See complete Problem 62
Square diagonal and Inscribed Circle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 62
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See complete Problem 62
Square diagonal and Inscribed Circle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Let A:(0,0),B:(0,s),C;(s,s) and D:(s,0) with s =side of square. Therefore,O:(s/2,s/2) & Arc BD: x^2+y^2=s^2 and the circle:(x-s/2)^2 + (y-s/2)^2 =s^2/4. E is:[(s(5 -V7)/8,s(V7+5)/8].
ReplyDeleteNow,CE^2=[(s-(s(5-V7)/8)]^2+[s-s(V7+5)/8)]^2
= s^2/2 or CE =s/V2
But BD^2 = 2s^2 or BD = V2s which gives the desired result viz. CE = BD/2
Ajit: ajitathle@gmail.com
http://henrik-geometry.webs.com/62.htm
ReplyDeleteLet AB = a so that OE = a/2 & OC = a/sqrt2
ReplyDeleteApply Apollonius to Tr. AEC
a^2 + CE^2 = 2 (a^2/2 + a^2/4) from which
CE = a/sqrt2 = OC
Sumith Peiris
Moratuwa
Sri Lanka
AE^2=AO.AC, Hence triangle ACE is similar to triangle AEO. We have CE/AC=OE/AE,Since OE/AE=1/2, We get CE=AC/2.
ReplyDelete