How do you calculate the volume of a pyramid without the height?

Question

I was solving a problem, and I was able to get to here, which was where I got stuck:

Original Problem: If it isn't possible, here is the original problem: Points $A, B$ and $C$ lie on Sphere $O$ with radius 20. If $AB=13, BC=14$, and $AC=15$, what is the shortest distance from $O$ to triangle $ABC$.

What I did: Take pyramid $ABCO$. The shortest distance from $O$ to $ABC$ is the height. So, I am trying to find the volume with the base lengths and slant heights.

Question: How do you calculate the volume of a pyramid whose base is a triangle of sides $13, 14, 15$ and with 3 slant heights of $20$?

Answer 1

First (as you ask it) obtain the volume $V$ whose square is given by the Cayley Menger determinant (http://mathworld.wolfram.com/Cayley-MengerDeterminant.html):

$$V^2=\dfrac{1}{288}\begin{vmatrix} 0&1&1&1&1\\ 1&0&D_{OA}^2&D_{OB}^2&D_{OC}^2\\ 1&D_{OA}^2&0&D_{AB}^2&D_{AC}^2\\ 1&D_{OA}^2&D_{AB}^2&0&D_{BC}^2\\ 1&D_{OA}^2&D_{BC}^2&D_{BC}^2&0 \end{vmatrix}=\dfrac{1}{288}\begin{vmatrix} 0&1&1&1&1\\ 1&0&20^2&20^2&20^2\\ 1&20^2&0&13^2&15^2\\ 1&20^2&13^2&0&14^2\\ 1&20^2&15^2&14^2&0 \end{vmatrix}=\dfrac{1}{288}75411000$$

Then compute the base area: ${\frak A}=\dfrac12\|\vec{AB}\times \vec{AC}\|$ (half the norm of the cross product). I leave you its computation.

At last, write $V=\dfrac13 h \frak{A}$, where $h$ is the looked for altitude, i.e, the distance from point $O$ to plane $ABC$.

Answer 2

Well, there is another way to find the distance from $O$ to the plane $ABC$, without looking for the volume. Since $O$ is the center of a sphere and the three points $A, B, C$ lie on that sphere, the orthogonal projection $O^*$ of the center $O$ on the plane $ABC$ is in fact the center of the circle superscribed around triangle $ABC$. Therefore, $AO_* = BO_* = CO_* = R_*$ is the circumradius of $ABC$. However, say triangle $AOO_*$ is right-angled with $AO = R$ the radius of the sphere and $AO_*=R_*$ the radius of the circumcircle of triangle $ABC$. Thus, by Pythagoras' theorem $$OO^* = \sqrt{AO^2 - AO_*^2} = \sqrt{R^2 - R_*^2}$$ Assume $BC = a, \, CA = b, \, AB = c$. Then one can calculate, by say using Heron's formula for the area of $ABC$ and then the law of sines, that the circumradius of $ABC$ is $$R_* = \frac{abc}{4 \,\text{Area}(ABC)} = \frac{ abc}{\sqrt{\big(a + b + c\big)\big(-a + b + c\big)\big(a - b + c\big)\big(a + b - c\big)}}$$ Thus, the distance you want to find is given by the formula $$\text{dist}\big(O, \, \text{plane}(ABC)\,\big) = OO^* = \sqrt{R^2 \, - \, \frac{ a^2b^2c^2}{\big(a + b + c\big)\big(-a + b + c\big)\big(a - b + c\big)\big(a + b - c\big)}}$$ When I plug your specific numbers $ R=20, \, c=13, \, a=14, \, b=15$ Wolfram Alpha produces the fairly cute number $$\text{dist}\big(O, \, \text{plane}(ABC)\,\big) = OO^* = \frac{15 \, \sqrt{95}}{8}$$