My Math Forum [ASK] Height of an Icosahedron

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 March 21st, 2016, 07:14 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry [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?
 March 22nd, 2016, 02:26 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2222 AO/AB = sin(72°)
 March 22nd, 2016, 03:51 AM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry What does it have to do with my question? Does the triangle ABO even have a right angle? Isn't it isosceles instead?
 March 22nd, 2016, 04:23 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2222 The height you seek is 2AO.
 March 22nd, 2016, 05:18 PM #5 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Where did the 72 degrees come from?
 March 22nd, 2016, 05:51 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2222 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°).
 March 22nd, 2016, 06:10 PM #7 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry 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.
 March 22nd, 2016, 06:28 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2222 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°).
 March 22nd, 2016, 06:31 PM #9 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry I think I understand. Might come back later if I meet a dead-end again.
April 6th, 2016, 07:02 PM   #10
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Quote:
 Originally Posted by skipjack 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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