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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11 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

    ReplyDelete
    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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  5. We can use trigonometry to find the length of segment AD, then we'll use the Pythagorean Theorem on triangle ABD to find the length of AB.

    Segment AD is adjacent to angle a. Segment DC, the measure of which we know, is opposite to that angle. So, we can use the tan trig ratio... tan(x) = 3/y. If you're confused as to where I'm going with this so far, stay tuned. Segment AD is opposite to angle 2a, and we know the measure of segment BD. We can once again use tan: tan(2x) = y/4.
    We are now left with two equations:
    tan(x) = 3/y
    tan(2x) = y/4
    If we apply the inverse of tan to both of them, then we are left with:
    x = tan^-1(3/y)
    2x = tan^-1(y/4)
    This means that the second one is two times the first (note the x), and we can combine them into one equation:
    2tan^-1(3/y) = tan^-1(y/4)
    If we plug this into our good friend Mathway.com, we find y to be approximately 5.745. That is the measure of segment AD. We now have the two legs of triangle ABD: 4 and 5.745. We find the hypotenuse by solving 4^2 + 5.745^2 = c^2. c is approximately 7.0004.
    AB = 7.0004.

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  6. Simple Solution
    In Triangle ACD, < ACD = 90 - @
    So in Triangle ABC, < BAC = 180 - 2@ - (90 - @) = 90 - @
    So < BAC = < BCA
    Hence BA = BC = 3 + 4 = 7

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Solution 3

    Extend CB to E such that AB = BE
    Then < BAE = < BEA = @ = < CAD (since < ABC = 2@)
    Hence Triangles ACD & AEC are similar (having angles @, < ACD)
    and so < CAE = < ADC = 90

    So B is the centre of Triangle ACE and AB = BC = 3+4 = 7

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete