2016 AMC 10A Problem 24

Call the quadrilateral $ABCD$ with $AB = BC = CD = 200$ and unknown side $AD = x$. Let $O$ be the center and $R = 200\sqrt2$ the radius, so $R^2 = 80000$.

Step 1: the shape is an isosceles trapezoid. Fold the picture along the line through $O$ that cuts $BC$ in half at a right angle. This fold swaps $B$ and $C$. Since $AB = CD$, it also swaps $A$ and $D$. So the figure is symmetric, and $BC$ is parallel to $AD$, with legs $AB = CD = 200$.

Step 2: the distance from $O$ to a side of length $200$. In triangle $AOB$, the two sides from $O$ are radii, so the line from $O$ to the midpoint of $AB$ is perpendicular to $AB$. By the Pythagorean theorem its length is
$$\sqrt{R^2 - 100^2} = \sqrt{80000 - 10000} = \sqrt{70000} = 100\sqrt7.$$
So triangle $AOB$ has area $\tfrac12 \cdot 200 \cdot 100\sqrt7 = 10000\sqrt7$.

Step 3: the diagonal $AC$. Point $B$ is $200$ from both $A$ and $C$, and $O$ is $R$ from both. So the line $OB$ is the perpendicular bisector of $AC$, and $AC$ is twice the distance from $A$ to line $OB$. That distance is the height of triangle $AOB$ onto side $OB$:
$$\text{height} = \frac{2 \cdot \text{area}}{OB} = \frac{20000\sqrt7}{200\sqrt2} = \frac{100\sqrt7}{\sqrt2}.$$
So
$$AC^2 = \left(\frac{200\sqrt7}{\sqrt2}\right)^2 = \frac{40000 \cdot 7}{2} = 140000.$$

Step 4: two right triangles on base $AD$. Drop a perpendicular from $C$ to line $AD$, meeting it at $K$. Because the trapezoid is symmetric, $KD = \tfrac{x - 200}{2}$ and $AK = \tfrac{x + 200}{2}$. The height $CK$ is shared by right triangles $CKD$ and $CKA$:
$$CK^2 = 200^2 - \left(\frac{x - 200}{2}\right)^2 = 140000 - \left(\frac{x + 200}{2}\right)^2.$$

Step 5: solve. Move the squares to one side:
$$\left(\frac{x + 200}{2}\right)^2 - \left(\frac{x - 200}{2}\right)^2 = 140000 - 40000 = 100000.$$
Expanding, the left side is $\tfrac{(x^2 + 400x + 40000) - (x^2 - 400x + 40000)}{4} = \tfrac{800x}{4} = 200x$. So $200x = 100000$ and $x = 500$.

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

Three equal sides in a circle force a symmetric shape, an isosceles trapezoid, so its two diagonals are equal and a single right-triangle comparison finds the last side. The radius only enters once, to get a diagonal. Step 5 is really Ptolemy's theorem (diagonal times diagonal equals the sum of the products of opposite sides) worked out by hand for a trapezoid.

With the law of cosines and Ptolemy's theorem. Each side of length $200$ sits opposite an angle $\theta$ at the center: $200^2 = 2R^2(1 - \cos\theta)$, so $\cos\theta = \tfrac34$. The diagonal $AC$ spans two such angles, and $\cos 2\theta = 2 \cdot \tfrac{9}{16} - 1 = \tfrac18$, so $AC^2 = 2R^2(1 - \tfrac18) = 140000$. The other diagonal $BD$ has the same length. Ptolemy's theorem for a quadrilateral in a circle says $AC \cdot BD = AB \cdot CD + BC \cdot AD$, so $140000 = 40000 + 200x$ and $x = 500$.