Friday, January 17, 2025

Geometry Problem 1589: Prove the Ratio of BD to OD is Square Root of Two in a 90° Circular Sector

Challenging Geometry Puzzle: Problem 1589. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1589

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4 comments:

  1. https://photos.app.goo.gl/h3uTtaoAeAgFZffF6

    Define point P and Q
    Due to symmetry we have
    Arc(AC)=Arc(PQ ) and Arc(BC)=Arc(BP)

    Angle CAD= 45=> Arc(BC)=Arc(BP)=45
    And Arc(AC)=Arc(PQ)=45

    AD become angle bisector of angle OAB
    So BD/OD= BA/OA= sqrt(2)

    ReplyDelete
  2. Let AC = DC = a, OD = b and radius = R

    So AD = V2.R

    OADC is a concyclic quadrilateral and so using Ptolemy

    ab + aR = V2.aR from which
    b = (V2-1).R
    and so R = (V2 + 1).b (using conjugate of V2 - 1)

    Now BD/OD = (R-b)/b = R/b - 1

    Therefore BD/OD = (V2 + 1) - 1 = V2

    (Using Trigonometry of course is a shortcut since we can see that < ACO = < ADO = 67.5 and tan 67.5 = V2+1)

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. 2nd Geometry Solution

    AC = CD so < ADC = 45

    Since AODC is concyclic , < AOC = < ADC = 45
    Since AO = CO, < ACO = 67.5 = < ADO
    Hence < OAD = 22.5 = < BAD
    So AD bisects < OAB

    Therefore BD / OD = AB / AO = V2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. 3rd Geomtery Solution using similar Triangles

    Let radius = R and AC = DC = a

    Triangles OCD and BAD are similar having angles 45 and 22.5 (see my solution 2 above)

    So BD / OD = V2a / a = V2R / R = V2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete