Can't say I follow your solution yet.
[SPOILER]Why is angle DGF 50 degrees?[/SPOILER]

[quote=philmoore;220028]Can't say I follow your solution yet.
[spoiler]Why is angle DGF 50 degrees?[/spoiler][/quote]
[spoiler]BGF = 20 (e.g. because BAF is 40 or AGB is 60) and DGB is 60/2[/spoiler]

David

[quote=Xyzzy;100082][COLOR=white].[/COLOR][/quote]

Click this post for the diagram.

[quote=philmoore;100095]Thanks for posting that, Mike. I'll try to explain what I found interesting. It was provoked by David's observation of reflection. Suppose we label the vertices consecutively as V[sub]1[/sub], V[sub]2[/sub], V[sub]3[/sub], ... V[sub]18[/sub]. Then what we notice is that the following line segments are all concurrent: V[sub]1[/sub]V[sub]7[/sub], V[sub]2[/sub]V[sub]9[/sub], V[sub]3[/sub]V[sub]12[/sub], and the reflections of the first two segments across the last: V[sub]4[/sub]V[sub]15[/sub] and V[sub]5[/sub]V[sub]17[/sub]. My guess is that there is some underlying symmetry that explains why these lines are concurrent, but I haven't done much research on it. Could something related to Pascal's theorem be at work here?[/quote]

When I looked at your diagram again, all became blindingly
obvious (which is why my "proof" above was a bit casual).
Let us call the centre A and the "multiple coincidences" v[sub]n[/sub].
I spotted that V[sub]-3[/sub]v[sub]-1[/sub]v[sub]1[/sub]V[sub]3[/sub] bisected AV[sub]0[/sub] perpendicularly.
The reason is obvious: V[sub]-3[/sub]V[sub]3[/sub]V[sub]9[/sub] is an equilateral triangle.

Talk about a penny dropping!

David

Note also that just as AV[sub]1[/sub] and V[sub]0[/sub]V[sub]7[/sub]
intersect on V[sub]-3[/sub]V[sub]3[/sub] (D in the original problem),
so do AV[sub]2[/sub] and V[sub]0[/sub]V[sub]5[/sub] (e.g. H or E)

David

Regular 18-agon
 

[quote=philmoore;220028]Can't say I follow your solution yet.
[spoiler]Why is angle DGF 50 degrees?[/spoiler][/quote]

I'm sure that you intended this as a friendly, tactful hint from
one busy teacher to another (less busy) that I was a bit slapdash.
I don't believe you couldn't work out the angle between
two diagonals of a regular 18-agon!

OK let's start with that.
For our purposes, sides are diagonals.

V[sub]0[/sub]V[sub]1[/sub] is // to V[sub]-1[/sub]V[sub]2[/sub] etc up to V[sub]-8[/sub]V[sub]9[/sub].
(Clock arithmetic of course).
Rotating through 20 degrees 8 times gives us 9*9 diagonals.

V[sub]-1[/sub]V[sub]1[/sub] is // to V[sub]-2[/sub]V[sub]2[/sub] etc up to V[sub]-8[/sub]V[sub]8[/sub].
Rotating through 20 degrees 8 times gives us 9*8 diagonals.

9*9 + 9*8 = 18C2

V[sub]0[/sub]V[sub]1[/sub]V[sub]-1[/sub] = 10 degrees.

[U]Now for my solution to the original problem:[/U]

Let the centre of the 18-agon be A. Let V[sub]0[/sub] be B and V[sub]1[/sub] be C.
V[sub]-3[/sub]V[sub]3[/sub] bisects AB perpendicularly.
(AV[sub]+/-3[/sub]B are equilateral).

V[sub]1[/sub]V[sub]0[/sub]V[sub]7[/sub] = 60
AV[sub]0[/sub]V[sub]7[/sub] = 20
V[sub]0[/sub]AV[sub]1[/sub] = 20
V[sub]0[/sub]V[sub]7[/sub] intersects AV[sub]1[/sub] on V[sub]-3[/sub]V[sub]3[/sub] at D.
Reflect V[sub]-3[/sub]V[sub]3[/sub] in AV[sub]1[/sub] to get V[sub]-1[/sub]V[sub]5[/sub] which passes through D. It intersects AV[sub]0[/sub]
at E?
Reflect V[sub]-1[/sub]V[sub]5[/sub] in AV[sub]0[/sub] to get V[sub]1[/sub]V[sub]-5[/sub].

Show that V[sub]-5[/sub]V[sub]1[/sub]V[sub]0[/sub] = 50 degrees,
confirming that E? = E.

Now find V[sub]0[/sub]DV[sub]-1[/sub]

David