AMC12 2009 A

To answer the question we are asked, it is enough to compute $AC^2$ for two different angles, preferably for both extremes ($45$ and $60$ degrees). You can use the law of cosines to do so. Alternately, it is enough to compute $AC^2$ for one of the extreme angles. In case it falls inside one of the given intervals, we are done. In case it falls on the boundary between two options, we also have to argue whether our $AC^2$ is the minimal or the maximal possible value of $AC^2$. Below we show a complete solution in which we also show that all possible values of $AC^2$ do indeed lie in the given interval. Let $C_1$ be the point the ship would reach if it turned $45^\circ$, and $C_2$ the point it would reach if it turned $60^\circ$. Obviously, $C_1$ is the furthest possible point from $A$, and $C_2$ is the closest possible point to $A$. Hence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$. We can find $AC_1^2$ and $AC_2^2$ as follows: Let $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$. The angles in the triangle $BD_1C_1$ are $45^\circ$, $45^\circ$, and $90^\circ$, hence $BD_1 = D_1C_1 = BC_1 / \sqrt 2$. Similarly, the angles in the triangle $BD_2C_2$ are $30^\circ$, $60^\circ$, and $90^\circ$, hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \sqrt 3 / 2$. Hence we get: \[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700\] \[AC_1^2 = AD_1^2 + D_1C_1^2 \]\[= (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 \]\[= 100 + 400/\sqrt 2 + 200 + 200 \]\[= 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800\] Therefore for any valid $C$ the value $AC^2$ is surely in the interval $\boxed{ \textbf{(D)}[700,800] }$. From the law of cosines, $500-400\cos120^\circ<AC^2<500-400\cos135^\circ\implies700<AC^2<500+200\sqrt{2}$. This is essentially the same solution as above. The answer is $\boxed{\textbf{(D)}}$.