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Source: St Petersburg Olympiad 2018, Grade 11, P7

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#1 h Jun 21, 2018, 1:08 PM • 2 Y

Y by Adventure10, Mango247

Points $A,B$ lies on the circle $S$ . Tangent lines to $S$ at $A$ and $B$ intersects at $C$ . $M$ -midpoint of $AB$ . Circle $S_1$ goes through $M,C$ and intersects $AB$ at $D$ and $S$ at $K$ and $L$ . Prove, that tangent lines to $S$ at $K$ and $L$ intersects at point on the segment $CD$ .

This post has been edited 2 times. Last edited by RagvaloD, Jun 22, 2018, 9:06 AM

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#2 h Jun 21, 2018, 1:13 PM • 2 Y

Y by Adventure10, Mango247

RagvaloD wrote:

tangent lines to $C$ at $K$ and $L$

Is this a typo?
If $C$ is replaced by $S$ , then here's my solution.
Since the inverse of $C$ in $S$ is the midpoint of $AB$ , so $(CD)$ is orthogonal to $S$ , and thus we have the conclusion, as the tangents meet at the midpoint of $CD$ .

This post has been edited 1 time. Last edited by WizardMath, Jun 21, 2018, 1:19 PM

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#3 h Jun 21, 2018, 1:20 PM • 3 Y

Y by omarius, Adventure10, Mango247

Yes, it was typo

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#4 h Nov 26, 2018, 8:41 AM • 1 Y

Y by Adventure10

Let $O$ be the center of circle $S$ , $P$ the intersection of the line $KL$ and $AB$ , and $Q$ the intersection of tangent lines to $S$ at $K$ and $L$ . Since $PA \cdot PB=PK \cdot PL=PD \cdot PM$ and M is the midpoint of $AB$ , it can be concluded that $(A,B;P,D)=-1$ or $P$ is on the polar of $D$ wrt $S$ . And from $P$ is on the polar of $C$ wrt $S$ , $CD$ is the polar of $P$ wrt $S$ . Polar of $Q$ wrt $S$ passes through $P$ ; therefore, polar of $P$ wrt $S$ also passes through $Q$ , i.e., $Q$ lies on $CD$ .

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#5 h Jun 24, 2019, 10:37 AM • 3 Y

Y by Delta0001, Adventure10, Mango247

St. Petersburg 2018 Grade 11 P7 wrote:

Points $A,B$ lies on the circle $S$ . Tangent lines to $S$ at $A$ and $B$ intersects at $C$ . $M$ -midpoint of $AB$ . Circle $S_1$ goes through $M,C$ and intersects $AB$ at $D$ and $S$ at $K$ and $L$ . Prove, that tangent lines to $S$ at $K$ and $L$ intersects at point on the segment $CD$ .

Solution: This problem is based on the Lemma:
Lemma: In an acute angled $\Delta ABC$ with $AB<AC$ with circumcenter $O$ and incenter $I$ . Let $AI$ $\cap$ $BC$ $=$ $D$ and $AI$ $\cap$ $\odot (ABC)$ $=$ $M_A$ . Let $OM_A$ $\cap$ $\odot (ABC)$ $=$ $M_{BC}$ . Let $M$ be the midpoint of $BC$ . Let $KM_A,$ $LM_A$ be the tangents from $M_A$ to $\odot (DMM_{BC})$ . Then, $M_A$ is the center of $\odot (BKLC)$
Proof: Firstly, observe, $\angle DMM_A=90^{\circ}=\angle DAM_{BC}$ $\implies$ $A$ $\in$ $\odot (DMM_{BC})$ . Now we want to show $BKCL$ cyclic with center $M_A$ . Let $DM_{BC}$ $\cap$ $\odot (ABC)$ $=$ $E$ . Then, $EDMM_A$ is cyclic. Hence, By Radical Axes Theorem, $AM_{BC}$ , $BC$ and $EM_A$ concur, say at $X$ . Let $F$ be the center of $\odot (ADMM_{BC})$ $\implies$ $FKEM_AL$ is cyclic. Hence, again By Radical Axes Theorem, $X$ lies on $KL$ $\implies$ $BKCL$ is cyclic. Now just notice, $\angle FKL= \angle KBL=\frac{1}{2} \angle KM_AL$ $\implies$ $M_A$ is the center of $\odot (BKCL)$ $\qquad \blacksquare$

Apply this Lemma, and the Original Problem Follows!

This post has been edited 2 times. Last edited by AlastorMoody, Jun 24, 2019, 10:41 AM

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