Simple Geom problem.

[philmoore]

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

[davieddy]

Originally Posted by philmoore

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

BGF = 20 (e.g. because BAF is 40 or AGB is 60) and DGB is 60/2

David

[davieddy]

Originally Posted by Xyzzy

.

Click this post for the diagram.

Originally Posted by philmoore

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 V₁, V₂, V₃, ... V₁₈. Then what we notice is that the following line segments are all concurrent: V₁V₇, V₂V₉, V₃V₁₂, and the reflections of the first two segments across the last: V₄V₁₅ and V₅V₁₇. 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?

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_n.
I spotted that V_-3v_-1v₁V₃ bisected AV₀ perpendicularly.
The reason is obvious: V_-3V₃V₉ is an equilateral triangle.

Talk about a penny dropping!

David

[davieddy]

Note also that just as AV₁ and V₀V₇
intersect on V_-3V₃ (D in the original problem),
so do AV₂ and V₀V₅ (e.g. H or E)

David

[davieddy]

Originally Posted by philmoore

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

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.

V₀V₁ is // to V_-1V₂ etc up to V_-8V₉.
(Clock arithmetic of course).
Rotating through 20 degrees 8 times gives us 9*9 diagonals.

V_-1V₁ is // to V_-2V₂ etc up to V_-8V₈.
Rotating through 20 degrees 8 times gives us 9*8 diagonals.

9*9 + 9*8 = 18C2

V₀V₁V_-1 = 10 degrees.

Now for my solution to the original problem:

Let the centre of the 18-agon be A. Let V₀ be B and V₁ be C.
V_-3V₃ bisects AB perpendicularly.
(AV_+/-3B are equilateral).

V₁V₀V₇ = 60
AV₀V₇ = 20
V₀AV₁ = 20
V₀V₇ intersects AV₁ on V_-3V₃ at D.
Reflect V_-3V₃ in AV₁ to get V_-1V₅ which passes through D. It intersects AV₀
at E?
Reflect V_-1V₅ in AV₀ to get V₁V_-5.

Show that V_-5V₁V₀ = 50 degrees,
confirming that E? = E.

Now find V₀DV_-1

David