Prove that some circles are concurrent.

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Prove that some circles are concurrent.

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Source: Moldova 2008 IMO-BMO Second TST Problem 3

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#1 h Mar 29, 2008, 2:47 PM • 1 Y

Y by Adventure10

Let $\omega$ be the circumcircle of $ABC$ and let $D$ be a fixed point on $BC$ , $D\neq B$ , $D\neq C$ . Let $X$ be a variable point on $(BC)$ , $X\neq D$ . Let $Y$ be the second intersection point of $AX$ and $\omega$ . Prove that the circumcircle of $XYD$ passes through a fixed point.

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¬[ƒ(Gabriel)³²¹º]¼

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#2 h Mar 29, 2008, 3:13 PM • 2 Y

Y by Adventure10, Mango247

w.l.o.g. $\beta \ge \gamma$ . Let E the second intersection between the circumcircle of XYD and $\omega$ . DE meets $\omega$ on F. So 'cause DXYE is cyclic $\angle FEY = \angle YAC + \angle BCA$ , so $\angle FEC = \angle BCA$ and then F is the symmetric of A w.r.t. the axis of BC. Then 'cause E is the intersection between FD and $\omega$ that are all fixed it's also fixed.

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The QuattoMaster 6000

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#3 h Mar 29, 2008, 4:12 PM • 1 Y

Y by Adventure10

freemind wrote:

Let $\omega$ be the circumcircle of $ABC$ and let $D$ be a fixed point on $BC$ , $D\neq B$ , $D\neq C$ . Let $X$ be a variable point on $(BC)$ , $X\neq D$ . Let $Y$ be the second intersection point of $AX$ and $\omega$ . Prove that the circumcircle of $XYD$ passes through a fixed point.

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#4 h Mar 29, 2008, 4:24 PM • 2 Y

Y by Adventure10, Mango247

Let $X_1$ be a fixed point on $BC$ and $X_{2}$ be any point on $BC$ ( $X_{1}, X_{2}$ are distinct from D). Let $AX_{i}$ cut the circumcenter of the triangle $ABC$ at $Y_{i}$ .
Let the circumcenter of the triangle $DX_{1}Y_1$ touch the circumcenter of the triangle $ABC$ at $T$ .

Then $\angle TDX_{1}= 180^{\circ}-\angle{TY_1A}=\angle{TBA}=180^{\circ}-\angle{TY_{2}A}$ , therefore the point $T$ belongs to the circumcenter of the triangle $DX_{2}Y_{2}$ , and since the points $X_{2}, Y_{2}$ are mobile, the problem is solved.

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#5 h Mar 29, 2008, 11:42 PM • 2 Y

Y by Adventure10, Mango247

Suppose that the diagram is as given (the other situations are the same). Let $AD\cap \omega = A,F$ . Let $\Gamma$ be the circle through $D$ , $F$ that is tangent to $BC$ . Clearly $\Gamma$ is fixed. Let $\Gamma\cap \omega = E,F$ (so $E$ is fixed). Now $\angle XYE = \angle AYE = \pi - \angle AFE = \pi - \angle AFD = \pi - \angle EDX$ so $DXYE$ is cyclic. Hence the circumcircle of $\triangle DXY$ passes through a fixed point, $E$ .

Attachments:

Fixed Point.pdf (15kb)

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#6 h Mar 30, 2008, 12:56 AM • 2 Y

Y by Adventure10, Mango247

For the record, the circumcircle of DXY passes through D, a fixed point

But we know what you mean.

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#7 h Sep 8, 2020, 5:15 PM

Y by

If $E=(DXY)\cap \omega$ and $X'$ is another variable point in $BC$ and $Y'$ the intersection of $AX'$ with $\omega$ we have,
$\angle X'DE=AYE=180-AY'E=180-X'Y'E$
So $X'Y'DE$ is cyclic and hence we are done

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