AMC 10 · 2024 · #14

Grade 8 probability

Archetype Compound Area by Decomposition / Difference

probability-basicarea-circlescoordinate-geometryfraction-arithmetic identify-subproblemsarea-differenceconvert-to-algebra ↑ Prerequisites: probability-basicarea-circlescoordinate-geometry

📏 Medium solution 💡 3 insights

Problem

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 $m + n$ ?

$\textbf{(A) }39 \qquad \textbf{(B) }71 \qquad \textbf{(C) }73 \qquad \textbf{(D) }75 \qquad \textbf{(E) }135 \qquad$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

135

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Toolkit + CCSS Solution

Understand

Restated: Region $B$ in the plane consists of all $(x, y)$ with $|x| + |y| \leq 8$. Region $T$ consists of all $(x, y)$ with $(x^2 + y^2 - 25)^2 \leq 49$. A dart lands uniformly at random in $B$. The probability that it lands in $T$ is $\tfrac{m}{n} \pi$ in lowest terms. Find $m + n$.

Givens: Dartboard: $B = \{(x,y) : |x| + |y| \leq 8\}$; Target: $T = \{(x,y) : (x^2 + y^2 - 25)^2 \leq 49\}$; Dart lands uniformly at random in $B$; $P(\text{lands in } T) = \tfrac{m}{n}\pi$ with $\gcd(m, n) = 1$; Answer choices: (A) $39$, (B) $71$, (C) $73$, (D) $75$, (E) $135$

Unknowns: $m + n$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

Both regions live in the plane and the answer is a ratio of areas, so the picture is everything. Tool #1 (Draw a Diagram) reveals what each inequality really is: $|x| + |y| \leq 8$ is a square (rotated $45^\circ$) with diagonals on the axes, and $(x^2 + y^2 - 25)^2 \leq 49$ is an annulus centered at the origin. Tool #7 (Identify Subproblems) splits the calculation in three clean parts — area of $B$ (rotated square), area of $T$ (annulus), and the ratio. Tool #13 (Convert to Algebra) tidies the target inequality: let $r^2 = x^2 + y^2$, then $(r^2 - 25)^2 \leq 49$ becomes $|r^2 - 25| \leq 7$, i.e. $18 \leq r^2 \leq 32$ — two concentric circles.

Execute — Answer: B

#1 Draw a Diagram 7.G.B.6 Step 1

Identify region $B$.
The inequality $|x| + |y| \leq 8$ is a square rotated $45^\circ$ with vertices on the axes at $(\pm 8, 0)$ and $(0, \pm 8)$, since on each line $\pm x \pm y = 8$ we get the four sides.
The two diagonals lie along the axes and each has length $16$, so Area$(B) = \tfrac{1}{2} d_1 d_2 = \tfrac{1}{2} \cdot 16 \cdot 16 = 128$.

$$\text{Area}(B) = \tfrac{1}{2}(16)(16) = 128$$

💡 An absolute-value sum is a tilted square — draw the four corner triangles and read off the area at a glance.

#13 Convert to Algebra 8.EE.A.2 Step 2

Unpack the target inequality.
Let $r^2 = x^2 + y^2$ (squared distance from origin).
The inequality $(r^2 - 25)^2 \leq 49$ is the same as $|r^2 - 25| \leq 7$, i.e.
$-7 \leq r^2 - 25 \leq 7$, which gives $18 \leq r^2 \leq 32$.

$$(r^2 - 25)^2 \leq 49 \iff 18 \leq r^2 \leq 32$$

💡 Squaring then bounding $\leq 49$ is just $|\cdot| \leq 7$ — Grade 8 "square root and squaring" reverse each other.

#7 Identify Subproblems 7.G.B.4 Step 3

Compute the area of the annulus $T$.
The region $18 \leq r^2 \leq 32$ is the ring between circles of radius-squared $18$ and $32$.
Its area is the outer disk minus the inner disk: $\pi (32) - \pi(18) = 14 \pi$.
(No need to take square roots — the area formula $\pi r^2$ uses $r^2$ directly.)

$$\text{Area}(T) = \pi(32) - \pi(18) = 14\pi$$

💡 Annulus $=$ big circle minus small circle. The area uses $r^2$, so the bounds $r^2 \in [18, 32]$ plug in directly.

#1 Draw a Diagram 8.G.B.7 Step 4

Check that $T \subseteq B$ so we can use Area$(T)$ directly.
The outermost circle of $T$ has $r^2 = 32$, so $r = \sqrt{32} \approx 5.66$.
The closest point of the boundary of $B$ to the origin is at distance $\tfrac{8}{\sqrt{2}} = 4\sqrt{2} \approx 5.66$.
Since $\sqrt{32} = 4\sqrt{2}$ exactly, the outer circle of $T$ is tangent to the four sides of $B$ from inside — so $T \subseteq B$ exactly, with the circle kissing the square's edges.

$$\sqrt{32} = 4\sqrt{2} = \text{dist(origin, side of }B) \;\Longrightarrow\; T \subseteq B$$

💡 The picture shows the annulus snugly inside the diamond — the outer circle just touches the diamond's sides. Without this check, we couldn't use Area$(T)$ as is.

#7 Identify Subproblems 6.RP.A.3 Step 5

Compute the probability as a ratio of areas and simplify.
$P = \tfrac{14\pi}{128} = \tfrac{7 \pi}{64}$.
So $m = 7$, $n = 64$.
$\gcd(7, 64) = 1$ ($7$ prime, $64 = 2^6$), so the fraction is in lowest terms.

$$\frac{m}{n}\pi = \frac{14\pi}{128} = \frac{7\pi}{64} \;\Longrightarrow\; m + n = 7 + 64 = 71 \;\Rightarrow\; \textbf{(B)}$$

💡 Geometric probability is just the area-ratio — reduce, then add the numerator and denominator.

[1] #1 7.G.B.6 Identify region $B$. The inequality $|x| + |y| \leq 8$ is a square rotated $45^\

[2] #13 8.EE.A.2 Unpack the target inequality. Let $r^2 = x^2 + y^2$ (squared distance from origi

[3] #7 7.G.B.4 Compute the area of the annulus $T$. The region $18 \leq r^2 \leq 32$ is the rin

[4] #1 8.G.B.7 Check that $T \subseteq B$ so we can use Area$(T)$ directly. The outermost circl

[5] #7 6.RP.A.3 Compute the probability as a ratio of areas and simplify. $P = \tfrac{14\pi}{128

Review

Reasonableness: The annulus has area $14\pi \approx 44$, and the square has area $128$, so the probability is about $44/128 \approx 0.34$. That fits a thick-ish ring covering a sizable chunk of the diamond — visually plausible. As a second check, the numbers $7$ and $64$ should be relatively prime, and they are. The other answer choices match other plausible-looking simplifications: (A) $39 = 7 + 32$ would come from forgetting to subtract the inner disk, (E) $135 = 14 + 121$ would come from squaring $\sqrt{32}$ incorrectly. Our (B) $= 71$ is the clean derivation.

Alternative: Tool #6 (Guess and Check) on simplification: $\tfrac{14\pi}{128}$ has obvious factor $2$. Divide top and bottom by $2$: $\tfrac{7\pi}{64}$. Check $\gcd(7, 64)$: $7$ is prime, doesn't divide $64$. Done. $7 + 64 = 71$. This is essentially Tool #3 (Eliminate) too — (A), (C), (D), (E) are all off by what looks like one common arithmetic slip each.

CCSS standards used (min grade 8)

7.G.B.6 Solve real-world problems involving area, surface area, and volume (Computing the area of the rotated square $B$ as $\tfrac{1}{2} d_1 d_2 = 128$ using the diagonal formula for a rhombus.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Reducing $(r^2 - 25)^2 \leq 49$ to $|r^2 - 25| \leq 7$, i.e. $18 \leq r^2 \leq 32$ — using the inverse-of-squaring move.)
7.G.B.4 Know the formulas for area and circumference of a circle (Computing the annulus area as $\pi(32) - \pi(18) = 14\pi$ using $A = \pi r^2$ on the two bounding circles.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Checking that $\sqrt{32} = 4\sqrt{2}$ equals the perpendicular distance $\tfrac{8}{\sqrt{2}}$ from the origin to a side of $B$, so $T \subseteq B$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Computing the geometric probability as the ratio Area$(T)$ / Area$(B) = 14\pi/128 = 7\pi/64$.)

⭐ This AMC 10 problem only needs Grade 8 square-root reasoning plus the Grade 7 circle-area formula that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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