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

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=== Solution 3 ===
 
=== Solution 3 ===
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Let <math>BC=x</math> and <math>CD=y+\sqrt{11}</math>. From Pythagorus with <math>AD</math>, we obtain <math>x^2+y^2=1001</math>. Since <math>AC</math> and <math>BD</math> are perpendicular diagonals of a quadrilateral, then <math>AB^2+CD^2=BC^2+AD^2</math>, so we have <cmath>\left(y+\sqrt{11}\right)^2+11=x^2+1001.</cmath> Substituting <math>x^2=1001-y^2</math> and simplifying yields <cmath>y^2+\sqrt{11}y-990=0,</cmath> and the quadratic formula gives <math>y=9\sqrt{11}</math>. Then from <math>x^2+y^2=1001</math>, we plug in <math>y</math> to find <math>x^2=\boxed{110}</math>.
+
Let <math>BC=x</math> and <math>CD=y+\sqrt{11}</math>. From Pythagoras with <math>AD</math>, we obtain <math>x^2+y^2=1001</math>. Since <math>AC</math> and <math>BD</math> are perpendicular diagonals of a quadrilateral, then <math>AB^2+CD^2=BC^2+AD^2</math>, so we have <cmath>\left(y+\sqrt{11}\right)^2+11=x^2+1001.</cmath> Substituting <math>x^2=1001-y^2</math> and simplifying yields <cmath>y^2+\sqrt{11}y-990=0,</cmath> and the quadratic formula gives <math>y=9\sqrt{11}</math>. Then from <math>x^2+y^2=1001</math>, we plug in <math>y</math> to find <math>x^2=\boxed{110}</math>.
 +
 
 +
 
 +
=== Solution 4 ===
 +
 
 +
Let <math>E</math> be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
 +
<cmath>\begin{align*}
 +
BC^2&=BE^2+CE^2 \\
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&=(AB^2-AE^2)+(CD^2-DE^2) \\
 +
&=CD^2+\sqrt{11}^2-(AE^2+DE^2) \\
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&=CD^2+11-AD^2 \\
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&=CD^2-990
 +
\end{align*}</cmath>
 +
Followed by dropping the perpendicular like in solution 1, we obtain system of equation
 +
<cmath>BC^2=CD^2-990</cmath>
 +
<cmath>BC^2+CD^2-2\sqrt{11}CD=990</cmath>
 +
Rearrange the first equation yields
 +
<cmath>BC^2-CD^2=990</cmath>
 +
Equating it with the second equation we have
 +
<cmath>BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD</cmath>
 +
Which gives <math>CD^2=\frac{BC^2}{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
  
 
== See also ==
 
== See also ==

Latest revision as of 18:35, 7 December 2019

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 \[BC^2-CD^2=990\] 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

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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