## powers are equal

For discussing Olympiad level Geometry Problems
photon
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### powers are equal

\$P\$ is inside \$\Delta ABC\$ such that \$AB,BC,CA\$ are tangents to the circumcircles of \$\Delta BPC,\Delta CPA,\Delta APB\$ respectively . Extended \$AP,BP,CP\$ cuts the circumcircle of \$\Delta ABC\$ at \$A',B',C'\$ .

Prove that power of \$A\$ (with respect to \$\odot B'PC\$ ),power of \$B\$ (wrt \$\odot C'PA\$ ),power of \$C\$ (wrt \$\odot A'PB\$ ) are equal .

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Nirjhor
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### Re: powers are equal

I might be wrong, but \$A,B,C\$ all lie on the circles they are meant to be respected to. So they have equal power \$0\$.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am

### Re: powers are equal

Nirjhor wrote: \$A,B,C\$ all lie on the circles they are meant to be respected to.
Not true.

First, prove that, \$AB',\$ etc. are tangent to \$\odot B'PC,\$ etc. Then, the powers of \$A, B, \$ and \$C\$ w.r.t. the given circles are equal to the squares of \$AB', BC', CA'\$. So, we'll be done if we can show \$AB'=BC'=CA'\$.

Now, for the proof.
\$P\$ is a Brocard Point of \$\triangle ABC\$. \$\therefore \angle ABP=\angle BCP=\angle CAP\$.
So,
\$\angle AB'P=\angle AB'B=\angle ACB=\angle ACP+\angle PCB=\$ \$\angle ACP+\angle PBA=\angle ACP+\angle B'BA=\angle ACP+\angle B'CA=\angle B'CP\$
\$\therefore AB'\$ is tangent to \$\odot B'PC\$. Similarly, \$BC', CA'\$ are tangent to \$\odot C'PA, \odot A'BP\$.
Finally, \$AB', BC', CA'\$ subtend equal angles on the circle. \$\therefore AB'= BC'= CA'\$, and, we're done.
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Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: powers are equal

Sorry I messed the construction up.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.