How many points $P$ such that $\angle APB=\angle BPC=\angle CPA $ are there?

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Given that $\triangle ABC$ is arbitrary. How many points $P$ such that $\angle APB=\angle BPC=\angle CPA $ are there?

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Hint
If $P$ is inside $\Delta ABC$, the angles must sum to $360^\circ$, if $P$ is outside, the angles can never be all equal.
Can you now prove that $P$ must be unique? (It even has a special name)