Iberoamerican Olympiad 2013 - Problem 2

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Iberoamerican Olympiad 2013 - Problem 2

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Source: http://oim2013.opm.org.pa/pdfs/examen_pt.pdf

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118 posts

#1 h Aug 13, 2014, 5:25 PM • 3 Y

Y by Davi-8191, Adventure10, Mango247

Let $X$ and $Y$ be the diameter's extremes of a circunference $\Gamma$ and $N$ be the midpoint of one of the arcs $XY$ of $\Gamma$ . Let $A$ and $B$ be two points on the segment $XY$ . The lines $NA$ and $NB$ cuts $\Gamma$ again in $C$ and $D$ , respectively. The tangents to $\Gamma$ at $C$ and at $D$ meets in $P$ . Let $M$ the the intersection point between $XY$ and $NP$ . Prove that $M$ is the midpoint of the segment $AB$ .

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bcp123

#2 h Aug 13, 2014, 5:46 PM • 3 Y

Y by Pluto1708, Derpy_Creeper, Adventure10

In order for this to be true, we don't need the information that $XY$ is a diameter.
$NP$ is $N$ -symmedian of $NCD$ so it's enough to prove that $AB$ is antiparallel to $CD$ .
$\angle NBA=\alpha$ and $\angle NYX=\beta\Rightarrow \angle DNY=\alpha-\beta\Rightarrow \angle YCD=\alpha-\beta$ . Since $N$ is midpoint of arc $XY$ , $\angle NCY=\beta$ so $\angle NCD=\alpha$ , as needed.

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sayantanchakraborty

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sayantanchakraborty

#3 h Oct 5, 2014, 12:10 PM • 3 Y

Y by Derpy_Creeper, Adventure10, Mango247

Hmm.... I used projectivity to kill the problem.

Let $J$ be the intersection of $NP$ with the circle.Then obciously $CDJN$ is a harmonic quadrilateral.Thus $NC,ND,NJ$ and the tangent at $N$ to the circle form a harmonic pencil.In other words $NA,NB,NM$ and the tangent at $N$ to the circle form a pencil.But note that the tangent at $N$ is parallel to diameter $XY$ since $N$ is the midpoint of arc $XY$ .So $M$ is the midpoint of $AB$ ,as desired.

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#4 h Nov 7, 2014, 8:32 PM • 4 Y

Y by matl, Derpy_Creeper, Adventure10, Mango247

Let $X'$ and $Y'$ be the intersections of the line $XY$ with the lines $PC$ and $PD$ , respectively.
Using Menelaus's theorem we get $\frac{X'C \cdot PN \cdot MA}{PC \cdot MN \cdot X'A} = 1 = \frac{Y'D \cdot PN \cdot MB}{PD \cdot MN \cdot Y'B} \implies \frac{X'C \cdot PD \cdot MA \cdot Y'B}{Y'D \cdot PC \cdot MB \cdot X'A} = 1$
Since $PC$ and $PD$ are tanget to $\Gamma$ , $PC=PD$ .
Also, $\angle X'AC = \angle NAO = 90^\circ - \angle CNO = 90^\circ - \angle NCO = \angle X'CA$ , then $X'C=X'A$ . Analogously $Y'D = Y'B$ .

Then the first result simply becomes $MA=MB$ .

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2312 posts

#5 h Nov 7, 2014, 10:40 PM • 9 Y

Y by danielwatson, matl, Pluto1708, primesarespecial, Derpy_Creeper, alexgsi, Adventure10, Mango247, AlexCenteno2007

My solution :

Let $\omega_A$ be a circle passing through $A, C$ and tangent to $XY.$
Let $\omega_B$ be a circle passing through $B, D$ and tangent to $XY.$

Since $NA \cdot NC=NB \cdot ND, PC^2=PD^2,$
so $NP$ is the radical axis of $\omega_A$ and $\omega_B$ $\Longrightarrow$ $MA^2=MB^2.$

Q.E.D

This post has been edited 1 time. Last edited by TelvCohl, Aug 17, 2016, 1:38 PM

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#6 h Aug 17, 2016, 1:34 PM • 1 Y

Y by Adventure10

through P we drawn ST//AB (S on NC and T on NB) we have m(XAC)=m(XCA) (because NX=NY) ==> m(PCS)=m(PSC)==> PS=PC,analogously PD=PT ==> PS=PT and combinate ST//AB.

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#7 h Feb 11, 2018, 7:35 PM • 1 Y

Y by Adventure10

........

This post has been edited 1 time. Last edited by Gregorio, May 10, 2018, 10:18 PM
Reason: ..

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jbaca

#8 h Apr 26, 2020, 11:26 PM • 1 Y

Y by AlexCenteno2007

Solution. By definition of $N$ , we get
$NA\cdot NC=NX^2=NY^2=NB\cdot ND$ thus $ABDC$ is cyclic. Since $NP$ is the $N$ -symmedian of $\bigtriangleup CND$ , we're done. $\blacksquare$

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34 posts

#9 h Oct 24, 2020, 3:08 PM • 1 Y

Y by Nawi

Using some trig:
We'll need the following lemma: Let $ABC$ a triangle and $AD$ a cevian. Then $\frac{\sin(\angle BAD)}{\sin(\angle DAC)} = \frac{BD}{DC}\frac{CA}{AB}$ .
Let $M'$ be the intersection between $CD$ and $PN$ , and let $Q$ the intersection between the circumference and $CD$ . We have that $\angle{ NAB} = \angle NDC$ since they both open the same arcs and $NX = NY$ . Similarly, $\angle NBA = \angle NCD$ and then triangles $NAB$ and $NDC$ are similar and therefore $\frac{AN}{BN} = \frac{ND}{NC}$ .

Observe that $\angle CDQ = \angle CNQ = \angle PCQ$ since $\angle CDP = \angle PCD = \angle CND$ , and similarly $\angle PND = \angle DCQ = \angle QDP$ . Call $\alpha = \sin(\angle CNP), \beta = \sin(\angle PND)$

Trig version of Ceva on $CDP \implies \frac{\sin(\angle DPM')}{\sin(\angle M'PC} \frac{\alpha^2}{\beta^2} = 1 \implies \frac{\sin(\angle DPM')}{\sin(\angle M'PC} = \frac{\beta^2}{\alpha^2} = \frac{M'D}{M'C}$ , where the last equality comes from using the lemma on triangle $CPD$ with cevian $PM'$

Now the lemma on triangle $CND$ and cevian $NM'$ gives $\frac{\alpha}{\beta} = \frac{CM'}{M'D}\frac{ND}{NC} \implies \frac{\beta}{\alpha} = \frac{ND}{NC} = \frac{AN}{BN}$ .

Finally, apply the lemma on $NAB$ with cevian $NM \implies \frac{\alpha}{\beta} = \frac{AM}{BM}\frac{BN}{AN} = \frac{AM}{BM}\frac{\alpha}{\beta} \implies \frac{AM}{BM} = 1$ , as desired

This post has been edited 1 time. Last edited by manifoldaora, Oct 24, 2020, 3:09 PM
Reason: some typos

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277 posts

#10 h Apr 7, 2021, 1:17 PM

Y by

Davi Medeiros wrote:

Let $X$ and $Y$ be the diameter's extremes of a circunference $\Gamma$ and $N$ be the midpoint of one of the arcs $XY$ of $\Gamma$ . Let $A$ and $B$ be two points on the segment $XY$ . The lines $NA$ and $NB$ cuts $\Gamma$ again in $C$ and $D$ , respectively. The tangents to $\Gamma$ at $C$ and at $D$ meets in $P$ . Let $M$ the the intersection point between $XY$ and $NP$ . Prove that $M$ is the midpoint of the segment $AB$ .

Posting for Storage. Have essentially the same solution as #8, but I spelled out the ending

Let $K$ be the midpoint of $DC$ and let the tangents at $N$ and $C$ intersect at a point $T$ and let the tangents at $N$ and $D$ intersect at a point $S$ . We know that the tangent through $N$ is parallel to $XY$ because $N$ is the midpoint of arc $XY$ .
By the tangent chord theorem, we know that $\angle BDC = \angle NDC = \angle CNT = \angle NAY = \angle NAB$ meaning that $\triangle NAB \sim \triangle NDC$ .
Moreover, by the Symmedian Lemma, $\angle KND = \angle PNC$ . Then by the Ratio Lemma and using the fact that $\triangle NAB \sim \triangle NDC$ $\frac{AM}{MB} = \frac{\sin \angle MAN}{\sin \angle MNB} \cdot \frac{AN}{BN} = \frac{\sin \angle DNK}{\sin \angle KNC} \cdot \frac{DN}{NC} = \frac{DK}{KC}$ where the last equality folows from the Converse of the Ratio Lemma. Clearly, by definition $\frac{DK}{KC} = 1$ meaning that $M$ is indeed the midpoint of $AB$ .

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JustKeepRunning

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JustKeepRunning

#11 h Jun 9, 2021, 11:10 AM • 1 Y

Y by LeYohan

Instat-killled by symmedians...

In fact, we will prove the result in general for any chord $XY$ . By definition, $NP$ is a symmedian in $\triangle NCD$ . If we can show that $AB$ is anti-parallel to $CD$ , then we know that the symmedian bisects all antiparallel lines, so we would be done. This is easy; just notice that $\angle XBN=\frac{\overarc{XN}+\overarc{YD}}{2}=\frac{\overarc{YN}+\overarc{YD}}{2}=\angle NCD,$ so we are done.

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#12 h Dec 24, 2022, 3:32 AM

Y by

As $N$ is midpoint of arc $XY$ of $\Gamma$ we have $\angle NYX = \angle NXY = 45^{\circ}$ . Then, $\angle NCD = \angle NCY + \angle YCD = 45^{\circ} + \angle YND = \angle NBX \Rightarrow ABDC$ is a cyclic quadrilateral.
So $\triangle NAB \sim \triangle NDC$ . Let $M'$ be the midpoint of $CD$ .
It's well know that $NP$ y $NM'$ are isogonals, then $\angle CNP = \angle ANM = \angle DNM'$ , therefore $M$ is to the $\triangle NAB$ as $M'$ is to the $\triangle NDC$ .
Thus $M$ is midpoint of $AB$ $\blacksquare$

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