# Six pyramids in a cube

This question is from the German mathematics competition Känguru der Mathematik. In this competition students have to solve 30 mathematical tasks like this in 90 minutes without calculator. Actually they are given 5 possible answers, but for this community a bit of additional challenge does not hurt.

You are given a cube with unknown edge length. A point $$M$$ inside the cube with unknown coordinates appears. Now imagine 6 pyramids inside the cube with each cube face being the base of one pyramid and $$M$$ being their common apex. The volumes of 5 of the 6 pyramids are 2, 5, 10, 11 and 14. What is the volume of the sixth pyramid?

• I'm sure I am missing something obvious. It appears to me that this is not even possible. Perhaps because I am assuming the pyramids have 4 equal triangles on their other faces? (From Wikipedia: "When unspecified, a pyramid is usually assumed to be a regular square pyramid, like the physical pyramid structures.") – user41655 Oct 24 '17 at 22:07
• @Rob try and picture the cube first, then picture a point in the cube, then picture a pyramid with the base on each side of the cube, and the vertex being the point that we picked. – phroureo Oct 24 '17 at 22:15
• @Rob: In school, we used "pyramid" to denote any "cone with a square base" (if that makes sense). If the tip of the cone was in the middle of the base (which is what you're referring to), we called it a "regular pyramid". I have a feeling that the question here is relying on the same interpretation. Your own quote even hints at this: "a pyramid is usually assumed to be a regular square pyramid" If it's an assumption, then it's not literally correct. Analogously, a quadrilateral is commonly assumed to have no concave corners, even though concave quadrilaterals do exist. – Flater Oct 25 '17 at 12:54
• @Rob: Example The three are all quadrilaterals, but unless otherwise specified, you can assume that people refer to the first shape when they say "a quadrilateral" – Flater Oct 25 '17 at 12:59

6

because the volume of a pyramid is proportional to its height, and we know that each pair of opposite pyramids together has the same total height. Therefore, all three pairs of pyramids from opposite sides have the same combined volume.

There is only one number we can make twice using the given values, and we have (2,14), (5,11), and (6,10).

The six pyramids each have the same base area $b$, and can be partitioned into three pairs whose bases are opposing sides of the cube. For each such pair of opposite pyramids, the two heights' sum is the cube's edge length $e$.

A pyramid's volume is $V = \frac{b h}{3}$, and thus $h =\frac{3 V}{b}$. Therefore, for a given pair with heights $h_A$ and $h_B$ and volumes $V_A$ and $V_B$,

$$h_a + h_b = e = \frac{3 V_A}{b} + \frac{3 V_B}{b}$$

which implies that

$$\frac{e b}{3} = V_A + V_B$$

Since $e$ and $b$ are the same for each pair of opposite pyramids, the total volume of each opposing pyramid pair is also constant.

We are given five of the six pyramid volumes, namely $2, 5, 10, 11, 14$. The six volumes must be able to be placed into three pairs such that each pair has the same total volume.

$(5,11)$ and $(2,14)$ is the only couple of pairs in which the sums are equal. This means the total volume of each opposing pair is 16 and the remaining pair of volumes is $(10,6)$.

• This is basically identical to Eric's answer. Make sure you check others' answers before posting. – boboquack Oct 25 '17 at 21:32
• @boboquack It is the same line of reasoning but the explanation here is significantly more in-depth and is potentially useful. – mherzl Oct 25 '17 at 21:46
• Welcome to Puzzling.SE @mherzl :) – ABcDexter Oct 26 '17 at 8:25