Difference between revisions of "2000 AIME II Problems/Problem 8"
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=== Solution 3 === | === 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>. | 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 === | ||
== See also == | == See also == |
Revision as of 18:22, 7 December 2019
Contents
[hide]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 .
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
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.