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Math Help - Geometry - Area

  1. #1
    Junior Member
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    Geometry - Area

    Hi,

    I've been working on this problem all night and I don't know what I'm doing wrong! I keep on going in circles, I think I'm either missing something or making it to hard. I'm assuming P can be anywhere...

    Problem:
    P is a point inside an equilateral triangle ABC. Perpendiculars PD, PE AD PF are dropped from P onto BC, CA and AB respectively Prove that [PAF]+[PBD]+[PCE]=[PAE]+[PCD]+[PBF]

    Thanks!
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Red face

    This argument is long and messy, but it works. I haven't been able to find a shorter proof.

    Suppose for convenience that the sides of the triangle are 1 unit in length. The first thing to notice is that the sum of the distances from P to the three sides is equal to √3/2.

    . . . . . . . . . . . . . . . . . . . .

    In this picture (gacked from here, where there is a full explanation), the area of the whole triangle is (1/2)base×height, and the height is √3/2. But this is equal to the sum of the three coloured triangles, whose bases are all equal to 1 and whose heights are h_1,\;h_2 and h_3. So h_1 + h_2 + h_3 = \sqrt3/2.

    Notice that the area of the triangle ABC is (1/2)base×height = √3/4.

    What we want to show here is that in the next picture (cunningly Photoshopped by me) the sum of the areas of the darker-coloured triangles is the same as the sum of the areas of the lighter-coloured triangles.

    . . . . . . . . . . . . . . . . . . . .

    Call the bases of the darker-coloured triangles b_1,\;b_2 and b_3. Then their total area is {\textstyle\frac12}(b_1h_1+b_2h_2+b_3h_3).

    The first main step is to find expressions for b_1,\;b_2 and b_3 in terms of h_1,\;h_2 and h_3.

    For b_2, the vertical height of the point E above the horizontal line BC is b_2\cos(\pi/3) = {\textstyle\frac{\sqrt3}2}b_2. But it is also equal to h_1+h_2\cos(\pi/6) = h_1+{\textstyle\frac12}h_2. That leads to the equation b_2 = {\textstyle\frac1{\sqrt3}}(2h_1+h_2).

    Similarly, the vertical height of F above the line BC is equal to √3/2 (the height of A) minus the vertical drop from A to F. It is also equal to the drop from F to P plus the drop from P to D. This leads to the equation b_3 = 1 - {\textstyle\frac1{\sqrt3}}(2h_1+h_2).

    For b_1, the distance DC is equal to 1-b_1 and also to {\textstyle\frac{\sqrt3}2}h_2 + {\textstyle\frac12}b_2 ={\textstyle\frac{\sqrt3}2}h_2 + {\textstyle\frac1{2\sqrt3}}(2h_1+h_2). This gives the equation b_1 = 1 - {\textstyle\frac1{\sqrt3}}(h_1+2h_2).

    Let A denote the sum of the areas of the darker-coloured triangles. Then

    . . . . . . . . . . . . . . . \begin{array}{rcl}2A&=& h_1b_1 +h_2b_2 + h_3b_3 \\<br />
&=& h_1(1 - {\textstyle\frac1{\sqrt3}}(h_1+2h_2)) + h_2{\textstyle\frac1{\sqrt3}}(2h_1+h_2) + h_3(1 - {\textstyle\frac1{\sqrt3}}(2h_1+h_2)) \\<br />
&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[-h_1^2 + h_2^2 - 2h_1h_3 - h_3^2\bigr] \\<br />
&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[h_2^2 - (h_1+h_3)^2\bigr] .\end{array}

    Now use the fact that h_1 + h_2 + h_3 = \sqrt3/2 to write this as

    . . . . . . . . . . . . . . . \begin{array}{rcl}2A&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[h_2^2 - ({\textstyle\frac{\sqrt3}2} - h_2)^2\bigr] \\<br />
&=& h_1 + h_3 + {\textstyle\frac1{\sqrt3}}\bigl[-{\textstyle\frac34} + \sqrt3h_2\bigr] \\<br />
&=& h_1 + h_2 + h_3 - {\textstyle\frac{\sqrt3}4} \\<br />
&=& \textstyle\frac{\sqrt3}2 - \textstyle\frac{\sqrt3}4 = \textstyle\frac{\sqrt3}4, \end{array}

    which is the area of the triangle ABC. Thus the sum of the areas of the darker-coloured triangles is half the area of the whole triangle, which is what we wanted to show.
    Last edited by Opalg; December 10th 2007 at 09:35 AM.
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