Two circles meet at points $A$ and $B$. A line through $B$ intersects the first circle again at $K$ and the second circle at $M$. A line parallel to $AM$ is tangent to the first circle at $Q$. The line $AQ$ intersects the second circle again at $R$. $(a)$ Prove that the tangent to the second circle at $R$ is parallel to $AK$. $(b)$ Prove that these two tangents meet on $KM$.
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Tags: trigonometry, geometry proposed, geometry
AK1024
07.11.2010 19:08
Let $KA$ meet the second circle again at $L$.The points $P$ and $S$ are shown in the diagram.
Let $\angle KQP=y$ and $\angle AQK=x$ ($x>y$).
Then we know $\angle KBQ=y$ and $\angle ABQ=x$ so that $\angle KBA=x-y$.
Also $\angle KAQ=y\implies \angle RAL=y\implies \angle SRL=y$.
Consider the power of the point $K$ wrt to the second circle. $\frac{KA}{KL}=\frac{KB}{KM}$ so that $\triangle KAB$ is similar to $\triangle KML$ i.e. $\angle KLM=\angle KBA=x-y$.
Since the $Q$-tangent is parallel to $AM$ we have $\angle QAM=x \implies \angle RLM=x$.
$\angle KBA=x-y \wedge \angle RLM =x \implies \angle RLA=y= \angle SRL \implies KA$ is parallel to $R$-tangent.
skytin
07.11.2010 19:45
(a) + (b) let tangent line to first circle at point Q and BM intersects at point X , easy to see that angle QXM = AMX = ARB , so QXRB is cyclic . use lemma on picture . Lemma for (AQB) and for (RBQ) , so tangent line from X to (RBQ) || KQ . use Lemma for circles (ABQ) and (ABR) , so KQ || RM use Lemma for (RBQ) and (ABR) tangent to RBQ from point X is parallel to RM , so RX is tangent to (ABR) and use Lemma for (ABQ) and (RBQ) , so AK || AX . done
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jayme
08.11.2010 13:34
Dear Mathlinkers, very nice application of the Reim's theorem and one of it converse, and of a converse of the pivot theorem. Sincerely Jean-Louis
sayantanchakraborty
12.05.2014 14:08
(a)$\angle{RQJ}=\angle{AKQ}=\angle{RAM},\angle{ARM}=\angle{ABK}=\angle{AQK} \Rightarrow \angle{KAQ}=\angle{AMR}=\angle{QRZ}\Rightarrow RZ \parallel AK$. (b)Use sine rule in $\triangle{LKR}$ and $\triangle{LQR}$ and a few steps...does the job.
Synthetic_Potato
10.03.2019 15:00
Let $Q'$ be the second intersection of $BQ$ with the second circle. Notice that $Q'R \parallel AM$ (since both are parallel to the tangent at $Q$ to the first circle. Now a simple angle chase yields $BQ\parallel$ tangent at $R$ to the second circle. Let the tangents at $Q,R$ meet at $S$. Notice that $BQRS$ is a cyclic quadrilateral and also $\angle KBR = \angle SQR$. Since $\angle SQR = \angle SBR$, $K,B,S$ are collinear. So the tangents at $Q, R$ to the respective circles meet on $KM$. $\blacksquare$.