Proposed Problem

See complete Problem 16 at:

http://gogeometry.com/problem/problem016.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, February 18, 2009

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

Labels:
angle,
cevian,
congruence,
perpendicular,
Problem 4

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It is possible to provide a proof using only elementary geometry!

ReplyDeleteHints:

Auxiliary lines, Perpendicular bisector, Congruence, Problem 4.

Since a plane geometry solution hasn't appeared so far here's a simple trigonometric solution:

ReplyDeleteBy 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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ReplyDeletehttp://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

ReplyDeleteThe solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJcDNRdVlFR3VKNlk

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

ReplyDeleteGreetings

Let CE=1 so that AC=2.

ReplyDeleteHence 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

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

ReplyDelete