Awesome Problems I've Solved

hi ^-^ i solve a lot of math, and wanted a place where i document fun, challenging, and elegant mathematical problems i’ve solved over time

---

2015 AMC 12B #25

Problem

A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \geq 1$, once the bee reaches $P_j$, she turns $30^\circ$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2016}$, she is exactly $a\sqrt{b}+c\sqrt{d}$ inches away from $P_0$, where $a$, $b$, $c$, and $d$ are positive integers and $b$ and $d$ are square-free. Find $a+b+c+d$.

Solution

Modeling the Flight with Complex Numbers

Represent the plane as the complex plane and place $P_0$ at the origin.

Let

$$\omega=e^{i\pi/6}.$$

Since the bee turns $30^\circ$ after each segment, the $k$-th flight segment has:

• Length $k$
• Direction angle $(k-1)\cdot 30^\circ$

Thus the displacement from $P_{k-1}$ to $P_k$ is

$$k\omega^{k-1}.$$

Therefore, the position of $P_{2016}$ relative to $P_0$ is

$$S=\sum_{k=1}^{2016}k\omega^{k-1}.$$

Our goal is to compute $|S|$.

Exploiting the Fact that $\omega^{12}=1$

Since

$$\omega^{12}=e^{2\pi i}=1,$$

and

$$2016=168\cdot 12,$$

we group the sum into $168$ blocks of $12$ consecutive terms:

$$S=\sum_{m=0}^{167}\sum_{t=1}^{12}(12m+t)\omega^{t-1}.$$

Expanding,

$$S=\sum_{m=0}^{167}\left(12m\sum_{t=1}^{12}\omega^{t-1}+\sum_{t=1}^{12}t\omega^{t-1}\right).$$

Because $\omega$ is a primitive $12$th root of unity,

$$\sum_{t=1}^{12}\omega^{t-1} = 1+\omega+\omega^2+\cdots+\omega^{11} = 0.$$

Hence all terms involving $12m$ vanish, giving

$$S = 168\sum_{t=1}^{12}t\omega^{t-1}.$$

Let

$$C=\sum_{t=1}^{12}t\omega^{t-1}.$$

Then

$$S=168C.$$

Evaluating the Root-of-Unity Sum

Consider the finite geometric series

$$1+x+x^2+\cdots+x^{11} = \frac{1-x^{12}}{1-x}.$$

Differentiate both sides:

$$\sum_{t=1}^{12}t\,x^{t-1} = \frac{-12x^{11}(1-x)+(1-x^{12})}{(1-x)^2}.$$

Substituting $x=\omega$ and using $\omega^{12}=1$ gives

$$C = \frac{-12\omega^{11}(1-\omega)}{(1-\omega)^2} = -\frac{12\omega^{11}}{1-\omega}.$$

Since $\omega^{11}=\omega^{-1}$,

$$C = -\frac{12\omega^{-1}}{1-\omega}.$$

Multiplying numerator and denominator by $\omega$,

$$C = -\frac{12}{\omega-\omega^2}.$$

Factoring $\omega$ from the denominator,

$$C = -\frac{12}{\omega(1-\omega)} = -\frac{12}{1-\omega}.$$

Therefore,

$$S = 168\left(-\frac{12}{1-\omega}\right) = -\frac{2016}{1-\omega}.$$

Computing $\frac{1}{1-\omega}$

Since

$$\omega = \cos 30^\circ+i\sin 30^\circ = \frac{\sqrt3}{2}+\frac{i}{2},$$

we have

$$1-\omega = \frac{2-\sqrt3}{2}-\frac{i}{2}.$$

A standard identity gives

$$\frac{1}{1-e^{i\theta}} = \frac12+\frac{i}{2}\cot\left(\frac{\theta}{2}\right).$$

Using $\theta=\frac{\pi}{6}$,

$$\frac{1}{1-\omega} = \frac12+\frac{i}{2}\cot 15^\circ.$$

Since

$$\cot 15^\circ=2+\sqrt3,$$

it follows that

$$\frac{1}{1-\omega} = \frac12+\frac{i}{2}(2+\sqrt3).$$

Therefore,

$$S = -2016\left(\frac12+\frac{i}{2}(2+\sqrt3)\right) = -1008-1008(2+\sqrt3)i.$$

Finding the Distance from the Origin

The distance from $P_0$ to $P_{2016}$ is

$$|S| = 1008\sqrt{1+(2+\sqrt3)^2}.$$

Compute the quantity inside the square root:

$$1+(2+\sqrt3)^2 = 1+4+4\sqrt3+3 = 8+4\sqrt3.$$

Factor:

$$8+4\sqrt3 = 4(2+\sqrt3).$$

Hence

$$|S| = 2016\sqrt{2+\sqrt3}.$$

Use the identity

$$2+\sqrt3 = \frac{(\sqrt3+1)^2}{2}.$$

Thus

$$\sqrt{2+\sqrt3} = \frac{\sqrt3+1}{\sqrt2} = \frac{\sqrt6+\sqrt2}{2}.$$

Substituting,

$$|S| = 2016\cdot\frac{\sqrt6+\sqrt2}{2} = 1008\sqrt6+1008\sqrt2.$$

Therefore,

$$a=1008,\quad b=6,\quad c=1008,\quad d=2.$$

Computing the Requested Sum

$$a+b+c+d = 1008+6+1008+2 = 2024.$$

Final Answer: 2024

---

2004 AIME I #12

Problem

Let $S$ be the set of ordered pairs $(x,y)$ such that $0<x\le1$, $0<y\le1$, and both

$$\left\lfloor \log_2\left(\frac{1}{x}\right)\right\rfloor$$

and

$$\left\lfloor \log_5\left(\frac{1}{y}\right)\right\rfloor$$

are even integers. Given that the area of the graph of $S$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Solution

Determining the Valid Values of $x$

Suppose

$$\left\lfloor \log_2\left(\frac{1}{x}\right)\right\rfloor=2k$$

for some integer $k\ge0$.

By the definition of the floor function,

$$2k\le\log_2\left(\frac{1}{x}\right)<2k+1.$$

Exponentiating base $2$ gives

$$2^{2k}\le\frac{1}{x}<2^{2k+1}.$$

Taking reciprocals reverses the inequalities:

$$2^{-(2k+1)}<x\le2^{-2k}.$$

Therefore, for each $k\ge0$, the valid interval for $x$ is

$$\left(2^{-(2k+1)},\,2^{-2k}\right].$$

The length of this interval is

$$2^{-2k}-2^{-(2k+1)} = 2^{-(2k+1)}.$$

Hence the total measure of all valid $x$-values is

$$\sum_{k=0}^{\infty}2^{-(2k+1)} = \frac{1}{2}\sum_{k=0}^{\infty}\left(\frac{1}{4}\right)^k.$$

Using the geometric series formula,

$$\frac{1}{2}\cdot\frac{1}{1-\frac{1}{4}} = \frac{1}{2}\cdot\frac{4}{3} = \frac{2}{3}.$$

Thus the total valid length in the $x$-direction is

$$\frac{2}{3}.$$

Determining the Valid Values of $y$

Similarly, suppose

$$\left\lfloor \log_5\left(\frac{1}{y}\right)\right\rfloor=2m$$

for some integer $m\ge0$.

Then

$$2m\le\log_5\left(\frac{1}{y}\right)<2m+1.$$

Exponentiating base $5$ yields

$$5^{2m}\le\frac{1}{y}<5^{2m+1}.$$

Taking reciprocals,

$$5^{-(2m+1)}<y\le5^{-2m}.$$

The length of this interval is

$$5^{-2m}-5^{-(2m+1)} = 4\cdot5^{-(2m+1)}.$$

Therefore the total measure of all valid $y$-values is

$$\sum_{m=0}^{\infty}4\cdot5^{-(2m+1)} = \frac{4}{5}\sum_{m=0}^{\infty}\left(\frac{1}{25}\right)^m.$$

Again applying the geometric series formula,

$$\frac{4}{5}\cdot\frac{1}{1-\frac{1}{25}} = \frac{4}{5}\cdot\frac{25}{24} = \frac{5}{6}.$$

Thus the total valid length in the $y$-direction is

$$\frac{5}{6}.$$

Computing the Area of $S$

The conditions on $x$ and $y$ are independent, so the area of $S$ is the product of the valid lengths in the two directions:

$$\text{Area}(S) = \frac{2}{3}\cdot\frac{5}{6} = \frac{10}{18} = \frac{5}{9}.$$

Hence

$$\frac{m}{n} = \frac{5}{9},$$

so

$$m=5, \qquad n=9.$$

Computing $m+n$

Therefore,

$$m+n = 5+9 = 14.$$

Final Answer: 14

---

2010 AIME I #2

Problem

For a positive integer $n$, let $\theta(n)$ denote the number of integers $0<x<2010$ such that

$$x^2-n$$

is divisible by $2010$.

Determine the remainder when

$$\sum_{n=0}^{2009} n\cdot\theta(n)$$

is divided by $2010$.

Solution

Interpreting the Sum

For each integer $x$ with $0<x<2010$, there is exactly one residue

$$n \in \{0,1,\dots,2009\}$$

such that

$$x^2 \equiv n \pmod{2010}.$$

The value $\theta(n)$ counts how many integers $x$ produce the residue $n$.

Therefore, each $x$ contributes its corresponding residue $n$ exactly once to the sum

$$\sum_{n=0}^{2009} n\cdot\theta(n).$$

Hence

$$\sum_{n=0}^{2009} n\cdot\theta(n) = \sum_{x=1}^{2009} (x^2 \bmod 2010).$$

Reducing Modulo $2010$

Since

$$x^2 \bmod 2010 \equiv x^2 \pmod{2010},$$

we have

$$\sum_{n=0}^{2009} n\cdot\theta(n) \equiv \sum_{x=1}^{2009} x^2 \pmod{2010}.$$

Thus we only need to compute

$$\sum_{x=1}^{2009} x^2.$$

Using the sum-of-squares formula,

$$\sum_{x=1}^{2009} x^2 = \frac{2009\cdot2010\cdot4019}{6}.$$

Since

$$\frac{2010}{6}=335,$$

this becomes

$$2009\cdot335\cdot4019.$$

Computing the Remainder

Working modulo $2010$,

$$2009 \equiv -1 \pmod{2010}$$

and

$$4019 \equiv -1 \pmod{2010}.$$

Therefore,

$$2009\cdot335\cdot4019 \equiv (-1)\cdot335\cdot(-1) = 335 \pmod{2010}.$$

So the required remainder is

$$335.$$

Final Answer: 335

---

2025 AMC 12A #1

Problem

Solution

Counting Lattice Points

Let the side lengths of $R$, $S$, and $T$ be $r$, $s$, and $t$, respectively.

Since each square is aligned with the lattice, a square of side length $m$ contains

$$(m+1)^2$$

lattice points.

Thus,

$$N(R)=(r+1)^2, \qquad N(S)=(s+1)^2.$$

We are given

$$(r+1)^2=\frac94(s+1)^2.$$

Since $r+1$ and $s+1$ are integers,

$$r+1=3k, \qquad s+1=2k$$

for some positive integer $k$.

Hence

$$r=3k-1, \qquad s=2k-1.$$

Counting Points in $R\cup S$

Since $R$ and $S$ meet along the $y$-axis and $r>s$, their intersection contains exactly

$$s+1$$

lattice points.

Therefore

$$N(R\cup S) = (r+1)^2+(s+1)^2-(s+1).$$

Substituting $r+1=3k$ and $s+1=2k$ gives

$$N(R\cup S) = 9k^2+4k^2-2k = 13k^2-2k.$$

Since $T$ contains one-fourth of the lattice points in $R\cup S$,

$$(t+1)^2 = \frac{13k^2-2k}{4}.$$

Using the Fraction Condition

Let $x_1$ be the width of $S\cap T$ and $x_2$ the width of $R\cap T$.

Then

$$t=x_1+x_2.$$

Since $T$ has height $t$, the intersections contain

$$N(S\cap T)=(x_1+1)(t+1),$$

and

$$N(R\cap T)=(x_2+1)(t+1)$$

lattice points.

The given ratio becomes

$$\frac{(x_1+1)(t+1)}{(s+1)^2} = 27\cdot \frac{(x_2+1)(t+1)}{(r+1)^2}.$$

Using

$$(r+1)^2=\frac94(s+1)^2,$$

we obtain

$$x_1+1 = 12(x_2+1).$$

Hence

$$x_1=12x_2+11.$$

Since

$$t=x_1+x_2,$$

it follows that

$$t=13x_2+11.$$

Let

$$X=13x_2+12=t+1.$$

Deriving a Pell Equation

Since

$$(t+1)^2 = \frac{13k^2-2k}{4},$$

and $k=\frac{s+1}{2}$, letting

$$M=s+1$$

gives

$$4X^2 = \frac{13M^2-4M}{4}.$$

Multiplying by $4$,

$$16X^2 = 13M^2-4M.$$

Rearranging,

$$13M^2-4M-16X^2=0.$$

Multiply by $13$:

$$169M^2-52M-208X^2=0.$$

Complete the square:

$$(13M-2)^2-4-208X^2=0.$$

Thus

$$(13M-2)^2-208X^2=4.$$

Dividing by $4$ gives

$$\left(\frac{13M-2}{2}\right)^2-52X^2=1.$$

Let

$$Y=\frac{13M-2}{2}.$$

Then

$$Y^2-52X^2=1.$$

This is a Pell equation.

Solving the Pell Equation

The fundamental positive solution of

$$Y^2-52X^2=1$$

is

$$(Y,X)=(649,90).$$

Since

$$X=t+1,$$

we obtain

$$t=89.$$

Also,

$$Y=\frac{13M-2}{2}=649,$$

so

$$13M-2=1298,$$

and therefore

$$M=100.$$

Hence

$$s=M-1=99.$$

Since

$$r+1=\frac32(s+1),$$

we have

$$r+1=150,$$

so

$$r=149.$$

Computing the Sum

Therefore

$$r+s+t = 149+99+89 = 337.$$

Final Answer: 337

---

HMMT 2025 Problem 3

Problem

A polynomial $P(x)$ is a base-$n$ polynomial if it is of the form

$$a_dx^d+a_{d-1}x^{d-1}+\cdots+a_1x+a_0,$$

where each $a_i$ is an integer between $0$ and $n-1$ inclusive and $a_d>0$.

Find the largest positive integer $n$ such that for any real number $c$, there exists at most one base-$n$ polynomial $P(x)$ for which

$$P(\sqrt2+\sqrt3)=c.$$

Solution

Reformulating the Uniqueness Condition

Let

$$\alpha=\sqrt2+\sqrt3.$$

Suppose two distinct base-$n$ polynomials $P(x)$ and $Q(x)$ satisfy

$$P(\alpha)=Q(\alpha).$$

Then their difference

$$D(x)=P(x)-Q(x)$$

is a nonzero polynomial satisfying

$$D(\alpha)=0.$$

Since the coefficients of $P$ and $Q$ lie in

$$\{0,1,\dots,n-1\},$$

the coefficients of $D$ all lie in

$$[-(n-1), n-1].$$

Therefore uniqueness holds exactly when every nonzero integer polynomial vanishing at $\alpha$ has at least one coefficient whose absolute value is at least $n$.

Our task is therefore to determine the smallest possible value of

$$\max |d_i|$$

among all nonzero integer polynomials

$$D(x)$$

satisfying

$$D(\alpha)=0.$$

Finding the Minimal Polynomial of $\alpha$

We compute:

$$\alpha^2 = (\sqrt2+\sqrt3)^2 = 5+2\sqrt6.$$

Hence

$$\alpha^2-5=2\sqrt6.$$

Squaring again gives

$$(\alpha^2-5)^2=24.$$

Expanding,

$$\alpha^4-10\alpha^2+25=24,$$

so

$$\alpha^4-10\alpha^2+1=0.$$

Thus the minimal polynomial of $\alpha$ is

$$m(x)=x^4-10x^2+1.$$

Any integer polynomial vanishing at $\alpha$ must therefore be a multiple of $m(x)$.

Producing a Small Multiple

Consider

$$(x^2+1)m(x).$$

Multiplying,

$$(x^2+1)(x^4-10x^2+1) = x^6-9x^4-9x^2+1.$$

The coefficients are

$$1, 0, -9, 0, -9, 0, 1.$$

Therefore there exists a nonzero polynomial vanishing at $\alpha$ whose coefficients all have absolute value at most $9$.

Consequently, uniqueness fails for

$$n=10,$$

because the coefficients lie in

$$[-9,9].$$

Thus

$$n\le 9.$$

Showing That 9 Is Best Possible

We now prove that every nonzero multiple of

$$m(x)=x^4-10x^2+1$$

has some coefficient whose absolute value is at least $9$.

Let

$$Q(x)=m(x)G(x),$$

where

$$G(x)=\sum g_i x^i$$

is a nonzero integer polynomial.

Let

$$M=\max |g_i|.$$

Choose an index $k$ such that

$$|g_k|=M.$$

The coefficient of $x^{k+2}$ in $Q(x)$ equals

$$q_{k+2} = g_{k+2}-10g_k+g_{k-2}.$$

Therefore

$$|q_{k+2}| \ge 10|g_k|-|g_{k+2}|-|g_{k-2}| \ge 10M-M-M = 8M.$$

Since $M\ge1$,

$$|q_{k+2}|\ge8.$$

If every coefficient of $Q$ had absolute value at most $8$, equality would have to hold throughout.

That forces

$$|g_{k+2}|=|g_{k-2}|=M$$

and all three terms must have the same sign.

Repeating the argument shows that infinitely many coefficients of $G$ would have magnitude $M$, impossible because $G$ has finite degree.

Hence no nonzero multiple of $m(x)$ can have all coefficients bounded by $8$.

Therefore every nonzero polynomial vanishing at $\alpha$ has a coefficient with absolute value at least

$$9.$$

Finishing the Argument

We have shown:

• There exists a nonzero polynomial vanishing at $\alpha$ with all coefficients bounded by $9$.
• No nonzero polynomial vanishing at $\alpha$ can have all coefficients bounded by $8$.

Therefore the smallest possible coefficient bound is exactly

$$9.$$

Thus uniqueness holds for base-$9$ polynomials and fails for base-$10$ polynomials.

Hence the largest possible value of $n$ is

$$\boxed{9}.$$

Final Answer: 9