1984 AMC 12 Problem 29

Question

Find the largest value of

\frac{y}{x}

for pairs of real numbers

(x,y)

which satisfy

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

3+2\sqrt2

2+\sqrt3

3\sqrt3

6

6+2\sqrt3

Interpret

\frac{y}{x}

as the slope of a line through the origin

At an extreme slope, the line

y=mx

is tangent to the circle

The circle lies entirely where

x\gt0,

so

\frac{y}{x}

is the slope

m

of the line

y=mx

through a point on it. At an extreme, this line is tangent. The distance from the center

(3,3)

to

mx-y=0

must equal

\sqrt6,

so

\frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt6.

Squaring and simplifying gives

m^2-6m+1=0,

hence

m=3\pm2\sqrt2.

The larger value is

3+2\sqrt2.

Therefore, the correct answer is A.

Problem 29 in other years