Proposed Problem

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Complete Problem 316

Level: High School, SAT Prep, College geometry

## Wednesday, July 8, 2009

### Problem 316: Circular segments and Inscribed Squares

Labels:
circular segment,
perpendicular,
side,
square

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Draw OQ parallel to AB meeting EF in Q. Thus, OQ=a/2 and QF=a-x. We've ON^2=(b/2)^2+(b+x)^2 & OF^2=(a/2)^2+(a-x)^2 but ON = OF being the radii of the circle. So,(b/2)^2+(b+x)^2=(a/2)^2+(a-x)^2 which, on simplification, directly yields x=5(a-b)/8

ReplyDeleteOG^2 = (a-x)^2 + a^2/4 and OM^2 = (x+b)^2 + b^2/4

ReplyDeleteSince the Radius = OG = OM, (a-x)^2 + a^2/4 = (x+b)^2 + b^2/4

Hence 2x(a+b) = (a^2 - b^2)/4

Dividing by (a+b), x = (5/8)(a+b)

Sumith Peiris

Moratuwa

Sri Lanka