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

9 comments:

  1. 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

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

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    2. 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

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    3. 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 ....

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

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  3. 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

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

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  5. 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

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