Friday, June 13, 2014

Geometry Problem 1021: Triangle, Cevian, Double Angle, Sides

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 1021.

Online Math: Geometry Problem 1021: Triangle, Cevian, Double Angle, Sides

Posted by Antonio Gutierrez at 2:21 PM
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11 comments:

  1. UnknownJune 15, 2014 at 5:17 AM

    Draw CH prenpedicular to AB.
    Let AH = x.
    By pythagorean theorem AC^2 - AH^2 = CB^2 - BH^2 ----> B^2 - x^2 = (b + c/2)^ - ( c- x)^2
    From short develop finally we get AH = x = 3c/8 - b/2
    Now draw isoceles triangle CDP (continue AP ....)
    Notice that:
    1) <A = <CDP + <CPD = 2*<APC

    ReplyDelete
    Replies
    1. Antonio GutierrezJune 15, 2014 at 7:19 AM

      D is on AP.

      Delete
    2. UnknownJune 15, 2014 at 9:01 AM

      Yes , sure but the solution is right......?

      Delete
    3. UnknownJune 15, 2014 at 9:08 AM

      I mean DH = PH = 3c/4 - x = 3c/4 - 3c/8 +b/2 = 3c/8 + b/2
      then DA = DH - x = 3c/8 + b/2 - 3c/8 + b/2 = b
      from here CA = DA = b ----> <CDP + <CPD = 2*<APC = <A

      Delete
    4. Antonio GutierrezJune 15, 2014 at 11:18 AM

      I understand the idea of your solution.
      Your step AH = x = 3c/8 - b/2 is OK
      but your step: "I mean DH = PH = 3c/4 - x = ..."
      should be: DH = AH then triangles AHC and DHC are congruent (SAS) then CD = b then ....

      Delete
  2. IftikharJune 18, 2014 at 1:08 AM

    Does this problem have a purely geometrical proof? I proved it using trigonometry.

    ReplyDelete
  3. UnknownOctober 6, 2014 at 11:41 PM

    Applying Stewart's theorem,(cp=x)
    b^2.c/4 + (b+c/2)^2.3c/4=c(x^2 +3c/4.c/4)
    =>b^2+(b+c/2)^2.3=4(x^2+3c^2/16)
    =>b^2+3b^2+3bc+3c^2/4=4x^2 +3c^2/4
    =>b^2+3/4.bc=x^2 =>x^2=b(b+3c/4)
    ,',In ΔAPC, x^2=b(b+3c/4). By triangle properties, <A=2<APC

    ReplyDelete
  4. AnonymousJanuary 3, 2015 at 3:36 PM

    See my synthetic solution at my page https://www.facebook.com/photo.php?fbid=784793084934933&set=a.784793394934902.1073741826.100002127454113&type=1&theater

    ReplyDelete
  5. AnonymousMarch 6, 2015 at 5:48 PM

    with C as center, radius b, make a circle O, circle O intercept AB at E, intercept CB at F;
    Extend BC to intercept circle O at G.
    CA=CE, /_ CAB = /_CEA=2x
    so BA*BE=BF*BG; BA=c,let BE=y, BF=1/2c, BG=2b+1/2c;
    so BF= b+1/4c,
    so PE=b,
    so trangle CEP is isoceles,
    so /_ CAB =2 /_CPA

    ReplyDelete
  6. Sumith PeirisMay 10, 2017 at 8:09 AM

    Extend PA to X such that XA = b. Let Y be the foot of the perpendicular from C to AP. Let AY = p.

    Then b^2 - p^2 = (b+c/2)^2 - (c-p)^2 from which p = 3c/8 - b/2 and PY = 3c/8+b/2

    So XY = PY and hence XC = CP and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisMay 10, 2017 at 9:01 AM

    Solution 2

    Mark Z on AP such that CZ = b
    Then AZ = 2p = 3c/4 - b (from my previous proof) and so PZ = b = ZC
    Hence < CZA = 2P = A

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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