Difference between revisions of "1999 AIME Problems/Problem 6"
| (4 intermediate revisions by 4 users not shown) | |||
| Line 2: | Line 2: | ||
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> | ||
| − | == Solution == | + | ==Solution== |
<cmath>\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ | <cmath>\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ | ||
B' = & (\sqrt {1800}, \sqrt {600})\\ | B' = & (\sqrt {1800}, \sqrt {600})\\ | ||
Latest revision as of 15:42, 20 July 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 
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. 
