{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": null, "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\n\n$$\n\\min (x, y)+z=\\min (x+z, y+z) .\n$$\n\nConsider 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 $L H S=R H S$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n1. [10]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "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 $=\\min \\left(x^{n}, y^{n}\\right)=\\min (n x, n y)=n x$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n2. [10]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "By a tropical polynomial we mean a function of the form\n\n$$\np(x)=a_{n} \\odot x^{n} \\oplus a_{n-1} \\odot x^{n-1} \\oplus \\cdots \\oplus a_{1} \\odot x \\oplus a_{0}\n$$\n\nwhere exponentiation is as defined in the previous problem.\nLet $p$ be a tropical polynomial. Prove that\n\n$$\np\\left(\\frac{x+y}{2}\\right) \\geq \\frac{p(x)+p(y)}{2}\n$$\n\nfor 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\n\n$$\n\\min \\left\\{x_{1}+y_{1}, x_{2}+y_{2}, \\ldots, x_{n}+y_{n}\\right\\} \\geq \\min \\left\\{x_{1}, x_{2}, \\ldots, x_{n}\\right\\}+\\min \\left\\{y_{1}, y_{2}, \\ldots, y_{n}\\right\\} .\n$$\n\nIndeed, 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 $\\min _{i}\\left\\{x_{i}+y_{i}\\right\\}=x_{m}+y_{m} \\geq \\min _{i} x_{i}+\\min _{i} y_{i}$.\nNow, let us write a tropical polynomial in a more familiar notation. We have\n\n$$\np(x)=\\min _{0 \\leq k \\leq n}\\left\\{a_{k}+k x\\right\\} .\n$$\n\nSo\n\n$$\n\\begin{aligned}\np\\left(\\frac{x+y}{2}\\right) & =\\min _{0 \\leq k \\leq n}\\left\\{a_{k}+k\\left(\\frac{x+y}{2}\\right)\\right\\} \\\\\n& =\\frac{1}{2} \\min _{0 \\leq k \\leq n}\\left\\{\\left(a_{k}+k x\\right)+\\left(a_{k}+k y\\right)\\right\\} \\\\\n& \\geq \\frac{1}{2}\\left(\\min _{0 \\leq k \\leq n}\\left\\{a_{k}+k x\\right\\}+\\min _{0 \\leq k \\leq n}\\left\\{a_{k}+k y\\right\\}\\right) \\\\\n& =\\frac{1}{2}(p(x)+p(y)) .\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n3. [35]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "(Fundamental Theorem of Algebra) Let $p$ be a tropical polynomial:\n\n$$\np(x)=a_{n} \\odot x^{n} \\oplus a_{n-1} \\odot x^{n-1} \\oplus \\cdots \\oplus a_{1} \\odot x \\oplus a_{0}, \\quad a_{n} \\neq \\infty\n$$\n\nProve that we can find $r_{1}, r_{2}, \\ldots, r_{n} \\in \\mathbb{R} \\cup\\{\\infty\\}$ so that\n\n$$\np(x)=a_{n} \\odot\\left(x \\oplus r_{1}\\right) \\odot\\left(x \\oplus r_{2}\\right) \\odot \\cdots \\odot\\left(x \\oplus r_{n}\\right)\n$$\n\nfor all $x$.", "solution": "Again, we have\n\n$$\np(x)=\\min _{0 \\leq k \\leq n}\\left\\{a_{k}+k x\\right\\} .\n$$\n\nSo 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 $y=p(x)$.\nSo $p(x)$ 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.\nNow, compare $p(x)$ with\n\n$$\n\\begin{aligned}\nq(x) & =a_{n} \\odot\\left(x \\oplus r_{1}\\right) \\odot\\left(x \\oplus r_{2}\\right) \\odot \\cdots \\odot\\left(x \\oplus r_{n}\\right) \\\\\n& =a_{n}+\\min \\left(x, r_{1}\\right)+\\min \\left(x, r_{2}\\right)+\\cdots+\\min \\left(x, r_{n}\\right) .\n\\end{aligned}\n$$\n\nFor $r_{k-1} \\leq x \\leq r_{k}$, 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.\n\n## Juggling [125]\n\nA juggling sequence of length $n$ is a sequence $j(\\cdot)$ of $n$ nonnegative integers, usually written as a string\n\n$$\nj(0) j(1) \\ldots j(n-1)\n$$\n\nsuch that the mapping $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ defined by\n\n$$\nf(t)=t+j(\\bar{t})\n$$\n\nis 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.\nFor 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.)\n\nNow 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.\n\nThe 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.\n![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-3.jpg?height=353&width=1001&top_left_y=1683&top_left_x=562)\n\nFigure 1: Juggling diagram of 441.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n4. [40]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "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 $t \\equiv 2(\\bmod 3)$.)", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n5. [10]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "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 $(0,1, \\ldots, n-1)$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n6. [40]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "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 $2^{n}-1$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n7. [30]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Prove that the number of balls $b$ in a juggling sequence $j(0) j(1) \\cdots j(n-1)$ is simply the average\n\n$$\nb=\\frac{j(0)+j(1)+\\cdots+j(n-1)}{n} .\n$$", "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\n\n$$\nb(M n-J) \\leq M(j(0)+j(1)+\\cdots+j(n-1)) \\leq b(M n+J)\n$$\n\nDividing by $M n$, we get\n\n$$\nb\\left(1-\\frac{J}{n M}\\right) \\leq \\frac{j(0)+j(1)+\\cdots+j(n-1)}{n} \\leq b\\left(1+\\frac{J}{n M}\\right)\n$$\n\nSince we can take $M$ to be arbitrarily large, we must have\n\n$$\nb=\\frac{j(0)+j(1)+\\cdots+j(n-1)}{n}\n$$\n\nas desired.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n8. [40]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "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 $f(0)=f(1)=2$.\n\n## Incircles [180]\n\nIn 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.\n![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-5.jpg?height=369&width=329&top_left_y=493&top_left_x=898)\n\nAt the end of this section you can find some terminology and theorems that may be helpful to you.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n9. [5]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "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:\n\n$$\na=y+z, \\quad b=x+z, \\quad c=x+y .\n$$\n\nSolving this system gives us\n\n$$\n\\begin{aligned}\n& A E=x=\\frac{b+c-a}{2}, \\\\\n& B F=y=\\frac{a+c-b}{2}, \\\\\n& C D=z=\\frac{a+b-c}{2}\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n10. [15]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "11", "problem_type": null, "exam": "HMMT", "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\n\n$$\n\\frac{B D}{D C} \\cdot \\frac{C E}{E A} \\cdot \\frac{A F}{F B}=1\n$$\n\nSince $A F=A E, B D=B F$, and $C D=C E$ (due to equal tangents), we see that the LHS is indeed 1.\n\nRemark: The point of concurrency is known as the Gergonne point.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n11. [15]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "12", "problem_type": null, "exam": "HMMT", "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 $A E F$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-6.jpg?height=554&width=560&top_left_y=238&top_left_x=826)\n\nSince $A E=A F$ 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.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n12. [35]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "13", "problem_type": null, "exam": "HMMT", "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\n![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-6.jpg?height=369&width=361&top_left_y=1282&top_left_x=931)\n\n$$\n\\begin{aligned}\n\\angle D A_{1} C_{1}+\\angle B_{1} C_{1} A_{1} & =\\frac{1}{2} \\angle D I C_{1}+\\frac{1}{2} \\angle B_{1} I F+\\frac{1}{2} \\angle F I A_{1} \\\\\n& =\\frac{1}{4}(\\angle E I D+\\angle D I F+\\angle F I E) \\\\\n& =\\frac{1}{4} \\cdot 360^{\\circ} \\\\\n& =90^{\\circ} .\n\\end{aligned}\n$$\n\nIt 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 $A_{1} B_{1} C_{1}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n13. [35]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "14", "problem_type": null, "exam": "HMMT", "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\n\n$$\nB X^{\\prime}-X^{\\prime} C=B Z-C Y=(E Y-C Y)-(F Z-B Z)=C E-B F=C D-B D .\n$$\n\nIt follows that $B X^{\\prime}=C D$, and thus $X^{\\prime}=X$. So the excircle touches $B C$ at $X$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-7.jpg?height=849&width=551&top_left_y=237&top_left_x=836)", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n14. [40]", "solution_match": "\nSolution: "}}
{"year": "2008", "tier": "T4", "problem_label": "15", "problem_type": null, "exam": "HMMT", "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.\n\n## Glossary and some possibly useful facts\n\n- 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.\n- 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.\n- The excircles of a triangle $A B C$ are the three circles on the exterior the triangle but tangent to all three lines $A B, B C, C A$.\n![](https://cdn.mathpix.com/cropped/2025_01_24_0c4d02e946d4192126fag-7.jpg?height=367&width=383&top_left_y=2034&top_left_x=920)\n- The orthocenter of a triangle is the point of concurrency of the three altitudes.\n- 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\n\n$$\n\\frac{B X}{X B} \\cdot \\frac{C Y}{Y A} \\cdot \\frac{A Z}{Z B}=1\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-team2-solutions.jsonl", "problem_match": "\n15. [40]", "solution_match": "\nSolution: "}}