IGO 2016 Medium/4
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
IGO 2016 Medium/4
4. Let $w$ be the circumcircle of rightangled 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 $\angle PKC = 90$.
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
 Thamim Zahin
 Posts: 98
 Joined: Wed Aug 03, 2016 5:42 pm
Re: IGO 2016 Medium/4
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, $\triangle CAB \sim \triangle CKP \Rightarrow \angle CAB= \angle CKP= 90^\circ$
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, $\triangle CAB \sim \triangle CKP \Rightarrow \angle CAB= \angle CKP= 90^\circ$
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I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.