All-Russian Olympiad 2010 grade 9 P-3

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Ovchinnikov Denis

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Ovchinnikov Denis

#1 h Sep 9, 2010, 9:59 AM • 1 Y

Y by Adventure10

Lines tangent to circle $O$ in points $A$ and $B$ , intersect in point $P$ . Point $Z$ is the center of $O$ . On the minor arc $AB$ , point $C$ is chosen not on the midpoint of the arc. Lines $AC$ and $PB$ intersect at point $D$ . Lines $BC$ and $AP$ intersect at point $E$ . Prove that the circumcentres of triangles $ACE$ , $BCD$ , and $PCZ$ are collinear.

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imfeyn

#2 h Sep 16, 2010, 8:32 AM • 1 Y

Y by Adventure10

Let's show circle $PAZB$ , circulcircle of $ACE$ , and circulcircle of $BCD$ tangent internally respectively.

$M$ is a midpoint of line segment $PZ$ and $O_1$ is circumcenter of triangle $ACE$ .
If two circles are tangent then their centers and tangent point are colinear. So, it is sufficient to show that $\angle PAO_1 = \angle PAM$ .
Let $\angle APZ = \theta$ , then $\angle PAM = \theta$ .
$\angle PAO_1 =\angle EAO_1 = \frac{\pi}{2} -\angle ACE = \frac{\pi}{2} -(\pi - \angle ACB)= \angle ACB -\frac{\pi}{2} = \frac{1}{2} \angle AZB - \frac{\pi}{2} = \theta = \angle PAM$
And we are done.
In the same way, we can show that circle $PAZB$ , and circulcircle of $BCD$ tangent internally.

From this we can find the radical point(let it be $F$ ) of circle $PAZB$ , circulcircle of $ACE$ , and circulcircle of $BCD$ and is a point which is intersection of line $PZ$ and tangent line of circle $PAZB$ at $A$ (or $B$ ).
$FZ \cdot FP$ : power of point $F$ with respect to circumcircle of $PCZ, PAZB$
= ${FA}^2$ : power of point $F$ with respect to circulcircle of $PCZB, ACE$
= ${FB}^2$ : power of point $F$ with respect to circulcircle of $PCZB, BDE$

Therefore $F$ is a radical point of circumcircles of $PCZ, ACE, BCD$ . and $C$ is a common point of three circle.

Three circumcircles of $PCZ, ACE, BCD$ pass $C$ and $FC$ is a common radical line.

And Center of three circumcircles of $PCZ, ACE, BCD$ pass $C$ and $FC$ are colinear!

This post has been edited 1 time. Last edited by imfeyn, Sep 16, 2010, 12:15 PM

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jayme

#3 h Sep 16, 2010, 11:40 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
nice proof...
To go more quickly, you can use a converse of the pivot theorem in order to prove the first step.
Sincerely
Jean-Louis

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#4 h Sep 16, 2010, 12:52 PM • 4 Y

Y by BakyX, MahdiTA, Adventure10, Mango247

see attached file

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Zhero

#5 h Nov 24, 2010, 3:39 AM • 2 Y

Y by Adventure10, Mango247

Let the circumcircles of $\triangle CAE$ and $\triangle CBD$ intersect at $T$ . We claim that $PCZT$ is cyclic. This will finish the problem, because the circumcircles of $CAE$ , $CBD$ , and $CPZ$ have $TC$ as a common chord, whence their centers all lie on the perpendicular bisector of $TC$ .

Perform an inversion about $P$ with arbitrary radius. We obtain two circles $\omega_1$ and $\omega_2$ intersecting at $C'$ and $P$ , with a common external tangent intersecting $\omega_1$ at $E'$ and $\omega_2$ at $D'$ . $T'$ is the intersection of lines $E'A'$ and $B'D'$ . $Z'$ is the reflection of $C'$ across $A'B'$ . We wish to show that $P'$ , $Z'$ , and $T'$ are collinear.

We will first show that quadrilateral $P'E'T'B'$ is cyclic. Let $\angle E'C'A' = D'C'B' = \theta$ . Then $\angle B'A'T' = \angle A'B'T' = \theta$ and $\angle A'T'B' = 180^{\circ} - 2 \theta$ . Furthermore, $\angle E'P'B' = \angle E'P'A' + \angle A'P'B' = \angle E'C'A' + \angle A'P'C' + \angle C'P'B'$ , which equals $\theta + \angle C'A'B' + \angle C'B'A' = \theta + 180^{\circ} - \angle A'C'B' = \theta + 180^{\circ} - \theta = 2\theta$ , so $\angle E'P'B' + \angle B'T'E' = 2 \theta + 180^{\circ} - 2 \theta = 180^{\circ}$ , so $P'E'T'B'$ is cyclic.

Note that $\angle A'Z'B' = \angle A'C'B' = 180 - \theta$ and $\angle A'P'B' = \theta$ , so $P'A'Z'B'$ is cyclic. Let $Z''$ be the intersection of $P'T'$ with the circumcircle of $P'A'B'$ . Then $\angle B'A'Z'' = \angle B'P'Z'' = \angle B'P'T' = \angle B'E'T' = \angle P'A'B' = \angle B'A'Z'$ . Hence, $Z'$ and $Z''$ both lie in the intersection of the circumcircle of $PA'B'$ and the line not passing through $C'$ that makes an angle of $\angle B'A'Z' = \angle B'A'Z''$ with $A'B'$ , whence $Z' = Z''$ . It follows that $A'$ , $Z'$ , and $T'$ are collinear, as desired.

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#6 h Nov 24, 2010, 10:32 PM • 4 Y

Y by Adventure10, Mango247, Mogmog8, and 1 other user

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#7 h Nov 30, 2010, 1:46 AM • 3 Y

Y by Fermat_Theorem, Adventure10, Mango247

Sorry, I was too tired to think in school today

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#8 h Jul 27, 2014, 9:25 PM • 1 Y

Y by Adventure10

Take inversion with any radius and with center C.Now,let X' be the image of X in this inversion.Now,it suffices to prove that the lines E'A',B'D' and P'Z' are concurrent.Now,by simple angle chase we obtain that A'Z'B',A'P'B' and A'SB'(S is the intersection of A'E' and B'D') are iscolles,so Z',P' and are collinear and we are finished.

This post has been edited 1 time. Last edited by junioragd, Aug 1, 2014, 7:13 PM

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jlammy

#9 h Jan 3, 2015, 1:27 PM • 2 Y

Y by Adventure10, Mango247

imfeyn wrote:

Let's show circle $PAZB$ , circulcircle of $ACE$ , and circulcircle of $BCD$ tangent internally respectively.

Another way to prove this is as follows:

$PAZC$ is cyclic with diameter $\overline{PZ}$ . Let $CD$ meet $(PAZC)$ at $Q$ . By Reim's theorem on $(ABC), (PAQB)$ , $BC \parallel PQ$ . By a converse of Reim's theorem on $(ACE), (PAQ)$ , these two circles are tangent at $A$ . The second result follows analogously, and we proceed as before.

I think jayme was a little terse: it took me some time to figure out that the application of the pivot theorem was to $\triangle ABC$ with $\triangle PBE$ as the reference triangle. The circle $(BBC)$ is defined similarly to a Brocard circle, being tangent to $\overline{PB}$ at $B$ .

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#10 h Nov 15, 2015, 4:13 AM • 1 Y

Y by Adventure10

This is a very nice illustration of https://www.awesomemath.org/wp-content/uploads/article_1_lema_coaxalitate.pdf

Let $(ACE)$ and $(BCD)$ meet $ZA$ and $ZB$ again respectively at $X,Y$ .

Now, by some angle-chasing we get that $XE \parallel PZ \parallel YD$ .

Now, to prove that $(PCZ),(ACE),(BCD)$ are coaxal it shall suffice to showing by this lemma that

$\frac{ZX}{PE}=\frac{ZY}{PD}$ which follows from the angle chase.

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suli

#11 h Nov 15, 2015, 6:15 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

Here's an epic inversion solution.

Let $M$ be the second intersection of circumcircles of $AEC$ and $CDB$ .

Invert with respect to $C$ . Then we obtain $AB$ as a common external tangent of circumcircles of $ACPE$ and $BCPD$ which intersect at $P$ and $C$ , which remains the intersection of $AD$ and $BE$ ; it is $M$ that becomes the intersection of $AE$ and $BD$ (not $P$ ). Also, $Z$ , which is the midpoint of a diameter in the original diagram, is inverted into the reflection of $C$ across line $AB$ .

Lemma. $M, Z, P$ are collinear.
Proof. Let $\angle ZAB = x$ and $\angle ZBA = y$ . Then $x = \angle CAB = \angle AEB$ and $y = \angle CBA = \angle ADB$ . Now notice that
$\angle ZAM = \angle BAM - x = \angle ECA - x = y$ and similarly $\angle ZBM = x$ , so $MA, MB$ are tangents to circumcircle of $ZAB$ . Thus $ZM$ is a symmedian of triangle $ZAB$ .

Let $PC$ intersect $AB$ at $N$ . By radical axis theorem, $NA^2 = NB^2$ , so $N$ is the midpoint of $AB$ . Now notice that
$\angle ANZ = \angle ANC = \angle NPB + \angle NBP = y + \angle NBP = \angle PBZ$ and
$\frac{AN}{NZ} = \frac{NB}{NC} = \frac{PB}{CB} = \frac{PB}{BZ},$ so triangles $ANZ$ and $PBZ$ are similar by SAS Similarity. Hence $\angle AZN = \angle PZB$ , so $PZ$ is a symmedian of $AZB$ as well. Thus $P, Z, M$ are collinear points on a symmedian of $AZB$ , proving the Lemma.

Returning to our original diagram, the Lemma shows that $P, C, M, Z$ are concyclic. Then it is immediate to see that the desired circumcenters all lie on the perpendicular bisector of their common chord $CM$ !

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