Geometry - Area

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.

. . . . . . . . . . . . . . . . . . . .http://www.qbyte.org/puzzles/p117s1.gif

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.

. . . . . . . . . . . . . . . . . . . .http://www.thelances.co.uk/imagesC/triangle.png

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

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

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

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

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.