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 $20$?

Why do you want to find the area? That doesn't seem to be clearly connected to what came before. — hmakholm left over Monica, Mar 18 '17 at 22:02
I wanted the volume so I could use the volume formula backwards to get the height. I'm not sure if that is the best way, but it was the first thing I thought of. — user406996, Mar 18 '17 at 22:05
I think what you really need to find is the radius of the circumcircle of the triangle. If you cut the sphere through the center and (say) A, perpendicular to the plane of the triangle, you will find a right triangle with hypotenuse $20$ and the legs being the circumradius and the height you're looking for. — hmakholm left over Monica, Mar 18 '17 at 22:06
Oh, I see! When you cut the sphere and take the cross section, you can see the right triangle! Thanks! — user406996, Mar 18 '17 at 22:09
If you google for 13-14-15 triangle, you will find some interesting numerical coincidences that should help you derive the circumradius. — hmakholm left over Monica, Mar 18 '17 at 22:10
I got the area is 84 and the circumradius is 65/8. So the answer would be $\sqrt{20^2-\frac{65}{8}^2}$? — user406996, Mar 18 '17 at 22:11
Why nobody looks at the definition of distance? $${\rm dist}(O,\triangle ABC)=\inf\limits_{x\in O,y\in\triangle ABC}{\rm dist}(x,y)$$ "Points $A$,$B$ and $C$ lie on Sphere $O$ ..." implies $A,B,C\in O$; $A,B,C \in \triangle ABC$, so the distance should be zero. — GNUSupporter 8964民主女神地下教會, Mar 18 '17 at 22:22
@GNUSupporter: Because that interpretation would make it an extremely unlikely problem. It is much more likely that the problem asks for the distance between the center of the sphere and the plane of the triangle, and has been garbled in the transcription here. — hmakholm left over Monica, Mar 18 '17 at 22:25

Jean Marie · Accepted Answer · 2017-03-18T22:49:21.273

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 $ABC$.

score 1 · Answer 2 · answered May 16 '17 at 04:08

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 $$\text{dist}\big(O, \, \text{plane}(ABC)\,\big) = OO^* = \frac{15 \, \sqrt{95}}{8}$$

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

2 Answers2