Polar Fox: Dude, did you get any permission to use a trademark?

Red Fox: What permission? This is a public service we are engaged in.

Polar Fox: Oh really? Still I say that logo has an owner. It says in its corner, see.

Red Fox: Go ahead; let those capitalist scumbags know about it.

Polar Fox: Oh come on Red, stop acting like a mild Chávesian socialist. You spent all your time in youtube and facebook.

Red Fox: Yeah, I am extending my base in the internet for a revolution.

Polar Fox: Yeah, sure. The only thing I see revolving is your chubby belly.

Red Fox: What permission? This is a public service we are engaged in.

Polar Fox: Oh really? Still I say that logo has an owner. It says in its corner, see.

Red Fox: Go ahead; let those capitalist scumbags know about it.

Polar Fox: Oh come on Red, stop acting like a mild Chávesian socialist. You spent all your time in youtube and facebook.

Red Fox: Yeah, I am extending my base in the internet for a revolution.

Polar Fox: Yeah, sure. The only thing I see revolving is your chubby belly.

Let the second and third circles be x^2+y^2=4 and

ReplyDeletex^2+y^2=9. Let the three vertices of the ET be (1,0),(x1,y1) &(x2,y2) with the side being = a. Then we can say:

x1^2+y1^2=9

x2^2+y2^2=4

(x1-x2)^2+(y1-y2)^2=(x1-1)^2+y1^2

(x1-x2)^2+(y1-y2)^2=(x2-1)^2+y2^2

a^2=(x1-x2)^2+(y1-y2)^2

These five eqns. with five unknowns can be simultaneously solved to obtain: x1 = 3/2, y1= 3V3/2, x2=-1, y2=V3 and a=V7 where V = square root.

Ajit

http://i38.tinypic.com/250kmdh.png

ReplyDeletehttp://i37.tinypic.com/o7txlg.gif

Let the corners of the triangle be A (on the smallest circle), B (on the second circle), and C (on the biggest circle). Let O be their common center. Let y be the side length of triangle ABC.

ReplyDelete1) Connect center O to three corners

2) Let the angles in triangle OAB be x1, x2, and x3 (BOA=x1, OAB=x2 ve ABO=x3)

3) The sides of OAB are 1, 2, and y. And x1 is the angle opposite of length y.

4) The sides of OAC are 1, y, and 3. And x2+60 is the angle opposite of length 3.

5) The sides of OBC are 2, 3, and y. And x3+60 is the angle opposite of length y.

6) We can write the cosine theorem in three triangles,

OAB: 1^2+2^2 – 2*1*2*cos(x1)=y^2

OAC: 1^2+y^2 – 2*1*y*cos(x2+60)=3^2

OBC: 2^2+y^2 – 2*2*y*cos(x3+60)=3^2

Also

x1+x2+x3=180

I have solved above four equations simultaneously with MS Excel Solver. Here are the answers:

x1 ~ 120

x2 ~ 40.8

x3 ~ 19.1

y ~ 2.6458 = sqrt(7)

Ozgur from Izmir

Let O be their common center of thee circles ,be ABC triangle whith AB = BC = CA = d and OA = 3 , OB = 2 ,OC =1.

ReplyDeleteThen OA.AB = OC.AB + OB.AC .( OA.AB =3d and

OC.AB + OB.AC = 1d + 2d = 3d) .

Then Ptolemy's theorem ==> (O,A,B,C) concyclic

and angle(O) + angle(A) = 180 degrrees ==> angle(O) =120 degrrees.

then (OBC) triangle ==> OB^2 + OC^2 + OB.OC = BC^2 or

2^2 + 1^2 +1.2 = d^2 ==> d^2 = 7 ==> d = sqrt(7).

Philip Pydnaios from Hellas.

==

Thank you Philip Pydnaios.

ReplyDelete