Tough Isosceles (Posted on 2005-07-20)

	Submitted by McWorter
Rating: 5.0000 (1 votes)
Solution:	(Hide)
Charlie's Solution If AD and AC are the same length, then triangle ADC is isosceles. Triangle ADO is also isosceles and has the common angle ADC with triangle ADC. Hence triangles ADC and ADO are similar. Thus angles DAC and AOD are equal. Now angles DAB and the central angle DOB subtend the same arc DB, so angle DAB is half that of DOB. But then angle AOD is also half that of DOB, making AOD one third of angle AOB; impossible to construct with staightedge and compass. My (now ugly) solution We show that angle AOC is one third of angle AOB, an impossible straightedge and compass construction. Let x be the angle AOC and y the angle OAB. Then angle ACD equals x+y. Since AC=AD, angle ADC=x+y. Since A and D are on the circumference of the circle, angle OAD=x+y. But angle OAD also equals angle CAD plus y, whence angle OAD=180-2(x+y)+y. Hence x+y=180-2(x+y)+y, or 3x=180-2y. But 180-2y is precisely the angle AOB, whence x trisects angle AOB.

	Subject	Author	Date
	re: solution	McWorter	2005-07-21 00:08:57
	solution	Charlie	2005-07-20 18:56:12
	Define Strictly	Bret	2005-07-20 17:31:21
	Arcs and Right Triangles	Bret	2005-07-20 17:21:19
	Conditional Solution	Bractals	2005-07-20 16:23:16
	Playing around with Geometer's Sketchpad....	Charlie	2005-07-20 15:50:33