## Thursday, November 20, 2014

### Geometry Problem 1062: Scalene Triangle, 60 degrees, Euler Line, Orthocenter, Circumcenter, Distance, Congruence, Transversal

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

Click the diagram below to enlarge it. 1. http://s30.postimg.org/73ub5x029/Problem_1062.png

1. Draw points O’, E, M, N and H’ per sketch.
Per the result of problem 761, triangle A1BC1 is equilateral

http://gogeometry.blogspot.com/2012/06/problem-761-scalene-triangle-60-degrees.html
Note that BO’ is an angle bisector of ∠ (A1BC1) and triangle O’OB is isosceles
We have ∠ (OO’E)= ∠ (OBE)= ∠ (EBH) => BE is and angle bisector of ∠ (OBH)
So BE ⊥OH => EO=EH and OC1=HA1
2. OH=OA1-HA1=OA1-OC1=(R-A1A2)-(R-C1C2)= C1C2- A1A2 where R is the radius of circle O

3. We have MN=1/2OH=BN-BM= ½ AC- 1/2BC => OH= AC- BC
4. Since A1BC1 is equilateral so OH= AC-BC= AC1-CA1

2. Minor correction of last 2 lines of my solution due to typo errors

3. We have MN=1/2.OH=BN-BM= ½ AB- 1/2BC => OH= AB- BC
4. A1BC1 is equilateral so OH= AB- BC= (BC1+AC1)-(BA1+CA1)= AC1-CA1 since BC1=BA1

Peter Tran

3. AOHC is cyclic because <AOC=<AHC=120
<BA1C1=<A1HC+<A1CH=<OAC+(90-60)=(90-60)+30=30+30=60, so triangle BC1A1 is equilateral and triangle A1HC is isosceles. By supplementary angles <C1HA=30, so triangle C1HA is isosceles as well.
<C1BO=<HBA1=90-<C, which makes C1O=HA1 (4).
OH=OC2-HA2=(C1C2+C1O)-(HA1+A1A2)=C1C2-A1A2 (1).
By isosceles triangles declared earlier, AC1-CA1=C1H-HA1=(C1O+OH)-HA1=OH (2).
Because BC1=BA1 due to equilateral triangle, AB-BC=(AC1+BC1)-(CA1+BA1)=AC1-CA1=OH (3).