AMC 10 · 2025 · #25

Grade 8 geometry-2d

Archetype Geometric Probability by Area └ Sub-archetype Area-Proportional Region Probability

geometric-probabilityperpendicular-bisectorcircular-sectorarea-difference casework ↑ Prerequisites: geometric-probabilitycircular-sector

📏 Long solution 💡 4 insights

Problem

A point P is dropped uniformly at random inside a square ABCD. Look at the triangle △ APB, whose three sides are AP, BP, and the fixed side AB. Find the probability that AP is neither the shortest nor the longest of these three sides. The probability has the form (a+bπ-c√(d))/e with gcd(a,b,c,e)=1 and d squarefree, and you report a+b+c+d+e.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Because P is uniform in the square, every probability here is just the area of the region of good points, so the whole problem becomes a geometry-of-regions question. Tool #7 (Identify Subproblems) is primary: "AP is the middle side" is not one condition but two disjoint ones, BP ＜ AP ＜ AB and AB ＜ AP ＜ BP, and each carves out its own region. Tool #4 (Introduce a Variable) puts coordinates on the square so each comparison of lengths becomes a clean condition on P=(x,y): comparing AP with BP becomes a vertical line, comparing AP with AB=1 becomes a circle. Tool #1 (Draw a Diagram) turns those conditions into a sector, a rectangle, and a right triangle whose areas we can read off. Tool #16 (Count the Complement) handles the second case, where the good region is the part of a rectangle OUTSIDE the circle, most easily found by subtracting the inside part. Adding the two case areas gives the probability.

1STEP 1

Turn probability into area

Uniform P makes probability equal area, so scale the square to side 1 with A=(0,1), B=(1,1) on the top edge.

A=(0,1), B=(1,1), AB=1, Pr(event)=[good region]

2STEP 2

"Middle side" splits into two cases

AP being the middle length splits into two disjoint cases: BP ＜ AP ＜ 1, or 1 ＜ AP ＜ BP.

Case 1: BP ＜ AP ＜ 1 Case 2: 1 ＜ AP ＜ BP

3STEP 3

Translate each comparison to a region

AP ＜ BP is the left half x ＜ 1/2, and AP ＜ 1 is inside the quarter circle of radius 1 centered at A.

AP ＜ BP⇔ x ＜ 1/2, AP ＜ 1⇔ x²+(y-1)² ＜ 1

4STEP 4

Case 1 area: sector minus a right triangle

Case 1 is the 60° sector ABN (area π/6) minus right triangle AMN (area √3/8), so it measures π/6-√3/8.

[Case 1]=π/6-√3/8

5STEP 5

Case 2 area: rectangle minus the circle's part

The left rectangle has area 1/2; removing the circle's left part π/12+√3/8 leaves 1/2-π/12-√3/8.

[Case 2]=1/2-(π/12+√3/8)=1/2-π/12-√3/8

6STEP 6

Add the cases and read off the integers

Adding gives 1/2+π/12-√3/4=(6+π-3√3)/12, so 6+1+3+3+12=25 — answer (A).

P=(6+π-3√3)/12 → a+b+c+d+e=25 (A)

Answer

Numerically P=(6+3.1416-5.196)/12≈3.946/12≈0.329, a sensible probability strictly between 0 and 1. The two pieces are also individually positive and believable: Case 1 =π/6-√3/8≈0.31 and Case 2 =1/2-π/12-√3/8≈0.02, so the near-A inside-the-circle region does most of the work while the thin outside-the-circle strip contributes a little. A structural check confirms the setup: in Case 1 the fixed side AB is the longest and in Case 2 it is the shortest, so in both cases AP genuinely sits in the middle. Finally gcd(6,1,3,12)=1 and 3 is squarefree, so the required form is met and a+b+c+d+e=25 is forced, matching (A).

💡Key takeaway

"Middle side" means between the other two, which is really two region cases; draw them as a pie-slice and a scooped rectangle, then add the areas.

Turn probability into area
"Middle side" splits into two cases
Translate each comparison to a region
Case 1 area: sector minus a right triangle
Case 2 area: rectangle minus the circle's part
Add the cases and read off the integers