## Sunday, December 23, 2012

### Problem 835: Isosceles Triangle, Double Angle, Triple Angle, Auxiliary Lines

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 835.

1. Result: 150

Solution: Let BC=AC=1 a:alfa, b:beta
From sinus theorem in BDC
DC=sina/sin(a+b)
Sinus theorem in ADC using DC we get
sina*sin(2a+3b)=sin2a*sin(a+b) since sin2a=2*sina*cosa we get
2*sin(a+b)*cosa=sin(2a+b)+sinb=sin(2a+3b)
sinb=sin(2a+3b)-sin(2a+b)=2*sinb*cos(2a+2b)=sinb
eliminating sinb(b can't be zero) we get
cos(2a+2b)=1/2 which means 2a+2b=60, a+b=30
So x=180-30=150 degrees.

1. And using elementary geometry?

2. http://img38.imageshack.us/img38/3597/problem835.png

Make drawing per attached sketch
1. Draw altitude CF and bisector AF of angle DAC
Make angle ACE= beta
2. Due to symmetric properties , CDE and AEDB are isosceles
And BE, AD and CF are concurrent at point G
3. In triangle AGC , GE is an angle bisector => angle BGF=angle AGF=angle AGE=angle CGE=60
In triangle ABC we have 60+ 4alpha+4.beta=180 => alpha+ beta=30
In triangle BDC x=180-alpha-beta= 150

1. not clear at all. Your proposition is to
1) Drop an altitude from C to the base, intercepting the base at F ( I think) that will bisect the vertex angle at C into 2 equal angles of 2beta each
2) The second construction I am not sure I understand. but it seems that you are proposing and angle bisector from A intersecting BC at E or intersecting DC At E.
the rest pointing to concurrency at G etc. I have no clue at all.
do you have a sketch of your construction?

3. Minor typo correction.
Line 3 of my solution should read as "1. Draw altitude CF of triangle ABC and bisector AE of angle DAC "
Peter Tran

4. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 25, 2012 at 9:04 AM

Fie P punctul de pe segmentul AD astfel incat m(: PC este mediatoarea segmetului AB si bisectoarea unghiului =>m(In ∆BPC
m(6α+6β=180 =>α+β=30 de unde in ∆BDC x=180-(α+β)=150

5. Let CX be the bisector of < ACB, X being on AD. Now thro' congruence or symmetry < AXC = < BXC. But D is the in centre of Tr. ACX, so XD bisects < BXC. It thus follows that < BXD = 60.
So 2 alpha + 2 beta + 120 = 180 in Tr. ADC, implying that alpha + beta = 30.
So from Tr. BDC x = 150

Sumith Peiris
Moratuwa
Sri Lanka