Saturday, February 18, 2023

Geometry Problem 1507: Understanding Triangle Geometry: Double Angles, Altitudes, and Measurements. Difficulty Level: High School.

Geometry Problem 1507. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1507: Understanding Triangle Geometry: Double Angles, Altitudes, and Measurements. Difficulty Level: High School.

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8 comments:

  1. According to the Diagram <B = 2@.....................Case 1
    According to the Data given <B = @...................Case 2

    Case 1:-
    Let BE bisect <B, E on AC
    Hence ABDE is concyclic and BE is perpendicular to AC
    So AB = BC = 3+4 = 7

    Case 2:-
    <CAD = <B = @
    So AC is tangential to circle ABD at A
    Hence AC = sqrt (3 X 7) = sqrt21 and so
    AD = sqrt (21 - 9) = sqrt12
    Therefore AB = sqrt(12 + 16) = sqrt 28 = 2.sqrt7

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. Thank you, Sumith, for your solutions and observation. To unify the idea, the graph has been modified to address Case 1.

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  2. Tr ABC => isosceles => AB=7 (the sketch will come soon)

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    Replies
    1. https://photos.app.goo.gl/QSZMRAiAmrBdUsbQA

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    2. https://photos.app.goo.gl/v4gnknaopkZNweet5

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  3. Thank you Peter for your prompt solution

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  4. 1st solution: Let CE be the C- altitude, H - orthocenter; BEHD being cyclic, <CHD=<B=2@, or <ACH=@, AB=BC.

    2nd solution: Take F reflection of C about D, AC is tangent to the circle AFB and from p.o.p. AC^2=CF.BC= 6.7 ; by Pithagoras AD^2=AC^2-BD^2=42-9=33 and AB^2=AD^2+BD^2=33+16=49, AB=BC=7

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