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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 2"

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== Solution ==
 
== Solution ==
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'''Solution 1'''
 
[[Completing the square]] to find a geometric interpretation,
 
[[Completing the square]] to find a geometric interpretation,
 
<math>x^2 + y^2 - 30x - 40y + 24^2 = 0 \Longleftrightarrow (x - 15)^2 + (y - 20)^2 = 7^2</math>
 
<math>x^2 + y^2 - 30x - 40y + 24^2 = 0 \Longleftrightarrow (x - 15)^2 + (y - 20)^2 = 7^2</math>
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<cmath>- 176m^2 + 600m - 351 = 0</cmath>  
 
<cmath>- 176m^2 + 600m - 351 = 0</cmath>  
 
Solving, <math>m = \frac {117}{44}</math> (or the extraneous root, <math>\frac {3}{4}</math>). Therefore <math>m + n = 117 + 44 = \boxed{161}</math>.
 
Solving, <math>m = \frac {117}{44}</math> (or the extraneous root, <math>\frac {3}{4}</math>). Therefore <math>m + n = 117 + 44 = \boxed{161}</math>.
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'''Solution 2'''
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We obtain the previous circle as explained above. Let <math>l</math> be the line that intersects the circle with largest slope. Draw the segment of length <math>25</math> from the center of the circle to the origin. Let <math>\angle A</math> be the angle determined by the segment just drawn and the x-axis, and <math>\angle B</math> be the angle determined by the segment just drawn and the y-axis.
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The slope of <math>l</math> is equal to <math>tan (A+B)</math>, or
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<math>\frac {tan A + tan B}{1 - tan A tan B}</math>
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= <math>\frac {\frac {4}{3} + \frac {7}{24}}{1 - \frac {4}{3} \frac {7}{24}}</math>
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= <math>\frac {117}{44}</math>, so the answer is <math>\boxed {161}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:48, 17 April 2009

Problem 2

If $x^2 + y^2 - 30x - 40y + 24^2 = 0$, then the largest possible value of $\frac{y}{x}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine $m + n$.

Solution

Solution 1 Completing the square to find a geometric interpretation, $x^2 + y^2 - 30x - 40y + 24^2 = 0 \Longleftrightarrow (x - 15)^2 + (y - 20)^2 = 7^2$

Consider the line through the circle, passing through the origin, $y = mx$. We want to maximise $\frac{y}{x} = m$. If the line $l_1$ passes through the circle, then we can steepen the line until it is tangent.

Pre2005 MockAIME 1-2.png

Therefore we must find the slope of the tangent, when the following simultaneous equations has just one solution: \[\begin{cases} y = mx \\ (x - 15)^2 + (y - 20)^2 = 7^2 \end{cases}\] Substituting, \[(x - 15)^2 + (mx - 20)^2 - 7^2 = 0\] If there is one solution, the discriminant must be $0$. Therefore \[- 176m^2 + 600m - 351 = 0\] Solving, $m = \frac {117}{44}$ (or the extraneous root, $\frac {3}{4}$). Therefore $m + n = 117 + 44 = \boxed{161}$.

Solution 2 We obtain the previous circle as explained above. Let $l$ be the line that intersects the circle with largest slope. Draw the segment of length $25$ from the center of the circle to the origin. Let $\angle A$ be the angle determined by the segment just drawn and the x-axis, and $\angle B$ be the angle determined by the segment just drawn and the y-axis.

The slope of $l$ is equal to $tan (A+B)$, or

$\frac {tan A + tan B}{1 - tan A tan B}$

= $\frac {\frac {4}{3} + \frac {7}{24}}{1 - \frac {4}{3} \frac {7}{24}}$

= $\frac {117}{44}$, so the answer is $\boxed {161}$.

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
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