Difference between revisions of "1998 AHSME Problems/Problem 23"
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The second curve becomes <math>(x-2)^2 + (y-6)^2 = 40+k</math>, which is a circle centered at <math>(2,6)</math> with radius <math>r=\sqrt{40+k}</math>. | The second curve becomes <math>(x-2)^2 + (y-6)^2 = 40+k</math>, which is a circle centered at <math>(2,6)</math> with radius <math>r=\sqrt{40+k}</math>. | ||
| − | The distance between the two centers is <math>5</math>, and therefore the two circles intersect | + | The distance between the two centers is <math>5</math>, and therefore the two circles intersect if <math>2\leq r \leq 12</math>. |
From <math>\sqrt{40+k} \geq 2</math> we get that <math>k\geq -36</math>. From <math>\sqrt{40+k}\leq 12</math> we get <math>k\leq 104</math>. | From <math>\sqrt{40+k} \geq 2</math> we get that <math>k\geq -36</math>. From <math>\sqrt{40+k}\leq 12</math> we get <math>k\leq 104</math>. | ||
Latest revision as of 08:12, 2 December 2018
Problem
The graphs of 
and 
intersect when 
satisfies 
, and for no other values of . Find 
.
Solution
Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form 
.
The first curve becomes 
, which is a circle centered at 
with radius 
.
The second curve becomes 
, which is a circle centered at 
with radius 
.
The distance between the two centers is 
, and therefore the two circles intersect if 
.
From 
we get that 
. From 
we get 
.
Therefore 
.
See also
| 1998 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
