[ASK] Height of an Icosahedron

[Monoxdifly]

In the icosahedron above, what is the proper way of determining the length of PP'?
My workmate thinks that it is twice the length of P to the midpoint of E'C' plus E'A, since she thinks that (and it does look) ACC'E' is a rectangle.
However, I think that E'A must not be the height of the rectangle, since the height of the rectangle should be the height of the triangle A'BB'. Also, I think the icosahedron's height is twice the length of P to the center of A'B'C'D'E' plus triangle A'BB', though I am not sure if the length of P to the center of A'B'C'D'E' equals the length of P to the midpoint of E'C'.
Which one of us is correct?

 

[Opalg]

In the icosahedron above, what is the proper way of determining the length of PP'?
My workmate thinks that it is twice the length of P to the midpoint of E'C' plus E'A, since she thinks that (and it does look) ACC'E' is a rectangle.
However, I think that E'A must not be the height of the rectangle, since the height of the rectangle should be the height of the triangle A'BB'. Also, I think the icosahedron's height is twice the length of P to the center of A'B'C'D'E' plus triangle A'BB', though I am not sure if the length of P to the center of A'B'C'D'E' equals the length of P to the midpoint of E'C'.
Which one of us is correct?

Imagine that you are looking vertically down on your icosahedron from somewhere vertically above the point $P$. The line $POP'$ will pass through the centre of each of the pentagons $ABCDE$ and $A'B'C'D'E'$, at the point marked by a black spot in the diagram below. Notice that neither of the lines $E'C'$, $AC$ meets the line $POP'.$ Also, $ACC'E'$ is indeed a rectangle, but the plane containing it is skew to the line $POP'$, and only meets it in the single point $O$ at the centre of the icosahedron.

View attachment 5441​

Finding the diameter of the icosahedron (in other words the distance $PP'$) is quite tricky. As far as I know, the easiest way to do this is to use coordinates. The 12 vertices of a regular icosahedron are given by all possible triples using the numbers $0,\pm1,\pm\tau$, where $\tau = \frac12\bigl(1+\sqrt5\bigr)$ is the golden ratio. See this link for an explanation. The length of each side of the icosahedron is then $\sqrt{1 + \tau^2 + (\tau-1)^2} = 2$, and the diameter is $2\sqrt{\tau + 2}.$

So if the icosahedron has sides of length $1$ then the diameter is $\sqrt{\tau + 2} = \sqrt{\frac12\bigl(5+\sqrt5\bigr)}.$

 

[Monoxdifly]

Do you imply that we can't determine the height without using coordinates?

 

[Opalg]

Do you imply that we can't determine the height without using coordinates?

No, I'm not saying that. I just think that it is likely to be difficult.

 

[Monoxdifly]

Another question:
I want to prove the formula of regular icosahedron's volume. The reference says that the volume is \(\displaystyle \frac{5\times(3+\sqrt{5})}{12}\times{a^3}\). For this, I need to split the icosahedron to 20 triangular pyramid with the same size. I assumed that the base area is \(\displaystyle \frac{1}{4}a^2\sqrt{3}\) and the pyramid's height is \(\displaystyle \frac{a}{2\sqrt{5-2\sqrt{5}}}\) (I assumed that the pyramid's height should be equal the distance from midpoints of A'B' to the center point of pentagon A'B'C'D'E'. However, no matter how hard I tried and changed the form, I was unable to reach the form \(\displaystyle \frac{5\times(3+\sqrt{5})}{12}\times{a^3}\). Is the pyramid's height not equal the distance from midpoints of A'B' to the center point of pentagon A'B'C'D'E'? If so, how to determine their distance?

 

[Opalg]

Another question:
I want to prove the formula of regular icosahedron's volume. The reference says that the volume is \(\displaystyle \frac{5\times(3+\sqrt{5})}{12}\times{a^3}\). For this, I need to split the icosahedron to 20 triangular pyramid with the same size. I assumed that the base area is \(\displaystyle \frac{1}{4}a^2\sqrt{3}\) and the pyramid's height is \(\displaystyle \frac{a}{2\sqrt{5-2\sqrt{5}}}\) (I assumed that the pyramid's height should be equal the distance from midpoints of A'B' to the center point of pentagon A'B'C'D'E'. However, no matter how hard I tried and changed the form, I was unable to reach the form \(\displaystyle \frac{5\times(3+\sqrt{5})}{12}\times{a^3}\). Is the pyramid's height not equal the distance from midpoints of A'B' to the center point of pentagon A'B'C'D'E'? If so, how to determine their distance?

You have $20$ pyramids. Each pyramid has one of the triangular faces of the icosahedron as its base, and the centre of the icosahedron (the point $O$ in your diagram) as its "top" vertex. The formula for the volume of a pyramid is that it is one-third times the base area times the height. So the volume of the icosahedron is given by $$V = 20 \times \frac13 \times \frac{\sqrt3a^2}4 \times h,$$ where $h$ is the height of the triangular pyramids, namely the distance from the point $O$ to the centre of one of the triangular faces of the icosahedron.

What you know so far is that the distance from $O$ to one of the vertices of the icosahedron is $\frac a2\sqrt{\tau + 2} = \frac a4\sqrt{10 + 2\sqrt5}.$ Within a face of the icosahedron, the distance from a vertex to the centre of the face is $\frac {\sqrt3a}3.$ You can now calculate $h$ by applying Pythagoras' theorem to a triangle with hypotenuse $\frac a4\sqrt{10 + 2\sqrt5}$, the other two sides being $h$ and $\frac {\sqrt3a}3.$ The calculation is $$h^2 = \frac{(10+2\sqrt5)a^2}{16} - \frac{a^2}3 = \frac{(14 + 6\sqrt5)a^2}{48}.$$ You now need a bit of algebraic expertise to notice that $14 + 6\sqrt5 = \bigl(3+\sqrt5\bigr)^2$. Using that, you can take the square root of the previous equation, to get $h = \dfrac{\bigl(3+\sqrt5\bigr)a}{4\sqrt3}.$ Plug that into the above formula for $V$ and you will find that \(\displaystyle V = \frac{5\bigl(3+\sqrt5\bigr)a^3}{12},\) just as the book says.

 

[Monoxdifly]

So it's down to coordinates again, then.

 

[Opalg]

So it's down to coordinates again, then.

(Nod) That's the only way I know how to do it.

 

[Monoxdifly]

(Nod) That's the only way I know how to do it.

I did it thanks to someone from My Math Forum who gave me some hints. If I recall he is also a member in this forum. His username is skipjack.

The distance from P to P' equals the distance from A to C'. We can find AC' by applying Pythagorean theorem on the rectangle ACC'E' with AC is the diagonal of pentagon ABCDE which can be found using trigonometry. Now that I have found the height, I don't need to know the distance between the two pentagons anymore.
'

 

[Monoxdifly]

ACC'E' is a rectangle, but the entire rectangle is "skewed" in the sense that you mean. Hence the angle ACC' is a right angle and Pythagoras can be used.

Note that sin(108°)/sin(36°) = sin(72°)/sin(36°) = 2sin(36°)cos(36°)/sin(36°) = 2cos(36°).

Thanks to this hint, I understand that cos36° = \(\displaystyle \frac{1+\sqrt{5}}{4}\), so 2cos36° = \(\displaystyle \frac{1+\sqrt{5}}{2}\).

Assuming the length of each triangle's side is a, using Pythagorean Theorem, we have:
\(\displaystyle AC'=\sqrt{(AC)^2+(CC')^2}\)
\(\displaystyle =\sqrt{\left(\frac{1+\sqrt{5}}{2}a\right)^2+(a)^2}\)
\(\displaystyle =\sqrt{\frac{4a^2}{4}+\frac{(1+\sqrt{5})^2a^2}{4}}\)
\(\displaystyle =\frac{1}{4}\sqrt{4a^2+(1+2\sqrt{5}+5)a^2}\)
\(\displaystyle =\frac{1}{4}\sqrt{4a^2+(6+2\sqrt{5})a^2}\)
\(\displaystyle =\frac{1}{4}\sqrt{4a^2+6a^2+2\sqrt{5}a^2}\)
\(\displaystyle =\frac{1}{4}\sqrt{10a^2+2\sqrt{5}a^2}\)
\(\displaystyle =\frac{a}{4}\sqrt{10+2\sqrt{5}}\)
Because AC' = PP', which is the icosahedron's height, the height of the icosahedron is \(\displaystyle \frac{a}{4}\sqrt{10+2\sqrt{5}}\).