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

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The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent addition,
 
The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent addition,
 
   
 
   
−
<cmath> \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{3}</cmath>
+
<cmath> \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}</cmath>
  
 
So <math>\boxed{A}</math> is the answer
 
So <math>\boxed{A}</math> is the answer

Revision as of 19:25, 8 March 2021

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

See Also

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

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