Question

In convex hexagon $ABCDEF,$ all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B,$ $\angle C,$ $\angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2} + 1).$ Find $AB.$

Answer

Solution:

The angles of a hexagon sum to $720^\circ,$ so each of the four congruent angles measures $\frac{720 - 2 \cdot 90}{4} = 135$ degrees. Let $AB = x.$ The diagonals $BF$ and $CE$ cut off the right isosceles triangles $FAB$ and $CDE,$ each with legs $x$ and hypotenuse $x\sqrt{2},$ and the $135^\circ$ angles guarantee that the remaining piece $BCEF$ is a rectangle with sides $x\sqrt{2}$ and $x.$

Hence the area is $2 \cdot \frac{1}{2}x^2 + x \cdot x\sqrt{2} = x^2(1 + \sqrt{2}) = 2116(\sqrt{2} + 1),$ so $x^2 = 2116$ and $AB = x = 46.$

2006 AIME II Problem 1

Solution:

Problem 1 in Other Years