AMC10 2012 B

Answer (A): Since 4 of the 6 segments have length $a$, some 3 of the points (call them $A$, $B$, and $C$) must form an equilateral triangle of side length $a$. The fourth point $D$ must be a distance $a$ from one of $A$, $B$, or $C$, and without loss of generality it can be assumed to be $A$. Thus $D$ lies on a circle of radius $a$ centered at $A$. The distance from $D$ to one of the other 2 points (which can be assumed to be $B$) is $2a$, so $BD$ is a diameter of this circle and therefore is the hypotenuse of right triangle $DCB$ with legs of lengths $a$ and $b$. Thus $b^2=(2a)^2-a^2=3a^2$, and the ratio of $b$ to $a$ is $\sqrt{3}$.