2016 AMC 10A Problem 24

Let the quadrilateral be $ABCD$ with $AB = BC = CD = 200$ and unknown $AD = x$, and let $O$ be the center, $R = 200\sqrt 2$. Equal chords subtend equal central angles, so $\angle AOB = \angle BOC = \angle COD = \theta$.

Step 1: the central angle. In isosceles triangle $AOB$, the law of cosines gives
$$200^2 = R^2 + R^2 - 2R^2\cos\theta = 160000(1 - \cos\theta),$$
so $1 - \cos\theta = \tfrac14$ and $\cos\theta = \tfrac34$.

Step 2: the diagonal. Chord $AC$ subtends $\angle AOC = 2\theta$, with $\cos 2\theta = 2\cos^2\theta - 1 = \tfrac{9}{8} - 1 = \tfrac18$. Then
$$AC^2 = 2R^2(1 - \cos 2\theta) = 160000\cdot\tfrac78 = 140000.$$
By symmetry $BD = AC$ (both cut off two of the equal chords), so $AC\cdot BD = 140000$.

Step 3: Ptolemy. For the cyclic quadrilateral $ABCD$,
$$AC\cdot BD = AB\cdot CD + BC\cdot AD \implies 140000 = 200\cdot 200 + 200x,$$
giving $200x = 100000$ and $x = 500$.

The answer is $\boxed{\textbf{(E)}\ 500}$.

Three equal chords force an isosceles trapezoid whose two diagonals are equal, and Ptolemy's theorem turns "product of diagonals" into a linear equation in the missing side. The radius enters only through one central angle, which the law of cosines extracts from a single equal side. Whenever a cyclic quadrilateral has known sides and radius, aim for the diagonals via central angles, then Ptolemy.

Trig: a chord subtending central angle $\phi$ has length $2R\sin(\phi/2)$. From $200 = 400\sqrt2\sin(\theta/2)$ we get $\sin(\theta/2) = \frac{\sqrt2}{4}$. The fourth side subtends $360^\circ - 3\theta$, so its length is $2R\sin(3\theta/2) = 400\sqrt2\left(3s - 4s^3\right)$ with $s = \frac{\sqrt2}{4}$, which equals $400\sqrt2\cdot\frac{5\sqrt2}{8} = 500$.

• Assuming the three equal chords subtend 90 degrees each (as if the radius were 100sqrt2), which makes the fourth side 200 and the figure a square.
• Applying Ptolemy with the wrong pairing of opposite sides, e.g. $AB\cdot BC + CD\cdot AD$, which are adjacent rather than opposite.