Let $\tau$ be a circle with radius $r$. Let $A$ and $B$ be distinct
points on $\tau$ such that $AB < \sqrt{3}r$. Let the circle with center $B$ and
radius $AB$ meet $\tau$ again at $C$. Let $P$ be the point inside $\tau$ such that
triangle $ABP$ is equilateral. Finally, let line $CP$ meet $\tau$ again at $Q$.
Prove that $PQ = r$.
Canada 2002 problem:2(Good)
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sourav das
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You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: Canada 2002 problem:2(Good)
This problem can be solved easily with complex numbers.
If we let the circle to be unit circle, then letting $A=a$ and $B=1$ , point $C$ becomes $\overline{a}= \frac 1 a$
Again, point $P$ becomes $a \omega^2+ \omega $ and then it is easy enough to find the point $m$ where $CP$ intersects the circle again and $|m-p|=1$.
If we let the circle to be unit circle, then letting $A=a$ and $B=1$ , point $C$ becomes $\overline{a}= \frac 1 a$
Again, point $P$ becomes $a \omega^2+ \omega $ and then it is easy enough to find the point $m$ where $CP$ intersects the circle again and $|m-p|=1$.
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