Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, June 13, 2016
Geometry Problem 1228: Three Tangent Circles, Midpoint, Major Arc, Common Tangent Line, Square
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AMO4 and BMO4 are congruent SSS so O4M bisects < AMB hence Tr.s AMD and BMC are congruent ASA (since AD//BC)
ReplyDeleteSo ABCD is a rectangle
Proving that ABCD is indeed a square appears to involve tedious equations
Lets assume circle O4 has radius R.
ReplyDeleteO4O3=R-R3. Let N be midpoint of AB, O4Nis perpendiclar to AB.
O4O3^2 = [O4N+R3]^2 + [AE-AN]^2
We get R^2-2R.R3=O4N^2+2O4N.R3+[AE-AN]^2
Also R^2=O4N^2+AN^2,
We get O4N+R=(AN^2-[AE-AN]^2)/2R3 or
O4N+R=AE.EB/2R3
Since AE=2sqrt(R1R3),EB=2sqrt(R2R3),AB=2sqrt(R1R2)
We get O4N+R=AB=CD hence ABCD is a square.
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