Difference between revisions of "2000 AIME II Problems/Problem 8"
Dgreenb801 (talk | contribs) m (→Solution) |
(→Solution 4) |
||
(12 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
== 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>. | + | 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 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 19:35, 7 December 2019
Contents
Problem
In trapezoid , leg
is perpendicular to bases
and
, and diagonals
and
are perpendicular. Given that
and
, find
.
Solution
Solution 1
Let be the height of the trapezoid, and let
. Since
, it follows that
, so
.
Let be the foot of the altitude from
to
. Then
, and
is a right triangle. By the Pythagorean Theorem,
The positive solution to this quadratic equation is .
![[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]](http://latex.artofproblemsolving.com/3/3/c/33cba5bae648539c2b01af0bfdabeeb2f45e237b.png)
Solution 2
Let . Dropping the altitude from
and using the Pythagorean Theorem tells us that
. Therefore, we know that vector
and vector
. Now we know that these vectors are perpendicular, so their dot product is 0.
As above, we can solve this quadratic to get the positve solution
.
Solution 3
Let and
. From Pythagoras with
, we obtain
. Since
and
are perpendicular diagonals of a quadrilateral, then
, so we have
Substituting
and simplifying yields
and the quadratic formula gives
. Then from
, we plug in
to find
.
Solution 4
Let be the intersection of the diagonals. Since the diagonals are perpendicular, applying the Pythagorean Theorem multiple times we have
Followed by dropping the perpendicular like in solution 1, we obtain system of equation
Rearrange the first equation yields
Equating it with the second equation we have
Which gives
.
Substituting into equation 1 obtains the quadratic in terms of
Solving the quadratic to obtain
.
~ 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.