## Wednesday, February 18, 2009

### Problem 13: Triangle, Angles, Congruence, Cevian

Proposed Problem

See complete Problem 13 at:
http://gogeometry.com/problem/problem013.htm

Level: High School, SAT Prep, College geometry

1. Solucion:
http://forogeometras.com/index.php?topic=884.0
http://img25.imageshack.us/img25/7149/rsolmarzo8.jpg

2. trazar DE//AB
el angulo EDC=4x
Los triangulos ABD y DCE son congruentes
entonces ABD=3x
en el triangulo ABC: 3x+4x+3x+5x=180
x=12º

atte:tino!!!

3. Dear Tino,
Why triangles ABD and DCE are congruent?

4. We've CD/sin(5x)=BC/sin(8x) and AB/sin(3x)=BC/sin(4x). Thus, sin(5x)/sin(8x)=sin(3x)/sin(4x) or
sin(5x)/2sin(4x)cos(4x)=sin(3x)/sin(4x) or
sin(5x)/2cos(4x)=sin(3x)
By inspection if x=10 then sin(5x)=cos(4x) and sin(3x)=1/2. Thus, x=10 is clearly a solution to the equation.
Ajit: ajitathle@gmail.com

5. This comment has been removed by a blog administrator.

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

7. Solution by cristian, my partnert >:)

http://youtu.be/neHBBXZiaRI

Go Go Geometry!

Geo-Greetings ^^!

9. 1. Draw E on AB such that BE = BD and F on CD such that DF = BD also. This forms 2 isosceles triangles BDE (with 2 6x angles) and BDF (with 2 4x angles).
2. By angle chasing <BFA = 4x = <BAD so tr. ABF is isosceles also and
BF = AB = CD.
3. By angle chasing <BDC = <EBF = 180 - 8x so tr BEF = tr BCD by SAS.
4. So <BEF = <CBD = 5x and by angle subtraction <DEF = x
then also <BFE = <BCD = 3x and by angle subtraction <EFD = x also.
5. So tr. EDF is also isosceles and DE = DF = BD = BE.
6. This implies tr. BDE is actually equilateral and 6x = 60. X = 10.

1. Btw: compare this construction to problem #9. They are the same.

10. I derived the following equation in the same way that Joe did in 2009:
sin(5x) / 2cos(4x) = sin(3x)
At this point, I rearranged the terms and apply the product-to-sum formula:
sin(5x) = 2cos(4x)sin(3x) = sin(7x) - sin(x)
Now, rearrange the terms again and apply the sum-to-product formula,
sin(x) = sin(7x) - sin(5x) = 2cos(6x)sin(x)
sin(x)(2cos(6x) - 1) = 0
This suggests either sin(x) = 0 or cos(6x) = 1/2.
The former is invalid (which would imply x is a multiple of 180°), whereas the latter suggest 6x = 60°, 300°, ..., that is x = 10°, 50°, ...
In order to keep ∠DBC = 5x under 180°, we must choose x = 10°.