RPO_1}$ collinear. First, $\angle RAC + \angle ACP = 90^{\circ}$ , and $\angle RAC = \angle RFC = 90-\angle FRC = 90-\angle FPC$ . So, $\angle ACP = \angle FPC$ . Because $\angle O_1 P C = \angle O_1 C P = 90 - \angle PCA = \angle RAC = \widehat{RC}$ and $\angle RPQ = \widehat{RQ}$ , $\angle QPF = \widehat{QF}$ , $\angle FPC = \widehat{FC}$ , we have $\angle RPQ + \angle QPF + \angle FPC + \angle CPO_1 = \widehat{RQ} + \widehat{QF} + \widehat{CF} + \widehat{RC} = 180^{\circ}$ , so $\overline{RPO_1}$ collinear, as desired.
Now, since $\overline{O_1 P R}$ collinear and $\measuredangle FPR = 90^{\circ}$ , we have $\measuredangle O_1 P F = 90 ^{\circ}$ , so $\mathrm{Pow}_\Delta O_1 = O_1 P^2 = O_1 C^2 = \mathrm{Pow}_\Pi O_1$ since $\measuredangle ACO_1 = 90^{\circ}$ . Thus, $O_1$ also lies on the radical axis of $\Delta$ and $\Pi$ . But this implies that $LO_1$ is the radical axis of $\Delta$ and $\Pi$ , so $LO_1 \perp DF$ , since $D$ is the center of $\Pi$ and $F$ is the center of $\Delta$ , which is what we wanted to show.

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#10 h Mar 2, 2025, 4:17 AM

Y by

Let $O_1$ be center of $\Omega$ and $O'$ be center of $\Gamma$ , let $RR'$ diameter on $\Gamma$ .
By Doble midbase and PoP it is enough to show that $LO_1 \perp DR'$ , notice that by the problem condition $CR' \parallel AB$ and by Reim's we have that $PQ \cap CR'=E$ lies on $\Omega$ , so by shooting Lemma we have $R'P$ tangent to $\Omega$ and so $R,P,O_1$ are colinear with $O_1P$ being tangent to $(R',R'P)$ but also notice $O_1C^2=O_1P^2$ and therefore $O_1$ lies on the radax of $(AC), (R',R'P)$ but also since $CPQA$ is cyclic we have by PoP that $L$ lies on radax of $(AC), (R',R'P)$ as well and therefore our perpendicularity is now trivially true thus we are done :cool:

:cool:

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