AIME 2000 II · 第 8 题

Solution

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. By the 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 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

$$\begin{aligned} BC^2&=BE^2+CE^2 \\ &=(AB^2-AE^2)+(CD^2-DE^2) \\ &=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\ &=CD^2+11-AD^2 \\ &=CD^2-990 \end{aligned}$$ 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

$$CD^2-BC^2=990$$ Equating it with the second equation we have

$$CD^2-BC^2=BC^2+CD^2-2\sqrt{11}CD$$ Which gives $CD=\frac{BC^2}{\sqrt{11}}$. 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 ~edits by fermat_sLastAMC