Thursday, April 23, 2009

Problem 283: Circular Sector 90 degrees, Semicircle, Circle inscribed, Radius

Proposed Problem

 Problem 283: Circular Sector 90 degrees, Semicircle, Circle inscribed, Radius.

See also: Complete Problem 283, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry


  1. With O as origin, the smallest circle with its centre at (r,h) is represented by (x-r)^2+(y-h)^2=r^2 while the circle with centre at C is:
    (x-R/2)^2+y^2=R^2/4. These two equations when solved simultaneously give: x=(2r+R)h^2 + or -2h^2(2rR-h^2)^(1/2)/(4(h^2+r^2)+R^2-4rR).
    For them to be touching each other at one point (and not intersect) we can conclude that h^2 =2rR. Like wise, the two circles given by:(x-r)^2+(y-h)^2=r^2 and x^2+y^2=R^2 are touching at a point if only if -r^2h^4+2(rRh)^2-r^2R^4+4r^4R^2 =0 where h^2 = 2rR gives us, -4r^4*R^2 + 4r^4*R^2 + 4r^3*R^3 - r^2*R^4 = 0 or 4r^3*R^3 = r^2*R^4 or r = R/4

  2. can someone give a better answer involved with circle geometry please!!! urgent!!!

  3. i need like a step by step way to complate this + reasoning.. i dont really understand the above comment... thanks :)

  4. Applying Euclid's Book II, 13th Proposition to Tr. OCD:(R/2+r)^2 =(R/2)^2+ OD^2 - rR. Now substitute OD = (R - r) in this equation to obtain: R = 4r,QED.

  5. If X is the foot of the perpendicular from D to OC, DX^2 =
    (R-r)^2 - r^2 = (r + R/2)^2 - (R/2 - r)^2

    Upon simplifying
    r = R/4

    Sumith Peiris
    Sri Lanka

  6. See the drawing

    F the tangent intersection of CirleD and CircleO
    E intersection of CircleD and OA
    G intersection of CircleD and CircleC
    Define the line L tangent of CircleD and CircleO at F
    F in CircleO => OF ⊥ L, F in CircleD => DF ⊥ L
    => O, D and F are aligned

    Define H intersection of ED and CircleD
    ED⊥OA => ED//OB
    => ∠FDH=∠DOC

    Draw semi circle with center O and define I intersection of OB and the semi circle
    F and I on CircleO => ∠FOB=2 ∠FIB
    F and E on CircleD => ∠FDH=2 ∠FED
    ∠FDH=∠FOB => ∠FIB = ∠FED and F, E , I are aligned

    D, G and C are aligned and DH//CO=> ∠DCO = ∠CDH
    => ∠GED = ∠GBC
    ED//BC => E, G and B are aligned

    OI=OB and OA⊥IB => ∠EIO = ∠EBO
    OB//ED =>∠EBO= ∠BED

    => ∠GED=∠FED
    =>∠GDH=2∠GED and ∠GCO=2∠GBO
    ∠GED=∠FED => ∠DCO=∠DOC
    => ODC is isosceles => DO=DC

    DO=R-r and DC=r+R/2
    Therefore r=R/4