The circumcenter

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The circumcenter

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Source: IMO Shortlist 1994, G3

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#1 h Oct 22, 2005, 9:58 PM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, Adventure10

A circle $C$ has two parallel tangents $L'$ and $L"$ . A circle $C'$ touches $L'$ at $A$ and $C$ at $X$ . A circle $C"$ touches $L"$ at $B$ , $C$ at $Y$ and $C'$ at $Z$ . The lines $AY$ and $BX$ meet at $Q$ . Show that $Q$ is the circumcenter of $XYZ$

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Tiks

#2 h Oct 23, 2005, 4:32 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

Nice problem but not hard.

Assume that C=(O,R),C'=(O_1,R_1) and C"=(O_2,R_2).OX=OY=R,O_1X=O_1Z=R_1 and O_2Y=O_2Z=R_2 =>
circumcircle of XYZ it is incircle of OO_1O_2.I am prove that BX prpendicular of OO_1 and AY prpendicular of OO_2=>Q is the circumcenter of XYZ.AY prpendicular of OO_2 <=>AO^2-(AO_2)^2=OY^2-(O_2Y)^2=R^2-(R_2)^2,AO^2=R^2+4RR_1 and (AO_2)^2=(2(RR_1)^1/2-2(RR_2)^1/2)^2+(2R-R_2)^2=>AO^2-(AO_2)^2=R^2-R_2^2<=> R^2=4R_1R_2 but this is easy question.

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#3 h Oct 24, 2005, 8:43 PM • 2 Y

Y by Adventure10, Mango247

In addition, I proved that:

${\overline {\underline |\ OZ\perp AB\ | }}$ (very nice !);

$\ 2\rho=\frac{R\sqrt R}{\sqrt R_1 +\sqrt R_2}$ , where $\rho$ is the inradius of the triangle $XYZ$ (easily);

$OZ\perp O_1O_2\Longleftrightarrow R_1=R_2$ (easily).

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Zhero

#4 h Aug 5, 2010, 3:21 AM • 2 Y

Y by Adventure10, Mango247

Let $C$ be tangent to $L'$ at $D$ . A homothety centered at $Y$ maps $C$ to $C''$ . Since $L'$ and $L''$ are parallel, $D$ is mapped to $B$ , so $D$ , $Y$ , and $B$ are collinear. Similarly, $A$ , $Z$ , and $B$ are collinear.

Let the common external tangents of $C$ and $C'$ intersect at $P$ . is the external homothety center of $C$ and $C'$ . $Y$ and $Z$ are the internal homothety centers of $C$ and $C''$ and $C'$ and $C''$ , respectively. Hence, by d'Alembert's theorem, $P$ , $Y$ , and $Z$ are collinear.

Let $PZ$ intersect $C'$ at $Y'$ . Since the homothety centered at $P$ maps $C'$ to $C$ , $\frac{PY'}{PA} = \frac{PY}{PD}$ , so $PY' \cdot PD = PA \cdot PY$ . By power of a point on $C'$ , $PA^2 = PY' \cdot PZ$ , so $PY' = \frac{PA^2}{PZ}$ . Substituting this into $PY' \cdot PD = PA \cdot PY$ , so $ADYZ$ is cyclic. Since $B = AZ \cap DY$ , $BY \cdot BD = BZ \cdot BA$ , so $B$ lies on the radical axis of $C$ and $C'$ . By symmetry, $A$ must lie on the radical axis of $C$ and $C'$ , so $Q$ must be the radical center of $C$ , $C'$ , and $C''$ . Since $QX$ , $QY$ , and $QZ$ are tangents and $Q$ has equal power with respect to each of these circles, $QX = QY = QZ$ , as desired.

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#5 h May 20, 2014, 1:22 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Strangely enough, I found G3 harder than G4.
Let $C$ touch $L'$ and $L''$ at $M$ and $N$ repectively.
Let the centre of $C$ be $O_1$ , the centre of $C'$ be $O_2$ and centre of $C''=O_3$ .
Join $O_1,X$ and $O_2,X$ as $AO_2\parallel MN\implies \angle AO_2X=\angle XO_1N$
$\implies \Delta XO_1N~\Delta AO_2X\implies \angle AXO_2~\angle O_1XN\implies A,X,N$ are collinear.
Similarly we can prove that $A,Z,B$ are collinear and $B,Y,M$ are collinear.
$\angle AZX=\angle MAN$ (alternate segment, $L'$ is tangent to $C'$ )
$\angle MAN=\angle ANM$ as $C$ is tangent to $L'$ and $L''$ at diametrically opposite points.
$\implies \angle AZX=\angle XNB\implies XZMN$ is cyclic, $XN\cap ZB=A\implies A$ lies on the radical axis of the circles $C$ and $C''$ , but as $C$ and $C''$ touch each other $\implies AY$ is the direct common tangent to $C$ and $C''$ .
Similarly, we can prove that $BX$ is the direct common tangent to $C$ and $C'$ .
$\implies AY\perp O_1O_3$ and $BY\perp O_1O_2$
$X,Y,Z$ are the points where the incircle of $\Delta O_1O_2O_3$ touches the sides.
This means that the circumcentre of $\Delta XYZ=$ the incentre of $\Delta O_1O_2O_3$
As $AY\perp O_1O_3\implies AY$ passes through the incentre of $\Delta O_1O_2O_3$
$BX\perp O_1O_2\implies BX$ passes through the incentre of $\Delta O_1O_2O_3$
$\implies AY\cap BX=\text{ the incentre of } \Delta O_1O_2O_3=\text{the circumcenter of} \Delta XYZ$

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