2024 AMC 10B Problem 14

P11-15~4 minPrint

Geometric Probability Circles Coordinate Geometry Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

A dartboard is the region $B$ in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8$ . A target $T$ is the region where $(x^2 + y^2 - 25)^2 \le 49$ . A dart is thrown and lands at a random point in B. The probability that the dart lands in $T$ can be expressed as $\frac{m}{n} \cdot \pi$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ?

Show solution without answering

Key insight

The board is a tilted square of area 128. The target is a ring of area 14 pi that fits exactly inside it.

Solution

Probability here means (area of target) divided by (area of board). So find each area.

The board.

Points with $|x| + |y| \le 8$ : on the axes this reaches $(8, 0)$ , $(0, 8)$ , $(-8, 0)$ , $(0, -8)$ . In the first quarter the edge is the line $x + y = 8$ , and the other quarters are mirror images. So the board is a square tilted on its corner.
Its two diagonals each have length $16$ . A square's area is half the product of its diagonals: $\frac12 \cdot 16 \cdot 16 = 128$ .

The target.

3. $(x^2 + y^2 - 25)^2 \le 49$ means the number $x^2 + y^2 - 25$ is between $-7$ and $7$ . Add $25$ :
$18 \le x^2 + y^2 \le 32.$
4. $x^2 + y^2$ is the squared distance from the center $(0,0)$ . So the target is the ring between the circle of radius $\sqrt{18}$ and the circle of radius $\sqrt{32}$ .
5. Area of the ring $=$ big circle minus small circle $= 32\pi - 18\pi = 14\pi$ .

Does the ring fit on the board?

The closest point of the board's edge to the center is the middle of an edge. For the edge from $(8,0)$ to $(0,8)$ that is $(4, 4)$ , at distance $\sqrt{16 + 16} = \sqrt{32}$ .
The outer circle has radius $\sqrt{32}$ as well, so it just touches the edges and the whole ring is on the board.

The probability.

$\dfrac{14\pi}{128} = \dfrac{7}{64}\pi$ . Since $7$ and $64$ share no factor, $m = 7$ , $n = 64$ , and $m + n = 71$ .

The answer is $\boxed{\textbf{(B)}\ 71}$ .

Why this works

A "random point" probability is a ratio of areas, so the real job is to recognize each shape. An equation with $|x| + |y|$ makes a tilted square, and one with $x^2 + y^2$ makes circles. Always check that the target region fits inside the board. If part of it hung over the edge, you would have to cut that part off.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The board $|x| + |y| \le 8$ is a square (a diamond) with vertices $(\pm 8, 0)$ and $(0, \pm 8)$ . Its diagonals both have length $16$ , so its area is $\frac12 \cdot 16 \cdot 16 = 128$ .

The target condition $(x^2 + y^2 - 25)^2 \le 49$ means $-7 \le x^2 + y^2 - 25 \le 7$ , i.e.
$18 \le x^2 + y^2 \le 32.$
This is the ring between circles of radii $\sqrt{18}$ and $\sqrt{32}$ centered at the origin, with area $\pi(32 - 18) = 14\pi$ .

Does the ring lie inside the board? The nearest points of the board's boundary to the origin are on the lines $\pm x \pm y = 8$ , at distance $\frac{8}{\sqrt2} = \sqrt{32}$ . So the outer circle is exactly inscribed in the diamond and the whole ring is inside.

The probability is $\frac{14\pi}{128} = \frac{7}{64}\pi$ , so $m + n = 7 + 64 = 71$ .

The answer is $\boxed{\textbf{(B)}\ 71}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 10: Geometric Probability
Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 12: Circles and Angles