SD is tangent to (QDM)

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SD is tangent to (QDM)

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Source: ELMO Shortlist 2013: Problem G13, by Ray Li

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6877 posts

#1 h Jul 23, 2013, 1:32 AM • 4 Y

Y by bcp123, megarnie, Adventure10, Mango247

In $\triangle ABC$ , $AB<AC$ . $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$ , respectively. $M$ is the midpoint of segment $BC$ , and $\omega$ is the circumcircle of $\triangle APD$ . Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$ . $QB$ meets $\omega$ again at $R$ , and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$ . Prove $SD$ is tangent to the circumcircle of $\triangle QDM$ .

Proposed by Ray Li

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#2 h Jul 23, 2013, 6:12 AM • 4 Y

Y by Pirkuliyev Rovsen, bcp123, DanDumitrescu, Adventure10

Since the cross ratio $(B,C,D,P)$ is harmonic, then $MB^2=MC^2=MD \cdot MP$ $\Longrightarrow$ circle $(M)$ with diameter $\overline{BC}$ is orthogonal to $(K) \equiv \omega$ $\Longrightarrow$ $Q \in (M).$ Inversion with center $B$ and power $BD \cdot BP$ takes $(K)$ into itself and carries $(M)$ into a line orthogonal to $(K),$ due to conformity. $R$ is the inverse of $Q,$ hence $RK$ is perpendicular bisector of $\overline{DP}$ $\Longrightarrow$ $\triangle SDP$ is S-isosceles $\Longrightarrow$ $\angle SDP=\angle SPD \equiv \angle QPD=\angle MQD$ $\Longrightarrow$ $SD$ is tangent to $\odot(QDM).$

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#3 h Jul 24, 2013, 10:14 AM • 4 Y

Y by bcp123, Pirkuliyev Rovsen, Adventure10, Mango247

Other solution, thanks to the same brilliant idea of Luis:

$Q$ belongs to Apollonius circle of $\Delta ABC$ , hence $QD, QP$ are bisectors of $\angle BQC$ which, being onto the circle of diameter $BC$ , is a right angle, consequently $\angle BQD=45^\circ$ and $R$ is the midpoint of the arc $DP$ of the circle $(ADP)$ and $\angle SDP=\angle SPD=\angle DQM$ , done.

Best regards,
sunken rock

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#4 h Aug 17, 2013, 6:17 PM • 2 Y

Y by Adventure10, Mango247

The Solution

Pretty standard problem.

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#5 h Apr 30, 2014, 3:53 AM • 1 Y

Y by Adventure10

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#6 h Jul 6, 2014, 4:23 PM • 1 Y

Y by Adventure10

Wow, G13!

To be honest I found the G4 more challenging than this.

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Let $\omega$ denote the circumcircle of $\triangle APD.$ By the angle bisector theorem, $\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{BP}{PC},$ so $\omega$ is the Appolonius circle of $B$ and $C$ with ratio $\dfrac{AB}{AC}.$ Since $Q$ lies on this circle too, $\dfrac{QB}{QC} = \dfrac{BD}{DC},$ implying $QD$ bisects $\angle BQC$ internally. Since $\angle PAD = 90^{\circ}$ and quadrilateral $APQD$ is cyclic, $\angle PQD = 90^{\circ},$ so $QP$ bisects $\angle BQC$ externally (this follows from the fact that the internal and external angle bisectors are perpendicular). Also, note that since $DA \perp AP,$ $AD$ is a diameter of $\omega.$

Now, note that $PB \times DC = PC \times BD,$ which implies $(P,D,B,C)$ is harmonic. Hence, $MC^2 = MB^2 = MD \cdot MP = MQ^2,$ where we have used the fact that $MQ^2 = MD \times MP$ which follows from PoP. Hence, $MB=MC=MQ,$ which implies that $M$ is the circumcenter of $\triangle BCQ.$ Since $M$ is the midpoint of $BC,$ this forces $\triangle BQC$ to be right angled at $Q.$ By our previous observations, we have that $\angle BQM = \angle QMC = \angle BQP = 45^{\circ}.$

Now, $\angle RPD = \angle PQR = 45^{\circ} = \angle RQD = \angle RDP,$ so $\triangle RPD$ is isoceles with $RP=RD.$ Since $AD$ is a diameter of $\omega$ and $R$ lies on the perpendicular bisector of $PD,$ $RS$ must bisect $PD,$ implying $RS$ is the perpendicular bisector of $PD.$ Hence, $SP = SD$ and $\angle SDP = \angle SPD.$ Since $MQ$ is tangent to $\omega,$ $\angle MQD= \angle QPD.$ It follows that $\angle SDP = \angle MQD,$ completing the proof. $\blacksquare$

This post has been edited 1 time. Last edited by AnonymousBunny, Jul 7, 2014, 4:55 PM

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#7 h Jul 7, 2014, 2:37 PM • 1 Y

Y by Adventure10

Obviously, the circle with centre $M$ through $B, C$ is orthogonal to the circle $APD$ , so $\angle BQC = \dfrac{\pi}{2}$ , thus $QB$ bisects $\angle PQD$ . So, $SDP$ is isosceles, or $\angle SDP = \angle SPD = \angle MQD$ ...

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sayantanchakraborty

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sayantanchakraborty

#8 h Dec 18, 2014, 3:02 PM • 1 Y

Y by Adventure10

Yes its just an easy cross ratio exercise.

$AP,AD,AB,AC$ being a harmonic pencil implies $(P,D;B,C)=-1 \implies MD \cdot MP=MB^2=MQ^2=MC^2 \implies QB \perp QC$ .Also $Q$ being a point on the Apollonius circle means $\angle{BQD}=\angle{{DQC}=45^{\circ} \implies PR=RM \implies \angle{DQM}=\angle{QPD}=\angle{SMP}}$ which implies the result.

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#9 h Nov 12, 2019, 1:12 PM • 2 Y

Y by Pakistan, Adventure10

TBH This was really very easy.

ELMO 2013 Shortlist G13 wrote:

In $\triangle ABC$ , $AB<AC$ . $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$ , respectively. $M$ is the midpoint of segment $BC$ , and $\omega$ is the circumcircle of $\triangle APD$ . Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$ . $QB$ meets $\omega$ again at $R$ , and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$ . Prove $SD$ is tangent to the circumcircle of $\triangle QDM$ .

Proposed by Ray Li

Solution:-

Claim 1:- $(P,D;B,C)=-1$ .
Just see that $\angle BAD=\angle DAC$ and $\angle PAD=90^\circ$ . So, $(P,D;B,C)=-1$ .

Claim 2:- $MB=MQ=MC$ .
As $(P,D;B,C)=-1$ and $MB=MC\implies MD.MP=MB^2=MC^2=MQ^2\implies \angle BQC=90^\circ$ .

Claim 3:- $R$ is the midpoint of $\widehat{PD}$ of $\odot(APD)$ .
As $(P,D;B,C)=-1\implies \angle PQB=\angle BQD$ as $\angle BQC=90^\circ$ . So, $R$ is the midpoint of $\widehat{PD}$ of $\odot(PQD)$ .

Claim 4:- $\angle SDQ=\angle QMD$ .
Let $\angle QPC=\theta\implies \angle QSD=2\theta\implies\angle QDS=90^\circ-2\theta$ . Now notice that $\angle PQB=45^\circ\implies\angle QBM=45^\circ-\theta\implies\angle QMD=90^\circ-2\theta=\angle QDS$ .

So, $SD$ is tangent to $\odot(QDM)$ . $\blacksquare$ .

This post has been edited 1 time. Last edited by amar_04, Nov 16, 2019, 10:01 PM

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#10 h Mar 19, 2020, 1:41 PM • 3 Y

Y by gamerrk1004, Aryan-23, SenatorPauline

Solution (with aops29)

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#11 h Apr 5, 2022, 8:04 PM

Y by

v_Enhance wrote:

In $\triangle ABC$ , $AB<AC$ . $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$ , respectively. $M$ is the midpoint of segment $BC$ , and $\omega$ is the circumcircle of $\triangle APD$ . Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$ . $QB$ meets $\omega$ again at $R$ , and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$ . Prove $SD$ is tangent to the circumcircle of $\triangle QDM$ .

Proposed by Ray Li

Wut? (did in around 5 mins coz i'm stupid, how could i forgot cross ratio+angle bisector stuff)
No need of even naming claims n'stuff, just note that $-1=(P, D; B, C)$ so since $\angle PQD=90$ we have that $\angle BQD=\angle DQC$ and since using MacLaurins theorem and PoP we have $MQ^2=MD \cdot MP=MB^2=MC^2$ so $M$ is center of $(BQC)$ meaning that $\angle BQC=90$ to $\angle PQB=\angle BQD=\angle DQC=45$ meaning that $PS=SD$ as $S$ lies on the perpbisector of $PD$ , so now by angle chase
$\angle DQM=\angle SPD=\angle SDP \implies SD \; \text{tangent to} \; (QDM)$ Thus we are done

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#12 h Sep 4, 2023, 6:55 PM • 1 Y

Y by centslordm

...

It is well-known that $MD\cdot MP=MB^2=MC^2$ , and by power of a point it also equals $MQ^2$ , hence $M$ is the center of $(BCQ)$ , i.e. $\angle BQC=90^\circ$ . Since $Q$ lies on the $A$ -apollonius circle $\omega$ of $\triangle ABC$ , it also follows that $\overline{QD}$ bisects $\angle BQC$ internally, hence $\angle BQD=\angle RQD=45^\circ$ . Since $\overline{PD}$ is a diameter of $\omega$ , it follows that $\overline{RS}$ bisects $\overline{DP}$ .

Now, we have $\angle SDP=\angle SPD=90^\circ-\angle BDQ$ and $\angle DQM=\angle BQM-\angle BQD=\angle BQM-45^\circ$ , so it suffices to show that $135^\circ=\angle BQM+\angle BDQ=\angle QBD+\angle BDQ$ , which is clear by looking at $\triangle BDQ$ . $\blacksquare$

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CT17

#13 h Sep 4, 2023, 10:13 PM

Y by

Note that $MQ^2 = MD\cdot MP = MB^2=MC^2$ so $Q$ lies on $(BC)$ and hence $R' = CQ\cap (PD)$ is diametrically opposite to $R$ . But by projecting through $Q$ onto $BC$ , $(PD;RR')$ is a harmonic bundle, so $RR'$ is actually the diameter perpendicular to $PD$ . Then we have

$\measuredangle SDM = \measuredangle DPS = \measuredangle DPQ = \measuredangle DQM$
as desired.

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