2005 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2005 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12B solutions, or check the answer key.

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Concepts:circletangent linecoordinate geometry

Difficulty rating: 1630

14.

A circle having center (0,k),(0, k), with k>6,k \gt 6, is tangent to the lines y=x,y = x, y=xy = -x and y=6.y = 6. What is the radius of this circle?

6266\sqrt{2} - 6

66

626\sqrt{2}

1212

6+626 + 6\sqrt{2}

Solution:

Since the circle is tangent to y=6y = 6 and its center (0,k)(0, k) is above that line, the radius is r=k6.r = k - 6.

The distance from (0,k)(0, k) to the line xy=0x - y = 0 is 0k2=k2,\dfrac{|0 - k|}{\sqrt2} = \dfrac{k}{\sqrt2}, and this must also equal r.r.

Setting k2=k6\dfrac{k}{\sqrt2} = k - 6 gives k=6221=62(2+1)=12+62.k = \dfrac{6\sqrt2}{\sqrt2 - 1} = 6\sqrt2\,(\sqrt2+1) = 12 + 6\sqrt2.

Then r=k6=6+62.r = k - 6 = 6 + 6\sqrt2.

Thus, the correct answer is E.

Problem 14 in Other Years