Thursday, November 21, 2013

Geometry Problem 936: Circle, Semicircle, Diameter, Tangent, Radius, Chord, Metric Relations

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

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Online Geometry Problem 936: Circle, Semicircle, Diameter, Tangent, Radius, Chord, Metric Relations

Posted by Antonio Gutierrez at 8:30 PM
Labels: , , , , , ,

5 comments:

  1. Jacob HA (EMK2000)November 22, 2013 at 6:35 AM

    Enlarge circle Q with center E, such that Q→O, (Homothetic transformation)
    let D→F, then EQO, COF and EDF are straight lines, with EF=7.

    Since then CF is the diameter of circle O, thus ∠CED=∠CEF=90°.

    Using Pythagoras theorem, we have
    CE² = 4²−3² = 7
    CF² = CE² + EF² = 7 + 49 = 56
    So OE = OF = √14

    Since ΔEQD~ΔEOF,
    x/√14 = 3/7
    x = (3/7)√14

    ReplyDelete
  2. AnonymousNovember 22, 2013 at 2:28 PM

    OC^2=DC*(DC+DE)/2
    QE=OC*DE/(DC+DE)=(DC*(DC+DE)/2)^(1/2)*DE/(DC+DE)
    =(DC/2)^(1/2)*DE/(DC+DE)^(1/2)
    =(2)^(1/2)*3/(7)^(1/2)
    =(14)^(1/2)*(3/7)

    ReplyDelete
  3. Ivan BazarovDecember 1, 2013 at 11:44 AM

    If <BDE=a then <OED=90-a and <EOD=2a-90. By circle theorem if <EOD=2a-90 then <BCE=a-45 and <ECO=(a-45)+45=a=<EDB making CEDO cyclic. That makes <DEC=90 and CE=sqrt7 by Pythagoras. If QD cuts CE at P then Q is cicrumcenter of PED so QD=1/2*DP.

    By angle bisector PE=3/sqrt7 so DP=6*sqrt(2/7) by Pythagoras. QD is half so QD=3*sqrt(2/7)

    ReplyDelete
  4. Sumith PeirisJanuary 30, 2019 at 10:28 AM

    Super!

    ReplyDelete
  5. Sumith PeirisFebruary 18, 2020 at 10:21 AM

    Trigonometry Solution

    Draw the Common Tangent at E, XEY

    If < XEC = p, then < COE = 2p = < OQD, so < QED = p = < XEC whence < CED = 90

    So Cos(2p) = 3/4. From Tr. QED, cos (p) = 3/2x

    Now cos (2p) = 2cos^2 (p) - 1
    Hence 9/(2x^2) - 1 = 3/4

    Solving, x = 3.sqrt(2/7)

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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