Geometry Problem. Post your solution in the comment box below.

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

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## Wednesday, March 11, 2015

### Problem 1096: Tangent Circles, Common Tangent, Chord, Radius, Center

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http://s9.postimg.org/h8iwun773/Pro_1096.png

ReplyDeleteDraw lines per sketch

We have ∠ ( D₃ O2 O₃)=∠ ( O₃OC)= 30 degrees

And OO3= R-r

OO2^2= (R-r)^2-4.r^2= R^2-2.R.r-3.r^2

OC^2=3/4.OO2^2= ¾(R^2-2.R.r-3.r^2)

CB^2=R^2-OC^2= ¼(R^2+6.R.r+9.r^2)

AB^2=4.CB^2=(R+3.r)^2= > AB= R+3.r

See revised sketch with more details below.

ReplyDeletehttp://s22.postimg.org/iruqw8anl/Pro_10961.png

Let OO2 = h and perpendicular from O to AB be p

ReplyDeleteThen < O1O2D1 = 30 and so p = sqrt3/2 h

Now AB^2 = 4(R^2- p^2)

and h^2 = (R-r)^2 -4r^2

So AB^2 = 4R^2 - 3h^2 = 4R^2 - 3(R-r)^2 - 12r^2 = (R+3r)^2

So AB = R + 3r

Sumith Peiris

Moratuwa

Sri Lanka