Monday, August 7, 2023

Geometry Problem 1550: Solving for Segment BD: An Angle Puzzle in Right Triangle ABC

Challenging Geometry Puzzle: Problem 1550. 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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Geometry Problem 1550: Solving for Segment BD: An Angle Puzzle in Right Triangle ABC

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8 comments:

  1. Let AD & AE extended meet circle ABC (with diameter AC) at P & Q respectively.
    Draw DM perpendicular to BP, M on BP

    BP = CQ (both subtend 20 degrees on the circumference) = y, say

    < ACB = < DBP = < DPB = 35 and so,
    Triangle BDP is isosceles and hence BM = y/2

    Triangles BDM & CEQ are similar (90,35,55) and so
    BD/BM = CE/CQ

    Therefore BD/(y/2) = 3/y which yields
    BD = 3/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Trigonometry Solution

    Let tan35 = t

    BD = c.tan20 = c/tan70 = c./(tan(2X35) =c.(1-t^2)/2t........(1)
    BE = c.tan35 = ct.......(2)
    BC = c.tan55 = c/tan35 = c/t .......(3)
    (3) - (2); CE = BC - BE = c.(1/t - t) from (2) & (3)
    So c.(1-t^2)/t = 3......(4)
    From (1) & (4) BD = 3/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. The trigonometric solution is beautiful because it is simple. The geometric solution is beautiful because it makes ingenious auxiliary constructions.

    ReplyDelete
  4. 2nd Geometry Solution - much simpler

    Let FA be perpendicular to AC, F on CB extended
    Let BD = x

    Now FD = DA = DC so DE = a - x - 3 and FB = a - 2x
    FB = BE = a - 2x = BD + DE = x + (a - 3 - x) = a - 3

    Hence 2x = 3 and x = 3/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Nice solution by creating the right triangle FAC.

    ReplyDelete
  6. Notice that AD=CD and, if ADCF is a rhombus, F is the circumcenter of triangle AEC, <FCE=<ADB, FC||=AD, G, projection of F onto CB is midpoint of EC, while triangles CFG and DAB are congruent, so CG=BD, done.

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