AMC 10 · 2023 · #19

Grade 7 geometry-2d

geometric-probabilityarea-rectanglessymmetry-argumentprobability-basic easier-related-problemidentify-subproblemssymmetry-argument ↑ Prerequisites: geometric-probabilityarea-rectangles

📏 Medium solution 💡 3 insights

Problem

Sonya the frog chooses a point uniformly at random lying within the square
$[0, 6]$ $\times$ $[0, 6]$ in the coordinate plane and hops to that point. She then randomly
chooses a distance uniformly at random from $[0, 1]$ and a direction uniformly at
random from {north, south, east, west}. All of her choices are independent. She now
hops the distance in the chosen direction. What is the probability that she lands
outside the square?

Pick an answer.

(A)

$frac{1}{6}$

(B)

$frac{1}{12}$

(C)

$frac{1}{4}$

(D)

$frac{1}{10}$

(E)

$frac{1}{9}$

AMC 10 2023 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.

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Toolkit + CCSS Solution

Understand

Restated: Sonya the frog lands at a uniformly random point $(X, Y)$ in the square $[0, 6] \times [0, 6]$, then picks a uniformly random distance $D$ in $[0, 1]$ and a uniformly random cardinal direction. She hops that distance in that direction. Find the probability that her landing spot is outside the square.

Givens: $(X, Y)$ uniform on $[0, 6] \times [0, 6]$; $D$ uniform on $[0, 1]$; Direction uniform on $\{$N, S, E, W$\}$, all independent; Hop changes one coordinate by $\pm D$ depending on the direction; Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{1}{12}$, (C) $\tfrac{1}{4}$, (D) $\tfrac{1}{10}$, (E) $\tfrac{1}{9}$

Unknowns: $P(\text{Sonya lands outside the }6\times 6\text{ square})$

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #1 Draw a Diagram, #7 Identify Subproblems

Direct computation across all four directions is messy, but symmetry collapses it. Tool #9 (Easier Problem): solve the conditional probability for ONE direction — say north — then use Tool #7 (Identify Subproblems) and the law of total probability with symmetry to get the full answer. Tool #1 (Diagram) does the heavy lifting: draw the $(Y, D)$ rectangle and shade $Y + D > 6$ — the favorable region is a small right triangle whose area is read off by inspection.

Execute — Answer: B

#9 Solve an Easier Related Problem 7.SP.C.7 Step 1

Exploit symmetry to reduce to one direction.
The square and the four cardinal directions are symmetric under $90^\circ$ rotation, so $P(\text{out}\mid\text{N}) = P(\text{out}\mid\text{S}) = P(\text{out}\mid\text{E}) = P(\text{out}\mid\text{W}) =: p$.
By the law of total probability $P(\text{out}) = \tfrac{1}{4}(p + p + p + p) = p$ — the overall probability equals the one-direction conditional probability.

$$P(\text{out}) = \sum_{\text{dir}} \tfrac{1}{4}\,P(\text{out}\mid\text{dir}) = p$$

💡 Four identical copies, each weighted $\tfrac{1}{4}$, just give the single copy back — symmetry turns the four-case problem into one case.

#7 Identify Subproblems 7.SP.C.7 Step 2

Reduce the one-direction problem to a $2$D geometric-probability question.
Conditional on hopping north, the new position is $(X, Y + D)$.
Since $X$ never changes and $D \le 1$, the $x$-coordinate stays in $[0, 6]$.
Sonya escapes the square iff $Y + D > 6$.
So $p = P(Y + D > 6)$ with $Y$ uniform on $[0, 6]$ and $D$ uniform on $[0, 1]$, independent.

$$p = P(Y + D > 6),\ Y \sim U[0, 6],\ D \sim U[0, 1]$$

💡 Only the $y$-coordinate can leave the square when hopping north — and only when start position is within $1$ of the top edge.

#1 Draw a Diagram 6.G.A.1 Step 3

Draw the $(Y, D)$ sample space as a $6 \times 1$ rectangle.
The line $Y + D = 6$ enters the rectangle at $(Y, D) = (5, 1)$ (top) and exits at $(6, 0)$ (right).
The favorable region $\{Y + D > 6\}$ is the small right triangle in the lower-right corner of the rectangle with vertices $(5, 1)$, $(6, 1)$, $(6, 0)$ — legs of length $1$ each.

$$\text{triangle area} = \tfrac{1}{2} \cdot 1 \cdot 1 = \tfrac{1}{2}$$

💡 Drawing the rectangle and the line $Y + D = 6$ makes the favorable region obvious — a right triangle with both legs equal to $1$.

#7 Identify Subproblems 7.SP.C.7 Step 4

Compute the geometric probability.
Total sample-space area is $6 \cdot 1 = 6$.
So $p = \text{favorable}/\text{total} = (\tfrac{1}{2})/6 = \tfrac{1}{12}$.
By step 1, the overall probability $P(\text{out}) = p = \tfrac{1}{12}$.

$$P(\text{out}) = \dfrac{1/2}{6} = \dfrac{1}{12} \;\Rightarrow\; \textbf{(B) }\dfrac{1}{12}$$

💡 When two variables are uniform and independent, probability becomes area — divide favorable area by total area.

[1] #9 7.SP.C.7 Exploit symmetry to reduce to one direction. The square and the four cardinal di

[2] #7 7.SP.C.7 Reduce the one-direction problem to a $2$D geometric-probability question. Condi

[3] #1 6.G.A.1 Draw the $(Y, D)$ sample space as a $6 \times 1$ rectangle. The line $Y + D = 6$

[4] #7 7.SP.C.7 Compute the geometric probability. Total sample-space area is $6 \cdot 1 = 6$. S

Review

Reasonableness: Sanity check the magnitude: the dangerous strip near each edge is only $1$ unit wide (since $D \le 1$), so the dangerous strip area is at most $4 \cdot (6 \cdot 1) = 24$ out of $36$ — but escape requires both being in the strip AND drawing a $D$ large enough, which trims it heavily. The answer $\tfrac{1}{12} \approx 0.083$ feels right — small but not tiny. Cross-check the triangle: vertices $(5,1), (6,1), (6,0)$, legs of length $1$, area $\tfrac{1}{2}$. Ratio $\tfrac{1/2}{6} = \tfrac{1}{12}$. Matches choice (B).

Alternative: Tool #13 (Convert to Algebra) using a direct integral: $p = \int_0^6 \int_0^1 \mathbf{1}_{y + d > 6}\,\tfrac{1}{6}\,\tfrac{1}{1}\,dd\,dy = \tfrac{1}{6}\int_5^6 (1 - (6 - y))\,dy = \tfrac{1}{6}\int_5^6 (y - 5)\,dy = \tfrac{1}{6} \cdot \tfrac{1}{2} = \tfrac{1}{12}$. Same answer (B).

CCSS standards used (min grade 7)

7.SP.C.7 Develop probability models and use them to find probabilities of events (Modeling $Y$ and $D$ as independent uniforms, using symmetry to reduce $4$ direction cases to $1$, and reading the geometric-probability ratio favorable-area/total-area.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Recognizing the favorable region $\{(Y, D) : Y + D > 6\} \cap [0,6] \times [0,1]$ as a right triangle with legs $1$ and area $\tfrac{1}{2}$.)

⭐ By rotational symmetry, the overall escape probability equals the probability of escaping when hopping in any one chosen direction (say north). For north, only the $y$-coordinate matters: shade the rectangle $[0,6] \times [0,1]$ of $(Y, D)$ pairs, and the escape region $Y + D > 6$ is a right triangle with legs $1$ and area $\tfrac{1}{2}$. Divide by the total area $6$ to get $\textbf{(B) }\tfrac{1}{12}$.