Express the area of the orange triangle in terms of the areas of the other colours.
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Solution
Orange = (purple-blue-green)/2.
Let G denote the area of the green triangle, B the blue, P the purple, and O the orange. Add G to O, and B to O, getting triangles G’ and B’ sharing a common base with P. Let x, y, z be the vectors from the left endpoint of the common base to the tips (opposite the common base) of the G, B, and P triangles respectively.
Notice that the right endpoint of the common base is now the centroid of those three tips, so its vector is the average (x+y+z)/3 of the three. The area of G’ is (1/2) (x+y+z)/3 X x, where X denotes the cross-product for vectors.
Similarly, the area of B’ is (1/2) (x+y+z)/3 X y, and the area of P is – (1/2) (x+y+z)/3 X z. Then the sum of areas G’+B’ minus P is (1/2) (x+y+z)/3 X x + (1/2) (x+y+z)/3 X y + (1/2) (x+y+z)/3 X z = (1/6) (x+y+z) X (x+y+z) = 0, because the cross-product of a vector with itself is always zero.
But G’ = G + O, and B’ = B + O, so that gives G+O+B+O = P, or 2O = P – B – G, and finally O = (1/2)(P-B-G).
2 replies on “The paper airplane”
Here’s a proof without vectors – but didn’t come up with this until after reading the above proof that does use vectors.
Incomprehensible text.