In [trapezoid](https://artofproblemsolving.com/wiki/index.php/Trapezoid) $ABCD$, leg $\overline{BC}$ is [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) to bases $\overline{AB}$ and $\overline{CD}$, and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001}$, find $BC^2$.
Solution 1
Let $x = BC$ be the height of the trapezoid, and let $y = CD$. Since $AC \perp BD$, it follows that $\triangle BAC \sim \triangle CBD$, so $\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}$.
Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE = x$, and $ADE$ is a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle). By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), 
\[x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0\]
The positive solution to this [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation) is $x^2 = \boxed{110}$.


Solution 2
Let $BC=x$. Dropping the altitude from $A$ and using the Pythagorean Theorem tells us that $CD=\sqrt{11}+\sqrt{1001-x^2}$. Therefore, we know that vector $\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle$ and vector $\vec{AC}=\langle-\sqrt{11},-x\rangle$. Now we know that these vectors are perpendicular, so their dot product is 0.\[\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0\]
\[(x^2-11)^2=11(1001-x^2)\]
\[x^4-11x^2-11\cdot 990=0.\]
As above, we can solve this quadratic to get the positve solution $BC^2=x^2=\boxed{110}$.

Solution 3
Let $BC=x$ and $CD=y+\sqrt{11}$. From Pythagoras with $AD$, we obtain $x^2+y^2=1001$. Since $AC$ and $BD$ are perpendicular diagonals of a quadrilateral, then $AB^2+CD^2=BC^2+AD^2$, so we have \[\left(y+\sqrt{11}\right)^2+11=x^2+1001.\] Substituting $x^2=1001-y^2$ and simplifying yields \[y^2+\sqrt{11}y-990=0,\] and the quadratic formula gives $y=9\sqrt{11}$. Then from $x^2+y^2=1001$, we plug in $y$ to find $x^2=\boxed{110}$.


Solution 4
Let $E$ be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
\begin{align*} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{align*}
Followed by dropping the perpendicular like in solution 1, we obtain system of equation
\[BC^2=CD^2-990\]
\[BC^2+CD^2-2\sqrt{11}CD=990\]
Rearrange the first equation yields
\[BC^2-CD^2=990\]
Equating it with the second equation we have
\[BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD\]
Which gives $CD^2=\frac{BC^2}{11}$.
Substituting into equation 1 obtains the quadratic in terms of $BC^2$
\[(BC^2)^2-11BC^2-11\cdot990=0\]
Solving the quadratic to obtain $BC^2=\boxed{110}$.
~ Nafer