The South African Mathematical Olympiad Third Round

Set $x=c-d$ and $y=b-d$. We have $$a-d=(c-d)+(a-c)=x+3y$$ and $$a-d=(b-d)+(a-b)=y+2x,$$ hence $x+3y=y+2x$, which implies $x=2y$. Now we get $a-d=5y$ and $$b-c=(b-d)-(c-d)=y-x=-y,$$ so $$\frac{a-d}{b-c}=-5.---\ast \ast Alternativesolution.\ast \ast Wearegiventhat$$ a - b = 2c - 2d, \text{ thus } a + 2d = b + 2c, $$and$$ a - c = 3b - 3d, \text{ thus } a + 3d = 3b + c. $$Multiplythefirstequationby4andthesecondoneby3,andsubtract:$$ a - d = 4(a + 2d) - 3(a + 3d) = 4(b + 2c) - 3(3b + c) = -5b + 5c, $$thus$$ \frac{a-d}{b-c} = -5.