Difference between revisions of "1984 AHSME Problems/Problem 29"

Latest revision as of 19:09, 12 September 2022

Problem

Find the largest value for $\frac{y}{x}$ for pairs of real numbers $(x, y)$ which satisfy $(x-3)^2+(y-3)^2=6$ .

$\mathrm{(A) \ }3+2\sqrt{2} \qquad \mathrm{(B) \ }2+\sqrt{3} \qquad \mathrm{(C) \ } 3\sqrt{3} \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 6+2\sqrt{3}$

Solution

Let $\frac{y}{x}=k$ , so that $y=kx$ . Substituting this into the given equation $(x-3)^2+(y-3)^2=6$ yields $(x-3)^2+(kx-3)^2=6$ . Multiplying this out and forming it into a quadratic yields $(1+k^2)x^2+(-6-6k)x+12=0$ .

We want $x$ to be a real number, so we must have the discriminant $\geq0$ . The discriminant is $(-6-6k)^2-4(1+k^2)(12)=36k^2+72k+36-48-48k^2=-12k^2+72k-12$ . Therefore, we must have $-12k^2+72k-12\geq0$ , or $k^2-6k+1\leq0$ . The roots of this quadratic, using the quadratic formula, are $3\pm2\sqrt{2}$ , so the quadratic can be factored as $(k-(3-2\sqrt{2}))(k-(3+2\sqrt{2}))\leq0$ . We can now separate this into $3$ cases:

Case 1: $k<3-2\sqrt{2}$ Then, both terms in the factored quadratic are negative, so the inequality doesn't hold.

Case 2: $3-2\sqrt{2}<k<3+2\sqrt{2}$ Then, the first term is positive and the second is negative and the second is positive, so the inequality holds.

Case 3: $k>3+2\sqrt{2}$ Then, both terms are positive, so the inequality doesn't hold.

Also, when $k=3-2\sqrt{2}$ or $k=3+2\sqrt{2}$ , the equality holds.

Therefore, we must have $3-2\sqrt{2}\leq k\leq3+2\sqrt{2}$ , and the maximum value of $k=\frac{y}{x}$ is ${3+2\sqrt{2}}, \boxed{\text{A}}$ .

Solution 2

The equation represents a circle of radius $\sqrt{6}$ centered at $A=(3,3)$ . To find the maximal $k$ with $y=kx$ is equivalent to finding the maximum slope of a line passing through the origin $O$ and intersecting the circle. The steepest such line is tangent to the circle at some point $X$ . We have $XA=\sqrt{6}$ , $OA=\sqrt{18}$ , $\angle OXA = 90$ because the line is tangent to the circle. Using the pythagorean theorem, we have $OX=\sqrt{12}$ .

The slope we are looking for is equivalent to $\tan (\theta + 45)$ where $\angle AOX = \theta$ . Using tangent addition,

$\tan (\theta + 45)= \frac{\tan \theta + \tan 45}{1-\tan\theta\tan 45} = \frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}$

So $\boxed{A}$ is the answer

Solution 3

Following the steps of Solution 2, we get $OX=\sqrt{12}$ . Let $X = (x, y)$ . Based on the distance to the origin and to the centre of the circle

$x^2 + y^2 = 12$

$(x - 3)^2 + (y - 3)^2 = 6$

Expanding the latter and substituting the former into the equation gives $30 - 6(x + y) = 6$ . $x + y = 4$ . We can square this and remove the 1st equation to get $2xy = 4$ , or $xy = 2$ .

Substitute $y = 4 - x$ to get

$x(4 - x) = 2$

$x^2 - 4x + 2 = 0$

$x = 2 \pm \sqrt{2}$

Since $x$ and $y$ are symmetric, we want $x$ to be the lower number to maximize $k$ , so $x = 2 - \sqrt{2}$ and $y = 2 + \sqrt{2}$ .

$\frac{y}{x} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}}$

$\frac{y}{x} = \frac{(2 + \sqrt{2})^2}{2}$

$\frac{y}{x} = 3 + 2\sqrt{2}$

@@ Line 48: / Line 48: @@
 <cmath>x^2 - 4x + 2 = 0</cmath>
-<math></math>x = 2 \pm \sqrt{2}<math>4
+<cmath>x = 2 \pm \sqrt{2}</cmath>
-Since </math>x<math> and </math>y<math> are symmetric, we want </math>x<math> to be the lower number to maximize </math>k<math>, so </math>x = 2 - \sqrt{2}<math> and </math>y = 2 + \sqrt{2}$.
+Since <math>x</math> and <math>y</math> are symmetric, we want <math>x</math> to be the lower number to maximize <math>k</math>, so <math>x = 2 - \sqrt{2}</math> and <math>y = 2 + \sqrt{2}</math>.
 <cmath>\frac{y}{x} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}}</cmath>

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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