Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, March 5, 2015
Geometry Problem 1093. Square, Circle, Tangent, Radius, Side, Sangaku, Japanese
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By construction, y = OO1 = 1-OB = 1-√2/2. x is to OB-y as y is to √2/2, hence x=3-2√2
ReplyDeleteLet R is the radius of circles O1 and O2
ReplyDeleteAlong diagonal AC we have AC= 2.R+2.R.sqrt(2)= sqrt(2)
So R= 1/(2.(1+sqrt2))
Along Diagonal BD we have BD= 2.R+2.x+ 2.x.sqrt(2)= sqrt(2)
Replace value of R in above we get x= 3-2.sqrt(2)
Call R the radius of the big circles
ReplyDeleteConsider the half diagonal AO: R(1+√2)=√2/2 thus R=(√2/2)/ (1+√2)
Consider the half diagonal BO:R+x(1+√2)=√2/2. Replacing with above found value of R:
(√2/2)/ (1+√2)+x(1+√2)=√2/2 thus √2/2 - (√2/2)/ (1+√2)= x(1+√2)
Developing the left hand expression
√2/2 - (√2/2)/ (1+√2)= 1/(1+√2)
Eqation becomes 1/(1+√2)= x(1+√2) and finally x=1/(1+√2)2
Numerically: R≈0.29 and x=0.17, both values fit well into the picture