IGO 2016 Medium/2
- Thamim Zahin
- Posts:98
- Joined:Wed Aug 03, 2016 5:42 pm
2. Let two circles $C_1$ and $C_2$ intersect in points $A$ and $B$. The tangent to $C_1$ at $A$ intersects $C_2$ in $P$ and the line $PB$ intersects $C_1$ for the second time in $Q$ (suppose that $Q$ is outside $C_2$). The tangent to $C_2$ from $Q$ intersects $C_1$ and $C_2$ in $C$ and $D$, respectively (The points $A$ and $D$ lie on different sides of the line $PQ$). Show that $AD$ is bisector of the angle $CAP$.
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/2
Extend the segment $QD$. Name it $C'$.
$\angle CQB=\angle BAC=\angle a$
$\angle BAD=\angle DPB=\angle b$ [cyclic quad]
$\angle DAP=\angle C'DP=\angle a$ [$DC'$ is tangent to $(DAP)$]
Now, notice that $\angle DAC=\angle b+\angle c$ and $\angle DAP=\angle a$
So, we have to proof that $\angle c+\angle b=\angle a$
Now, in triangle $\triangle DPQ$, $DC'$ is extension of $QD$.
So, $\angle c+\angle b=\angle a$
Or. $\angle CAD=\angle PAD$
$[Done]$
$\angle CQB=\angle BAC=\angle a$
$\angle BAD=\angle DPB=\angle b$ [cyclic quad]
$\angle DAP=\angle C'DP=\angle a$ [$DC'$ is tangent to $(DAP)$]
Now, notice that $\angle DAC=\angle b+\angle c$ and $\angle DAP=\angle a$
So, we have to proof that $\angle c+\angle b=\angle a$
Now, in triangle $\triangle DPQ$, $DC'$ is extension of $QD$.
So, $\angle c+\angle b=\angle a$
Or. $\angle CAD=\angle PAD$
$[Done]$
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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.