## Friday, September 2, 2011

### Problem 664. Triangle, 60 Degrees, Orthocenter, Incenter, Circumcenter, Angles

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

Click the figure below to see the complete problem 664.

1. http://img855.imageshack.us/img855/8060/problem664.png
Per the result of problem 662 we have:
5 points A,D,E,F and C are con-cyclic
Let O is the center of pentagon ADEFC ( see picture)
Since tri. AFC is isosceles >> (FAC)=30
Since (BAC)=76 >> (DAF)=16, (DOF)=32
So (DEF)=1/2*(360-32)= 164
Peter Tran

2. Sketch of the Solution:

FD = R cos 60° = R/2
BD = 2FD = R = BF
BE bisects DBF
A = 76°, B = 60°, C = 44°
∠ABD = 14°, ∠DBE = ∠FBE = 30° - 14° = 16°
DF = 2DN (where N is the 9-point centre and midpoint of DF)
=2R sin 16°
DE = EF = √(R² - 2Rr)
r = 4R sin A/2. sin B/2. sin C/2
=2R sin 38°. sin 22°
=R(cos 16° - cos 60°)
=R(cos 16° - 1/2)
DE² = R² - 2R² (cos 16° - 1/2)
= 2 R² - 2 R² cos 16°
= 4 R² sin² 8°
DE = EF = 2R sin 8°
2.DE.EF.cos x = DE² + EF² - DF²
8 R²sin²8°.cos x = 8R² sin²8°- 4R²sin²16°
2cos x sin² 8° = 2sin²8° - sin² 16°
cos x = 1 – [sin²16°/ 2sin²8°]
= 1 – 2cos²8° = - cos 16°
x = 180° – 16° = 164°

3. Refer Peter Tran's Solution:
A, D, E, F and C are con-cyclic
So ∠AED = ∠ACD = 90° - A = 90° - 76° = 14°,
and ∠CEF = ∠CAF = 90° - 60° = 30°
∠AEC = ∠AFC = 2∠ABC = 120°
Adding x = DEF = 14° + 30° + 120° = 164°

4. Ref:Prob 664
More generally.
x = C + 120°