## Wednesday, February 18, 2009

### Problem 16: Triangle, Cevian, Perpendicular, Angles, Congruence

Proposed Problem See complete Problem 16 at:
http://gogeometry.com/problem/problem016.htm

Level: High School, SAT Prep, College geometry

1. It is possible to provide a proof using only elementary geometry!
Hints:
Auxiliary lines, Perpendicular bisector, Congruence, Problem 4.

2. Since a plane geometry solution hasn't appeared so far here's a simple trigonometric solution:
By applying sine rule to Tr. ABC, we get
sin(7x)/sin(3x) = AC/BC. But AC = 2CE; hence
sin(7x)/sin(3x)= 2CE/BC. Now CE/BC = cos(2x) from Tr. BCE. Thus, sin(7x)/sin(3x)=2cos(2x) or sin(7x)/cos(2x)= 2sin(3x). By inspection if x=10 then sin(7x)=sin(70)=cos(20)=cos(2x) while sin(3x)=sin(30)= 1/2 which in turn makes LHS = RHS. Therefore, x = 10 is a solution to our equation.
Ajit: ajitathle@gmail.com

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4. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

5. 6. Video solution http://youtu.be/3luEqMUwzXM (in spanish :P)

Greetings

7. Let CE=1 so that AC=2.
Hence BC= (1 / cos(2x))
In △ABC,
sin(180-7x)/AC = sin(3x)/BC
sin(7x)/2 = sin(3x)cos(2x)
sin(7x) =2sin(3x)cos(2x)
sin(7x) = sin(5x) + sin(x) (trigonometric identity)
sin(x) = sin(7x) - sin(5x)
sin(x) = 2cos(6x)sin(x) (trigonometric identity)
0 = 2cos(6x)sin(x)-sin(x)
0 = sin(x) (2cos(6x)-1)
sin(x)=0 or (2cos(6x)-1)=0
x=0 (rejected) or 2cos(6x)=1
cos(6x)=0.5
6x=60
x=10

8. Extend CE to F such that BF || AC. Then FBC is isosceles with base FC=AC and base angles 2x. Extend AC to G such that CG=CB. Then BFCG is parallelogram and BG=FC=AC. Therefore, triangle ABG and its cevian BC display the same configuration as Problem 5, and therefore x=10