AMC 12A 2025 Problems & Solutions | Step-by-Step Math Hints

The AMC 12A 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) 4:30

Problem 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?

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)=4

Final Answer

(B) 4

Problem 3

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?

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} = 52

Hint 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 = 7

Hint 6

From (5) in Hint 4:

x = 2(7) + 14 = 28

Final Answer

(A) 28

Problem 4

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?

Hints & Final Answer

Hint 1

Use casework

Hint 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

Therefore, the fourth statement must be true. So, we found one valid case.

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) 1

Problem 5

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).

Concentric squares with alternating shading

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) + \cdots

Hint 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)

\tfrac34

Problem 6

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 7

In a certain alien world, the maximum running speed v of an organism is dependent on its number of toes n and number of eyes m. The relationship can be expressed as $v = k n^a m^b$ centimeters per hour, where k, a, and b are integer constants. In a population where all organisms have 5 toes, $\log v = 4 + 2 \log m$ ; and in a population where all organisms have 25 eyes, $\log v = 4 + 4 \log n$ , where the logarithms are base 10. What is $k + a + b$ ?

Hints & Final Answer

Hint 1

\log(ab) = \log a + \log b,\quad \log(a^n) = n\log a

Hint 2

For 5 toes:

\begin{aligned} \log v &= 4 + 2\log m \\ \Rightarrow \log(k5^a m^b) &= 4 + \log m^2 = \log(10^4 m^2) \\ \Rightarrow k5^a m^b &= 10^4 m^2 \end{aligned}

Since

k,a

are constants but

m

can vary,

b = 2

Hint 3

Similarly, for 25 eyes:

\log v = 4 + 4\log n \Rightarrow \log(k n^a 25^2) = 4 + 4\log n = \log(10^4 n^4)

Hence

a = 4

Hint 4

From Hints 2 and 3:

\begin{aligned} k 5^4 m^2 &= 10^4 m^2 \\ \Rightarrow k &= \frac{10^4}{5^4} = 16 \end{aligned}

k + a + b = 16 + 4 + 2 = 22

Final Answer

Problem 8

Pentagon ABCDE is inscribed in a circle, and $\angle BEC = \angle CED = 30^\circ.$ Let line AC and line BD intersect at point F, and suppose that $AB = 9$ and $AD = 24.$ What is $BF?$

$\dfrac{57}{11}$
$\dfrac{59}{11}$
$\dfrac{60}{11}$
$\dfrac{61}{11}$
$\dfrac{63}{11}$

Hints & Final Answer

Hint 1

Use the Angle Bisector Theorem.

Hint 1 figure

Hint 2

\text{Using the Angle Bisector Theorem in } \triangle ABD:

\frac{BF}{DF} = \frac{AB}{AD} = \frac{9}{24} = \frac{3}{8}

Hint 3 figure

Hint 3

Find $BD.$

Hint 4

Using the Law of Cosines in $\triangle ABD:$

BD^2 = AB^2 + AD^2 - 2(AB)(AD)\cos 60^\circ

= 9^2 + 24^2 - 2(9)(24)\left(\tfrac{1}{2}\right) = 441 \Rightarrow BD = 21

Hint 5

From Hints 2 and 4:

\frac{BF}{DF} = \frac{3}{8} \Rightarrow \frac{BF}{BD} = \frac{3}{3+8} = \frac{3}{11}

\Rightarrow BF = 21\left(\frac{3}{11}\right) = \frac{63}{11}

Final Answer

(E)

\dfrac{63}{11}

Problem 9

Let w be the complex number $2+i$ , where $i=\sqrt{-1}$ . What real number r has the property that r, w, and $w^2$ are three collinear points in the complex plane?

$\tfrac{3}{4}$
1
$\tfrac{7}{5}$
$\tfrac{3}{2}$
$\tfrac{5}{3}$

Hints & Final Answer

Hint 1

w = 2 + i

w^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i

Hint 2

Hint 2 diagram

Hint 3

Treat complex numbers as points:

w \rightarrow (2,1), \quad w^2 \rightarrow (3,4)

The line through them has slope

m = \frac{4 - 1}{3 - 2} = 3

Equation through

(2,1)

y - 1 = 3(x - 2) \Rightarrow y = 3x - 5

Hint 4

The real number

r

lies on the real axis (imaginary part 0), so set

y = 0

0 = 3r - 5 \Rightarrow r = \tfrac{5}{3}

Final Answer

(E)

\tfrac{5}{3}

Problem 10

In the figure shown below, major arc $\overset{\frown}{AD}$ and minor arc $\overset{\frown}{BC}$ have the same center, O. Also, A lies between O and B, and D lies between O and C. Major arc $\overset{\frown}{AD}$ , minor arc $\overset{\frown}{BC}$ , and each of the two segments $AB$ and $CD$ have length $2\pi$ . What is the distance from $O$ to $A$ ?

Problem 10 figure

1
$1 - \pi + \sqrt{\pi^2 + 1}$
$\tfrac{\pi}{2}$
$\tfrac{\sqrt{\pi^2 + 1}}{2}$
2

Hints & Final Answer

Hint 1

Hint 1 figure

\begin{align*} \overset{\frown}{AD}: &\quad \frac{360 - \alpha}{360} \cdot 2\pi r = 2\pi \quad (1)\\ \overset{\frown}{BC}: &\quad \frac{\alpha}{360} \cdot 2\pi(2\pi + r) = 2\pi \quad (2) \end{align*}

Hint 2

From (1):

\dfrac{360 - \alpha}{360} \cdot r = 1 \quad (3)

From (2):

\dfrac{\alpha}{360}(2\pi + r) = 1 \Rightarrow \alpha = \dfrac{360}{2\pi + r} \quad (4)

Hint 3

Substitute (3) and (4):

\left(360 - \frac{360}{2\pi + r}\right)r = 360

Simplify:

\left(1 - \frac{1}{2\pi + r}\right)r = 1 \Rightarrow (2\pi + r - 1)r = 2\pi + r

Hint 4

r^2 + (2\pi - 2)r - 2\pi = 0

Hint 5

Use the quadratic formula:

r = \frac{2 - 2\pi \pm \sqrt{(2\pi - 2)^2 + 8\pi}}{2}

Hint 6

Simplify:

r = 1 - \pi \pm \sqrt{\pi^2 + 1}

Since

r > 0

, we take the positive root:

r = 1 - \pi + \sqrt{\pi^2 + 1}

Final Answer

(B)

1 - \pi + \sqrt{\pi^2 + 1}

Problem 11

The orthocenter of a triangle is the concurrent intersection of the three (possibly extended) altitudes. What is the sum of the coordinates of the orthocenter of the triangle whose vertices are $A(2,31)$ , $B(8,27)$ , and $C(18,27)$ ?

5
17
$10 + 4\sqrt{17} + 2\sqrt{13}$
$\tfrac{113}{3}$
54

Hints & Final Answer

Hint 1

If you also draw $\triangle ABC$ as an acute triangle, the calculation remains the same. So, there is no need to draw an accurate figure for this problem.

Hint 1 figure

Hint 2

The orthocenter

H

is the intersection of the altitudes

AE

and

BD

. Hence,

H = AE \cap BD.

Hint 3

Compute slopes:

S_{AC} = \frac{31 - 27}{2 - 18} = \frac{4}{-16} = -\tfrac{1}{4}

So the slope of altitude

BD

is the negative reciprocal:

S_{BD} = 4

Equation of

BD

y - 27 = 4(x - 8)

Hint 4

Since

S_{BC} = 0

BC

is horizontal, so altitude

AE

is vertical:

x = 2

Hint 5

Substituting

x = 2

into

y - 27 = 4(x - 8)

y - 27 = 4(2 - 8) = -24 \Rightarrow y = 3

H(2,3)

Final Answer

(A) 5

Problem 12

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

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 13

Let $C = \{1, 2, 3, \ldots, 13\}$ . Let $N$ be the greatest integer such that there exists a subset of $C$ with $N$ elements that does not contain five consecutive integers. Suppose $N$ integers are chosen at random from $C$ without replacement. What is the probability that the chosen elements do not include five consecutive integers?

$\tfrac{3}{130}$
$\tfrac{3}{143}$
$\tfrac{5}{143}$
$\tfrac{1}{26}$
$\tfrac{5}{78}$

Hints & Final Answer

Hint 1

\{1,2,3,4,5\}, \{6,7,8,9,10\}

From each set, we should remove at least one element

\Rightarrow N \le 11

Hint 2

Consider the subset

\{1,2,3,4,6,7,8,9,11,12,13\}

. By removing 5 and 10, there are no five consecutive integers.

\Rightarrow N \ge 11

Hint 3

From Hints 1 and 2,

\Rightarrow N = 11

Hint 4

N_{\text{total}} = \binom{13}{11} \times 11!

Hint 5

Let the two numbers that are removed be

x

and

y

, where

x < y

1, 2, \ldots, x-1; \, x; \, x+1, \ldots, y-1; \, y; \, y+1, \ldots, 13

To avoid having five consecutive integers:

x - 1 \le 4 \Rightarrow x \le 5

y - 1 - x \le 4 \Rightarrow y - x \le 5

13 - y \le 4 \Rightarrow 9 \le y

Hint 6

Possible pairs

(x, y)

are:

(5,9), \quad (5,10), \quad (4,9)

Hint 7

N_{\text{favorable}} = 3 \times 11!

Hint 8

From Hints 4 and 7:

P = \frac{3 \times 11!}{\binom{13}{11} \times 11!} = \frac{3}{\binom{13}{11}} = \frac{3}{13 \times 6} = \frac{1}{26}

Final Answer

(D)

\tfrac{1}{26}

Problem 14

Points F, G, and H are collinear with G between F and H. The ellipse with foci at G and H is internally tangent to the ellipse with foci at F and G, as shown below.

Problem 14 figure

The two ellipses have the same eccentricity, e, and the ratio of their areas is 2025. (Recall that the eccentricity of an ellipse is e=\frac{c}{a}, where c is the distance from the center to a focus, and 2a is the length of the major axis.) What is e?

$\tfrac{3}{5}$
$\tfrac{16}{25}$
$\tfrac{4}{5}$
$\tfrac{22}{23}$
$\tfrac{44}{45}$

Hints & Final Answer

Hint 1

Hint 1 figure

c^2=a^2-b^2,\qquad e=\frac{c}{a}

Hint 2

Since

e_1=e_2

\frac{c_1}{a_1}=\frac{c_2}{a_2} \ \Rightarrow\ \frac{c_1}{c_2}=\frac{a_1}{a_2}=k

Now,

b_1^2=a_1^2-c_1^2=(a_2k)^2-(c_2k)^2 =k^2(a_2^2-c_2^2)=k^2 b_2^2 \ \Rightarrow\ b_1=kb_2

Hence,

\frac{c_1}{c_2}=\frac{a_1}{a_2}=\frac{b_1}{b_2}=k

Therefore, the two ellipses are similar.

Hint 3

\frac{\text{Area}_1}{\text{Area}_2}=k^2=2025 \ \Rightarrow\ k=45

Hint 4

Hint 4 figure

c_1+c_2+a_2=a_1

Using

a_1=45a_2

and

c_1=45c_2

45c_2+c_2+a_2=45a_2 \ \Rightarrow\ 46c_2=44a_2 \ \Rightarrow\ \frac{c_2}{a_2}=\frac{44}{46}=\frac{22}{23}

Final Answer

Since both ellipses have the same eccentricity

e=\frac{c}{a}

e=\frac{22}{23}

(D)

Problem 15

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}?

Hints & Final Answer

Hint 1

Use small examples

Hint 2

{11,12,…,20} works ⇒ Answer ≥ 10

Hint 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

Problem 16

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 $BP$ ?

Hints & Final Answer

Hint 1

Hint 1 diagram

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

\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

\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

Hint 7 diagram

Final Answer

(D) 21

Problem 17

The polynomial $(z+i)(z+2i)(z+3i)+10$ has three roots in the complex plane, where $i=\sqrt{-1}$ . What is the area of the triangle formed by these three roots?

Hints & Final Answer

Hint 1

Let

w=z+2i

Hint 2

(z+i)(z+2i)(z+3i)+10=0 \Rightarrow (w-i)(w)(w+i)+10=0 \Rightarrow w(w^2+1)+10=0 \Rightarrow w^3+w+10=0

Hint 3

w^3+w+10=(w+2)(w^2-2w+5)=0

Hint 4

Use 21. Quadratic Formula.

Hint 5

w=-2,\quad \frac{2\pm\sqrt{4-20}}{2}= -2,\ 1\pm 2i

Hint 6

z+2i=-2,\,1\pm2i \Rightarrow z=-2-2i,\ 1,\ 1-4i

Hint 7

Use the Shoelace Theorem.

Hint 8

Points:

(-2,-2),\ (1,0),\ (1,-4)

\text{Area}=\frac12\left|(-2)(0)+(1)(-4)+(1)(-2)-\big((-2)(1)+0(1)+(-4)(-2)\big)\right| =\frac12\left|-6-6\right|=6

Final Answer

(A) 6

Problem 18

How many ordered triples $(x,y,z)$ of different positive integers less than or equal to 8 satisfy $xy>z$ , $xz>y$ , and $yz>x$ ?

Hints & Final Answer

Hint 1

WLOG, assume

x < y < z

. We will multiply by

3!

at the end to account for all possible orderings. then

xz > y

and

yz > x

, so we should just focus on

xy > z

Hint 2

Use casework.

Hint 3

Base your casework on

z

Hint 4

z\ge 3

Hint 5

Case 1:

z=3

No valid cases.

Hint 6

Case 2:

z=4

Then

x=2,\,y=3 \rightarrow

1 case

Hint 7

Case 3:

z=5

x,y\in\{1,2,3,4\}\rightarrow

3 cases

Hint 8

Case 4:

z=6

x,y\in\{1,2,3,4,5\}:

\binom{5}{2}-(4+1)=10-5=5

\rightarrow

5 cases

Hint 9

Case 5:

z=7

x,y\in\{1,2,3,4,5,6\}:

\binom{6}{2}-(5+1)=15-6=9

\rightarrow

9 cases

Hint 10

Case 6:

z=8

x,y\in\{1,2,3,4,5,6,7\}:

\binom{7}{2}-(6+2)=21-8=13

\rightarrow

13 cases

Hint 11

Total

=3!\times(0+1+3+5+9+13)=6\times31=186

Final Answer

Problem 19

Let a, b, and c be the roots of the polynomial $x^3+kx+1$ . What is the sum

$a^3b^2 + a^2b^3 + b^3c^2 + b^2c^3 + c^3a^2 + c^2a^3 \, ?$

$-k$
$-k+1$
1
$k-1$
$k$

Hints & Final Answer

Hint 1

Use 30. Vieta’s Formula.

Hint 2

a+b+c=0,\qquad ab+ac+bc=k,\qquad abc=-1

Hint 3

Use Newton’s sums.

Hint 4

Let

T=a^3b^2+a^2b^3+b^3c^2+b^2c^3+c^3a^2+c^2a^3 =a^3(b^2+c^2)+b^3(a^2+c^2)+c^3(a^2+b^2).

Hint 5

Let

s_m=a^m+b^m+c^m

. Then

T=a^3(s_2-a^2)+b^3(s_2-b^2)+c^3(s_2-c^2) \Rightarrow T=s_2(a^3+b^3+c^3)-(a^5+b^5+c^5) \Rightarrow T=s_2s_3-s_5.

Hint 6

Given the cubic equation

x^3+kx+1=0

. By Newton’s sums:

s_m + k s_{m-2} + s_{m-3} = 0 \Rightarrow s_m = -k s_{m-2} - s_{m-3}\quad (1)

We know:

s_0 = a^0+b^0+c^0 = 1+1+1 = 3 \quad (2)

s_1 = a+b+c = 0 \quad (3)

s_2 = a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) \Rightarrow s_2 = 0^2 - 2(k) = -2k \quad (4)

Hint 7

From (1), (2), (3), and (4):

s_3 = -k s_1 - s_0 = -k(0) - 3 = -3 \quad (5)

s_4 = -k s_2 - s_1 = -k(-2k) - 0 = 2k^2

s_5 = -k s_3 - s_2 = -k(-3) - (-2k) = 5k \quad (6)

Hint 8

From Hint 5, (4), (5), and (6):

T=s_2s_3 - s_5 = (-2k)(-3) - 5k = k

Final Answer

(E)

k

Problem 20

The base of the pentahedron shown below is a $13\times8$ rectangle, and its lateral faces are two isosceles triangles with base of length $8$ and congruent sides of length $13$ , and two isosceles trapezoids with bases of length $7$ and $13$ and nonparallel sides of length $13$ .

Problem 20 figure

Hints & Final Answer

Hint 1

Let

x = \dfrac{13 - 7}{2} = 3

. Then:

\begin{align*} AM^2 &= 13^2 - 4^2 = 153 \Rightarrow AM = \sqrt{153}\\ h^2 &= AM^2 - x^2 = 153 - 9 = 144 \Rightarrow h = 12 \end{align*}

Hint 1 figure

Hint 2

You can split the pentahedron into a triangular prism and two pyramids.

Hint 2 figure

Hint 3

The volume is given by:

V = 2\left(\dfrac{12\times3\times8}{3}\right) + \dfrac{12\times8}{2}\times7

Simplify:

V = 192 + 336 = 528

Final Answer

(C)

528

Problem 21

There is a unique ordered triple (a,k,m) of nonnegative integers such that

$\frac{4^{a}+4^{a+k}+4^{a+2k}+\cdots+4^{a+mk}} {2^{a}+2^{a+k}+2^{a+2k}+\cdots+2^{a+mk}} \;=\; 964.$

What is $a+k+m$ ?

Hints & Final Answer

Hint 1

Use 4. Finite geometric series.

Hint 2

Start from

4^{a}+4^{a+k}+\cdots+4^{a+mk} \;=\; 964\bigl(2^{a}+2^{a+k}+\cdots+2^{a+mk}\bigr).

Factor and apply geometric sums:

4^{a}\!\left(1+4^{k}+(4^{k})^{2}+\cdots+(4^{k})^{m}\right) = 964\cdot 2^{a}\!\left(1+2^{k}+(2^{k})^{2}+\cdots+(2^{k})^{m}\right),

2^{a}\cdot \frac{(4^{k})^{m+1}-1}{4^{k}-1} \;=\; 964\left(\frac{(2^{k})^{m+1}-1}{2^{k}-1}\right).

Hint 3

Use difference of squares:

4^{k}-1=(2^{k})^{2}-1=(2^{k}-1)(2^{k}+1)

Hint 4

From Hints 2 and 3,

2^{a}\cdot \frac{\bigl(2^{k(m+1)}\bigr)^{2}-1}{(2^{k})^{2}-1} \;=\; 964\left(\frac{2^{k(m+1)}-1}{2^{k}-1}\right) \;\Longrightarrow\; 2^{a}\cdot\frac{2^{k(m+1)}+1}{2^{k}+1}=964,

hence

2^{a}\Bigl(2^{k(m+1)}+1\Bigr)=964\,(2^{k}+1).

Hint 5

k=0

, then

2^{a}(1+1)=964\cdot2

, impossible. So

k\ge1

Hint 6

Note that

964=4\cdot241

. Separate even and odd parts:

2^{a}\bigl((2^{k})^{m+1}+1\bigr) = 4\cdot241\,(2^{k}+1).

Even part:

2^{a}=4 \;\Rightarrow\; a=2 \quad (1)

Odd part:

(2^{k})^{m+1}+1=241(2^{k}+1).

Hint 7

From the odd part,

2^{k(m+1)}+1=241\cdot2^{k}+241 \;\Longrightarrow\; 2^{k(m+1)}-241\cdot2^{k}=240=16\cdot15.

Hence

2^{k}\bigl(2^{mk}-241\bigr)=240 \;\Rightarrow\; 2^{k}=16 \;\Rightarrow\; k=4 \quad (2),

and

2^{mk}-241=15 \;\Rightarrow\; 2^{mk}=256=2^{8} \;\Rightarrow\; mk=8 \;\Rightarrow\; m=2 \quad (3).

Hint 8

From (1), (2), and (3):

a+k+m \;=\; 2+4+2 \;=\; 8.

Final Answer

(A) 8

Problem 22

Three real numbers are chosen independently and uniformly at random between 0 and 1. What is the probability that the greatest of these three numbers is greater than 2 times each of the other two numbers? (In other words, if the chosen numbers are $a \ge b \ge c$ , then $a > 2b$ .)

$\tfrac{1}{12}$
$\tfrac{1}{9}$
$\tfrac{1}{8}$
$\tfrac{1}{6}$
$\tfrac{1}{4}$

Hints & Final Answer

Hint 1

WLOG, assume

a \ge b \ge c

, and at the end multiply by

3!

Hint 2

Use integrals to compute the probability over the ordered region.

Hint 3

For the ordered region we need

a>2b

b>c>0

. Thus, for each

a\in[0,1]

, we have

b\in[0,\tfrac{a}{2}]

and

c\in[0,b]

. We need to evaluate

\int_{0}^{1}\int_{0}^{a/2}\int_{0}^{b} \mathrm{d}c\,\mathrm{d}b\,\mathrm{d}a.

Hint 4

Compute the integral:

\begin{aligned} \int_{0}^{1}\int_{0}^{a/2}\int_{0}^{b} \mathrm{d}c\,\mathrm{d}b\,\mathrm{d}a &= \int_{0}^{1}\int_{0}^{a/2} b\,\mathrm{d}b\,\mathrm{d}a \\ &= \int_{0}^{1}\left[\frac{b^{2}}{2}\right]_{0}^{a/2}\,\mathrm{d}a = \int_{0}^{1}\frac{a^{2}}{8}\,\mathrm{d}a = \left[\frac{a^{3}}{24}\right]_{0}^{1} = \frac{1}{24}. \end{aligned}

Hint 5

Multiply by

3!

to account for all orderings:

3!\times \frac{1}{24}=\frac{1}{4}.

Final Answer

(E)

\tfrac{1}{4}

Problem 23

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\; 9

Hint 5

Case 2:

n=2 \;\Rightarrow\; 9\times 8=72

Hint 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 1

Hint 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} =9841

Final Answer

Problem 24

A circle of radius r is surrounded by 12 circles of radius 1, externally tangent to the central circle and sequentially tangent to each other, as shown. Then r can be written as $\sqrt{a}+\sqrt{b}+c$ , where $a,b,c$ are integers. What is $a+b+c$ ?

Problem 24 figure

Hints & Final Answer

Hint 1

Since there are 12 small circles around the big one, the central angle between adjacent tangency points is

\dfrac{360^\circ}{12}=30^\circ.

Hint 1 figure

Hint 2

Use the Law of Cosines in the triangle formed by the center of the big circle and the centers of two adjacent small circles.

Hint 3

Let the distance from the big center to a small center be

r+1

. Then with included angle

30^\circ

2^2=(r+1)^2+(r+1)^2-2(r+1)(r+1)\cos30^\circ,

4=(r+1)^2\!\left(2-\sqrt3\right) \quad\Longrightarrow\quad (r+1)^2=\frac{4}{\,2-\sqrt3\,} =4\!\left(\frac{2+\sqrt3}{(2-\sqrt3)(2+\sqrt3)}\right) =4(2+\sqrt3). \tag{1}

Hint 4

Note that

(1+\sqrt3)^2=1+2\sqrt3+3=4+2\sqrt3. \tag{2}

Hint 5

From (1) and (2),

(r+1)^2=4(2+\sqrt3)=2(4+2\sqrt3)=2(1+\sqrt3)^2,

hence

r+1=\sqrt2\,(1+\sqrt3)=\sqrt2+\sqrt6 \quad\Longrightarrow\quad r=\sqrt2+\sqrt6-1.

Therefore

a=2,\ b=6,\ c=-1

, and

a+b+c=2+6-1=7

Final Answer

(C)

7

Problem 25

Polynomials $P(x)$ and $Q(x)$ each have degree $3$ and leading coefficient $1$ , and their roots are all elements of $\{1,2,3,4,5\}$ . The function $f(x)=\dfrac{P(x)}{Q(x)}$ has the property that there exist real numbers $a < b < c < d$ such that the set of all real numbers $x$ for which $f(x)\le0$ consists of the closed interval $[a,b]$ together with the open interval $(c,d)$ . How many functions $f(x)$ are possible?

Hints & Final Answer

Hint 1

Domain differences matter when canceling factors in rational functions.

g(x)=\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-3)} \ne \frac{(x-1)(x-2)}{(x-4)(x-5)} = h(x).

The two are not equal because

3\notin\mathrm{Dom}(g(x))

, but

3\in\mathrm{Dom}(h(x))

.However, if we define

g_2(x)=\frac{(x-1)(x-2)(x-4)}{(x-4)(x-5)(x-4)},\qquad h_2(x)=\frac{(x-1)(x-2)}{(x-4)(x-5)},

then

g_2(x)=h_2(x)

Hint 2

The sign pattern for

f(x)

alternates across

a,b,c,d

Sign chart for f(x)

Hint 3

Since

f(x)=\dfrac{P(x)}{Q(x)}

a,b

must be the roots of

P(x)

, and

c,d

must be the roots of

Q(x)

Hint 4

Let the shared factor be

(x-e)

. Then

f(x)=\frac{(x-a)(x-b)(x-e)}{(x-c)(x-d)(x-e)}.

Hint 5

Use casework on possible

(a,b,c,d)

from

\{1,2,3,4,5\}

with

a < b < c < d

, and check valid

e

Hint 6

Let

(a,b,c,d)=(1,2,3,4)

. Then

e\in\{3,4,5\}

. From Hint 1,

e=3

and

e=4

yield the same

f(x)

. ⇒ 2 distinct cases.

Hint 6 figure

Hint 7

(a,b,c,d)=(1,2,3,5)\Rightarrow e=3,5

⇒ 1 case.

Hint 8

(a,b,c,d)=(1,2,4,5)\Rightarrow e=3,4,5

e=4

and

e=5

are the same ⇒ 2 cases.

Hint 9

(a,b,c,d)=(1,3,4,5)\Rightarrow e=4,5

⇒ 1 case.

Hint 10

(a,b,c,d)=(2,3,4,5)\Rightarrow e=1,4,5

e=4

and

e=5

same ⇒ 2 cases.

Hint 11

Total cases:

2+1+2+1+2=8.

Final Answer

8 (which is not among the given choices)

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