Difference between revisions of "1999 AIME Problems/Problem 6"
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A transformation of the first [[quadrant]] of the [[coordinate plane]] maps each point <math>(x,y)</math> to the point <math>(\sqrt{x},\sqrt{y}).</math> The [[vertex|vertices]] of [[quadrilateral]] <math>ABCD</math> are <math>A=(900,300), B=(1800,600), C=(600,1800),</math> and <math>D=(300,900).</math> Let <math>k_{}</math> be the area of the region enclosed by the image of quadrilateral <math>ABCD.</math> Find the greatest integer that does not exceed <math>k_{}.</math> | A transformation of the first [[quadrant]] of the [[coordinate plane]] maps each point <math>(x,y)</math> to the point <math>(\sqrt{x},\sqrt{y}).</math> The [[vertex|vertices]] of [[quadrilateral]] <math>ABCD</math> are <math>A=(900,300), B=(1800,600), C=(600,1800),</math> and <math>D=(300,900).</math> Let <math>k_{}</math> be the area of the region enclosed by the image of quadrilateral <math>ABCD.</math> Find the greatest integer that does not exceed <math>k_{}.</math> | ||
+ | The below solution may not be accurate, see talk page for detail and to discuss. | ||
== Solution == | == Solution == | ||
<cmath>\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ | <cmath>\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ |
Revision as of 14:40, 25 April 2021
Problem
A transformation of the first quadrant of the coordinate plane maps each point to the point The vertices of quadrilateral are and Let be the area of the region enclosed by the image of quadrilateral Find the greatest integer that does not exceed
The below solution may not be accurate, see talk page for detail and to discuss.
Solution
First we see that lines passing through and have equations and , respectively. Looking at the points above, we see the equations for and are and , or, after manipulation and , respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines.
Now take a look at and , which have the equations and . The image equations hence are and , respectively, which are the equations for circles.
To find the area between the circles (actually, parts of the circles), we need to figure out the angle of the arc. This could be done by . So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = . Hence the answer is .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.