Mock AIME 1 Pre 2005 Problems/Problem 2

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.

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 $\boxed {161}$ .

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by Problem 1	Followed by Problem 3
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Mock AIME 1 Pre 2005 Problems/Problem 2

Problem 2

Solution

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