Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Solution 4)
 
(15 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
A container in the shape of a right circular cone is 12 inches tall and its base has a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the liquid is <math>m - n\sqrt [3]{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>.
+
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 ==
−
{{solution}}
+
 
 +
=== Solution 1 ===
 +
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>.
 +
 
 +
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]],
 +
 
 +
<cmath>x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0</cmath>
 +
 
 +
The positive solution to this [[quadratic equation]] is <math>x^2 = \boxed{110}</math>.
 +
 
 +
<center><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></center>
 +
 
 +
=== Solution 2 ===
 +
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>
 +
<cmath>(x^2-11)^2=11(1001-x^2)</cmath>
 +
<cmath>x^4-11x^2-11\cdot 990=0.</cmath>
 +
As above, we can solve this quadratic to get the positve solution <math>BC^2=x^2=\boxed{110}</math>.
 +
 
 +
=== Solution 3 ===
 +
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 \\
 +
&=(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*}</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 ==
 
{{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]]
 +
{{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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_8&oldid=112675"