## Monday, May 19, 2008

### Elearn Geometry Problem 46

See complete Problem 46
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

1. x=30 degrees is also a solution to the problem

2. Solucion:
http://img8.imageshack.us/img8/9236/rsolmarzo7.jpg

3. In Tr. DBC, DC/DB = sin(3x+30)/sin(30) = 2sin(3x+30) While Tr. ADC gives,DC/AD =sin(2x)/sin(x)
2sin(3x+30)=2sin(x)cos(x)/sin(x) or
sin(3x+30)=sin(90-x) or 3x+30=90-x or x=15
Alternatively, sin(3x+30)=sin(90-x) may mean
sin(180 -(3x+30))=sin(90-x) giving us,
180 -(3x+30) = 90- x giving us x=30
Svvv is very correct in saying that 15 deg. and 30 deg. are both acceptable solutions to this problem.
Ajit: ajitathle@gmail.com

New condition: angle B is an obtuse angle.

5. Dear Antonio

This problem is proving very difficult to crack thro Euclidean geometry alone.

I tried to drop altitudes from B to DC and AC and work through the cyclic quad
Secondly tried to find point M on DC such that D is the centre of ABM Tr.
Thirdly found that if u find N on DC such that Tr. ANC is isoceles I found that < DBN is 30

But I cannot clinch it

Are any of my approaches right?

1. Dear Sumith,
Some ideas: equilateral triangle, isosceles triangle, congruence ....

2. Let BE = h be an altitude of triangle ABC and let CD = d

So BD = DA = DE = CE = c/2

CD is a median of triangle ABC, so
d^2 = a^2/2 +b^2/2 -c^2/4 ....(1)

Since triangles BDC and ADC are equal in area, easily
1/2 da/2 = 1/2 bh/2 hence d = bh/a ....(2)

Using Pythagoras h^2 = c^2 - (b-c/2)^2 = a^2 - (c/2)^2....(3)

Eliminating d and h between these 3 equations,
a^2/2 + b^2/2 - c^2/4 = (b^2/a^2)(a^2 - c^2/4).

Simplifying
a^/2 - c^2/4 = b^2/2 - b^2c^2/(4a^2) which gives,
2(a^2 - b^2) = (c^2/a^2)(a^2 -b^2)

Since a NOT = b, a^2 = (c/2)c
So BC^2 = BD. BA and BC is therefore a tangent to circle ADC.

Hence 2x = 30 and x = 15

Sumith Peiris
Moratuwa
Sri Lanka