Green vs. Blue

Yet another Catriona Shearer puzzle.

Green vs. Blue

Is more of this design green or blue (and by how much)?


I was hoping I’d find a clever way to make the answer drop right out, but nope, I ended up mucking about with lengths of sides of similar triangles. Right triangles fortunately, and you don’t need any hypotenuses or angles, so that’s something.

Add some lines and label vertices:


and I’ve shown some congruent angles.

No line segment lengths are given, so let’s use symbols: define AB = t = BC = CD = DA = GH, the last coming from the fact triangles ADG and FHG have to be congruent. Define BE = s = DG = FH = CH.

What we have to find is the difference between the blue area, which is ADG + DJF, and the green area, which is ABE + ECJ. But ADG and ABE are congruent, so we can cross them off and we just need DJF-ECJ.

ECJ is similar to ABE, and EC=t-s, so CJ = BE\times EC/AB = s(t-s)/t. The area of ECJ is s(t-s)^2/(2t).

DH is t-s, so the area of triangle DHF is s(t-s)/2.

HJ = s-s(t-s)/t = s^2/t so the area of triangle FHJ is s^3/(2t). Then the area of triangle DJF is s(t-s)/2+s^3/(2t).


\displaystyle {DJF-ECJ =s(t-s)/2+s^3/(2t) -s(t-s)^2/(2t)}

\displaystyle{= (st-s^2+s^3/t-st+2s^2-s^3/t)/2 = s^2/2}.

Independent of t! Mysteriously. I wish I could find a clever way to see that.

Anyway, the area of triangle GDF is s\times s/2 = 5. Mysteriously the same asDJF-ECJ! So the answer is, the blue area is larger by 5.

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