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Let us have square pyramid $N$ which has all edges of equal length. Point A is the apex of the pyramid. A point $P$ is randomly and uniformly chosen in the interior of the pyramid. The probability that $P$ is closer to $A$ than any other of the four vertices of the pyramid is equal to $\frac mn$ for relatively prime positive integers $m$ and $n$. Find $m+n$.

First, I thought that we could create a pyramid, half the height of pyramid $N$. This would fit inside pyramid $N$ and essentially create a frustum and new pyramid. However, this would not work, because we would only be reaching a majority of the cases. Help is much needed. Can someone guide me through the solution?

A Piercing Arrow
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1 Answers1

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Given a square pyramid $\mathcal{P}$ with all sides equal, the shape $\mathcal{S}$ where the points in $\mathcal{P}$ closer to apex than vertices is an unequal trapezohedron. It can be constructed as the intersection of two square pyramids.

Let's say the side of $\mathcal{P}$ is $2$. we can choose a coordinate system so that the apex $A$ and other vertices $B, C, D, E$ are located at $$A = (0,0,\sqrt{2}),\;B = (-1,-1,0),\; C = (1,-1,0),\; D = ( 1,1,0),\; E = (-1,1,0)$$ The shape $\mathcal{S}$ looks like this:

points closer to apex than vertices

As one can see, $\mathcal{S}$ have $8$ faces and it is the convex hull of following 10 points:

  • the apex $A$,
  • midpoints of 4 edges adjacent to $A$: $$m_B = \frac{A+B}{2}, m_C = \frac{A+C}{2}, m_D = \frac{A+D}{2}, m_E = \frac{A+E}{2}$$
  • centroids of 4 faces adjacent to $A$: $$c_{BC} = \frac{A+B+C}{3}, c_{CD} = \frac{A+C+D}{3}, c_{DE} = \frac{A+D+E}{3}, c_{EB} = \frac{A+E+B}{3}$$
  • the origin $O = (0,0,0)$.

Let $\mathcal{T}_1$ and $\mathcal{T}_2$ be the tetrahedra formed by the 4-tuples $(O, A, B, C)$ and $(O, X, m_B, c_{BC})$ respectively. It is easy to see $$\verb/Volume/(\mathcal{P}) = 4\verb/Volume/(\mathcal{T}_1) \quad\text{ and }\quad \verb/Volume/(\mathcal{S}) = 8\verb/Volume/(\mathcal{T}_2)$$ Together with following formula for their area, $$\begin{align} \verb/Volume/(\mathcal{T}_1) &= \frac16\left|\vec{A}\cdot(\vec{B}\times\vec{C})\right|\\ \verb/Volume/(\mathcal{T}_2) &= \frac16\left| \vec{A} \cdot \left( \frac{\vec{A}+\vec{B}}{2} \times \frac{\vec{A}+\vec{B}+\vec{C}}{3} \right) \right| = \frac{1}{36}\left|\vec{A}\cdot(\vec{B}\times\vec{C})\right| = \frac16 \mathcal{T}_1 \end{align}, $$ we find $$\verb/Volume/(\mathcal{S}) = (8)(\frac16)(\frac14)\verb/Volume/(\mathcal{P}) = \frac13 \verb/Volume/(\mathcal{P})$$ This means $\frac{m}{n} = \frac13 \implies m + n = 4$.

achille hui
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  • Nice solution. However, wouldn't a symmetry argument suffice? If you consider two such pyramids strapped together to make a regular octahedron, you can split the solid into 6 identical trapezohedra etc – Plato Dec 24 '17 at 16:03
  • @Plato yes, that argument will work. I also reached that conclusion a few hours ago but I didn't bother to include it into the answer. – achille hui Dec 24 '17 at 16:40