This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 , construct its center O using only a compass. Solution: Let r be the (unknown) radius of . Take a point A on , and draw a circle 1 with center A and radius R between r/ 2 and 2 r ; call the two intersection points with 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 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 . Proof: By construction, A,B,B lie on , 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 ....
View
Full
Document
 Summer '06
 Karigiannis
 Math, Geometry

Click to edit the document details