Sunday, October 29, 2023

Geometry Problem 1558: Three Equilateral Triangles

Challenging Geometry Puzzle: Problem 1558. 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 1558: Three Equilateral Triangles

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1 comment:

  1. Perpendiculars from C1, C and C2 to B1B2 (extended to R) meet B1B2 at P, Q and R respectively.
    Let C1B1= a and CB2 = b.
    Let B1P = t1, PB = t2, BQ = t3, QB2 = t4 and B2R = t5.
    Let C1P = u and C2R = v

    C1PRC2 is a trapezoid with C1P//CQ//C2R, so by connecting C2P and applying the midpoint theorem,
    CQ = (u+v)/2 ..........(1) and
    t2 + t3 = t4 + t5 ......(2)
    Now it is given that t1 + t2 = t3 + t4....... (3)
    From (2) and (3), t3 = (t1 + t5) / 2..........(4)

    Now consider Right Triangles B1C1P, BCQ and B2C2R
    CQ = (u + v)/2
    BQ = t3 = (t1+ t5)/2
    and the included angle is 90

    So BC = (a + b)/2
    Similarly we can show that CA = (a+b)/2 and AB = (a+b)/2

    Therefore Triangle ABC has BC = CA = AB and is thus equilateral

    Sumith Peiris
    Sri Lanka