{"year": "2014", "tier": "T3", "problem_label": "1", "problem_type": null, "exam": "USAJMO", "problem": "Let $a, b, c$ be real numbers greater than or equal to 1 . Prove that  $$ \\min \\left(\\frac{10 a^{2}-5 a+1}{b^{2}-5 b+10}, \\frac{10 b^{2}-5 b+1}{c^{2}-5 c+10}, \\frac{10 c^{2}-5 c+1}{a^{2}-5 a+10}\\right) \\leq a b c $$", "solution": "  Notice that  $$ \\frac{10 a^{2}-5 a+1}{a^{2}-5 a+10} \\leq a^{3} $$  since it rearranges to $(a-1)^{5} \\geq 0$. Cyclically multiply to get  $$ \\prod_{\\mathrm{cyc}}\\left(\\frac{10 a^{2}-5 a+1}{b^{2}-5 b+10}\\right) \\leq(a b c)^{3} $$  and the minimum is at most the geometric mean.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2014-notes.jsonl"}}
{"year": "2014", "tier": "T3", "problem_label": "2", "problem_type": null, "exam": "USAJMO", "problem": "Let $\\triangle A B C$ be a non-equilateral, acute triangle with $\\angle A=60^{\\circ}$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\\triangle A B C$, respectively. (a) Prove that line $O H$ intersects both segments $A B$ and $A C$ at two points $P$ and $Q$, respectively. (b) Denote by $s$ and $t$ the respective areas of triangle $A P Q$ and quadrilateral $B P Q C$. Determine the range of possible values for $s / t$.", "solution": "  We begin with some synthetic work. Let $I$ denote the incenter, and recall (\"fact 5\") that the arc midpoint $M$ is the center of $(B I C)$, which we denote by $\\gamma$.  Now we have that  $$ \\angle B O C=\\angle B I C=\\angle B H C=120^{\\circ} . $$  Since all three centers lie inside $A B C$ (as it was acute), and hence on the opposite side of $\\overline{B C}$ as $M$, it follows that $O, I, H$ lie on minor arc $B C$ of $\\gamma$.  We note this implies (a) already, as line $O H$ meets line $B C$ outside of segment $B C$. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-04.jpg?height=806&width=552&top_left_y=1339&top_left_x=752)  Claim - Triangle $A P Q$ is equilateral with side length $\\frac{b+c}{3}$.   Finally, since $\\angle P B H=30^{\\circ}$, and $\\angle B P H=120^{\\circ}$, it follows that $\\triangle B P H$ is isosceles and $B P=P H$. Similarly, $C Q=Q H$. So $b+c=A P+B P+A Q+Q C=A P+A Q+P Q$ as needed.  Finally, we turn to the boring task of extracting the numerical answer. We have  $$ \\frac{s}{s+t}=\\frac{[A P Q]}{[A B C]}=\\frac{\\frac{\\sqrt{3}}{4}\\left(\\frac{b+c}{3}\\right)^{2}}{\\frac{\\sqrt{3}}{4} b c}=\\frac{b^{2}+2 b c+c^{2}}{9 b c}=\\frac{1}{9}\\left(2+\\frac{b}{c}+\\frac{c}{b}\\right) . $$  So the problem is reduced to analyzing the behavior of $b / c$. For this, we imagine fixing $\\Gamma$ the circumcircle of $A B C$, as well as the points $B$ and $C$. Then as we vary $A$ along the \"topmost\" arc of measure $120^{\\circ}$, we find $b / c$ is monotonic with values $1 / 2$ and 2 at endpoints, and by continuity all values $b / c \\in(1 / 2,2)$ can be achieved.  So  $$ \\frac{1}{2}<\\frac{b}{c}<2 \\Longrightarrow 4 / 9<\\frac{s}{s+t}<1 / 2 \\Longrightarrow 4 / 5<\\frac{s}{t}<1 $$  as needed.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2014-notes.jsonl"}}
{"year": "2014", "tier": "T3", "problem_label": "3", "problem_type": null, "exam": "USAJMO", "problem": "Find all $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that  $$ x f(2 f(y)-x)+y^{2} f(2 x-f(y))=\\frac{f(x)^{2}}{x}+f(y f(y)) $$  for all $x, y \\in \\mathbb{Z}$ such that $x \\neq 0$.", "solution": "  The answer is $f(x) \\equiv 0$ and $f(x) \\equiv x^{2}$. Check that these work.  $$ x f(2 f(0)-x)=\\frac{f(x)^{2}}{x}+f(0) . $$  The nicest part of the problem is the following step:  $$ \\text { Claim - We have } f(0)=0 $$   $$ f(0)=0 $$  Claim - We have $f(x) \\in\\left\\{0, x^{2}\\right\\}$ for each individual $x$.  $$ x^{2} f(-x)=f(x)^{2} $$  holds for all nonzero $x$, but also for $x=0$. Let us now check that $f$ is an even function. In the above, we may also derive $f(-x)^{2}=x^{2} f(x)$. If $f(x) \\neq f(-x)$ (and hence $x \\neq 0$ ), then subtracting the above and factoring implies that $f(x)+f(-x)=-x^{2}$; we can then obtain by substituting the relation  $$ \\left[f(x)+\\frac{1}{2} x^{2}\\right]^{2}=-\\frac{3}{4} x^{4}<0 $$  which is impossible. This means $f(x)^{2}=x^{2} f(x)$, thus  $$ f(x) \\in\\left\\{0, x^{2}\\right\\} \\quad \\forall x . $$  Now suppose there exists a nonzero integer $t$ with $f(t)=0$. We will prove that $f(x) \\equiv 0$. Put $y=t$ in the given to obtain that  $$ t^{2} f(2 x)=0 $$  for any integer $x \\neq 0$, and hence conclude that $f(2 \\mathbb{Z}) \\equiv 0$. Then selecting $x=2 k \\neq 0$ in the given implies that  $$ y^{2} f(4 k-f(y))=f(y f(y)) $$  Assume for contradiction that $f(m)=m^{2}$ now for some odd $m \\neq 0$. Evidently  $$ m^{2} f\\left(4 k-m^{2}\\right)=f\\left(m^{3}\\right) $$  If $f\\left(m^{3}\\right) \\neq 0$ this forces $f\\left(4 k-m^{2}\\right) \\neq 0$, and hence $m^{2}\\left(4 k-m^{2}\\right)^{2}=m^{6}$ for arbitrary $k \\neq 0$, which is clearly absurd. That means  $$ f\\left(4 k-m^{2}\\right)=f\\left(m^{2}-4 k\\right)=f\\left(m^{3}\\right)=0 $$  for each $k \\neq 0$. Since $m$ is odd, $m^{2} \\equiv 1(\\bmod 4)$, and so $f(n)=0$ for all $n$ other than $\\pm m^{2}$ (since we cannot select $\\left.k=0\\right)$.  Now $f(m)=m^{2}$ means that $m= \\pm 1$. Hence either $f(x) \\equiv 0$ or  $$ f(x)= \\begin{cases}1 & x= \\pm 1 \\\\ 0 & \\text { otherwise }\\end{cases} $$  To show that the latter fails, we simply take $x=5$ and $y=1$ in the given.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2014-notes.jsonl"}}
{"year": "2014", "tier": "T3", "problem_label": "4", "problem_type": null, "exam": "USAJMO", "problem": "Let $b \\geq 2$ be a fixed integer, and let $s_{b}(n)$ denote the sum of the base- $b$ digits of $n$. Show that there are infinitely many positive integers that cannot be represented in the from $n+s_{b}(n)$ where $n$ is a positive integer.", "solution": "  For brevity let $f(n)=n+s_{b}(n)$. Select any integer $M$. Observe that $f(x) \\geq b^{2 M}$ for any $x \\geq b^{2 M}$, but also $f\\left(b^{2 M}-k\\right) \\geq b^{2 M}$ for $k=1,2, \\ldots, M$, since the base- $b$ expansion of $b^{2 M}-k$ will start out with at least $M$ digits $b-1$.  Thus $f$ omits at least $M$ values in $\\left[1, b^{2 M}\\right]$ for any $M$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2014-notes.jsonl"}}
{"year": "2014", "tier": "T3", "problem_label": "5", "problem_type": null, "exam": "USAJMO", "problem": "Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In her move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.", "solution": "  The answer is $k=6$. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-09.jpg?height=500&width=561&top_left_y=1372&top_left_x=753)  【 Example of a strategy for $A$ when $k=5$. We describe a winning strategy for $A$ explicitly. Note that after $B$ 's first turn there is one counter, so then $A$ may create an equilateral triangle, and hence after $B$ 's second turn there are two consecutive counters. Then, on her third turn, $A$ places a pair of counters two spaces away on the same line. Label the two inner cells $x$ and $y$ as shown below. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-09.jpg?height=295&width=500&top_left_y=2168&top_left_x=778)  Now it is $B$ 's turn to move; in order to avoid losing immediately, he must remove either $x$ or $y$. Then on any subsequent turn, $A$ can replace $x$ or $y$ (whichever was removed) and add one more adjacent counter. This continues until either $x$ or $y$ has all its neighbors filled (we ask $A$ to do so in such a way that she avoids filling in the two central cells between $x$ and $y$ as long as possible).  So, let's say without loss of generality (by symmetry) that $x$ is completely surrounded by tokens. Again, $B$ must choose to remove $x$ (or $A$ wins on her next turn). After $x$ is removed by $B$, consider the following figure. ![](https://cdn.mathpix.com/cropped/2024_11_19_3739490e48f1a6399d35g-10.jpg?height=398&width=612&top_left_y=503&top_left_x=728)  We let $A$ play in the two marked green cells. Then, regardless of what move $B$ plays, one of the two choices of moves marked in red lets $A$ win. Thus, we have described a winning strategy when $k=5$ for $A$.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2014-notes.jsonl"}}
{"year": "2014", "tier": "T3", "problem_label": "6", "problem_type": null, "exam": "USAJMO", "problem": "Let $A B C$ be a triangle with incenter $I$, incircle $\\gamma$ and circumcircle $\\Gamma$. Let $M, N, P$ be the midpoints of $\\overline{B C}, \\overline{C A}, \\overline{A B}$ and let $E, F$ be the tangency points of $\\gamma$ with $\\overline{C A}$ and $\\overline{A B}$, respectively. Let $U, V$ be the intersections of line $E F$ with line $M N$ and line $M P$, respectively, and let $X$ be the midpoint of $\\operatorname{arc} B A C$ of $\\Gamma$. (a) Prove that $I$ lies on ray $C V$. (b) Prove that line $X I$ bisects $\\overline{U V}$.", "solution": "  The fact that $I=\\overline{B U} \\cap \\overline{C V}$ and is Lemma 1.45 from EGMO. As for (b), we note: Claim - Line $I X$ is a symmedian of $\\triangle I B C$.  Since $B V U C$ is cyclic with diagonals intersecting at $I$, and $I X$ is symmedian of $\\triangle I B C$, it is median of $\\triangle I U V$, as needed.", "metadata": {"resource_path": "USAJMO/segmented/en-JMO-2014-notes.jsonl"}}