2005 AIME I Problems/Problem 14

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Problem

Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$ .

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 $A$ 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 $\boxed{936}$ .

2005 AIME I (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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