2012 AMC 10B Problem 21

Call the two points that are $2a$ apart $P$ and $Q$, and the other two points $R$ and $S$. The other five distances, $PR$, $PS$, $QR$, $QS$ and $RS$, are $a, a, a, a$ and $b$ in some order.

1. A useful fact. If a point $X$ is at distance $a$ from $P$ and at distance $a$ from $Q$, where $PQ = 2a$, then going from $P$ to $X$ to $Q$ is exactly as long as going straight from $P$ to $Q$. A detour is always longer than the straight path unless there is no detour. So $X$ lies on segment $PQ$, right in the middle.
2. Could $RS$ be the $b$? Then $PR, PS, QR, QS$ would all be $a$. By the fact, $R$ is the midpoint of $PQ$ and so is $S$. Then $R$ and $S$ are the same point, but the four points are supposed to be different. So $RS$ is not $b$; it equals $a$.
3. So $b$ is one of $PR, PS, QR, QS$. The other three of these equal $a$. The one odd distance touches only one of the points $R$, $S$, so the other point is at distance $a$ from both $P$ and $Q$. Call that point $R$. So $PR = QR = a$, and by the fact $R$ is the midpoint of $PQ$.
4. We can also name things so that $PS = a$ and $QS = b$ (otherwise swap the names $P$ and $Q$).
5. Now $PR = a$, $RS = a$ and $PS = a$. Triangle $PRS$ is equilateral, so the angle at $R$ between $RP$ and $RS$ is $60^\circ$.
6. Find $QS$. Drop a straight line from $S$ down to line $PQ$; in an equilateral triangle it lands at the middle of $PR$. Call that point $M$.
- The height $SM$ of an equilateral triangle with side $a$ is $\frac{\sqrt3}{2}a$.
- $MQ = MR + RQ = \frac{a}{2} + a = \frac{3a}{2}$.
7. Triangle $SMQ$ has a right angle at $M$, so
$$QS^2 = \left(\tfrac{3a}{2}\right)^2 + \left(\tfrac{\sqrt3}{2}a\right)^2 = \tfrac{9a^2}{4} + \tfrac{3a^2}{4} = 3a^2.$$
So $b = QS = a\sqrt3$, and $b : a = \sqrt3$.

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

When one distance equals the sum of two others, the three points must lie on one straight line; that is the "no shortcut" idea from step 1. It pins down a point exactly. Start with the most demanding length (here $2a$) and the rest of the picture is forced: a straight line of three points plus an equilateral triangle on top.

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 $\boxed{\textbf{(A)}\ \sqrt3}$.