1984 AHSME Problems/Problem 29

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 ${3+2\sqrt{2}}, \boxed{\text{A}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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1984 AHSME Problems/Problem 29

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