Tuesday, November 13, 2018

Geometry Problem 1400: Triangle with three rectangles on the sides, Perpendicular Bisectors, Concurrency

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1400: Triangle with three rectangles on the sides, Perpendicular Bisectors, Concurrency, Tutoring.

4 comments:

  1. Let KN & LP meet at O

    Let OB = u, OA = v & OC = w
    Let OE = OJ = a, OD = OJ = b, OH = c & OF = d

    Using Apollonius for Tr. OBE & ODJ,

    2.OP^2 + 2.BP^2 =
    a^2 + u^2 = b^2 + v^2 ...(1)

    Similarly

    b^2 + w^2 = d^2 + u^2 ...(2) and
    c^2 + v^2 = a^2 + w^2 ....(3)

    Adding (1), (2), (3) gives c = d

    So MQ must pass thro O

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. It is obvious that Triangles BGD, CFH, AJE makes one triangle if the opposite sides of the rectangles (that are not side of the original triangle) are overlapped. In this new triangle, the perpendicular bisectors overlap the KN, LP and MO rays because the sides of the triangle are only pushed away from the intersetion point without changing their direction and length. Because, the perpendicular bisectors of the triangle concur at one point, these rays also concur at one point.

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