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Used more standard formatting (as a result, the diff looks more extensive than it really is - use view "Side-by-side Markdown" to compare).
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Peter Mortensen
Peter Mortensen
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This one, via $($via Proof Without Words$)$Proof Without Words, is wonderful, but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

  Explanation: ![enter image description here][2]

Enter image description here

Set the radius to be $1$, then 
$$HK=2HI=2\cos\frac{\pi}{7}$$

Explanation:

$$AC=2AB=2\cos\frac{3\pi}{7}$$

Enter image description here

$$DG=2DF=-2\cos\frac{5\pi}{7}$$

Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$

So

So

$$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

$$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

This one $($via Proof Without Words$)$ is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

  Explanation: ![enter image description here][2]
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

This one, via Proof Without Words, is wonderful, but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it.

Enter image description here

 

Explanation:

Enter image description here

Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$

So

$$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

added 4 characters in body
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MR_BD
MR_BD
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This one ($($via Proof Without Words)$)$ is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

Explanation: ![enter image description here][2]
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

This one (via Proof Without Words) is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

Explanation: ![enter image description here][2]
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

This one $($via Proof Without Words$)$ is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

Explanation: ![enter image description here][2]
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$
added 5 characters in body
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Vim
Vim
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This one (via Proof Without Words) is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

Explanation: ![enter image description here][2] 
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

This one (via Proof Without Words) is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

Explanation: ![enter image description here][2] Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$

This one (via Proof Without Words) is wonderful but not immediately obvious. Ponder on it and you'll find out how fantastic it is when you get it. enter image description here

Explanation: ![enter image description here][2] 
Set the radius to be $1$, then $$HK=2HI=2\cos\frac{\pi}{7}$$ $$AC=2AB=2\cos\frac{3\pi}{7}$$ $$DG=2DF=-2\cos\frac{5\pi}{7}$$ So $$\begin{align} 2(\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7})&=HK+AC-DG\\ &=HK-(DG-AC)\\ &=HK-(DG-DE)\\ &=HK-EG\\ &=HK-JK\\ &=HJ\\ &=LO\\ &=1 \end{align}$$
added 417 characters in body
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Vim
Vim
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Vim
Vim
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  • 90
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