Saturday, July 8, 2023

Geometry Problem 1547: Tangents' Dance: Exploring B-to-AC Distance in a Circle's Grasp

Geometry Problem 1547. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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Geometry Problem 1547: Tangents' Dance: Exploring B-to-AC Distance in a Circle's Grasp

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2 comments:

  1. Let BD cut the circle at E. Let F be the midpoint of BE. Let BF = y and OF = z

    OFDC is a Rectangle. Triangles OBF & ABD are similar

    So 4/5 = y/(5-z) = z/(y+4) which gives z = 4(y+4)/5
    Substituting for z, y/(5 - 4(y+4)/5)) = 4/5

    Simplifying y = 36/41

    Therefore BD = 4 + 36/41 = 4.89 approximately

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. From right triangle OBA, tan(angleBAO) = 4/5, so BAO = atan(4/5)
    Now, BAC = 2 BAO and from right triangle BDA,
    BD = 5 sin(BAC) = 5 sin(2*atan(4/5) = 4.878

    Christos

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