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Heptagon diagonal intersection (Posted on 2025-05-13) Difficulty: 3 of 5
Let ABCDEFG be a regular heptagon. Find the measure of the smaller angle of intersection between AD and CF.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 2 of 2 |
I decided to generalize a bit and let ABCDEF be six consecutive vertices of a regular polygon of N sides.  Let the intersection of AD and CF be J.

Each side of the polygon has an associated arc whose arc length is 2pi/N radians.
Arc AC and DF each subtend two of those arcs, so the are each 4pi/N radians.
Angles DAF and CFA have a measure of half the subtended arcs DF and AC, so angles DAF and CFA are each 2pi/N radians.

Then triangle AJF has two angles of 2pi/N radians each.  The remaining angle, AJF is then (1-4/N)*pi radians.
The supplementary angle of AJF may also be the smallest.  That angle is 4pi/N radians.

For the specific problem we want the smaller of 4pi/N verses (1-4/N)*pi where N=7.  4pi/7 > (1-4/7)*pi = 3pi/7, so the measure of the smaller angle is 3pi/7 radians.

  Posted by Brian Smith on 2025-05-14 13:10:42
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