Emergency problems #1

Emergency problems #1 - MATH 348 Solved Emergency Problems...

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Unformatted text preview: MATH 348: Solved Emergency Problems, Week 1 Leonid Chindelevitch July 13, 2008 1 Tuesday, July 8, 2008 1. Problem: Given a circle ω , construct its center O using only a compass. Solution: Let r be the (unknown) radius of ω . Take a point A on ω , and draw a circle ω 1 with center A and radius R between r/ 2 and 2 r ; call the two intersection points with ω B and B , respectively. Draw two circles with centers B and B and radius AB = AB , respectively. Call C their intersection point different from A . Draw a circle with center C and radius AC , call its two intersection points with ω 1 D and D , respectively (so that B and D lie on the same side of AC ). Finally, draw two circles with centers D and D and radius AD = AD , respectively. Call X their intersection point different from A . We claim that X is the center O of ω . Proof: By construction, A,B,B lie on ω , so OA = OB = OB = r . Again by construction, AB = AB = BC = BC = R . Since all three points A,O,C are equidistant from B and B , they lie on the perpendic- ular bisector of BB . The isoceles triangles AOB and ABC are similar (because they have a common angle ∠ OAB = ∠ BAC at the base), so AO AB = AB AC = ⇒ AC = AB 2 AO = R 2 r . By construction again, we have AD = AD = DX = D X = R , and CA = CD = CD = R 2 r . By the same argument as above, we see that triangles ACD and ADX are similar, so AC AD = AD AX = ⇒ AX = AD 2 AC = R 2 R 2 /r = r . But X also lies on the perpendicular bisector of BB , so X = O ....
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Emergency problems #1 - MATH 348 Solved Emergency Problems...

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