Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 2"
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Consider the line through the circle, passing through the origin, <math>y = mx</math>. We want to maximise <math>\frac{y}{x} = m</math>. If the line <math>l_1</math> passes through the circle, then we can steepen the line until it is tangent. | Consider the line through the circle, passing through the origin, <math>y = mx</math>. We want to maximise <math>\frac{y}{x} = m</math>. If the line <math>l_1</math> passes through the circle, then we can steepen the line until it is tangent. | ||
| − | [[Image:Pre2005 1 | + | [[Image:Pre2005 MockAIME 1-2.png|center]] |
Therefore we must find the slope of the tangent, when the following simultaneous equations has just one solution: | Therefore we must find the slope of the tangent, when the following simultaneous equations has just one solution: | ||
Revision as of 16:15, 21 March 2008
Problem 2
If 
, then the largest possible value of 
can be written as 
, where 
and 
are relatively prime positive integers. Determine 
.
Solution
Completing the square to find a geometric interpretation,
Consider the line through the circle, passing through the origin, 
. We want to maximise 
. If the line 
passes through the circle, then we can steepen the line until it is tangent.
Therefore we must find the slope of the tangent, when the following simultaneous equations has just one solution:
![\[\begin{cases} y = mx \\ (x - 15)^2 + (y - 20)^2 = 7^2 \end{cases}\]](http://latex.artofproblemsolving.com/1/8/5/18566482abcdfdebbb8e4c517a85dc199a79b3d1.png)
Substituting,
![\[(x - 15)^2 + (mx - 20)^2 - 7^2 = 0\]](http://latex.artofproblemsolving.com/8/0/b/80bdd2226d893b09e9ddd7da542b86b182966e76.png)
If there is one solution, the discriminant must be 
. Therefore
![\[- 176m^2 + 600m - 351 = 0\]](http://latex.artofproblemsolving.com/3/a/6/3a695fa4bd5dd9fbefb0f984f387780efc0fd4ee.png)
Solving, 
(or the extraneous root, 
). Therefore 
.
See also
| Mock AIME 1 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
