sol2 - Solutions toExam 2: MTG 3214 * November 4, 2008 Name...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 dont 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....
View Full Document

Ask a homework question - tutors are online