AMC 10 · 2024 · #14
Grade 8 probabilityProblem
A dartboard is the region B in the coordinate plane consisting of points (x,y) such that ∣x∣+∣y∣≤8. A target T is the region where (x2+y2−25)2≤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 nm⋅π, where m and n are relatively prime positive integers. What is m+n?
Pick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $135$
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
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$(B) = \tfrac{1}{2} d_1 d_2 = \tfrac{1}{2} \cdot 16 \cdot 16 = 128$.
💡 An absolute-value sum is a tilted square — draw the four corner triangles and read off the area at a glance.
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 $18 \leq r^2 \leq 32$.
💡 Squaring then bounding $\leq 49$ is just $|\cdot| \leq 7$ — Grade 8 "square root and squaring" reverse each other.
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.)
💡 Annulus $=$ big circle minus small circle. The area uses $r^2$, so the bounds $r^2 \in [18, 32]$ plug in directly.
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.
💡 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.
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.
💡 Geometric probability is just the area-ratio — reduce, then add the numerator and denominator.
7.G.B.6 Identify region $B$. The inequality $|x| + |y| \leq 8$ is a square rotated $45^\ 8.EE.A.2 Unpack the target inequality. Let $r^2 = x^2 + y^2$ (squared distance from origi 7.G.B.4 Compute the area of the annulus $T$. The region $18 \leq r^2 \leq 32$ is the rin 8.G.B.7 Check that $T \subseteq B$ so we can use Area$(T)$ directly. The outermost circl 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.6Solve 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.2Use 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.4Know 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.7Apply 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.3Use 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!
⭐ This AMC 10 problem only needs Grade 8 square-root reasoning plus the Grade 7 circle-area formula that you already know!
More like this
Same archetype — closest grade level first.
