## Saturday, October 4, 2014

### Geometry Problem 1049: Hexagon inscribed, Circle, Circumcircle, Congruence, Angles

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

Click the figure below to view the complete problem 1049. 1. http://s21.postimg.org/4qgvbi7bb/pro_1049.png

Connect OGt
Since ODEG and OAFG are cyclic quadrilaterals
We have x= ∠ (EGt)+ ∠ (FGt)= ∠ (ODE)+ ∠ (OAF)
= 90- ½∠ (DOE)+90-1/2∠ (OAF)=180-1/2(∠DOE+∠ OAF)
But ∠ (DOE)+ ∠ (AOF)= 360-60-60-26-60= 152
So x=180-.5*152= 104

2. Ang.AOB=ang.COD=ang.EOF=60
ang.DOE+angFOA=152(=180-28)

x=360-ang.EGO-ang.FGO
ang.EGO=180-ang.EDO, ang.EDO=90-ang.DOE/2
ang.FGO=180-ang.FOA, ang.FOA=90-ang.FOA/2

Result: x=104degree

3. Ang.ODG=a ---> ang.EGO=180-a, and ang.DOE=180-2a
Ang.AOB=ang.COD=ang.EOF=60 ---> ang.FOA=360-60*3-28-(180-2a)=2a-28
---> ang OAF=(180-(2a-28))/2=104-a ---> ang.FGO=76+a
x=360-ang.EGO-ang.FGO=360-(180-a)-(76+a)=360-180-76=104

x=104deg.

4. Denote <DOE + <AOF by y
AB, CD, EF each subtends 60 deg at O. They sum upto 180 deg.
BC, DE, AF subtend angles whose sum is y + 28deg.
Thus y + 28 + 180 = 360, so y = 152
The ray OG divides angle x into 2 angles which are equal to ODE and OAF (Property of cyclic quadrilaterals).
ΔODE being isosceles, <ODE = <OED = 90 deg - ½ <DOE
Similarly <OAF = <OFA = 90 deg - ½ <AOF
So x = 180 deg - ½ y = 180 deg - 76 deg = 104 deg

5. Problem 1049:
Solution sent by Fukuchan Thanks.
Hi, Antonio,

See
attached picture here.

Ans. x=104 degree.

From Fukuchan

6. Denote the sum <DOE + <AOF by y
AB, CD, EF each subtends 60 deg at O. They sum to 180 deg.
BC, DE, AF subtend at O, angles whose sum is y + 28.
Thus y + 28 deg+ 180 deg = 360 deg, so y = 152 deg
The ray O through G divides angle x into 2 angles which are equal to <ODE and <OAF (Property of cyclic quadrilaterals).
ΔODE being isosceles, <ODE = <OED = 90 deg- ½ <DOE
Similarly <OAF = <OFA = 90 deg - ½ <AOF
So x = 180 deg - ½ y = 180 deg - 76 deg= 104 deg

7. Problem 1049:
More generally ∠ EGF = 90 deg + (½) ∠BOC

8. 