# Green vs. Blue

Yet another Catriona Shearer puzzle.

14.
Green vs. Blue Is more of this design green or blue (and by how much)?

Spoiler!

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 $s(t-s)/2+s^3/(2t)$.

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

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 as $DJF-ECJ$! So the answer is, the blue area is larger by 5.