Difference between revisions of "2000 AIME II Problems/Problem 8"

Latest revision as of 14:33, 20 January 2025

Problem

In trapezoid $ABCD$ , leg $\overline{BC}$ is 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

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 $x^2 = \boxed{110}$ .

$[asy] size(200); pathpen = linewidth(0.7); pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D); D(MP("A",A,(2,.5))--MP("B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7)); MP("\sqrt{11}",(A+B)/2,N);MP("\sqrt{1001}",(A+D)/2,NE);MP("\sqrt{1001}",(A+D)/2,NE);MP("x",(B+C)/2,W);MP("y",(D+C)/2);D(rightanglemark(B,IP(A--C,B--D),C,20)); [/asy]$

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 $CD^2-BC^2=990$ 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

@@ Line 47: / Line 47: @@
 Equating it with the second equation we have
 <cmath>CD^2-BC^2=BC^2+CD^2-2\sqrt{11}CD</cmath>
-Which gives <math>CD^2=\frac{BC^2}{11}</math>.
+Which gives <math>CD=\frac{BC^2}{\sqrt{11}}</math>.
 Substituting into equation 1 obtains the quadratic in terms of <math>BC^2</math>
 <cmath>(BC^2)^2-11BC^2-11\cdot990=0</cmath>
 Solving the quadratic to obtain <math>BC^2=\boxed{110}</math>.
-~ Nafer
+~Nafer ~edits by fermat_sLastAMC
 == See also ==

2000 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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