Friday, June 26, 2009

Problem 310: Circle Inscribed in a Semicircle, 45 degree Angle

Proposed Problem
Click the figure below to see the complete problem 310.

 Problem 310: Circle Inscribed in a Semicircle, Angle.
See also:
Complete Problem 310
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry


  1. With O as the origin we have A:(-R,0) and B:(R,0). Let D be (p,0) and inner circle have a radius r; hence C is (p,r). Since the two circles touch each other at E we can show that R^2-p^2=2rR, by solving the two equations simultaneously and setting the discriminant=0 to obtain a single point as the solution. Now if we substitute r=(R^2-p^2)/2R in the equation above we can determine E as [2pR^2/(R^2+p^2),R(R^2-p^2)/(R^2+p^2)]. Further by the distance formula, AE^2 =4pR^3/(R^2+p^2)+ 2R^2 while BE^2= 2R^2 - 4pR^3/(R^2+p^2) Hence, AE^2/BE^2 = (R+p)^2/(R-p)^2 = AD^2/DB^2 since AD=R+p and DB=R-p. In other words, AE/BE = AD/DB or D divides AB in the ratio of the two other sides of triangle ABE. Hence we can say DE biscts angle AEB which 90 deg since AB is a diameter of the larger circle. Hence angle AED=45 deg.
    QED. Ajit:

  2. Forgot to mention that the equations of the two circles above are: x^2 + y^2 = R^2 and (x-p)^2 +(y-r)^2 = r^2. Also, the discriminant referred to is: -p^6 + 2(R^2)(p^4)-(R^4)(p^2) + (2rRp)^2 which when equated to 0 gives us: R^2- p^2 = 2rR.
    If there's another easier way to obtain this condition, will someone please show me the same?

  3. Dear Ajit,
    Another way:
    1. Isosceles triangle DCE; therefore...
    2. Isosceles triangle AOE; therefore...
    3. Right triangle ODC; therefore...


  4. Yes, indeed!
    O,C & E are collinear and OC=R-r while OD=p and CD=r which gives us (R-r)^2 = p^2 + r^2 from where R^2 - p^2 = 2Rr.
    Thanks, Antonio.

  5. This problem can be done with plane geometry!
    Auxiliary Construction Hint:
    Draw tangent line at E; extend Ray ED and Arc EA turning counterclockwise till they intersect at F; and finally connect AF and BE.

  6. And furthermore, connect D and the intersection of AE and circle C (call the intersection G).

  7. Draw full circle diameter AB and a radius OF perpendicular to diameter AB ( F on extended portion of the circle)
    DE intersect OF at F’ . We will prove F’ coincide with F and angle AED face 90 degrees arc.
    We have CD perpendicular to AB and points O, C and E are collinear ( circle C tangent to dia. AB and circle O)
    Since CD //OF’ so triangles CDE similar to triangle OF’E ( Case AA)
    And triangle OF’E is isosceles
    So OF’=OE=OF= radius of circle O and point F’ coincide with F and angle AED=45

    Peter Tran

  8. Let < OAE = < OEB = p. Then < EOB = 2p.
    Now O, C, E are collinear, DC is perp. To OB and < DCE is isoceles.
    Hence < CDE = alpha- p and < EDB = alpha + p. Hence < CDB = 2. Alpha = 90, so alpha = 45.

    Sumith Peiris
    Sri Lanka

  9. Problem 310
    The points O, C, E are collinear.Let <OEA=<OAE=x, <CED=<CDE=y. But CD, AE are perpendicular in BD , EB respectively.Then <BDE=2x+y=90-y or 2x+2y=90 so
    x+y=45.But <AED=x+y=45.

  10. Very Simply,
    DC = CE so < OCD = 2 X < OED ....(1)
    OE = OA so < EOD = 2 X < AEO ....(2)

    (1) + (2) < OCD + < EOD = 90 = 2 X (< OED + < AED) = 2 Alpha

    Sumith Peiris
    Sri Lanka