Thursday, April 2, 2009

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

Proposed Problem

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

See complete Problem 276 at:

Level: High School, SAT Prep, College geometry


  1. If we've two circles of radii r1 & r2 that are tangent to each other then:
    (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

  2. Apply Pythagoras to Tr. ADM where.M is the centre of the circle with radius x.

    (a + x)^2 = (a-x)^2 + a^2. from which it follows that x = a/16.

    Sumith Peiris
    Sri Lanka

  3. you mean : (a + x)^2 = (a-x)^2 + a^2/4


  5. See the drawing

    Without 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