AMC 10 · 2025 · #25
Grade 8 geometry-2dA point P is chosen at random inside square ABCD. The probability that AP is neither the shortest nor the longest side of △APB can be written as ea+bπ−cd, where a,b,c,d, and e are positive integers, gcd(a,b,c,e)=1, and d is not divisible by the square of a prime. What is a+b+c+d+e?
Pick an answer.
AMC 10 2025 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: A point $P$ is dropped uniformly at random inside a square $ABCD$. Look at the triangle $\triangle APB$, 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 $\dfrac{a+b\pi-c\sqrt{d}}{e}$ with $\gcd(a,b,c,e)=1$ and $d$ squarefree, and you report $a+b+c+d+e$.
Givens: $P$ is chosen uniformly at random from the interior of square $ABCD$; $\triangle APB$ has sides $AP$, $BP$, and $AB$, where $AB$ is a full side of the square; The target probability is written $\dfrac{a+b\pi-c\sqrt{d}}{e}$ with $a,b,c,d,e$ positive integers, $\gcd(a,b,c,e)=1$, and $d$ not divisible by any perfect square; Answer choices: (A) 25, (B) 26, (C) 27, (D) 28, (E) 29
Unknowns: The value of $a+b+c+d+e$, which first requires finding the probability in the form $\dfrac{a+b\pi-c\sqrt{d}}{e}$
Understand
Restated: A point $P$ is dropped uniformly at random inside a square $ABCD$. Look at the triangle $\triangle APB$, 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 $\dfrac{a+b\pi-c\sqrt{d}}{e}$ with $\gcd(a,b,c,e)=1$ and $d$ squarefree, and you report $a+b+c+d+e$.
Givens: $P$ is chosen uniformly at random from the interior of square $ABCD$; $\triangle APB$ has sides $AP$, $BP$, and $AB$, where $AB$ is a full side of the square; The target probability is written $\dfrac{a+b\pi-c\sqrt{d}}{e}$ with $a,b,c,d,e$ positive integers, $\gcd(a,b,c,e)=1$, and $d$ not divisible by any perfect square; Answer choices: (A) 25, (B) 26, (C) 27, (D) 28, (E) 29
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable, #16 Change Focus / Count the Complement
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.
Execute — Answer: A
7.SP.C.7 Step 1 Turn probability into area
- Since $P$ is uniform inside the square, the probability of any event equals the fraction of the square's area where it holds.
- Scale the square to side length 1 so its area is 1 and "probability" literally equals "area of the good region." Place coordinates with $A=(0,1)$ and $B=(1,1)$ on the top edge, $C=(1,0)$, $D=(0,0)$; then $AB=1$ is fixed and $P=(x,y)$ ranges over the unit square.
💡 For a uniform point, chance is just the share of the area that works.
6.NS.C.7 Step 2 "Middle side" splits into two cases
- Among the three lengths $AP$, $BP$, $AB=1$, saying $AP$ is neither shortest nor longest means $AP$ is the middle value, i.e.
- it lies strictly between the other two.
- A number can sit in the middle in exactly two ways, depending on which of $BP$ and $AB$ is larger.
- This gives two disjoint cases whose areas we add.
💡 Being the middle of three numbers means being between the smallest and largest.
8.G.B.8 Step 3 Translate each comparison to a region
- Using $AP^2=x^2+(y-1)^2$ and $BP^2=(x-1)^2+(y-1)^2$, the comparison $AP<BP$ simplifies: $x^2<(x-1)^2$ gives $x<\tfrac12$.
- So $AP<BP$ (closer to $A$) is the left half $x<\tfrac12$, and $AP>BP$ is the right half $x>\tfrac12$ — split by the perpendicular bisector of $AB$.
- Next, $AP<1$ means $x^2+(y-1)^2<1$, i.e.
- $P$ is inside the quarter circle of radius 1 centered at $A$ (it runs from $B$ to $D$).
- So Case 1 = right of the divider and inside the circle; Case 2 = left of the divider and outside the circle.
💡 Closer-to-$A$ is one side of a line; shorter-than-a-side is inside a circle.
7.G.B.4 Step 4 Case 1 area: sector minus a right triangle
- Let $M=(\tfrac12,1)$ mark the top of the divider and let $N=(\tfrac12,1-\tfrac{\sqrt3}{2})$ be where the divider $x=\tfrac12$ meets the circle, so $AN=1$.
- In right triangle $AMN$ the leg $AM=\tfrac12$ is exactly half the hypotenuse $AN=1$, which is the 30-60-90 signature, so $\angle BAN=60^\circ$; and by the Pythagorean theorem $MN=\sqrt{1-\tfrac14}=\tfrac{\sqrt3}{2}$.
- The Case-1 region (right of the divider, inside the circle) is exactly the $60^\circ$ sector $ABN$ with the left sliver $\triangle AMN$ removed.
- Sector $ABN=\tfrac{60}{360}\pi(1)^2=\tfrac{\pi}{6}$ and $[\triangle AMN]=\tfrac12\cdot\tfrac12\cdot\tfrac{\sqrt3}{2}=\tfrac{\sqrt3}{8}$.
💡 Take the pie-slice up to the divider, then trim off the triangle poking past it.
7.G.B.4 Step 5 Case 2 area: rectangle minus the circle's part
- Case 2 lives in the left rectangle $x\in[0,\tfrac12],\,y\in[0,1]$ (area $\tfrac12$) and must lie OUTSIDE the circle, so it is easier to subtract the inside part.
- The whole quarter circle has area $\tfrac{\pi}{4}$.
- Its right portion (inside the circle, $x>\tfrac12$) is exactly the Case-1 region, $\tfrac{\pi}{6}-\tfrac{\sqrt3}{8}$, so its left portion is $\tfrac{\pi}{4}-\left(\tfrac{\pi}{6}-\tfrac{\sqrt3}{8}\right)=\tfrac{\pi}{12}+\tfrac{\sqrt3}{8}$.
- Removing that from the rectangle gives the good region.
💡 The good part is the rectangle with the circle's slice scooped out, so subtract the slice.
7.NS.A.3 Step 6 Add the cases and read off the integers
- Add the two disjoint areas.
- The $\sqrt3$ terms combine: $-\tfrac{\sqrt3}{8}-\tfrac{\sqrt3}{8}=-\tfrac{\sqrt3}{4}$, and the $\pi$ terms give $\tfrac{\pi}{6}-\tfrac{\pi}{12}=\tfrac{\pi}{12}$.
- So the probability is $\tfrac12+\tfrac{\pi}{12}-\tfrac{\sqrt3}{4}=\tfrac{6+\pi-3\sqrt3}{12}$.
- Matching $\tfrac{a+b\pi-c\sqrt d}{e}$ gives $a=6,\,b=1,\,c=3,\,d=3,\,e=12$ (indeed $\gcd(6,1,3,12)=1$ and $3$ is squarefree), so $a+b+c+d+e=6+1+3+3+12=25$.
- The answer is (A).
💡 Two disjoint regions just add, and the answer form tells you which integers to name.
7.SP.C.7 Since $P$ is uniform inside the square, the probability of any event equals the 6.NS.C.7 Among the three lengths $AP$, $BP$, $AB=1$, saying $AP$ is neither shortest nor 8.G.B.8 Using $AP^2=x^2+(y-1)^2$ and $BP^2=(x-1)^2+(y-1)^2$, the comparison $AP<BP$ simp 7.G.B.4 Let $M=(\tfrac12,1)$ mark the top of the divider and let $N=(\tfrac12,1-\tfrac{\ 7.G.B.4 Case 2 lives in the left rectangle $x\in[0,\tfrac12],\,y\in[0,1]$ (area $\tfrac1 7.NS.A.3 Add the two disjoint areas. The $\sqrt3$ terms combine: $-\tfrac{\sqrt3}{8}-\tfr Review
Reasonableness: Numerically $P=\tfrac{6+3.1416-5.196}{12}\approx\tfrac{3.946}{12}\approx0.329$, a sensible probability strictly between 0 and 1. The two pieces are also individually positive and believable: Case 1 $=\tfrac{\pi}{6}-\tfrac{\sqrt3}{8}\approx0.31$ and Case 2 $=\tfrac12-\tfrac{\pi}{12}-\tfrac{\sqrt3}{8}\approx0.02$, 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).
Alternative: Instead of geometry-by-regions, integrate directly. With $u=1-y$, Case 1 is $\int_{1/2}^{1}\sqrt{1-x^2}\,dx=\tfrac{\pi}{6}-\tfrac{\sqrt3}{8}$, and Case 2 is $\tfrac12-\int_{0}^{1/2}\sqrt{1-x^2}\,dx=\tfrac12-\tfrac{\pi}{12}-\tfrac{\sqrt3}{8}$; the two agree with the sector/rectangle computation and sum to $\tfrac{6+\pi-3\sqrt3}{12}$. A common trap is to compute only Case 1 (thinking the triangle must be near $A$) and forget Case 2, where $AP$ is middle but longer than $AB$; that would undercount and give a wrong, too-small probability.
CCSS standards used (min grade 8)
7.SP.C.7Develop a probability model and use it to find probabilities of events (Turning the uniform random point into an area model, so the probability equals the area of the good region in the unit square.)6.NS.C.7Understand ordering and absolute value of rational numbers (Reading "neither shortest nor longest" as "$AP$ is the middle of three lengths" and splitting it into the two ordered cases $BP<AP<1$ and $1<AP<BP$.)8.G.B.8Apply the Pythagorean theorem to find the distance between two points in a coordinate system (Writing $AP^2=x^2+(y-1)^2$ and $BP^2=(x-1)^2+(y-1)^2$ to convert $AP<BP$ into $x<\tfrac12$ and $AP<1$ into the quarter-circle $x^2+(y-1)^2<1$.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the $60^\circ$ sector area $\tfrac{\pi}{6}$, the quarter-circle area $\tfrac{\pi}{4}$, and the left/right splits of the circular region.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $MN=\sqrt{1-\tfrac14}=\tfrac{\sqrt3}{2}$ and recognizing the 30-60-90 triangle $AMN$ that fixes the $60^\circ$ central angle.)7.NS.A.3Solve real-world and mathematical problems involving the four operations with rational numbers (Adding the two case areas, combining the $\pi$ and $\sqrt3$ terms, and reading off $a,b,c,d,e$ to get $a+b+c+d+e=25$.)
⭐ "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.
⭐ "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.
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