Proposed Problem
See complete Problem 276 at:
gogeometry.com/problem/p276_square_circle_90_degree_arc.htm
Level: High School, SAT Prep, College geometry
Thursday, April 2, 2009
Problem 276: Square, 90 degree Arcs, Circle, Radius
Subscribe to:
Post Comments (Atom)
If we've two circles of radii r1 & r2 that are tangent to each other then:
ReplyDelete(common tangent)^2 = (r1+r2)^2 -(r1+r2)^2
= 4r1r2
In this case, the common tangent = a/2 while r1 =x and r2 =a. Thus, (a/2)^2 = 4xa or a^2/4 =4xa
or x=a/16
QED
Ajit: ajitathle@gmail.com
Apply Pythagoras to Tr. ADM where.M is the centre of the circle with radius x.
ReplyDelete(a + x)^2 = (a-x)^2 + a^2. from which it follows that x = a/16.
Sumith Peiris
Moratuwa
Sri Lanka
you mean : (a + x)^2 = (a-x)^2 + a^2/4
ReplyDeleteYes. Sorry for the mistake
Deletehttps://photos.app.goo.gl/jjyMxzP3bdYFvRGy6
ReplyDeleteSee the drawing
ReplyDeleteWithout using Pythagoras!
Define G the center of CircleG with radius x
Define the point E as the only point of intersection between CircleG and CircleA => A, E and G are aligned
Define K in BC such as KE is tangent to CircleA
BC is tangent of circleA on B =>CBE intercepts arc BE of circleA =>2 ∠CBE= ∠BAE
EK is tangent of circleA on E =>KEB intercepts arc BE of circleA
=> ∠KEB= ∠KBE => BKE is isosceles in K => KB=KE and ∠CKE= 2 ∠KBE
Define F on BC such as F belongs to CircleG => GF⊥BC
=> KB=KE and KE⊥GE and KF⊥FG and GF=GE=x
KF is tangent to CircleG and KE is tangent to CircleG => KE=KF
=> ΔKFG is congruent to ΔKEG (SAS)
=> ∠FKG= ∠EKG => ∠FKE= 2 ∠FKG => ∠FKG= ∠KBE
ΔAEK is congruent to ΔABK (SAS) => ∠ABK= ∠FKG
=>ΔABK is similar to KFG => AB/BK=KF/FG
By symmetry F is in the middle of BC, BK=KF => BK=AB/4=a/4
AB/BK=4=KF/FG =>KF=4FG, KF=AB/4=>KF=AB/16
Therefore x=a/16