Wednesday, January 15, 2025

Geometry Problem 1588: Unlocking the Tangent: Prove the Fifth of the Radius in a Sector with Semicircle

Challenging Geometry Puzzle: Problem 1588. 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 1588

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Posted by Antonio Gutierrez at 7:07 PM
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2 comments:

  1. Sumith PeirisJanuary 16, 2025 at 4:15 AM

    Trigonometry Solution

    Let < OAQ = @ so < OAC = 2@ (since Tr.s OAQ & CAQ are congruent SSS)

    tan @ = 1/2 = t (say)

    Triangle AOD is isosceles, so

    AD = 2r.cos2@ = 2r. (1-t^2)/(1+t^2) = 2r. (1-1/4)/(1+1/4) = 6r/5

    Therefore CD = AD - AC = 6r/5 - r = r/5

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Sumith PeirisJanuary 16, 2025 at 7:30 AM

    Geometry Solution

    Let U be the mid point of isosceles triangle OAC
    Let AQ & OC intersect at V

    S(OAQ) = 1/2 X r X r/2 = 1/2 X AQ X OV
    So OV = (r^2 / 2) / (V5 X r/2 ) = r /V5
    OC = 2.r/V5 & AV = 2.r/V5

    2.S(OAC) = OU.r = AV.OC
    Hence OU = (2.r/V5).(2.r/V5) / r = 4r/5

    So AU = 3r/5
    and CU = r - 3r/5 = 2r/5

    Therefore CD = 3r/5 - 2r/5 = r/5

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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