problems-ex2

problems-ex2 - Problem 1.6.4 on p.18 Referring to Fig 1.6A...

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

Solutions to a couple of problems: MTG 3214 October 29, 2008 Problem 1.8.5 on p.22 (assume problem 1.6.4 on p.18 has been solved; I’ll do this later) First a remark: The statement could have been more precise. For instance: “the 9 point circle cuts the sides at angles | B C | , | A B | and | A C | , seen from the points A , C and B respectively”. An equivalent statement is: “the 9 point circle cuts the sides at angles 2 | B C | , 2 | A B | and 2 | A C | , seen from its center N ”. I will prove the last statement. Refer to Fig. 1.8B. The chord DA is seen from N under angle: DNA = 2 DKA ( since D,K,A’ are on circle with center N) = 2 HKN ( same angles ) = 2 HAO ( since HKN ∼ △ HAO, reason below) = 2 | B C | ( by problem 1.8.5) , There holds that HKN ∼ △ HAO by SAS, since AH = 2 KH , the angles at H are the same, and HO = 2 HN
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . Problem 1.6.4 on p.18 Referring to Fig. 1.6A, we need to show that ≺ HAO = | B − C | . We have A = ≺ BAC = ≺ BAD + ≺ HAO + ≺ OAC. (1) We know that ≺ BAD = 90 o − B ( consider the right-angled △ BAD ) (2) and that ≺ OAC = ≺ OCA ( △ AOC is isosceles because O is circumcenter) = 90 o − B (3) where the last equality holds for a similar reason that in Fig. 1.6.A, ≺ OBC = α = 90 o − A holds. Plugging (2) , (3) in (1), and solving for ≺ HAO , yields: ≺ HAO = A − 180 o + 2 B = B − C, which is | B − C | since here B > C (if B were < C instead, you would ±nd that ≺ HAO = C − B , which is of course equal to | B − C | ). * Instructor: Patrick De Leenheer. 1...
View Full Document

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.

Ask a homework question - tutors are online