The AMC 10A 2025 exam was held on November 5th, 2025. Below you’ll find all problems, detailed hints, and final answers as soon as they’re released.
Problem 1
Andy and Betsy both live in Mathville. Andy leaves Mathville on his bicycle at 1:30 traveling due north at a steady 8 miles per hour. Betsy leaves on her bicycle from the same point at 2:30, traveling due east at a steady 12 miles per hour. At what time will they be exactly the same distance from their common starting point?
- 3:30
- 3:45
- 4:00
- 4:15
- 4:30
Hints & Final Answer
Hint 1
Use v = \frac{d}{t}Hint 2
d_{Andy} = d_{Betsy}Hint 3
v_{Andy} \cdot t_{Andy} = v_{Betsy} \cdot t_{Betsy}Hint 4
8t = 12(t - 1)Hint 5
8t = 12t - 12 \Rightarrow 4t = 12 \Rightarrow t = 3 \text{ hours}Final Answer
(E) 4:30Problem 2
A box contains 10 pounds of a nut mix that is 50 percent peanuts, 20 percent cashews, and 30 percent almonds. A second nut mix containing 20 percent peanuts, 40 percent cashews, and 40 percent almonds is added to the box resulting in a new nut mix that is 40 percent peanuts. How many pounds of cashews are now in the box?
- 3.5
- 4
- 4.5
- 5
- 6
Hints & Final Answer
Hint 1
| Peanuts | Cashews | Almonds | |
|---|---|---|---|
| Box 1 | 5 | 2 | 3 |
| Box 2 | 20x | 40x | 40x |
| New Box | 40% |
Hint 2
\frac{5+20x}{5+2+3+20x+40x+40x}=40\%Hint 3
\frac{5+20x}{10+100x}=0.4 \;\Rightarrow\; 5+20x=4+40x \;\Rightarrow\; 20x=1 \;\Rightarrow\; x=\frac{1}{20}Hint 4
2+40x = 2 + 40\left(\frac{1}{20}\right)=4Final Answer
(B) 4Problem 3
How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length 2025?
- 2025
- 2026
- 3012
- 3037
- 4050
Hints & Final Answer
Hint 1
There are two cases.Hint 2
Use the Triangle Inequality.Hint 3
Case 1:

Since 2025 is the longest side, a \le 2025.
By the Triangle Inequality: 2a > 2025.
Since a must be a positive integer, 1013 \le a \le 2025.
Hint 4
Case 2:

Since 2025 is the longest side, a ≤ 2025.
By the Triangle Inequality: 4050 > a.
Since a must be a positive integer, 1 \le a ≤ 2025.
Hint 5
We have counted the equilateral triangle with side length 2025 in both Case 1 and Case 2.Hint 6
From Hints 3 and 4:1013 + 2025 - 1 = (2025 - 1013 + 1) + (2025) - 1 = 3037
Final Answer
(D) 3037Problem 4
A team of students is going to compete against a team of teachers in a trivia contest. The total number of students and teachers is 15. Ash, a cousin of one of the students, wants to join the contest. If Ash plays with the students, the average age on that team will increase from 12 to 14. If Ash plays with the teachers, the average age on that team will decrease from 55 to 52. How old is Ash?
- 28
- 29
- 30
- 32
- 33
Hints & Final Answer
Hint 1
Assume there are a students and 15 - a teachers.Hint 2
If the age of Ash is x, and the sum of the ages of students and teachers are s and t respectively, then:\frac{s}{a} = 12, \quad \frac{x + s}{a + 1} = 14
Hint 3
\frac{t}{15 - a} = 55, \quad \frac{t + x}{16 - a} = 52Hint 4
\begin{aligned} s &= 12a \quad &(1) \\ x + s &= 14a + 14 \quad &(2) \\ t &= 55(15 - a) \quad &(3) \\ t + x &= 52(16 - a) \quad &(4) \end{aligned} From (1) and (2): x = 2a + 14 \quad (5) Substitute (3), (4), and (5): 55(15 - a) + 2a + 14 = 52(16 - a)Hint 5
55(16 - 1) - 55a + 2a + 14 = 52 \times 16 - 52a Simplify: 16(55 - 52) - 55 + 14 = 55a - 2a - 52a 48 - 41 = a \quad \Rightarrow \quad a = 7Hint 6
From (5) in Hint 4: x = 2(7) + 14 = 28Final Answer
(A) 28Problem 5
Consider the sequence of positive integers
1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, …
What is the 2025th term in this sequence?
- 5
- 15
- 16
- 44
- 45
Hints & Final Answer
Hint 1
1;\quad 2,1;\quad 2,3,2,1;\quad 2,3,4,3,2,1;\quad 2,3,4,5,4,3,2,1;\ \ldotsHint 2
The n^{\text{th}} part (n\ge 2) starts from 2, increases up to n, then decreases back to 1. Hence, the length of the n^{\text{th}} part is 2(n-1).Hint 3
If 2025 is in the m^{\text{th}} part, then 1+2\cdot1+2\cdot2+\cdots+2\cdot(m-2)\;\lt\;2025\;\le\;1+2\cdot1+2\cdot2+\cdots+2\cdot(m-1)Hint 4
1+(m-2)(m-1)\;\lt\;2025\;\le\;1+(m-1)m Since 2025=45^{2}, we get m=46.Hint 5
For the 46th part: 2,3,4,\ldots,45,46,45,44,\ldots,1 We want the 2025th number from the beginning: 2025-\big(1+2\cdot1+2\cdot2+\cdots+2\cdot44\big)=2025-(1+44\cdot45)=44 Therefore, we should find the 44th number in the 46th part, which is 45.Final Answer
(E) 45Problem 6
In an equilateral triangle each interior angle is trisected by a pair of rays. The intersection of the interiors of the middle \(20^\circ\)-angle at each vertex is the interior of a convex hexagon. What is the degree measure of the smallest angle of this hexagon?
- 80
- 90
- 100
- 110
- 120
Hints & Final Answer
Hint 1

Hint 2

Hint 3

Hint 4

Final Answer
(C) 100Problem 7
Suppose a and b are real numbers. When the polynomial x^3 + x^2 + ax + b is divided by x - 1, the remainder is 4. When the polynomial is divided by x - 2, the remainder is 6. What is b - a?
- 14
- 15
- 16
- 17
- 18
Hints & Final Answer
Hint 1
Use 32. Remainder of Polynomial DivisionHint 2
x - 1 = 0 \Rightarrow x = 1 \Rightarrow 1^3 + 1^2 + a + b = 4 \Rightarrow a + b = 2Hint 3
x - 2 = 0 \Rightarrow x = 2 \Rightarrow 2^3 + 2^2 + 2a + b = 6 \Rightarrow 2a + b = -6Hint 4
From the two equations: \begin{cases} a + b = 2 \\ 2a + b = -6 \end{cases} \Rightarrow a = -8,\; b = 10 Therefore, b - a = 18.Final Answer
(E) 18Problem 8
Agnes writes the following four statements on a blank piece of paper.
- At least one of these statements is true.
- At least two of these statements are true.
- At least two of these statements are false.
- At least one of these statements is false.
Each statement is either true or false. How many false statements did Agnes write on the paper?
- 0
- 1
- 2
- 3
- 4
Hints & Final Answer
Hint 1
Use caseworkHint 2
Use casework based on the second statement.Hint 3
Case 1: The second statement is true. Then the first statement must also be true.
- At least one of these statements is true. — T
- At least two of these statements are true. — T
- At least two of these statements are false.
- At least one of these statements is false.
Hint 4
Case 1–1: The third statement is true.
- At least one of these statements is true. — T
- At least two of these statements are true. — T
- At least two of these statements are false. — T
- At least one of these statements is false.
Then, we can have at most one false statement, which contradicts the truth of the third statement. ✗
Hint 5
Case 1–2: The third statement is false.
- At least one of these statements is true. — T
- At least two of these statements are true. — T
- At least two of these statements are false. — F
- At least one of these statements is false. — T
Hint 6
Case 2: The second statement is false. Then the fourth statement is true, and the first statement is also true. However, both possibilities for the third statement contradict themselves. ✗
- At least one of these statements is true. — T
- At least two of these statements are true. — F
- At least two of these statements are false.
- At least one of these statements is false. — T
Final Answer
(B) 1Problem 9
Let f(x)=100x^3-300x^2+200x. For how many real numbers a does the graph of y=f(x-a) pass through the point (1,25)?
- 1
- 2
- 3
- 4
- more than 4
Hints & Final Answer
Hint 1
25=f(1-a)Hint 2
100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25Hint 3
Divide both sides by 25: 4(1-a)^3 - 12(1-a)^2 + 8(1-a) = 1 Expand: 4(1 - 3a + 3a^2 - a^3) - 12(1 - 2a + a^2) + 8(1 - a) = 1Hint 4
Simplify to a cubic in a: 0 = 4a^3 - 4a + 1Hint 5
Let g(a)=4a^3 - 4a + 1. We don’t need the exact roots, only how many real roots there are.Hint 6
Use Intermediate Value Theorem (IVT):Hint 7
Evaluate g(a) at some points:- g(-2)=-32+8+1=-23<0
- g(-1)=-4+4+1=1>0
- g\!\left(\tfrac12\right)=\tfrac12-2+1=-\tfrac12<0
- g(1)=4-4+1=1>0
Hint 8
Use the Fundamental Theorem of Algebra (real consequence):A cubic polynomial with real coefficients has at most three real roots.
Hint 9
Therefore, g(a) can have at most three real roots.Hint 10
From Hints 6 and 7, there is one real root in each interval:(-2,-1),\quad(-1,\tfrac12),\quad(\tfrac12,1).
Hint 11
From Hints 9 and 10, g(a) has exactly three real roots.Final Answer
(C) 3Problem 10
A semicircle has diameter AB and chord CD of length 16 parallel to AB. A smaller semicircle with diameter on AB and tangent to CD is cut from the larger semicircle, as shown below.

What is the area of the resulting figure, shown shaded?
- 16\pi
- 24\pi
- 32\pi
- 48\pi
- 64\pi
Hints & Final Answer
Hint 1

Hint 2
By using the Pythagorean Theorem in \triangle OMD: R^2=r^2+8^2 \;\;\Rightarrow\;\; R^2-r^2=64Hint 3
\text{Area}_{\text{shaded}} = \tfrac12\pi\!\left(R^2-r^2\right) = \tfrac12\pi\cdot 64 = 32\piFinal Answer
(C) 32\piProblem 11
The sequence 1,x,y,z is arithmetic. The sequence 1,p,q,z is geometric. Both sequences are strictly increasing and contain only integers, and z is as small as possible. What is the value of x+y+z+p+q?
- 66
- 91
- 103
- 132
- 149
Hints & Final Answer
Hint 1
1,x,y,z:\; 1,\,1+d,\,1+2d,\,1+3d \quad (d\in\mathbb{Z}^+)Hint 2
1,p,q,z:\; 1,\,r,\,r^2,\,r^3 \quad (r\in\mathbb{Z}^+,\,r\ge2)Hint 3
1+3d=z=r^3Hint 4
r \equiv 1 \pmod{3}Hint 5
So, the first r that we should check is r=4.Hint 6
1+3d=z=r^3=64 \;\Rightarrow\; d=21Hint 7
Arithmetic sequence:
1,\,x,\,y,\,z \;=\; 1,\,1+d,\,1+2d,\,1+3d = 1,\,22,\,43,\,64Geometric sequence:
1,\,p,\,q,\,z \;=\; 1,\,r,\,r^2,\,r^3 = 1,\,4,\,16,\,64Sum:
x + y + z + p + q = 22 + 43 + 64 + 4 + 16 = 149Final Answer
(E) 149Problem 12
Carlos uses a 4-digit passcode to unlock his computer. In his passcode, exactly one digit is even, exactly one (possibly different) digit is prime, and no digit is 0. How many 4-digit passcodes satisfy these conditions?
- 176
- 192
- 432
- 464
- 608
Hints & Final Answer
Hint 1
Use casework.Hint 2
Base your casework on whether the even digit is prime or not.Hint 3
Case 1: the even digit is 2. There are 4 places to position the 2. The other three digits must be odd and not prime, so each can only be 1 or 9. 4 \times 2^3 = 32Hint 4
Case 2: the even digit is 4, 6, or 8. There are 3 choices for the even number and 4 ways to place it. We must select one odd prime (3,5,7) and choose its place among the remaining 3 positions: 3 \times 3. The other 2 digits can only be 1 or 9: 2^2. Overall: 3 \times 4 \times 3 \times 3 \times 2^2 \;=\; 432Final Answer
Total = 32 + 432 = 464 (D) 464Problem 13
In the figure below, the outside square contains infinitely many squares, each of them with the same center and sides parallel to the outside square. The ratio of the side length of a square to the side length of the next inner square is k, where 0 < k < 1. The spaces between squares are alternately shaded as shown in the figure (which is not necessarily drawn to scale).

The area of the shaded portion of the figure is 64% of the area of the original square. What is k?
- \tfrac35
- \tfrac{16}{25}
- \tfrac23
- \tfrac34
- \tfrac45
Hints & Final Answer
Hint 1
Use 4. Infinite geometric series.Hint 2
64\%\cdot a^2 = \big(a^2-(ak)^2\big) + \big((ak^2)^2-(ak^3)^2\big) + \big((ak^4)^2-(ak^5)^2\big) + \cdotsHint 3
1-k^2+k^4-k^6+k^8-k^{10}-\cdots = 1+(-k^2)+(-k^2)^2+(-k^2)^3+\cdots = \frac{1}{1+k^2}Hint 4
\frac{1}{1+k^2}=\frac{64}{100}=\frac{16}{25} \;\Rightarrow\; 25=16+16k^2 \;\Rightarrow\; k^2=\frac{9}{16} \;\Rightarrow\; k=\pm\frac34 Since 0 < k < 1, k=\tfrac34.Final Answer
(D) \tfrac34Problem 14
Six chairs are arranged around a round table. Two students and two teachers randomly select four of the chairs to sit in. What is the probability that the two students will sit in two adjacent chairs and the two teachers will also sit in two adjacent chairs?
- \frac{1}{6}
- \frac{1}{5}
- \frac{2}{9}
- \frac{3}{13}
- \frac{1}{4}
Hints & Final Answer
Hint 1
N(\text{total})=\binom{6}{2}\binom{4}{2}Hint 2
N(\text{favorable}) = 6 \times 3
Hint 3
\text{Probability} \;=\; \frac{6\times 3}{\binom{6}{2}\binom{4}{2}} \;=\; \frac{6\times 3}{15\times 6} \;=\; \frac{1}{5}Final Answer
(B) \dfrac{1}{5}Problem 15
In the figure below, ABEF is a rectangle, \overline{AD}\perp\overline{DE}, AF=7, AB=1, and AD=5.

What is the area of \triangle ABC?
- \frac{3}{8}
- \frac{4}{9}
- \frac{1}{8}\sqrt{13}
- \frac{7}{15}
- \frac{1}{8}\sqrt{15}
Hints & Final Answer
Hint 1
Use similar triangles.Hint 2
\triangle ABC \sim \triangle EDC\frac{x}{7 - y} = \frac{y}{5 - x}

Hint 3
x^2 = y^2 + 1Hint 4
From Hints 2 and 3: \frac{\sqrt{y^2+1}}{7-y} = \frac{y}{5-\sqrt{y^2+1}}Hint 5
\begin{aligned} 5\sqrt{y^2+1} - y^2 - 1 &= 7y - y^2 \\ 5\sqrt{y^2+1} &= 7y + 1 \\ 25(y^2+1) &= (7y+1)^2 \\ 25y^2 + 25 &= 49y^2 + 14y + 1 \\ 0 &= 24y^2 + 14y - 24 \\ 0 &= 12y^2 + 7y - 12 \\ y &= \frac{-7 \pm \sqrt{625}}{24} = \frac{-7 \pm 25}{24} \\ y &= \frac{3}{4} \text{ (since } y>0\text{)} \end{aligned}Hint 6
[ABC] = \frac{y \times 1}{2} = \frac{3}{8}Final Answer
(A) \frac{3}{8}Problem 16
There are three jars. Each of three coins is placed in one of the three jars, chosen at random and independently of the placements of the other coins. What is the expected number of coins in a jar with the most coins?
- \frac{4}{3}
- \frac{13}{9}
- \frac{5}{3}
- \frac{17}{9}
- 2
Hints & Final Answer
Hint 1
N(\text{total}) = 3^3Hint 2
Use casework.Hint 3
Use casework based on the number of coins in the jar with the most coins, = X.Hint 4
Expected value: E(X) = \sum xP(X=x) = 1\cdot P(X=1) + 2\cdot P(X=2) + 3\cdot P(X=3)Hint 5
Case 1: X=3
Hint 6
Case 2: X=2
Hint 7
Case 3: X=1
Hint 8
From Hints 4, 5, 6, and 7: E(X) = \frac{1}{3} + \frac{4}{3} + \frac{2}{9} = \frac{17}{9}Final Answer
(D) \frac{17}{9}Problem 17
Let N be the unique positive integer such that dividing 273436 by N leaves a remainder of 16 and dividing 272760 by N leaves a remainder of 15. What is the tens digit of N?
- 0
- 1
- 2
- 3
- 4
Hints & Final Answer
Hint 1
273436 \equiv 16 \pmod{N} ⟹ N \mid 273420Hint 2
272760 \equiv 15 \pmod{N} ⟹ N \mid 272745Hint 3
If a \mid b and a \mid c, then a \mid \gcd(b,c)Hint 4
From Hints 1, 2, and 3: N \mid \gcd(273420, 272745)Hint 5
If a \ge b, \gcd(a,b) = \gcd(\text{remainder of } a \div b,\, b)Hint 6
\begin{aligned} \gcd(273420, 272745) &= \gcd(675, 272745) \\ &= \gcd(675, 45) \\ &= 45 \end{aligned}Hint 7
Since N is greater than the remainders (16 and 15), N = 45Final Answer
(E) 4Problem 18
The harmonic mean of a collection of numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers in the collection. For example, the harmonic mean of 4, 4, and 5 is
\frac{1}{\frac{1}{3}\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{5}\right)} = \frac{30}{7}
What is the harmonic mean of all the real roots of the 4050th degree polynomial
\prod_{k=1}^{2025} (kx^2 - 4x - 3) = (x^2 - 4x - 3)(2x^2 - 4x - 3)(3x^2 - 4x - 3)\cdots(2025x^2 - 4x - 3)?
- -\frac{5}{3}
- -\frac{3}{2}
- -\frac{6}{5}
- -\frac{5}{6}
- -\frac{2}{3}
Hints & Final Answer
Hint 1
Use Vieta’s Formula.Hint 2
If r_1 and r_2 are the roots of kx^2 - 4x - 3, then r_1 + r_2 = \frac{4}{k}, \quad r_1 r_2 = -\frac{3}{k} Therefore, \frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{\frac{4}{k}}{-\frac{3}{k}} = -\frac{4}{3}Hint 3
The total of 4050 roots gives \frac{4050}{\frac{1}{r_1} + \frac{1}{r_2} + \cdots + \frac{1}{r_{4050}}} = \frac{4050}{2025 \times \left(-\frac{4}{3}\right)} = \frac{2}{-\frac{4}{3}} = -\frac{3}{2}Final Answer
(B) -\frac{3}{2}Problem 19
An array of numbers is constructed beginning with the numbers -1\quad 3\quad 1 in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row begins and ends with -1 and 1, respectively.

If the process continues, one of the rows will sum to 12,288. In that row, what is the third number from the left?
- -29
- -21
- -14
- -8
- -3
Hints & Final Answer
Hint 1
since each element is duplicate in the next row, the sum of each row is twice the sum of previous row:

Hint 2
S_n = 3\times2^{n-1}Hint 3
S_n = 12288 \Rightarrow 3\times2^{n-1}=12288 \Rightarrow 2^{n-1}=4096 \Rightarrow 2^{n-1}=2^{12} \Rightarrow n-1=12 \Rightarrow n=13
Hint 4
Find the pattern of the third number in each row:

Hint 5

Hint 6
The third number in the 13th row = 1 + (3 + 2 + 1 + 0 + (-1) + (-2) + \cdots + (-8)) = 1 + (-4 -5 -6 -7 -8) = -29
Final Answer
(A) -29Problem 20
A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and g > 0 meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as \dfrac{a\sqrt{b}-c}{d}, where a, b, c, and d are positive integers, b is not divisible by the square of any prime, and d is relatively prime to the greatest common divisor of a and c. What is a + b + c + d?
- 119
- 120
- 121
- 122
- 123
Hints & Final Answer
Hint 1

Hint 2

Hint 3
The equation of the circle: (x-20)^2 + (y-15)^2 = 100 The equation of line DG: y = mxHint 4
\begin{aligned} (x-20)^2+(mx-15)^2 &= 100 \\ \Rightarrow\; x^2-40x+400 + m^2x^2 - 30mx + 225 &= 100 \\ \Rightarrow\; (m^2+1)x^2 + (-40-30m)x + 525 &= 0 \end{aligned}Hint 5
Since the circle is tangent to the line, the quadratic equation should have exactly one root.Hint 6
Discriminant D=\Delta=0Hint 7
\begin{aligned} (-40-30m)^2 - 4(m^2+1)(525) &= 0 \\ \Rightarrow\; (4+3m)^2 - 21(m^2+1) &= 0 \\ \Rightarrow\; 16 + 24m + 9m^2 - 21m^2 - 21 &= 0 \\ \Rightarrow\; 12m^2 - 24m + 5 &= 0 \end{aligned}Hint 8
Use 21. Quadratic Formula.Hint 9
\begin{aligned} m &= \frac{24 \pm \sqrt{24^2 - 4\cdot12\cdot5}}{2\cdot12} = \frac{6 \pm \sqrt{21}}{6} \\ &\text{Since } m<1,\; m=\dfrac{6-\sqrt{21}}{6} \end{aligned}Hint 10
Equation of the line: y=mx At point G(40,\,15-g): \begin{aligned} 15-g &= \left(\frac{6-\sqrt{21}}{6}\right)40 = \frac{120-20\sqrt{21}}{3} \\ \Rightarrow\; g &= 15 - \frac{120-20\sqrt{21}}{3} = \frac{20\sqrt{21}-75}{3} \end{aligned}Final Answer
(A) 119Problem 21
A set of numbers is called sum-free if whenever x and y are (not necessarily distinct) elements of the set, x + y is not an element of the set. For example, {1, 4, 6} and the empty set are sum-free, but {2, 4, 5} is not. What is the greatest possible number of elements in a sum-free subset of {1, 2, 3, …, 20}?
- 8
- 9
- 10
- 11
- 12
Hints & Final Answer
Hint 1
Use small examplesHint 2
{11,12,…,20} works ⇒ Answer ≥ 10Hint 3
By using inductions, prove that we can select at most n numbers from {1,2,…,2n} to satisfy the property of the main problem.Final Answer
(C) 10Problem 22
A circle of radius r is surrounded by three circles, whose radii are 1, 2, and 3, all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below.

What is r ?
- \tfrac14
- \tfrac{6}{23}
- \tfrac{3}{11}
- \tfrac{5}{17}
- \tfrac{3}{10}
Hints & Final Answer
Hint 1

Hint 2
Use coordinate geometry.Hint 3

Hint 4
From (1) and (2): (y-3)^2-y^2=2r+3 \;\Rightarrow\; 9-6y=2r+3 \;\Rightarrow\; y=\frac{6-2r}{6}=\frac{3-r}{3} \quad (4)Hint 5
From (1) and (3): (x-4)^2-x^2=4r+8 \;\Rightarrow\; 16-8x=4r+8 \;\Rightarrow\; x=\frac{8-4r}{8}=\frac{2-r}{2} \quad (5)Hint 6
From (1), (4), and (5): \left(\frac{3-r}{3}\right)^2+\left(\frac{2-r}{2}\right)^2=1+r^2+2r \Rightarrow\; 4(3-r)^2+9(2-r)^2=36(1+r^2+2r) \Rightarrow\; 23r^2+132r-36=0Hint 7
Use 21. Quadratic Formula: r=\frac{-132\pm\sqrt{132^2-4\cdot23\cdot(-36)}}{2\cdot23} =\frac{-66\pm6\sqrt{121+23}}{23} \Rightarrow\; r=\frac{6}{23}Final Answer
(B) \tfrac{6}{23}Problem 23
Triangle \triangle ABC has side lengths AB=80, BC=45, and AC=75. The bisector of \angle B and the altitude to side \overline{AB} intersect at point P. What is BP?
- 18
- 19
- 20
- 21
- 22
Hints & Final Answer
Hint 1

Hint 2
Find cos B, BE, cos (B/2), and BP.Hint 3
\text{a}=45,\quad \text{b}=75,\quad \text{c}=80 By using the law of Cosines: \begin{aligned} \cos B &= \frac{a^2+c^2-b^2}{2ac} = \frac{45^2+80^2-75^2}{2(45)(80)} = \frac{9^2+16^2-15^2}{2(9)(16)} \\ &= \frac{81+256-225}{2(9)(16)} = \frac{112}{2(9)(16)} = \frac{7}{18} \end{aligned}Hint 4
In \triangle BCE: \cos B=\frac{BE}{a}\;\Rightarrow\; BE=a\cos B=(45)\!\left(\frac{7}{18}\right)=\frac{35}{2}Hint 5
Use 38. Trigonometric Identities to find cos(B/2)Hint 6
\cos^2\!\left(\frac{B}{2}\right) =\frac{1+\cos B}{2} =\frac{1+\frac{7}{18}}{2} =\frac{25}{36} \;\Rightarrow\; \cos\!\left(\frac{B}{2}\right)=\frac{5}{6}Hint 7
In \triangle BPE: \cos\!\left(\frac{B}{2}\right)=\frac{BE}{BP} \;\Rightarrow\; BP=\frac{BE}{\cos(B/2)} From Hints 4 and 6: BP=\frac{\frac{35}{2}}{\frac{5}{6}}=\frac{35}{2}\times\frac{6}{5}=21

Final Answer
(D) 21Problem 24
Call a positive integer fair if no digit is used more than once, it has no 0s, and no digit is adjacent to two greater digits. For example, 196, 23, and 12463 are fair, but 1546, 320, and 34321 are not fair. How many fair positive integers are there?
- 511
- 2584
- 9841
- 17711
- 19682
Hints & Final Answer
Hint 1
Use casework.Hint 2
Base your casework on the number of digits of the fair number.Hint 3
Let n = the number of digits of the fair number.Hint 4
Case 1: n=1 \;\Rightarrow\; 9Hint 5
Case 2: n=2 \;\Rightarrow\; 9\times 8=72Hint 6
Case 3: n=3 We should select 3 different digits from 1 to 9, which is \binom{9}{3}. WLOG, assume we selected 1, 2, and 3. We can’t place 1 in the middle position since no digit is adjacent to two greater digits. So, for 1 there are 2 ways to place it, then for 2 and 3, there are 2 and 1 ways respectively. \Rightarrow\; \binom{9}{3}\times 2 \times 2 \times 1Hint 7
Case 4: n=4 (Similarly) \binom{9}{4}\times 2 \times 2 \times 2 \times 1 \;=\; \binom{9}{4}\,2^{3}Hint 8
Case k: n=k \binom{9}{k}\,2^{\,k-1}Hint 9
Use Binomial Theorem.Hint 10
(1+2)^9 = \binom{9}{0}2^{0} + \binom{9}{1}2^{1} + \binom{9}{2}2^{2} + \cdots + \binom{9}{9}2^{9}Hint 11
\text{Ans}=\binom{9}{1}2^{0}+\binom{9}{2}2^{1}+\binom{9}{3}2^{2}+\cdots+\binom{9}{9}2^{8}Hint 12
(1+2)^9=\binom{9}{0}2^{0}+2\cdot\text{Ans} \;\Rightarrow\; \text{Ans}=\frac{3^{9}-1}{2} =\frac{19683-1}{2} =\frac{19682}{2} =9841Final Answer
(C) 9841Problem 25
A 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 \dfrac{a+b\pi-c\sqrt{d}}{e}, where a,b,c,d, and e are positive integers, \gcd(a,b,c,d,e)=1, and d is not divisible by the square of a prime. What is a+b+c+d+e?
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Hints & Final Answer
Hint 1
Use casework.Hint 2
Case1: AB<AP<BP Case2: BP<AP<ABHint 3
Calculate P(AB<AP).Hint 4
WLOG, assume that the side of the square is 1. P(AB<AP)=1-\frac{\pi}{4}

Hint 5
P(AP<BP)=\frac{1}{2}

Hint 6
P(AB<AP<BP)=[APMD]-[\text{Sector }ADP]

Hint 7
Given AP=1,\; AE=\tfrac12, \angle APE=30^\circ \Rightarrow \angle EAP=60^\circ \Rightarrow \angle DAP=30^\circ Then EP=\tfrac{\sqrt3}{2}\Rightarrow PM=1-\tfrac{\sqrt3}{2} \begin{aligned} P(AB<AP<BP) &= [APMD]-[\text{Sector }ADP]\\ &= \frac{(1+1-\frac{\sqrt3}{2})(\frac12)}{2}-\frac{30}{360}\pi(1)^2\\ &= \frac12-\frac{\sqrt3}{8}-\frac{\pi}{12} \end{aligned}

Hint 8
\begin{aligned} P(BP<AP<AB) &= [\text{Sector }APB]-[AEP]\\ &= \frac{60^\circ}{360^\circ}\pi(1)^2-\frac12\cdot\frac{\sqrt3}{2}\cdot\frac12\\ &= \frac{\pi}{6}-\frac{\sqrt3}{8} \end{aligned}

