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Math Help - Point in triangle...

  1. #1
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    Point in triangle...

    Triangle ABC with BC=10, AC=11 and AB = 12.

    Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.

    What are lengths od AP, BP abd CP?

    Work so far:
    area of ABC = 33SQRT(39) / 4
    triangle APB : height from P = (33SQRT(39)/2) / 36
    triangle APC : height from P = (33SQRT(39)/2) / 33
    triangle BPC : height from P = (33SQRT(39)/2) / 30

    From that, how would you calculate lengths of AP, BP and CP?
    I keep going around in circles! Thanks.
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  2. #2
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    Quote Originally Posted by Wilmer View Post
    Triangle ABC with BC=10, AC=11 and AB = 12.

    Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.

    What are lengths od AP, BP abd CP?

    Work so far:
    area of ABC = 33SQRT(39) / 4
    triangle APB : height from P = (33SQRT(39)/2) / 36
    triangle APC : height from P = (33SQRT(39)/2) / 33
    triangle BPC : height from P = (33SQRT(39)/2) / 30

    From that, how would you calculate lengths of AP, BP and CP?
    I keep going around in circles! Thanks.
    Maybe this will help

    Median (geometry) - Wikipedia, the free encyclopedia (subheading Equal-area division)
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  3. #3
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    Quote Originally Posted by Wilmer View Post
    Triangle ABC with BC=10, AC=11 and AB = 12.

    Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.

    What are lengths od AP, BP abd CP?

    Work so far:
    area of ABC = 33SQRT(39) / 4
    triangle APB : height from P = (33SQRT(39)/2) / 36
    triangle APC : height from P = (33SQRT(39)/2) / 33
    triangle BPC : height from P = (33SQRT(39)/2) / 30

    From that, how would you calculate lengths of AP, BP and CP?
    I keep going around in circles! Thanks.
    Alternatively, you could play with co-ordinate geometry.
    Placing B at the origin, C is at (10,0).
    Then locate the co-ordinates of A.
    You've worked out the triangle areas.
    You've got the height of P from the line [BC].
    Then P lies on the line y=h.
    The intersection of the line parallel to [AB] with appropriate perpendicular distance from [AB], with y=h,
    or the intersection of the line parallel to [AC] with y=h, correct distance from [AC]
    locates P.
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  4. #4
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    Had a closer look at this; turns out easy enough.

    We have triangle ABC with BC = a =10, AC = b = 11 and AB = c = 12 ; point P inside.

    STEP#1:
    Make B(0,0) and A(12,0) ; and C(g,h)
    Get the area k.
    h = 2k / c
    g = SQRT(a^2 - h^2)

    STEP#2:
    Make P(x,y)
    x = (c + g) / 3
    y = h / 3
    NOTE: since 3 inside triangles equal in area, then P is the centroid!

    STEP#3:
    AP = SQRT((c - x)^2 + y^2) ; ~6.9121
    BP = SQRT(x^2 + y^2) ; ~6.3857
    CP = SQRT((x - g)^2 + (h - y)^2) ; ~5.7542

    Thanks for the hints fellows.
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  5. #5
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    Note that  P is the centroid of the triangle , to calculate  AP , BP ,CP . We can find the length of the medians first , then obtain the answers by multiplying  \frac{2}{3} .


    If we let  M be the mid-point of  AB

    By Apollonius' Theorem ,  AC^2 + BC^2 = 2( AM^2 + CM^2)

    or  CM^2 = \frac{a^2 + b^2 }{2} - \frac{c^2}{4}

    Then substitute the values of  a,b,c ...
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  6. #6
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    Ahhhh YA! Thanks loads, SP.
    I tattooed Apollonius' Theorem on my wrist
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  7. #7
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    Here's an integer case: triangle sides 127-131-158 ; lengths to P: 87-85-68
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