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Let $\Delta ABC$ be a triangle with orthocenter $H$ and altitudes $AD, BE, CF$. Perform an inversion around $C$ with radius $\sqrt{CH \cdot CF}$. Where do the six points each go?

I'm having trouble with the wording of the statement. Does it mean inversion about the circle centered at $C$ with radius $\sqrt{CH \cdot CF}$? Rest, I can solve the matter myself.

Mathejunior
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  • What about inversion? – Dietrich Burde Jan 09 '18 at 19:47
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    Yes, it does seem that the inversion about the circle centered at $C$ with radius $\sqrt{CH\cdot CF}$ maps $A$ to $E$ and vice versa, $B$ to $D$ and vice versa, and $F$ to $H$ and vice versa... I've commonly seen people omitting mention of the circle and just mentioning the centre. –  Jan 09 '18 at 19:54
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    Inversion around $C$ whith radius $\sqrt{CH \cdot CF}$ means that the point $X$ will be inverted in the point $X'$ such that $XC\cdot X'C=XX'^2-(\sqrt{CH \cdot CF }^2$ – Ricardo Largaespada Jan 09 '18 at 20:01
  • There is also inversion in a point $C$, which sends $X$ to $C-\vec{CX}$. In 2D, this is also a 180 rotation around $C$. – mr_e_man Sep 09 '18 at 14:51

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