Right Pyramid question In a Right pyramid, that is one with all edges equal to each other and a height which goes to the center of the geometric figure which serves as it's base, can we say that if that figure is an isosceles triangle, then the height of the pyramid intersects the base in the point where the perpendicular bisectors of the triangle meet? 
 A: When nothing clever comes to mind, calculate. Insight may come later. 
Let the coordinates of the two base points of our triangle be $(-a,0,0)$ and $(a,0,0)$. Let the other vertex be at $(0,b,0)$. Let the "tip" of the pyramid be at $T=(0,y,z)$. Then the equal distance condition says that
$$a^2+y^2+z^2=(y-b)^2+z^2,$$
which means that $z$ is irrelevant (and in geometric hindsight this is obvious), and $2by=b^2-a^2$. 
Let $T_0$ be the point where the perpendicular from $T$ to the $x$-$y$ plane meets the plane. Then $T_0$ has coordinates $\left(0, \frac{b^2-a^2}{2b},0\right)$. 
Now let's see whether the perpendicular bisectors of the sides of the bottom triangle meet at $T_0$.
We only need to worry about the perpendicular bisector of the side that joins $(a,0,0)$ and $(0,b,0)$. 
For convenience, since we are working in the $x$-$y$ plane, we drop the third coordinate. 
The perpendicular bisector of the side that joins $(a,0,0)$ and $(0,b,0)$ goes through $(a/2,b/2)$ and has slope $\frac{a}{b}$. 
So let us find the slope of the line that joins $(a/2,b/2)$ to $T_0=(0,(b^2-a^2)/2b)$.
Calculate. It turns out to be $\frac{a}{b}$, exactly the right thing, so we are finished. 
Added: And now in hindsight it is obvious. Ditch the coordinates. The projection $T_0$ of $T$ onto the base is equidistant from the three vertices of the base triangle. The point equidistant from the three vertices of any triangle is the point where the perpendicular bisectors of the sides meet. 
