exam: HMMT

problem: (Distributive law) Prove that $(x \oplus y) \odot z=x \odot z \oplus y \odot z$ for all $x, y, z \in \mathbb{R} \cup\{\infty\}$.

solution:
This is equivalent to proving that

$$
\min (x, y)+z=\min (x+z, y+z) .
$$

Consider two cases. If $x \leq y$, then $L H S=x+z$ and $R H S=x+z$. If $x>y$, then $L H S=y+z$ and $R H S=y+z$. It follows that $L H S=R H S$.

solution_match: Solution:


problem: (Freshman's Dream) Let $z^{n}$ denote $z \odot z \odot z \odot \cdots \odot z$ with $z$ appearing $n$ times. Prove that $(x \oplus y)^{n}=x^{n} \oplus y^{n}$ for all $x, y \in \mathbb{R} \cup\{\infty\}$ and positive integer $n$.

solution: Without loss of generality, suppose that $x \leq y$, then $L H S=\min (x, y)^{n}=x^{n}=$ $n x$, and RHS $=\min \left(x^{n}, y^{n}\right)=\min (n x, n y)=n x$.


problem:
By a tropical polynomial we mean a function of the form

$$
p(x)=a_{n} \odot x^{n} \oplus a_{n-1} \odot x^{n-1} \oplus \cdots \oplus a_{1} \odot x \oplus a_{0}
$$

where exponentiation is as defined in the previous problem.
Let $p$ be a tropical polynomial. Prove that

$$
p\left(\frac{x+y}{2}\right) \geq \frac{p(x)+p(y)}{2}
$$

for all $x, y \in \mathbb{R} \cup\{\infty\}$. (This means that all tropical polynomials are concave.)

solution:
First, note that for any $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$, we have

$$
\min \left\{x_{1}+y_{1}, x_{2}+y_{2}, \ldots, x_{n}+y_{n}\right\} \geq \min \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}+\min \left\{y_{1}, y_{2}, \ldots, y_{n}\right\} .
$$

Indeed, suppose that $x_{m}+y_{m}=\min _{i}\left\{x_{i}+y_{i}\right\}$, then $x_{m} \geq \min _{i} x_{i}$ and $y_{m} \geq \min _{i} y_{i}$, and so $\min _{i}\left\{x_{i}+y_{i}\right\}=x_{m}+y_{m} \geq \min _{i} x_{i}+\min _{i} y_{i}$.
Now, let us write a tropical polynomial in a more familiar notation. We have

$$
p(x)=\min _{0 \leq k \leq n}\left\{a_{k}+k x\right\} .
$$

So

$$
\begin{aligned}
p\left(\frac{x+y}{2}\right) & =\min _{0 \leq k \leq n}\left\{a_{k}+k\left(\frac{x+y}{2}\right)\right\} \\
& =\frac{1}{2} \min _{0 \leq k \leq n}\left\{\left(a_{k}+k x\right)+\left(a_{k}+k y\right)\right\} \\
& \geq \frac{1}{2}\left(\min _{0 \leq k \leq n}\left\{a_{k}+k x\right\}+\min _{0 \leq k \leq n}\left\{a_{k}+k y\right\}\right) \\
& =\frac{1}{2}(p(x)+p(y)) .
\end{aligned}
$$


problem:
(Fundamental Theorem of Algebra) Let $p$ be a tropical polynomial:

$$
p(x)=a_{n} \odot x^{n} \oplus a_{n-1} \odot x^{n-1} \oplus \cdots \oplus a_{1} \odot x \oplus a_{0}, \quad a_{n} \neq \infty
$$

Prove that we can find $r_{1}, r_{2}, \ldots, r_{n} \in \mathbb{R} \cup\{\infty\}$ so that

$$
p(x)=a_{n} \odot\left(x \oplus r_{1}\right) \odot\left(x \oplus r_{2}\right) \odot \cdots \odot\left(x \oplus r_{n}\right)
$$

for all $x$.

solution:
Again, we have

$$
p(x)=\min _{0 \leq k \leq n}\left\{a_{k}+k x\right\} .
$$

So the graph of $y=p(x)$ can be drawn as follows: first, draw all the lines $y=a_{k}+k x$, $k=0,1, \ldots, n$, then trace out the lowest broken line, which then is the graph of $y=p(x)$.
So $p(x)$ is piecewise linear and continuous, and has slopes from the set $\{0,1,2, \ldots, n\}$. We know from the previous problem that $p(x)$ is concave, and so its slope must be decreasing (this can also be observed simply from the drawing of the graph of $y=p(x)$ ). Then, let $r_{k}$ denote the $x$-coordinate of the leftmost kink such that the slope of the graph is less than $k$ to the right of this kink. Then, $r_{n} \leq r_{n-1} \leq \cdots \leq r_{1}$, and for $r_{k-1} \leq x \leq r_{k}$, the graph of $p$ is linear with slope $k$. Note that is if possible that $r_{k-1}=r_{k}$, if no segment of $p$ has slope $k$. Also, since $a_{n} \neq \infty$, the leftmost piece of $p(x)$ must have slope $n$, and thus $r_{n}$ exists, and thus all $r_{i}$ exist.
Now, compare $p(x)$ with

$$
\begin{aligned}
q(x) & =a_{n} \odot\left(x \oplus r_{1}\right) \odot\left(x \oplus r_{2}\right) \odot \cdots \odot\left(x \oplus r_{n}\right) \\
& =a_{n}+\min \left(x, r_{1}\right)+\min \left(x, r_{2}\right)+\cdots+\min \left(x, r_{n}\right) .
\end{aligned}
$$

For $r_{k-1} \leq x \leq r_{k}$, the slope of $q(x)$ is $k$, and for $x \leq r_{n}$ the slope of $q$ is $n$ and for $x \geq r_{1}$ the slope of $q$ is 0 . So $q$ is piecewise linear, and of course it is continuous. It follows that the graph of $q$ coincides with that of $p$ up to a translation. By taking any $x<r_{n}$, we see that $q(x)=a_{n}+n x=p(x)$, we see that the graphs of $p$ and $q$ coincide, and thus they must be the same function.

## Juggling [125]

A juggling sequence of length $n$ is a sequence $j(\cdot)$ of $n$ nonnegative integers, usually written as a string

$$
j(0) j(1) \ldots j(n-1)
$$

such that the mapping $f: \mathbb{Z} \rightarrow \mathbb{Z}$ defined by

$$
f(t)=t+j(\bar{t})
$$

is a permutation of the integers. Here $\bar{t}$ denotes the remainder of $t$ when divided by $n$. In this case, we say that $f$ is the corresponding juggling pattern.
For a juggling pattern $f$ (or its corresponding juggling sequence), we say that it has $b$ balls if the permutation induces $b$ infinite orbits on the set of integers. Equivalently, $b$ is the maximum number such that we can find a set of $b$ integers $\left\{t_{1}, t_{2}, \ldots, t_{b}\right\}$ so that the sets $\left\{t_{i}, f\left(t_{i}\right), f\left(f\left(t_{i}\right)\right), f\left(f\left(f\left(t_{i}\right)\right)\right), \ldots\right\}$ are all infinite and mutually disjoint (i.e. non-overlapping) for $i=1,2, \ldots, b$. (This definition will become clear in a second.)

Now is probably a good time to pause and think about what all this has to do with juggling. Imagine that we are juggling a number of balls, and at time $t$, we toss a ball from our hand up to a height $j(\bar{t})$. This ball stays up in the air for $j(\bar{t})$ units of time, so that it comes back to our hand at time $f(t)=t+j(\bar{t})$. Then, the juggling pattern presents a simplified model of how balls are juggled (for instance, we ignore information such as which hand we use to toss the ball). A throw height of 0 (i.e., $j(\bar{t})=0$ and $f(t)=t$ ) represents that no thrown takes place at time $t$, which could correspond to an empty hand. Then, $b$ is simply the minimum number of balls needed to carry out the juggling.

The following graphical representation may be helpful to you. On a horizontal line, an curve is drawn from $t$ to $f(t)$. For instance, the following diagram depicts the juggling sequence 441 (or the juggling sequences 414 and 144). Then $b$ is simply the number of contiguous "paths" drawn, which is 3 in this case.
![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-3.jpg?height=353&width=1001&top_left_y=1683&top_left_x=562)

Figure 1: Juggling diagram of 441.


problem: Prove that 572 is not a juggling sequence.

solution: We are given $j(0)=5, j(1)=7$ and $j(2)=2$. So $f(3)=3+j(0)=8$ and $f(1)=1+j(1)=8$. Thus $f(3)=f(1)$ and so $f$ is not a permutation of $\mathbb{Z}$, and hence 572 is not a juggling pattern. (In other words, there is a "collision" at times $t \equiv 2(\bmod 3)$.)


problem: Suppose that $j(0) j(1) \cdots j(n-1)$ is a valid juggling sequence. For $i=0,1, \ldots, n-1$, Let $a_{i}$ denote the remainder of $j(i)+i$ when divided by $n$. Prove that $\left(a_{0}, a_{1}, \ldots, a_{n-1}\right)$ is a permutation of $(0,1, \ldots, n-1)$.

solution: Suppose that $a_{i}=j(i)+i-b_{i} n$, where $b_{i}$ is an integer. Note that $f\left(i-b_{i} n\right)=$ $i-b_{i} n+j(i)=a_{i}$. Since $\left\{i-b_{i} n \mid i=0,1, \ldots, n-1\right\}$ contains $n$ distinct integers (as their residue $\bmod n$ are all distinct), and $f$ is a permutation, we see that after applying the map $f$, the resulting set $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$ is a set of $n$ distinct integers. Since $0 \leq a_{i}<n$ from definition, we see that ( $a_{0}, a_{1}, \ldots, a_{n-1}$ ) is a permutation of $(0,1, \ldots, n-1)$.


problem: Determine the number of juggling sequences of length $n$ with exactly 1 ball.

solution: $2^{n}-1$. Solution: With 1 ball, we simply need to decide at times should the ball land in our hand. That is, we need to choose a non-empty subset of $\{0,1,2, \ldots, n-1\}$ where the ball lands. It follows that the answer is $2^{n}-1$.

solution_match: Answer:


problem:
Prove that the number of balls $b$ in a juggling sequence $j(0) j(1) \cdots j(n-1)$ is simply the average

$$
b=\frac{j(0)+j(1)+\cdots+j(n-1)}{n} .
$$

solution:
Consider the corresponding juggling diagram. Say the length of an curve from $t$ to $f(t)$ is $f(t)-t$. Let us draw only the curves whose left endpoint lies inside $[0, M n-1]$. For every single ball, the sum of the lengths of the arrows drawn corresponding to that ball is between $M n-J$ and $M n+J$, where $J=\max \{j(0), j(1), \ldots, j(n-1)\}$. It follows that the sum of the lengths of the arrows drawn is between $b(M n-J)$ and $b(M n+J)$. Since the arrow drawn at $t$ has length $j(\bar{t})$, the sum of the lengths of the arrows drawn is $M(j(0)+j(1)+\cdots+j(n-1))$. It follows that

$$
b(M n-J) \leq M(j(0)+j(1)+\cdots+j(n-1)) \leq b(M n+J)
$$

Dividing by $M n$, we get

$$
b\left(1-\frac{J}{n M}\right) \leq \frac{j(0)+j(1)+\cdots+j(n-1)}{n} \leq b\left(1+\frac{J}{n M}\right)
$$

Since we can take $M$ to be arbitrarily large, we must have

$$
b=\frac{j(0)+j(1)+\cdots+j(n-1)}{n}
$$

as desired.


problem: Show that the converse of the previous statement is false by providing a non-juggling sequence $j(0) j(1) j(2)$ of length 3 where the average $\frac{1}{3}(j(0)+j(1)+j(2))$ is an integer. Show that your example works.

solution:
One such example is 210 . This is not a juggling sequence since $f(0)=f(1)=2$.

## Incircles [180]

In the following problems, $A B C$ is a triangle with incenter $I$. Let $D, E, F$ denote the points where the incircle of $A B C$ touches sides $B C, C A, A B$, respectively.
![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-5.jpg?height=369&width=329&top_left_y=493&top_left_x=898)

At the end of this section you can find some terminology and theorems that may be helpful to you.


problem: Let $a, b, c$ denote the side lengths of $B C, C A, A B$. Find the lengths of $A E, B F, C D$ in terms of $a, b, c$.

solution:
Let $x=A E=A F, y=B D=B F, z=C D=C E$. Since $B C=B D+C D$, we have $a=x+y$. Similarly with the other sides, we arrive at the following system of equations:

$$
a=y+z, \quad b=x+z, \quad c=x+y .
$$

Solving this system gives us

$$
\begin{aligned}
& A E=x=\frac{b+c-a}{2}, \\
& B F=y=\frac{a+c-b}{2}, \\
& C D=z=\frac{a+b-c}{2}
\end{aligned}
$$


problem: Show that lines $A D, B E, C F$ pass through a common point.

solution:
Using Ceva's theorem on triangle $A B C$, we see that it suffices to show that

$$
\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1
$$

Since $A F=A E, B D=B F$, and $C D=C E$ (due to equal tangents), we see that the LHS is indeed 1.

Remark: The point of concurrency is known as the Gergonne point.


problem: Show that the incenter of triangle $A E F$ lies on the incircle of $A B C$.

solution:
Let segment $A I$ meet the incircle at $A_{1}$. Let us show that $A_{1}$ is the incenter of $A E F$.
![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-6.jpg?height=554&width=560&top_left_y=238&top_left_x=826)

Since $A E=A F$ and $A A^{\prime}$ is the angle bisector of $\angle E A F$, we find that $A_{1} E=A_{1} F$. Using tangent-chord, we see that $\angle A F A_{1}=\angle A_{1} E F=\angle A_{1} F E$. Therefore, $A_{1}$ lies on the angle bisector of $\angle A F E$. Since $A_{1}$ also lies on the angle bisector of $\angle E A F, A_{1}$ must be the incenter of $A E F$, as desired.


problem: Let $A_{1}, B_{1}, C_{1}$ be the incenters of triangle $A E F, B D F, C D E$, respectively. Show that $A_{1} D, B_{1} E, C_{1} F$ all pass through the orthocenter of $A_{1} B_{1} C_{1}$.

solution:
Using the result from the previous problem, we see that $A_{1}, B_{1}, C_{1}$ are respectively the midpoints of the $\operatorname{arc} F E, F D, D F$ of the incircle. We have
![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-6.jpg?height=369&width=361&top_left_y=1282&top_left_x=931)

$$
\begin{aligned}
\angle D A_{1} C_{1}+\angle B_{1} C_{1} A_{1} & =\frac{1}{2} \angle D I C_{1}+\frac{1}{2} \angle B_{1} I F+\frac{1}{2} \angle F I A_{1} \\
& =\frac{1}{4}(\angle E I D+\angle D I F+\angle F I E) \\
& =\frac{1}{4} \cdot 360^{\circ} \\
& =90^{\circ} .
\end{aligned}
$$

It follows that $A_{1} D$ is perpendicular to $B_{1} C_{1}$, and thus $A_{1} D$ passes through the orthocenter of $A_{1} B_{1} C_{1}$. Similarly, $A_{1} D, B_{1} E, C_{1} F$ all pass through the orthocenter of $A_{1} B_{1} C_{1}$.


problem: Let $X$ be the point on side $B C$ such that $B X=C D$. Show that the excircle $A B C$ opposite of vertex $A$ touches segment $B C$ at $X$.

solution:
Let the excircle touch lines $B C, A C$ and $A B$ at $X^{\prime}, Y$ and $Z$, respectively. Using the equal tangent property repeatedly, we have

$$
B X^{\prime}-X^{\prime} C=B Z-C Y=(E Y-C Y)-(F Z-B Z)=C E-B F=C D-B D .
$$

It follows that $B X^{\prime}=C D$, and thus $X^{\prime}=X$. So the excircle touches $B C$ at $X$.
![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-7.jpg?height=849&width=551&top_left_y=237&top_left_x=836)


problem: Let $X$ be as in the previous problem. Let $T$ be the point diametrically opposite to $D$ on on the incircle of $A B C$. Show that $A, T, X$ are collinear.

solution:
Consider a dilation centered at $A$ that carries the incircle to the excircle. This dilation must send the diameter $D T$ to some the diameter of excircle that is perpendicular to $B C$. The only such diameter is the one goes through $X$. It follows that $T$ gets carried to $X$. Therefore, $A, T, X$ are collinear.

## Glossary and some possibly useful facts

- A set of points is collinear if they lie on a common line. A set of lines is concurrent if they pass through a common point.
- Given $A B C$ a triangle, the three angle bisectors are concurrent at the incenter of the triangle. The incenter is the center of the incircle, which is the unique circle inscribed in $A B C$, tangent to all three sides.
- The excircles of a triangle $A B C$ are the three circles on the exterior the triangle but tangent to all three lines $A B, B C, C A$.
![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-7.jpg?height=367&width=383&top_left_y=2034&top_left_x=920)
- The orthocenter of a triangle is the point of concurrency of the three altitudes.
- Ceva's theorem states that given $A B C$ a triangle, and points $X, Y, Z$ on sides $B C, C A, A B$, respectively, the lines $A X, B Y, C Z$ are concurrent if and only if

$$
\frac{B X}{X B} \cdot \frac{C Y}{Y A} \cdot \frac{A Z}{Z B}=1
$$