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March 21st, 2016, 07:14 PM   #1
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[ASK] Height of an Icosahedron


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?
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March 22nd, 2016, 02:26 AM   #2
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AO/AB = sin(72°)
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March 22nd, 2016, 03:51 AM   #3
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What does it have to do with my question? Does the triangle ABO even have a right angle? Isn't it isosceles instead?
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March 22nd, 2016, 04:23 PM   #4
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The height you seek is 2AO.
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March 22nd, 2016, 05:18 PM   #5
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Where did the 72 degrees come from?
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March 22nd, 2016, 05:51 PM   #6
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2AO = AC', which is a diagonal of the rectangle ACC'E'.

After using a little trigonometry to find AC, which is a diagonal of the regular pentagon ABCDE, AC' can be found by use of Pythagoras.

The result can be simplified to 2AB·sin(72°).
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March 22nd, 2016, 06:10 PM   #7
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I got $\displaystyle AC=\frac{\sin108^\circ}{\sin36^\circ}AB$. Also, is the triangle ACC' even a right triangle? Shouldn't the angle ACC' be $\displaystyle 150^\circ$ since the line CC' is actually skewed?

Last edited by skipjack; March 22nd, 2016 at 06:19 PM.
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March 22nd, 2016, 06:28 PM   #8
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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°).
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March 22nd, 2016, 06:31 PM   #9
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I think I understand. Might come back later if I meet a dead-end again.
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April 6th, 2016, 07:02 PM   #10
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Quote:
Originally Posted by skipjack View Post
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 $\displaystyle \frac{a}{4}\sqrt{10+2\sqrt{5}}$.
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