2012 AMC 10B Problem 21

Let $P$ and $Q$ be the two points at distance $2a$, and $R$, $S$ the other two. The remaining five segments $PR, PS, QR, QS, RS$ have lengths $a, a, a, a, b$.

If $RS = b$, then $PR = QR = a$, so $PR + RQ = 2a = PQ$: by the equality case of the triangle inequality, $R$ is the midpoint of $PQ$. The same argument makes $S$ the midpoint too, so $R = S$, contradicting distinctness.

Hence $RS = a$, and $b$ is one of the segments from $P$ or $Q$; by relabeling, let $PR = QR = PS = a$ and $QS = b$. As above, $R$ is the midpoint of $PQ$. Then $PS = SR = RP = a$, so triangle $PRS$ is equilateral and $\angle PRS = 60^\circ$, giving $\angle SRQ = 120^\circ$.

In triangle $SRQ$ we have $SR = RQ = a$ with included angle $120^\circ$. Drop the perpendicular from $S$ to line $PQ$; its foot $M$ is the midpoint of $PR$, so $SM = \frac{\sqrt3}{2}a$ and $MQ = \frac{a}{2} + a = \frac{3a}{2}$. Then
$$QS^2 = \left(\tfrac{3a}{2}\right)^2 + \left(\tfrac{\sqrt3 a}{2}\right)^2 = \tfrac{9a^2}{4} + \tfrac{3a^2}{4} = 3a^2, \qquad b = a\sqrt3.$$

The answer is $\boxed{\textbf{(A)}\ \sqrt3}$.

A distance equal to the sum of two others is a collinearity statement (the degenerate triangle inequality), and that rigidly places a point. Once three collinear points are fixed, the four equal distances leave the last point only one place to go: the apex of an equilateral triangle. Reasoning from the most constraining lengths first turns a "find the configuration" problem into a forced construction.

• Guessing 2 from a sketch of a rhombus or parallelogram without checking that all six distances really are a, a, a, a, 2a, b.
• Trying to make $PQ$ the long diagonal of a rhombus of side $a$; a rhombus of side $a$ has diagonals shorter than $2a$, so that figure degenerates.