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Source: Mexico 2003

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#1 h Mar 2, 2007, 11:53 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

$A, B, C$ are collinear with $B$ betweeen $A$ and $C$ . $K_{1}$ is the circle with diameter $AB$ , and $K_{2}$ is the circle with diameter $BC$ . Another circle touches $AC$ at $B$ and meets $K_{1}$ again at $P$ and $K_{2}$ again at $Q$ . The line $PQ$ meets $K_{1}$ again at $R$ and $K_{2}$ again at $S$ . Show that the lines $AR$ and $CS$ meet on the perpendicular to $AC$ at $B$ .

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#2 h Mar 2, 2007, 3:39 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

Let $AP$ and $CQ$ meet in $M$ . Since $\measuredangle APB = BQC = 90^{\circ}$ , then $\measuredangle BPM = BQM = 90^{\circ}$ , and so $BM$ is a diameter of the circle touching $AC$ .
Since $BS \bot CS$ and $\measuredangle BPM = 90^{\circ}$ , then $BS || AP$ , and so $AP \bot CS$ .
For the same reason $CQ \bot AR$ , and so $M$ is the orthocenter of $\Delta AXC$ where $X = AR \cap CS$ .
Then $MB$ is the altitude through $X$ of $\Delta AXC$ .

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#3 h Mar 2, 2007, 5:32 PM • 2 Y

Y by Adventure10, Mango247

Andreas wrote:

Since $BS \bot CS$ and $\measuredangle BPM = 90^{\circ}$ , then $BS || AP$ , and so $AP \bot CS$

Adreas , kindly sorry , but does this prove immediately that $AP \parallel BS$ ? Or I can not see something ?

Note

I prove that $\angle ABP= \angle AMB = \angle PMB = \angle PQB = \angle SQB = \angle SCB$ , so $PB \parallel CS$ , so $CS \perp AM$ , since $\angle APB = 90$

Babis

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#4 h Mar 2, 2007, 5:42 PM • 1 Y

Y by Adventure10

Well, correct me if I'm wrong, a convex quadrilateral is a parallelogram iff its opposite angles are congruent.

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#5 h Mar 2, 2007, 5:55 PM • 1 Y

Y by Adventure10

Andreas wrote:

Well, correct me if I'm wrong, a convex quadrilateral is a parallelogram iff its opposite angles are congruent.

Mmm ! Which quadrilateral do you mean ? ( I did not you are wrong , but I ask you to explain this point. Probably , I did not get it

.Again sorry . I'll try to prove that $\angle PBS = 90$ . Is this obvious ?.Because this was my original question ! Today it is not my lucky day !

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#6 h Mar 2, 2007, 6:08 PM • 2 Y

Y by Adventure10, Mango247

Let $AP \cap CS = K$ . Then the quadrilateral is $PBSK$ . But you are right, we don't know if it's convex. So your proof works better.

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#7 h Mar 2, 2007, 6:36 PM • 2 Y

Y by Adventure10, Mango247

Andreas wrote:

Let $AP \cap CS = K$ . Then the quadrilateral is $PBSK$ . But you are right, we don't know if it's convex. So your proof works better.

Andeas , the problem was this : In $PBSK$ we have only two right angles , so we can not decide that this is a parallelogram.And my problem was exactly to prove that one more angle is $90$ .I could not see this at once , so I asked you to help. Now I think that your nice solution is complete !
Thanks -
Babis

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#8 h Mar 3, 2007, 5:25 AM • 1 Y

Y by Adventure10

We have $\angle QSC=\angle QBC=\angle QPB$ (because of tangent) $=\angle RAB$ so $A,R,S,C$ are cyclic
This means $XR*XA=XS*XC$ so $X$ is on the radical axis of circle $K1$ and $K2$
So $XB$ and $AC$ are perpendicular $q.e.d$

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