# sol2 - Solutions toExam 2 MTG 3214 November 4 2008 Name...

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Unformatted text preview: Solutions toExam 2: MTG 3214 * November 4, 2008 Name : Student ID : This is a closed book exam and the use of calculators is not allowed. 1. Prove: The feet of the perpendiculars from a point to the sides of a triangle are collinear if and only if the point lies on the circumcircle. Solution : This is Theorem 2.51. 2. Prove that the nine point circle of the triangle formed by connecting the excenters of a given triangle ABC, is the circumcircle of triangle ABC. Solution : (This is problem 1 . 8 . 3 which was a monthly HW problem) Consider Fig 1.4B on p.12. The 9 point circle of 4 I a I b I c must certainly contain the 3 feet of its altitudes. I claim that these feet are A,B and C . To prove this, notice that the internal and external bisector of any angle are always perpendicular. (convince yourself by drawing a picture if you don’t see this!). On the other hand, there is only 1 circle that contains the 3 points A,B and C , namely the circumcircle of 4 ABC . Consequently, this circumcircle must be the 9 point circle of....
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## This note was uploaded on 09/12/2011 for the course MAP 4305 taught by Professor Deleenheer during the Summer '06 term at University of Florida.

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