Sunday, October 23, 2011

Problem 680: Concentric Circles, Radii, Chords, Perpendicular, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 680.

Online Geometry Problem 680: Concentric Circles, Radii, Chords, Perpendicular, Metric Relations

Posted by Antonio Gutierrez at 3:29 PM
Labels: , , , ,

9 comments:

  1. HenkieOctober 24, 2011 at 7:28 AM

    CD = b-a
    ED is diameter of the small circle (Thales).
    So by Pythagoras: (2r)² =(b-a)² + c², giving a² + b² + c² = 4r² + 2ab (result 1)

    Let F be the orthogonal projection of O to AB.
    AF = ½(a+b).
    Triangle DEC is similar to triangle DOF, so OF = ½c.
    Pythagoras in triangle OAF:
    R² = (½(a+b))² + (½c)², giving a² + b² + c² = 4R² - 2ab (result 2)
    Adding results 1 and 2 gives:
    2(a² + b² + c²) = 4r² + 4R²
    So: a² + b² + c² = 2(r² + R²)
    QED

    ReplyDelete
  2. Peter TranOctober 24, 2011 at 9:19 AM

    http://img24.imageshack.us/img24/8664/problem680.png
    Let F and G are projections of O over CE and AB
    F and G are the midpoints of CE and AB
    We have OF^2=r^2-(c/2)^2 and OG^2=R^2-(a+b)^2/4
    OF^2+OG^2=r^2=r^2-c^2/4+R^2-a^2/4-b^2/4-a.b/2
    or a^2+b^2+c^2=4R^2-2.a.b (1)
    Note that a.b= -Power of point C to Outer circle , radius R
    So a.b= -(OC^2-R^2)= R^2-r^2
    Replace this value in equation (1) we will get a^2+b^2+c^2=2(R^2+r^2)
    Peter Tran

    ReplyDelete
  3. AjitOctober 25, 2011 at 9:13 PM

    Just want to prove that a*b=R^2-r^2 in a slightly different way. Draw a line thru' A & O meeting the inner circle in P & Q resply. Now AP*AQ=AC*AD or (R-r)*(R+r)=a*b since AD=BC=b. Hence, R^2-r^2 =a*b. And (2r)^2=(b-a)^2+c^2 from rt. angled Tr. ECD. Hence etc.
    Ajit

    ReplyDelete
  4. PravinOctober 26, 2011 at 7:31 PM

    Let M be the midpoint of CD (AB also).

    Let CD = 2d.

    AM = MD implies a + d = b - d, 2d = b - a.

    EOD is a diameter.

    Easy to note

    4r^2 = c^2 + 4d^2 = c^2 + (b - a)^2

    Also

    R^2 - r^2 = AM^2 - CM^2

    = (a + d)^2 - d^2 = a^2 + 2ad

    = ab.

    Hence

    2(R^2 + r^2)

    = 2(R^2 - r^2) + 4r^2

    = 2ab + c^2 + (b - a)^2

    = a^2 + b^2 + c^2

    ReplyDelete
  5. UnknownMarch 5, 2013 at 5:32 AM

    draw a perpendicular to AB from center intersecting AB at F
    therefore by theorem AF=BF=a+b/2
    CF=AF-CA=b-a/2
    draw OG perpendiculr to EC from center meeting EC at G
    therefore EG=GC=OF=c/2 (OGCF is rectangle)
    in tri(OCF) by phythagors theorem (Pt)
    4r sq = c sq + a sq -2ab + b sq .......... 1
    in tri(OAF) by Pt
    4R sq = c sq + b sq+ a sq +2ab ...............2
    1 + 2
    we get the required result

    ReplyDelete
  6. PravinMarch 6, 2013 at 9:21 AM

    Draw ON perp to AB. ON bisects AB (resp CD)
    AD = AN + ND = NB + NC = BC = b
    DE is a diameter.
    In triangle AED, AO is median.
    So AE^2 + AD^2 = 2(AO^2 + OE^2)
    (a^2 + c^2) + b^2 = 2(R^2 + r^2) etc

    ReplyDelete

Subscribe to: Post Comments (Atom)