## Thursday, June 7, 2012

### Problem 761: Scalene Triangle, 60 degrees, Euler Line, Equilateral Triangle

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 761 details.

1. http://img835.imageshack.us/img835/6848/problem761.png

Draw additional lines per attached sketch
Note that ∆ABN is equilateral and BG/BM=BQ/BS= 2/3 => QG⊥BQ
And HG=2.GO and HQ=2.QR …. (Property of Euler line)
OR=RQ. √3=1/3 . HR. √3
So tan(x)= OR/HR= 1/√3 => x=30

2. Let X and Y be the midpoints of AC, AB
Let L, M be the feet of the altitudes from B, C
1/2 = cos 60° = AL/AB = AL/c. So AL = c/2 = AY
1/2 = cos 60° = AM/AC = AM/b. So AM = b/2 = AX
Follows LX = AX - AL = (b - c)/2 ; MY = AM - AY = (b - c)/2
So LX = MY = d say
Next ∆ACM is right angled at M and ∠ACM = 30°
(i.e.)∠LCH = 30°
So 1/2 = sin 30° = HL/CH, HL = CH/2 = OY by a well known property
Similar to HL = OY, we have HM = OX
For convenience denote OX = x, OY = y
Next HL ∥ OX implies LE/XE = HL/OX = y/x, 1 + LX/XE = y/x,
LX/XE = (y - x)/x,
(y - x).XE = x. LX = x.d
Similarly
HM ∥ OY implies DY/DM = OY/HM = y/x, 1 + MY/DM = y/x,
MY/DM = (y - x)/x,
(y - x).DM = x. MY = x.d
Thus XE = DM
Hence finally,
AD = AM + MD = b/2 + MD = AX + XE = AE and so ∆ADE is equilateral
(minor typos, if any may be excepted)

3. Since m(A) = 60 => m(COB) = 120
m(CHB) = 180-(B-30)-(C-30) = 120
So BHOC are concyclic
Therefore m(OHC) = m(OBC) = 30 ( Since triangle OBC is isosceles triangle with OBC=OCB=30 and COB =120)
Also m(HCA) = 30 (extend CH to meet AB at P and PCA is a 30-60-90 triangle)
So triangle EHC is an isosceles triangle with m(EHC) = m(OHC) = 30 = m(HCE)
So m(HEC) = 120
=> m(HEA) = 60
Therefore DAE is equilateral

1. Can you please explain how BHOC is concyclic?