2012 AMC 10B Problem 21

P21-25~5 minPrint

Triangles: Area & Pythagorean Angles & Polygons Add construction lines/points (drop altitudes, extend segments, connect centers)Split into exhaustive cases and handle each

Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$ , $a$ , $a$ , $a$ , $2a$ , and $b$ . What is the ratio of $b$ to $a$ ?

Show solution without answering

Key insight

Two points 2a apart, each exactly a from a third point, force that third point to be the midpoint. The fourth point then makes an equilateral triangle.

Solution

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 2: Angles
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Geometry, Chapter 19: Problem Solving Strategies in Geometry (§19.1)