AIME 2005 I · 第 14 题

Solution

Solution 1

Consider a point $E$ such that $AE$ is perpendicular to $BD$, $AE$ intersects $BD$, and $AE = BD$. E will be on the same side of the square as point $C$.

Let the coordinates of $E$ be $(x_E,y_E)$. Since $AE$ is perpendicular to $BD$, and $AE = BD$, we have $9 - 7 = x_E - 0$ and $10 - ( - 4) = 12 - y_E$ The coordinates of $E$ are thus $(2, - 2)$.

Now, since $E$ and $C$ are on the same side, we find the slope of the sides going through $E$ and $C$ to be $\frac { - 2 - 0}{2 - 8} = \frac {1}{3}$. Because the other two sides are perpendicular, the slope of the sides going through $B$ and $D$ are now $- 3$.

Let $A_1,B_1,C_1,D_1$ be the vertices of the square so that $A_1B_1$ contains point $A$, $B_1C_1$ contains point $B$, and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find $2$ vertices of the square, then apply the distance formula.

We find the coordinates of $C_1$ to be $(12.5,1.5)$ and the coordinates of $D_1$ to be $( - 0.7, - 2.9)$. Applying the distance formula, the side length of our square is $\sqrt {\left( \frac {44}{10} \right)^2 + \left( \frac {132}{10} \right)^2} = \frac {44}{\sqrt {10}}$.

Hence, the area of the square is $K = \frac {44^2}{10}$. The remainder when $10K$ is divided by $1000$ is $936$.

Solution 2

Let $(a,b)$ denote a normal vector of the side containing $A$. Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$, $ax+by=8a$, $bx-ay=10b-9a$, and $bx-ay=-4b-7a$. The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$, hence $8a+0b-12b=-4b-7a-10b+9a$, or $-3a=b$. We can take $a=-1$ and $b=3$. So the side of the square is $\frac{44}{\sqrt{10}}$, the area is $K=\frac{1936}{10}$, and the answer to the problem is $\boxed{936}$.

Solution 3

Let $\begin{bmatrix} a \\ b \end{bmatrix}$ be the unit vector parallel to one side of the square. The unit vector parallel to the perpendicular side is then $\begin{bmatrix} b \\ -a \end{bmatrix}$. The dot product of a diagonal of quadrilateral $ABCD$ with the corresponding unit vector gives the side length. Since the side lengths of the square are equal, $\overrightarrow{DB}\cdot \begin{bmatrix} a \\ b \end{bmatrix}= \overrightarrow{AC}\cdot \begin{bmatrix} b \\ -a \end{bmatrix}$. Plugging in $\overrightarrow{AC}=\begin{bmatrix} 8 \\ -12 \end{bmatrix}$ and $\overrightarrow{DB}=\begin{bmatrix} 14 \\ 2 \end{bmatrix}$ yields $a=3b$. Since $\begin{bmatrix} a \\ b \end{bmatrix}$ is a unit vector, it is equal to $\begin{bmatrix} \frac{3}{\sqrt{10}} \\ \frac{1}{\sqrt{10}} \end{bmatrix}$. Taking the dot product with $\overrightarrow{DB}$ gives the side length, which is $\frac {44}{\sqrt {10}}$, so the area is $\frac {44^2}{10}$, and the answer is $\boxed{936}$.