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Source: 2015 JBMO Shortlist G2

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30629 posts

#1 h Oct 8, 2017, 10:25 AM • 3 Y

Y by Adventure10, Mango247, lian_the_noob12

The point ${P}$ is outside the circle ${\Omega}$ . Two tangent lines, passing from the point ${P}$ touch the circle ${\Omega}$ at the points ${A}$ and ${B}$ . The median ${AM \left(M\in BP\right)}$ intersects the circle ${\Omega}$ at the point ${C}$ and the line ${PC}$ intersects again the circle ${\Omega}$ at the point ${D}$ . Prove that the lines ${AD}$ and ${BP}$ are parallel.

(Moldova)

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475 posts

#2 h Oct 8, 2017, 10:44 AM • 2 Y

Y by tenplusten, Adventure10

suffices to prove (P,B;M,AD\capBP)=-1
<=> (A,B;C,D)=-1, which is true.

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32 posts

#3 h Mar 11, 2018, 8:58 PM • 1 Y

Y by Adventure10

Because $MB$ is tanget to circle $\Omega$ it is well know that: $MB^2 = MC\cdot MA$ but because $MB=MP$ we have:
$MP^2 = MC\cdot MA$
Now this means that triangles $\triangle MPC$ and $\triangle MPA$ are similar. From there, we got:
$\angle MPC =\angle PAM$
But, line $PA$ is tanget to the circle $\Omega$ so: $\angle PAM= \angle ADP$ .
Now : $\angle MPC= \angle ADP$ so the lines $AD$ and $BP$ are parallel.

This post has been edited 3 times. Last edited by Bergo1305, Mar 11, 2018, 8:59 PM

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308 posts

#4 h Aug 25, 2018, 3:38 AM • 2 Y

Y by Adventure10, Mango247

In this proof, let $(ABC$ ) be the circumcircle of $\triangle ABC$ and Pow $_\omega (X)$ be the power of point $X$ with respect to circle $\omega$ .

We start out with a Lemma.

Lemma: Given $\triangle ABP$ with $M$ on the midpoint of $PB$ and $MB$ is tangent to $(ACB)$ , then $MP$ is tangent to $(ACP)$ .

Proof of Lemma: Let $\Omega=(ACB)$ and $\omega=(ACP)$ . Because $M$ lies on the Radical Axis $AC$ of $\Omega$ and $\omega$ , Pow $_\Omega (M)=$ Pow $_\omega (M).$ Assume for the sake of contradiction that $MP$ is not tangent to $\omega$ , or $MP$ intersects $\omega$ at a second time at $P'$ . Because $MP^2=MB^2$ and Pow $_\Omega (M)=MB^2$ , then Pow $_\omega (M)=MP^2$ . But this is a contradiction as Pow $_\omega (M)=MP\cdot MP' \neq MP\cdot MP=MP^2$ as $P$ and $P'$ are unique. Thus, $MP$ is tangent to $(ACP).$ $\Box$

By the Lemma, $\angle BPC=\angle PAC.$ Because $PA$ is tangent to $\Omega$ , $\angle PAC=\angle CDA$ . Hence, $\angle ADC=\angle PAC=\angle CPM$ , proving that $AB\parallel PB$ . $\blacksquare$

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#5 h Nov 26, 2018, 7:14 PM • 1 Y

Y by Adventure10

Since, $\Delta MPC \sim \Delta MAP$ and $\Delta PAC \sim PDA \implies \angle CPM =\angle PAM= \angle PDA \implies \boxed{AD||BP}$

Something obvious but interesting is that: $PC=2MC$

This post has been edited 2 times. Last edited by AlastorMoody, Nov 26, 2018, 7:23 PM

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#6 h Feb 2, 2019, 8:14 PM • 2 Y

Y by Adventure10, Mango247

Define $\mathcal{P_1}_{||AD}$ as the point at infinity along $AD$ ,
$-1 =(A,B;C,D) \overset{A}{=} (P,B;M, AD \cap BP) \implies AD \cap BP=\mathcal{P_1}_{||AD} \implies AD||BP$

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1051 posts

#7 h May 26, 2020, 4:01 AM

Y by

Here's a power of a point solution.

Claim: $\triangle MPC \sim \triangle MAP$

Proof: $PM^2 = BM^2 = CM \cdot AM,$ so $\frac{MP}{MC} = \frac{MA}{MP}.$

Notice that because $PM = \frac{AP}{2}, PC = 2CM.$ Now $PC \cdot PD = PA^2 = PB^2 = 4MB^2 = 4(MC \cdot MA)$ . But $PC = 2MC,$ so $PD = 2AM.$ Now $\frac{AC}{CD} = \frac{AM-CM}{PD-CD} = \frac{1}{2},$ so $\triangle ACD \sim \triangle MCP,$ and so $\angle MPC = \angle CDA = \angle PDA,$ as desired.

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#9 h Sep 1, 2020, 3:05 PM

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Since $PB$ is tangent to $\Omega$ and $AC$ intersects $BP$ in its midpoint we have that $(PCA)$ is tangent with $PB$ and so,
$\angle CDA= \angle CAP=\angle BPC=\angle BPD$ hence done.

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#10 h Feb 7, 2021, 2:54 PM

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Solution. It is easy to see that $(ACP)$ is tangent with $PB$ at point $P\implies C$ is the $A$ -Humpty Point of $\triangle ABP$ , $\angle PCB=180-\angle A\implies \angle PBA=\angle A=\angle DCP=\angle DAB.\blacksquare$

This post has been edited 1 time. Last edited by geometry6, Feb 7, 2021, 2:55 PM

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#12 h Jan 1, 2023, 12:46 PM

Y by

first by pow(M) MB^2=MC*MA=CP^2 because M is midpoint.so BP is tangent to (ACP) which means (1) <CAP=<CPB.PA is tangent to circle so <CAP=<ADC <=> (1) <ADP=<BPD and we are done.

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lian_the_noob12

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#13 h Feb 22, 2024, 10:17 PM

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$\color{red} \textbf{Geo Marabot Solve 4}$

$-1=(D,C ; AB \cap PD,P) \overset{A}{=} (AD\cap BP,M;B,P) \implies AD\cap BP \equiv \mathcal{P_1} \implies AD \parallel BP \blacksquare$

This post has been edited 4 times. Last edited by lian_the_noob12, Feb 22, 2024, 10:19 PM

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#14 h Mar 6, 2025, 10:25 PM

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Proof: It is clear that BP is tangent to (ACP) at point P. It follows that <DPB = <CPB = <CAP = <CDA = <PDA and we are done.

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