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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Posted by Antonio Gutierrez at 10:06 AM
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2 comments:

  1. Peter TranJanuary 17, 2025 at 7:32 PM

    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)

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  2. Sumith PeirisJanuary 17, 2025 at 8:25 PM

    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

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