{"year": "2002", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $a_{1}, a_{2}, a_{3}, \\ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.\n\nLet\n\n$$\nA_{n}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}\n$$\n\nProve that\n\n$$\na_{1}!a_{2}!\\ldots a_{n}!\\geq\\left(\\left\\lfloor A_{n}\\right\\rfloor!\\right)^{n}\n$$\n\nwhere $\\left\\lfloor A_{n}\\right\\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \\times 2 \\times \\cdots \\times a$ for $a \\geq 1$ (and $0!=1$ ). When does equality hold?", "solution": "Assume without loss of generality that $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n} \\geq 0$, and let $s=\\left\\lfloor A_{n}\\right\\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \\geq s \\geq a_{k+1}$.\n\nOur inequality is equivalent to proving that\n\n$$\n\\frac{a_{1}!}{s!} \\cdot \\frac{a_{2}!}{s!} \\cdot \\ldots \\cdot \\frac{a_{k}!}{s!} \\geq \\frac{s!}{a_{k+1}!} \\cdot \\frac{s!}{a_{k+2}!} \\cdot \\ldots \\cdot \\frac{s!}{a_{n}!}\n$$\n\nNow for $i=1,2, \\ldots, k, a_{i}!/ s!$ is the product of $a_{i}-s$ factors. For example, 9!/5! $=9 \\cdot 8 \\cdot 7 \\cdot 6$. The left side of inequality (1) therefore is the product of $A=a_{1}+a_{2}+\\cdots+a_{k}-k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B=(n-k) s-$ $\\left(a_{k+1}+a_{k+2}+\\cdots+a_{n}\\right)$ factors, all of which are at most $s$. Since $\\sum_{i=1}^{n} a_{i}=n A_{n} \\geq n s, A \\geq B$. This proves the inequality. [ 5 marks to here.]\n\nEquality in (1) holds if and only if either:\n(i) $A=B=0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1}=a_{2}=\\cdots=a_{n}$; or\n(ii) $a_{1}=1$ and $s=0$, that is, the only factors on either side of (1) are 1 's, which occurs if and only if $a_{i} \\in\\{0,1\\}$ for all $i$. [2 marks for both (i) and (ii), no marks for (i) only.]", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n1. ", "solution_match": "# Solution 1."}}
{"year": "2002", "tier": "T1", "problem_label": "1", "problem_type": null, "exam": "APMO", "problem": "Let $a_{1}, a_{2}, a_{3}, \\ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.\n\nLet\n\n$$\nA_{n}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}\n$$\n\nProve that\n\n$$\na_{1}!a_{2}!\\ldots a_{n}!\\geq\\left(\\left\\lfloor A_{n}\\right\\rfloor!\\right)^{n}\n$$\n\nwhere $\\left\\lfloor A_{n}\\right\\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \\times 2 \\times \\cdots \\times a$ for $a \\geq 1$ (and $0!=1$ ). When does equality hold?", "solution": "Assume without loss of generality that $0 \\leq a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n}$. Let $d=a_{n}-a_{1}$ and $m=\\left|\\left\\{i: a_{i}=a_{1}\\right\\}\\right|$. Our proof is by induction on $d$.\n\nWe first do the case $d=a_{n}-a_{1}=0$ or 1 separately. Then $a_{1}=a_{2}=\\cdots=a_{m}=a$ and $a_{m+1}=\\cdots=a_{n}=a+1$ for some $1 \\leq m \\leq n$ and $a \\geq 0$. In this case we have $\\left\\lfloor A_{n}\\right\\rfloor=a$, so the inequality to be proven is just $a_{1}!a_{2}!\\ldots a_{n}!\\geq(a!)^{n}$, which is obvious. Equality holds if and only if either $m=n$, that is, $a_{1}=a_{2}=\\cdots=a_{n}=a$; or if $a=0$, that is, $a_{1}=\\cdots=a_{m}=0$ and $a_{m+1}=\\cdots=a_{n}=1$. [ 2 marks to here.]\n\nSo assume that $d=a_{n}-a_{1} \\geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence\n\n$$\na_{1}+1, a_{2}, a_{3}, \\ldots, a_{n-1}, a_{n}-1\n$$\n\nthough not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n}>a_{1}+1$,\n\n$$\n\\begin{aligned}\na_{1}!a_{2}!\\ldots a_{n}! & =\\left(a_{1}+1\\right)!a_{2}!\\ldots a_{n-1}!\\left(a_{n}-1\\right)!\\cdot \\frac{a_{n}}{a_{1}+1} \\\\\n& \\geq\\left(\\left\\lfloor A_{n}\\right\\rfloor!\\right)^{n} \\cdot \\frac{a_{n}}{a_{1}+1} \\\\\n& >\\left(\\left\\lfloor A_{n}\\right\\rfloor!\\right)^{n}\n\\end{aligned}\n$$\n\nwhich completes the proof. Equality cannot hold in this case.", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n1. ", "solution_match": "\nSolution 2."}}
{"year": "2002", "tier": "T1", "problem_label": "2", "problem_type": null, "exam": "APMO", "problem": "Find all positive integers $a$ and $b$ such that\n\n$$\n\\frac{a^{2}+b}{b^{2}-a} \\text { and } \\frac{b^{2}+a}{a^{2}-b}\n$$\n\nare both integers.", "solution": "By the symmetry of the problem, we may suppose that $a \\leq b$. Notice that $b^{2}-a \\geq 0$, so that if $\\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \\geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \\geq 0$. Since $a, b>0$, we must have $a \\geq b-1$. [3 marks to here.] We therefore have two cases:\n\nCase 1: $a=b$. Substituting, we have\n\n$$\n\\frac{a^{2}+a}{a^{2}-a}=\\frac{a+1}{a-1}=1+\\frac{2}{a-1},\n$$\n\nwhich is an integer if and only if $(a-1) \\mid 2$. As $a>0$, the only possible values are $a-1=1$ or 2. Hence, $(a, b)=(2,2)$ or $(3,3)$. [1 mark.]\n\nCase 2: $a=b-1$. Substituting, we have\n\n$$\n\\frac{b^{2}+a}{a^{2}-b}=\\frac{(a+1)^{2}+a}{a^{2}-(a+1)}=\\frac{a^{2}+3 a+1}{a^{2}-a-1}=1+\\frac{4 a+2}{a^{2}-a-1} .\n$$\n\nOnce again, notice that $4 a+2>0$, and hence, for $\\frac{4 a+2}{a^{2}-a-1}$ to be an integer, we must have $4 a+2 \\geq a^{2}-a-1$, that is, $a^{2}-5 a-3 \\leq 0$. Hence, since $a$ is an integer, we can bound $a$ by $1 \\leq \\bar{a} \\leq 5$. Checking all the ordered pairs $(a, b)=(1,2),(2,3), \\ldots,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions. [3 marks.]\n\nThus, the ordered pairs that work are\n\n$$\n(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)\n$$\n\nwhere the last two pairs follow by symmetry. [2 marks if these solutions are found without proof that there are no others.]", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n2. ", "solution_match": "\nSolution."}}
{"year": "2002", "tier": "T1", "problem_label": "3", "problem_type": null, "exam": "APMO", "problem": "Let $A B C$ be an equilateral triangle. Let $P$ be a point on the side $A C$ and $Q$ be a point on the side $A B$ so that both triangles $A B P$ and $A C Q$ are acute. Let $R$ be the orthocentre of triangle $A B P$ and $S$ be the orthocentre of triangle $A C Q$. Let $T$ be the point common to the segments $B P$ and $C Q$. Find all possible values of $\\angle C B P$ and $\\angle B C Q$ such that triangle $T R S$ is equilateral.", "solution": "We are going to show that this can only happen when\n\n$$\n\\angle C B P=\\angle B C Q=15^{\\circ} .\n$$\n\nLemma. If $\\angle C B P>\\angle B C Q$, then $R T>S T$.\nProof. Let $A D, B E$ and $C F$ be the altitudes of triangle $A B C$ concurrent at its centre $G$. Then $P$ lies on $C E, Q$ lies on $B F$, and thus $T$ lies in triangle $B D G$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_f9f048d50025eb9b7941g-3.jpg?height=590&width=679&top_left_y=968&top_left_x=636)\n\nNote that -\n\n$$\n\\angle F A S=\\angle F C Q=30^{\\circ}-\\angle B C Q>30^{\\circ}-\\angle C B P=\\angle E B P=\\angle E A R\n$$\n\nSince $A F=A E$, we have $F S>E R$ so that\n\n$$\nG S=G F-F S<G E-E R=G R .\n$$\n\nLet $T_{x}$ be the projection of $T$ onto $B C$ and $T_{y}$ be the projection of $T$ onto $A D$, and similarly for $R$ and $S$. We have\n\n$$\nR_{x} T_{x}=D R_{x}+D T_{x}>\\left|D S_{x}-D T_{x}\\right|=S_{x} T_{x}\n$$\n\nand\n\n$$\nR_{y} T_{y}=G R_{y}+G T_{y}>G S_{y}+G T_{y}=S_{y} T_{y}\n$$\n\nIt follows that $R T>S T$.\n[1 mark for stating the Lemma, 3 marks for proving it.]\nThus, if $\\triangle T R S$ is equilateral, we must have $\\angle C B P=\\angle B C Q$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_f9f048d50025eb9b7941g-4.jpg?height=589&width=663&top_left_y=291&top_left_x=639)\n\nIt is clear from the symmetry of the figure that $T R=T S$, so $\\triangle T R S$ is equilateral if and only if $\\angle R T A=30^{\\circ}$. Now, as $B R$ is an altitude of the triangle $A B C, \\angle R B A=30^{\\circ}$. So $\\triangle T R S$ is equilateral if and only if $R T B A$ is a cyclic quadrilateral. Therefore, $\\triangle T R S$ is equilateral if and only if $\\angle T B R=\\angle T A R$. But\n\n$$\n\\begin{aligned}\n90^{\\circ} & =\\angle T B A+\\angle B A R \\\\\n& =(\\angle T B R+\\angle R B A)+(\\angle B A T+\\angle T A R) \\\\\n& =\\left(\\angle T B R+30^{\\circ}\\right)+\\left(30^{\\circ}+\\angle T A R\\right)\n\\end{aligned}\n$$\n\nand so\n\n$$\n30^{\\circ}=\\angle T A R+\\angle T B R\n$$\n\nBut these angles must be equal, so $\\angle T A R=\\angle T B R=15^{\\circ}$. Therefore $\\angle C B P=\\angle B C Q=15^{\\circ}$. [3 marks for finishing the proof with the assumption that $\\angle C B P=\\angle B C Q$.]", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n3. ", "solution_match": "# Solution."}}
{"year": "2002", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $x, y, z$ be positive numbers such that\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1\n$$\n\nShow that\n\n$$\n\\sqrt{x+y z}+\\sqrt{y+z x}+\\sqrt{z+x y} \\geq \\sqrt{x y z}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z}\n$$", "solution": "$$\n\\begin{aligned}\n\\sum_{\\text {cyclic }} \\sqrt{x+y z} & =\\sqrt{x y z} \\sum_{\\text {cyclic }} \\sqrt{\\frac{1}{x}+\\frac{1}{y z}} \\\\\n& =\\sqrt{x y z} \\sum_{\\text {cyclic }} \\sqrt{\\frac{1}{x}\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)+\\frac{1}{y z}} \\quad[1 \\text { mark. }] \\\\\n& =\\sqrt{x y z} \\sum_{\\text {cyclic }} \\sqrt{\\left(\\frac{1}{x}+\\frac{1}{y}\\right)\\left(\\frac{1}{x}+\\frac{1}{z}\\right)} \\quad[1 \\text { mark.] }\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\n& =\\sqrt{x y z} \\sum_{\\text {cyclic }} \\sqrt{\\left(\\frac{1}{x}+\\frac{1}{\\sqrt{y z}}\\right)^{2}+\\frac{(\\sqrt{y}-\\sqrt{z})^{2}}{x y z}} \\quad[2 \\text { marks. }] \\\\\n& \\geq \\sqrt{x y z} \\sum_{\\text {cyclic }}\\left(\\frac{1}{x}+\\frac{1}{\\sqrt{y z}}\\right) \\quad[1 \\text { mark. }] \\\\\n& =\\sqrt{x y z}\\left(1+\\sum_{\\text {cyclic }} \\frac{1}{\\sqrt{y z}}\\right) \\quad[1 \\text { mark. }] \\\\\n& =\\sqrt{x y z}+\\sum_{\\text {cyclic }} \\sqrt{x} \\quad[1 \\text { mark. }]\n\\end{aligned}\n$$\n\nNote. It is easy to check that equality holds if and only if $x=y=z=3$.", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution 1."}}
{"year": "2002", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $x, y, z$ be positive numbers such that\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1\n$$\n\nShow that\n\n$$\n\\sqrt{x+y z}+\\sqrt{y+z x}+\\sqrt{z+x y} \\geq \\sqrt{x y z}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z}\n$$", "solution": "Squaring both sides of the given inequality, we obtain\n\n$$\n\\begin{aligned}\n& \\sum_{\\text {cyclic }} x+\\sum_{\\text {cyclic }} y z+2 \\sum_{\\text {cyclic }} \\sqrt{x+y z} \\sqrt{y+z x} \\\\\n& \\quad \\geq x y z+2 \\sqrt{x y z} \\sum_{\\text {cyclic }} \\sqrt{x}+\\sum_{\\text {cyclic }} x+2 \\sum_{\\text {cyclic }} \\sqrt{x y} \\quad \\text { [1 mark.] }\n\\end{aligned}\n$$\n\nIt follows from the given condition $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$ that $x y z=\\sum_{\\text {cyclic }} x y$. Therefore, the given inequality is equivalent to\n\n$$\n\\sum_{\\text {cyclic }} \\sqrt{x+y z} \\sqrt{y+z x} \\geq \\sqrt{x y z} \\sum_{\\text {cyclic }} \\sqrt{x}+\\sum_{\\text {cyclic }} \\sqrt{x y} . \\quad[2 \\text { marks.] }\n$$\n\nUsing the Cauchy-Schwarz inequality [or just $x^{2}+y^{2} \\geq 2 x y$ ], we see that\n\n$$\n(x+y z)(y+z x) \\geq\\left(\\sqrt{x y}+\\sqrt{x y z^{2}}\\right)^{2}, \\quad[1 \\text { mark. }]\n$$\n\nor\n\n$$\n\\sqrt{x+y z} \\sqrt{y+z x} \\geq \\sqrt{x y}+\\sqrt{z} \\sqrt{x y z} . \\quad[1 \\text { mark. }]\n$$\n\nTaking the cyclic sum of this inequality over $x, y$ and $z$, we get the desired inequality. [2 marks.]", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution 2."}}
{"year": "2002", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $x, y, z$ be positive numbers such that\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1\n$$\n\nShow that\n\n$$\n\\sqrt{x+y z}+\\sqrt{y+z x}+\\sqrt{z+x y} \\geq \\sqrt{x y z}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z}\n$$", "solution": "This is another way of presenting the idea in the first solution.\nUsing the condition $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$ and the AM-GM inequality, we have\n\n$$\n\\begin{aligned}\nx+y z-\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{x}\\right)^{2} & =y z\\left(1-\\frac{1}{x}\\right)-2 \\sqrt{y z} \\\\\n& =y z\\left(\\frac{1}{y}+\\frac{1}{z}\\right)-2 \\sqrt{y z}=y+z-2 \\sqrt{y z} \\geq 0\n\\end{aligned}\n$$\n\nwhich gives\n\n$$\n\\sqrt{x+y z} \\geq \\sqrt{\\frac{y z}{x}}+\\sqrt{x} . \\quad[3 \\text { marks. }]\n$$\n\nSimilarly, we have\n\n$$\n\\sqrt{y+z x} \\geq \\sqrt{\\frac{z x}{y}}+\\sqrt{y} \\text { and } \\sqrt{z+x y} \\geq \\sqrt{\\frac{x y}{z}}+\\sqrt{z}\n$$\n\nAddition yields\n\n$$\n\\sqrt{x+y z}+\\sqrt{y+z x}+\\sqrt{z+x y} \\geq \\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z}\n$$\n\n[2 marks.] Using the condition $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$ again, we have\n\n$$\n\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}=\\sqrt{x y z}\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)=\\sqrt{x y z}, \\quad[1 \\text { mark. }]\n$$\n\nand thus\n\n$$\n\\sqrt{x+y z}+\\sqrt{y+z x}+\\sqrt{z+x y} \\geq \\sqrt{x y z}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z} . \\quad[1 \\text { mark. }]\n$$", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n4. ", "solution_match": "\nSolution 3."}}
{"year": "2002", "tier": "T1", "problem_label": "4", "problem_type": null, "exam": "APMO", "problem": "Let $x, y, z$ be positive numbers such that\n\n$$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1\n$$\n\nShow that\n\n$$\n\\sqrt{x+y z}+\\sqrt{y+z x}+\\sqrt{z+x y} \\geq \\sqrt{x y z}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z}\n$$", "solution": "This is also another way of presenting the idea in the first solution.\nWe make the substitution $a=\\frac{1}{x}, b=\\frac{1}{y}, c=\\frac{1}{z}$. Then it is enough to show that\n\n$$\n\\sqrt{\\frac{1}{a}+\\frac{1}{b c}}+\\sqrt{\\frac{1}{b}+\\frac{1}{c a}}+\\sqrt{\\frac{1}{c}+\\frac{1}{a b}} \\geq \\sqrt{\\frac{1}{a b c}}+\\sqrt{\\frac{1}{a}}+\\sqrt{\\frac{1}{b}}+\\sqrt{\\frac{1}{c}},\n$$\n\nwhere $a+b+c=1$. Multiplying this inequality by $\\sqrt{a b c}$, we find that it can be written\n\n$$\n\\sqrt{a+b c}+\\sqrt{b+c a}+\\sqrt{c+a b} \\geq 1+\\sqrt{b c}+\\sqrt{c a}+\\sqrt{a b} . \\quad[1 \\text { mark. }]\n$$\n\nThis is equivalent to\n\n$$\n\\begin{aligned}\n& \\sqrt{a(a+b+c)+b c}+\\sqrt{b(a+b+c)+c a}+\\sqrt{c(a+b+c)+a b} \\\\\n& \\geq a+b+c+\\sqrt{b c}+\\sqrt{c a}+\\sqrt{a b}, \\quad[1 \\text { mark. }]\n\\end{aligned}\n$$\n\nwhich in turn is equivalent to\n\n$$\n\\sqrt{(a+b)(a+c)}+\\sqrt{(b+c)(b+a)}+\\sqrt{(c+a)(c+b)} \\geq a+b+c+\\sqrt{b c}+\\sqrt{c a}+\\sqrt{a b}\n$$\n\n[1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c \\geq 2 \\sqrt{b c}$ ), we have\n\n$$\n\\left[(\\sqrt{a})^{2}+(\\sqrt{b})^{2}\\right]\\left[(\\sqrt{a})^{2}+(\\sqrt{c})^{2}\\right] \\geq(\\sqrt{a} \\sqrt{a}+\\sqrt{b} \\sqrt{c})^{2}\n$$\n\nor\n\n$$\n\\sqrt{(a+b)(a+c)} \\geq a+\\sqrt{b c} . \\quad[2 \\text { marks. }]\n$$\n\nTaking the cyclic sum of this inequality over $a, b, c$, we get the desired inequality. [2 marks.]", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n4. ", "solution_match": "# Solution 4."}}
{"year": "2002", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Let R denote the set of all real numbers. Find all functions $f$ from R to R satisfying:\n(i) there are only finitely many $s$ in R such that $f(s)=0$, and\n(ii) $f\\left(x^{4}+y\\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in $\\mathbf{R}$.", "solution": "The only such function is the identity function on $R$.\nSetting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find\n\n$$\nf(y)=f(f(y))\n$$\n\n[1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from (ii) that $f\\left(x^{4}+y\\right)=$ $x^{3} f(x)+f(y)$ for all $x, y \\in \\mathbf{R}$. Set $y=0$ to obtain\n\n$$\nf\\left(x^{4}\\right)=x^{3} f(x)\n$$\n\nfor all $x \\in \\mathrm{R}$, and so\n\n$$\nf\\left(x^{4}+y\\right)=f\\left(x^{4}\\right)+f(y)\n$$\n\nfor all $x, y \\in \\mathbf{R}$. The functional equation (3) suggests that $f$ is additive, that is, $f(a+b)=$ $f(a)+f(b)$ for all $a, b \\in \\mathbf{R}$. [1 mark.] We now show this.\n\nFirst assume that $a \\geq 0$ and $b \\in R$. It follows from (3) that\n\n$$\nf(a+b)=f\\left(\\left(a^{1 / 4}\\right)^{4}+b\\right)=f\\left(\\left(a^{1 / 4}\\right)^{4}\\right)+f(b)=f(a)+f(b)\n$$\n\nWe next note that $f$ is an odd function, since from (2)\n\n$$\nf(-x)=\\frac{f\\left(x^{4}\\right)}{(-x)^{3}}=\\frac{f\\left(x^{4}\\right)}{-x^{3}}=-f(x), \\quad x \\neq 0\n$$\n\nSince $f$ is odd, we have that, for $a<0$ and $b \\in R$,\n\n$$\n\\begin{aligned}\nf(a+b) & =-f((-a)+(-b))=-(f(-a)+f(-b)) \\\\\n& =-(-f(a)-f(b))=f(a)+f(b)\n\\end{aligned}\n$$\n\nTherefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \\in R$. [2 mers.].]\nWe now show that $\\{s \\in \\mathrm{R} \\mid f(s)=0\\}=\\{0\\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \\in \\mathrm{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \\in\\{s \\in R \\mid f(s)=0\\}$ for all $n \\in \\mathbf{N}$. However, this is a contradiction to the given condition (i). [1 mark.]\n\nIt's now easy to check that $f$ is one-to-one. Assume that $f(a)=f^{\\prime}(b)$ for some $a, b \\in \\mathbb{P}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \\in\\{s \\in$ $\\mathbf{R} \\mid f(s)=0\\}=\\{0\\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \\in \\mathbf{R}$. [1 mark.] This completes the proof.", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n5. ", "solution_match": "# Solution 1."}}
{"year": "2002", "tier": "T1", "problem_label": "5", "problem_type": null, "exam": "APMO", "problem": "Let R denote the set of all real numbers. Find all functions $f$ from R to R satisfying:\n(i) there are only finitely many $s$ in R such that $f(s)=0$, and\n(ii) $f\\left(x^{4}+y\\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in $\\mathbf{R}$.", "solution": "Again, the only such function is the identity function on R .\nAs in Solution 1, we first show that $f(f(y))=f(y), f(0)=0$, and $f\\left(x^{4}\\right)=x^{3} f(x)$. [2 marks.] From the latter follows\n\n$$\nf(x)=0 \\Longrightarrow f\\left(x^{4}\\right)=0\n$$\n\nand from condition (i) we get that $f(x)=0$ only possibly for $x \\in\\{0,1,-1\\}$. [1 mark.]\nNext we prove\n\n$$\nf(a)=b \\Longrightarrow f(\\sqrt[4]{|a-b|})=0\n$$\n\nThis is clear if $a=b$. If $a>b$ then\n\n$$\n\\begin{aligned}\nf(a) & =f((a-b)+b)=(a-b)^{3 / 4} f(\\sqrt[4]{a-b})+f(f(b)) \\\\\n& =(a-b)^{3 / 4} f(\\sqrt[4]{a-b})+f(b) \\\\\n& =(a-b)^{3 / 4} f(\\sqrt[4]{a-b})+f(f(a)) \\\\\n& =(a-b)^{3 / 4} f(\\sqrt[4]{a-b})+f(a),\n\\end{aligned}\n$$\n\nso $(a-b)^{3 / 4} f(\\sqrt[4]{a-b})=0$ which means $f(\\sqrt[4]{|a-b|})=0$. If $a<b$ we get similarly\n\n$$\n\\begin{aligned}\nf(b) & =f((b-a)+a)=(b-a)^{3 / 4} f(\\sqrt[4]{b-a})+f(f(a)) \\\\\n& =(b-a)^{3 / 4} f(\\sqrt[4]{b-a})+f(b)\n\\end{aligned}\n$$\n\nand again $f(\\sqrt[4]{|a-b|})=0$. [2 marks.]\nThus $f(a)=b \\Longrightarrow|a-b| \\in\\{0,1\\}$. Suppose that $f(x)=x+b$ for some $x$, where $|b|=1$. Then from $f\\left(x^{4}\\right)=x^{3} f(x)$ and $f\\left(x^{4}\\right)=x^{4}+a$ for some $|a| \\leq 1$ we get $x^{3}=a / b$, so $|x| \\leq 1$. Thus $f(x)=x$ for all $x$ except possibly $x= \\pm 1$. [ 1 mark.] But for example,\n\n$$\nf(1)=f\\left(2^{4}-15\\right)=2^{3} f(2)+f(f(-15))=2^{3} \\cdot 2-15=1\n$$\n\nand\n\n$$\nf(-1)=f\\left(2^{4}-17\\right)=2^{3} f(2)+f(f(-17))=2^{3} \\cdot 2-17=-1\n$$\n\n[1 mark.] This finishes the proof.", "metadata": {"resource_path": "APMO/segmented/en-apmo2002_sol.jsonl", "problem_match": "\n5. ", "solution_match": "# Solution 2."}}