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IGO 2016 Medium/4

Posted: Tue Jan 10, 2017 4:03 pm
by Thamim Zahin
4. Let $w$ be the circumcircle of right-angled triangle $ABC (\angle A = 90)$. Tangent to $w$ at point $A$ intersects the line $BC$ in point $P$. Suppose that $M$ is the midpoint of (the smaller) arc $AB$, and $PM$ intersects $w$ for the second time in $Q$. Tangent to $w$ at point $Q$ intersects $AC$ in $K$. Prove that $\angle PKC = 90$.

Re: IGO 2016 Medium/4

Posted: Mon Jun 05, 2017 10:23 pm
by Thamim Zahin
Suppose $AB<AC$ (the solution is same if $AB>AC$)
Here, $\triangle KAQ \sim \triangle KCQ \Rightarrow \frac{KA}{KC}=\frac{[KAQ]}{[KQC]}=(\frac{AQ}{QC})^2$

And, $\triangle PAB \sim \triangle PAC \Rightarrow \frac{PB}{PC}=\frac{[APB]}{[APC]}=(\frac{AB}{AC})^2$

We also have, $\triangle PQA \sim \triangle PAM \Rightarrow \frac{AQ}{AM}=\frac{PA}{PM} ...(i)$

And, $\triangle PBM \sim \triangle PQC \Rightarrow \frac{BM}{QC}=\frac{PM}{PC} ...(ii)$

By multiplying $(i)$ and $(ii)$ $\Rightarrow \frac{AQ}{AM}\times\frac{BM}{QC}=\frac{PA}{PM}\times\frac{PM}{PC}\Rightarrow\frac{AQ}{QC}=\frac{PA}{PC}$ [because $MA=MB$ ]

It is known that $\triangle PAB \sim \triangle PCA \Rightarrow \frac{PA}{PC}=\frac{AB}{AC}$
So, $\frac{AQ}{QC}=\frac{AB}{AC} \Rightarrow (\frac{AQ}{QC})^2=(\frac{AB}{AC})^2 \Rightarrow \frac{KA}{KC}=\frac{PB}{PC}$

So, $\triangle CAB \sim \triangle CKP \Rightarrow \angle CAB= \angle CKP= 90^\circ$