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

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## Thursday, April 2, 2009

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Problem 276: Square, 90 degree Arcs, Circle, Radius

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Proposed Problem

See complete Problem 276 at:

gogeometry.com/problem/p276_square_circle_90_degree_arc.htm

Level: High School, SAT Prep, College geometry

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

ReplyDeletedrawingWithout 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