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From: rlh@mit-eddie.UUCP (Roger L. Hale)
Newsgroups: net.math
Subject: Re: 80-80-20 Isosceles Triangle Problem
Message-ID: <831@mit-eddie.UUCP>
Date: Tue, 18-Oct-83 20:54:39 EDT
Article-I.D.: mit-eddi.831
Posted: Tue Oct 18 20:54:39 1983
Date-Received: Thu, 20-Oct-83 01:45:17 EDT
References: <439@5941ux.UUCP>
Organization: MIT, Cambridge, MA
Lines: 32

To restate the isosceles triangle problem submitted some time ago,

Given an isosceles triangle with angle A = 20 degrees
				 angle B = angle C = 80 degrees,

construct segment AD along AB such that length AD = length BC.

Find angle ADC.
___________________________________________________________________

One way comes to mind which worked for me in less than five minutes
(suggested by past playing with cyclotomic numbers, that is
more or less vector sums of the radii of regular polygons):

Draw a nonagon (9-gon), label the corners A through I.
Then isosceles triangle ABC above is triangle FAB.

Construct segment FX along FA with length = length AB = length FG.

Angle XFG = angle AFG = 60 degrees, since arc AG = 120.

So triangle XFG is equilateral;

Segment BX passes through the center, since B is opposite side FG;

Angle XBF = 10 degrees;

Angle BXF = 180 - 20 - 10 = 150 degrees, 
	quod erat inveniendum.

				Roger at MIT-DSPG
			     or ...!decvax!genrad!mit-eddie!rlh