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: 20,613 Thanks: 2071 
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: 20,613 Thanks: 2071 
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: 20,613 Thanks: 2071 
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: 20,613 Thanks: 2071 
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 deadend again.

April 6th, 2016, 07:02 PM  #10  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry  Quote:
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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