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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