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

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== Problem ==
 
== Problem ==
In trapezoid <math>ABCD</math>, leg <math>\overline{BC}</math> is perpendicular to bases <math>\overline{AB}</math> and <math>\overline{CD}</math>, and diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
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In [[trapezoid]] <math>ABCD</math>, leg <math>\overline{BC}</math> is [[perpendicular]] to bases <math>\overline{AB}</math> and <math>\overline{CD}</math>, and diagonals <math>\overline{AC}</math> and <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
  
 
== Solution ==
 
== Solution ==
Call the height of the trapezoid a. Triangles BAC and CBD are similar since AC is perpendicular to BD. Using similar triangles, <math>CD=a^2/11</math>. Call the foot of the altitude from A to CD H. We know AH=a, but AHD is a right triangle, so by the Pythagorean Theorem, <math>a=\sqrt{1001-(a^2/\sqrt{11}-\sqrt{11})^2}</math>. Letting <math>a^2=n</math> (which is what we're trying to find) and simplifying yields the equation <math>n^2-11n-10890=0</math>, and the positive solution is 110.
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=== Solution 1 ===
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Let <math>x = BC</math> be the height of the trapezoid, and let <math>y = CD</math>. Since <math>AC \perp BD</math>, it follows that <math>\triangle BAC \sim \triangle CBD</math>, so <math>\frac{x}{\sqrt{11}} = \frac{y}{x} \Longrightarrow x^2 = y\sqrt{11}</math>.
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Let <math>E</math> be the foot of the altitude from <math>A</math> to <math>\overline{CD}</math>. Then <math>AE = x</math>, and <math>ADE</math> is a [[right triangle]]. By the [[Pythagorean Theorem]],
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<cmath>x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0</cmath>
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The positive solution to this [[quadratic equation]] is <math>x^2 = \boxed{110}</math>.
 +
 
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<center><asy>
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size(200); pathpen = linewidth(0.7);
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pair C=(0,0),B=(0,110^.5),A=(11^.5,B.y),D=(10*11^.5,0),E=foot(A,C,D);
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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));
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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));
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</asy></center>
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=== Solution 2 ===
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Let <math>BC=x</math>. Dropping the altitude from <math>A</math> and using the Pythagorean Theorem tells us that <math>CD=\sqrt{11}+\sqrt{1001-x^2}</math>. Therefore, we know that vector <math>\vec{BD}=\langle \sqrt{11}+\sqrt{1001-x^2},-x\rangle</math> and vector <math>\vec{AC}=\langle-\sqrt{11},-x\rangle</math>. Now we know that these vectors are perpendicular, so their dot product is 0.<cmath>\vec{BD}\cdot \vec{AC}=-11-\sqrt{11(1001-x^2)}+x^2=0</cmath>
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<cmath>(x^2-11)^2=11(1001-x^2)</cmath>
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<cmath>x^4-11x^2-11\cdot 990=0.</cmath>
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As above, we can solve this quadratic to get the positve solution <math>BC^2=x^2=\boxed{110}</math>.
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=== Solution 3 ===
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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>.
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=== Solution 4 ===
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Let <math>E</math> be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
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<cmath>\begin{align*}
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BC^2&=BE^2+CE^2 \
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&=(AB^2-AE^2)+(CD^2-DE^2) \
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&=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
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\end{align*}</cmath>
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Followed by dropping the perpendicular like in solution 1, we obtain system of equation
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<cmath>BC^2=CD^2-990</cmath>
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<cmath>BC^2+CD^2-2\sqrt{11}CD=990</cmath>
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Rearrange the first equation yields
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<cmath>BC^2-CD^2=990</cmath>
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Equating it with the second equation we have
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<cmath>BC^2-CD^2=BC^2+CD^2-2\sqrt{11}CD</cmath>
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Which gives <math>CD^2=\frac{BC^2}{11}</math>.
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Substituting into equation 1 obtains the quadratic in terms of <math>BC^2</math>
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<cmath>(BC^2)^2-11BC^2-11\cdot990=0</cmath>
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Solving the quadratic to obtain <math>BC^2=\boxed{110}</math>.
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~ Nafer
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=II|num-b=7|num-a=9}}
 
{{AIME box|year=2000|n=II|num-b=7|num-a=9}}
 +
 +
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

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 (ProblemsAnswer KeyResources)
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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