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Suppose that the lateral faces $VAB, VBC,$ and $VCA$ of triangular pyramid $VABC$ all have the same height drawn from $V$. Let $F$ be the point in plane $ABC$ that is closest to $V$, so that $VF$ is the altitude of the pyramid. Show that $F$ is one of the special points of triangle $ABC$.

I've been thinking about this problem for a couple of days, and I think one of my main concerns is that I'm not actually trying to prove the right thing. I think that $F$ is the incenter of the triangle, but I could be wrong and it's actually the centroid. Consider a circle with center $V$ and height $h$ (let $h$ be the height of each lateral face). Then the intersection of this sphere with the plane containing $ABC$ forms an incircle of the triangle $ABC$. I think $F$ is the center of this circle, but I'm not sure how to justify it. I know it has something to do with the heights all being equal, but I can't convince myself that $ABC$ must be equilateral and I'm not sure what else would work. Any help would be really appreciated!

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  • $\begingroup$ Velcome to our site! $\endgroup$ – kjetil b halvorsen Jul 2 '14 at 15:50
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Let $F_A,F_B,F_C$ be the projections of $F$ on the sides of $ABC$. Since the angle between any lateral face and the plane where $ABC$ lies is always the same ($\arcsin\frac{VF}{h_{VAB}}$), it follows that $FF_A=FF_B=FF_C$ (since $F_A,F_B,F_C$ are also the feet of the altitudes from $V$ in $VAB,VBC,VCA$), so $F$ is the incenter of $ABC$, and the inradius of $ABC$ equals $\sqrt{h^2-VF^2}$.

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