Prove that some circles are concurrent.

freemind

Prove that some circles are concurrent.

**Posted:** Sat Mar 29, 2008 4:47 pm

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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.

**Posted:** Sat Mar 29, 2008 5:13 pm

The QuattoMaster 6000

#3

Re: Prove that some circles are concurrent.

**Posted:** Sat Mar 29, 2008 6:12 pm

andyciup

#4

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.

**Posted:** Sat Mar 29, 2008 6:24 pm

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Altheman

#5

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, $E$ .

**Posted:** Sun Mar 30, 2008 1:42 am

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kops723

#6

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

But we know what you mean.

**Posted:** Sun Mar 30, 2008 2:56 am