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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 $\displaystyle \frac{a}{4}\sqrt{10+2\sqrt{5}}$. Tags height, icosahedron Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post foothill Geometry 6 October 31st, 2014 10:19 PM Shamieh Algebra 5 August 27th, 2013 04:03 AM Telu Algebra 15 June 27th, 2013 01:36 AM chicagobearsconvulse Calculus 1 April 28th, 2009 05:11 AM symmetry Algebra 1 September 26th, 2007 11:51 AM

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