Geometry Problem

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

Click the figure below to see the complete problem 664.

## Friday, September 2, 2011

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

Labels:
60 degrees,
angle,
circle,
circumcenter,
incenter,
orthocenter,
triangle

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http://img855.imageshack.us/img855/8060/problem664.png

ReplyDeletePer the result of problem 662 we have:

angle(ADC)=(AEC)=(AFC)=120

5 points A,D,E,F and C are con-cyclic

Let O is the center of pentagon ADEFC ( see picture)

Since (ABC)=60 >> (BAD)=30

Since tri. AFC is isosceles >> (FAC)=30

Since (BAC)=76 >> (DAF)=16, (DOF)=32

So (DEF)=1/2*(360-32)= 164

Peter Tran

Sketch of the Solution:

ReplyDeleteFD = 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°

Refer Peter Tran's Solution:

ReplyDeleteA, 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°

Ref:Prob 664

ReplyDeleteMore generally.

x = C + 120°