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math.json

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+[
+ {
+  "instruction":"Compute$$ \\int_{0}^{\\pi} \\frac{x \\sin x}{1+\\sin ^{2} x} d x . $$",
+  "response":"We use the example from the introduction for the particular function $f(x)=\\frac{x}{1+x^{2}}$ to transform the integral into$$ \\pi \\int_{0}^{\\frac{\\pi}{2}} \\frac{\\sin x}{1+\\sin ^{2} x} d x . $$This is the same as$$ \\pi \\int_{0}^{\\frac{\\pi}{2}}-\\frac{d(\\cos x)}{2-\\cos ^{2} x}, $$which with the substitution $t=\\cos x$ becomes$$ \\pi \\int_{0}^{1} \\frac{1}{2-t^{2}} d t=\\left.\\frac{\\pi}{2 \\sqrt{2}} \\ln \\frac{\\sqrt{2}+t}{\\sqrt{2}-t}\\right|_{0} ^{1}=\\frac{\\pi}{2 \\sqrt{2}} \\ln \\frac{\\sqrt{2}+1}{\\sqrt{2}-1} . $$"
+ },
+ {
+  "instruction":"Compute up to two decimal places the number$$ \\sqrt{1+2 \\sqrt{1+2 \\sqrt{1+\\cdots+2 \\sqrt{1+2 \\sqrt{1969}}}}} $$where the expression contains 1969 square roots.",
+  "response":"Let$$ x_{n}=\\sqrt{1+2 \\sqrt{1+2 \\sqrt{1+\\cdots+2 \\sqrt{1+2 \\sqrt{1969}}}}} $$with the expression containing $n$ square root signs. Note that$$ x_{1}-(1+\\sqrt{2})=\\sqrt{1969}-(1+\\sqrt{2})<50 . $$Also, since $\\sqrt{1+2(1+\\sqrt{2})}=1+\\sqrt{2}$, we have $$ \\begin{aligned} x_{n+1}-(1+\\sqrt{2})=& \\sqrt{1+2 x_{n}}-\\sqrt{1+2(1+\\sqrt{2})}=\\frac{2\\left(x_{n}-(1-\\sqrt{2})\\right)}{\\sqrt{1+2 x_{n}}+\\sqrt{1+2(1+\\sqrt{2})}} \\\\ &<\\frac{x_{n}-(1+\\sqrt{2})}{1+\\sqrt{2}} . \\end{aligned} $$From here we deduce that$$ x_{1969}-(1+\\sqrt{2})<\\frac{50}{(1+\\sqrt{2})^{1968}}<10^{-3}, $$and the approximation of $x_{1969}$ with two decimal places coincides with that of $1+\\sqrt{2}=$ 2.41. This argument proves also that the limit of the sequence is $1+\\sqrt{2}$."
+ },
+ {
+  "instruction":"Let $M_{2 \\times 2}$ be the vector space of all real $2 \\times 2$ matrices. Let\n\n$$\nA=\\left(\\begin{array}{cc}\n1 & 2 \\\\\n-1 & 3\n\\end{array}\\right) \\quad B=\\left(\\begin{array}{cc}\n2 & 1 \\\\\n0 & 4\n\\end{array}\\right)\n$$\n\nand define a linear transformation $L: M_{2 \\times 2} \\rightarrow M_{2 \\times 2}$ by $L(X)=A X B$. Compute the trace of $L$.\n\n",
+  "response":" Identify $M_{2 \\times 2}$ with $\\mathbb{R}^{4}$ via\n\n$$\n\\left(\\begin{array}{ll}\na & b \\\\nc & d\n\\end{array}\\right) \\leftrightarrow\\left(\\begin{array}{l}\na \\\\nb \\\\nc \\\\nd\n\\end{array}\\right)\n$$\n\nand decompose $L$ into the multiplication of two linear transformations,\n\n$$\nM_{2 \\times 2} \\simeq \\mathbb{R}^{4} \\stackrel{L_{A}}{\\longrightarrow} \\mathbb{R}^{4} \\stackrel{L_{B}}{\\longrightarrow} \\mathbb{R}^{4} \\simeq M_{2 \\times 2}\n$$\n\nwhere $L_{A}(X)=A X$ and $L_{B}(X)=X B$. is\n\nThe matrices of these two linear transformations on the canonical basis of $\\mathbb{R}^{4}$\n\n$$\nL_{A}=\\left(\\begin{array}{rrrr}\n1 & 0 & 2 & 0 \\\\n0 & 1 & 0 & 2 \\\\n-1 & 0 & 3 & 0 \\\\n0 & -1 & 0 & 3\n\\end{array}\\right) \\quad \\text { and } L_{B}=\\left(\\begin{array}{llll}\n2 & 0 & 0 & 0 \\\\n1 & 4 & 0 & 0 \\\\n0 & 0 & 2 & 0 \\\\n0 & 0 & 1 & 4\n\\end{array}\\right)\n$$\n\nthen $\\operatorname{det} L=\\operatorname{det} L_{A} \\cdot \\operatorname{det} L_{B}=(9+6+2(2+3)) \\cdot(2 \\cdot 32)=2^{6} \\cdot 5^{2}$, and to compute the trace of $L$, we only need the diagonal elements of $L_{A} \\cdot L_{B}$, that is,\n\n$$\n\\operatorname{tr} L=2+4+6+12=24 \\text {. }\n$$\n\n"
+ },
+ {
+  "instruction":"Define\n\n$$\nF(x)=\\int_{\\sin x}^{\\cos x} e^{\\left(t^{2}+x t\\right)} d t .\n$$\n\nCompute $F^{\\prime}(0)$.\n\n",
+  "response":" Let\n\n$$\nG(u, v, x)=\\int_{v}^{u} e^{t^{2}+x t} d t .\n$$\n\nThen $F(x)=G(\\cos x, \\sin x, x)$, so\n\n$$\n\\begin{aligned}\nF^{\\prime}(x) &=\\frac{\\partial G}{\\partial u} \\frac{\\partial u}{\\partial x}+\\frac{\\partial G}{\\partial v} \\frac{\\partial v}{\\partial x}+\\frac{\\partial G}{\\partial x} \\\\n&=e^{u^{2}+x u}(-\\sin x)-e^{\\left(v^{2}+x v\\right)} \\cos x+\\int_{v}^{u} t e^{t^{2}+x t} d t\n\\end{aligned}\n$$\n\nand\n\n$$\nF^{\\prime}(0)=-1+\\int_{0}^{1} t e^{t^{2}} d t=\\frac{1}{2}(e-3) .\n$$\n\n"
+ },
+ {
+  "instruction":"What is the probability that the sum of two randomly chosen numbers in the interval $[0,1]$ does not exceed 1 and their product does not exceed $\\frac{2}{9}$?",
+  "response":"Let $x$ and $y$ be the two numbers. The set of all possible outcomes is the unit square$$ D=\\{(x, y) \\mid 0 \\leq x \\leq 1,0 \\leq y \\leq 1\\} . $$The favorable cases consist of the region$$ D_{f}=\\left\\{(x, y) \\in D \\mid x+y \\leq 1, x y \\leq \\frac{2}{9}\\right\\} . $$This is the set of points that lie below both the line $f(x)=1-x$ and the hyperbola $g(x)=\\frac{2}{9 x}$. equal toThe required probability is $P=\\frac{\\operatorname{Area}\\left(D_{f}\\right)}{\\operatorname{Area}(D)}$. The area of $D$ is 1 . The area of $D_{f}$ is$$ \\int_{0}^{1} \\min (f(x), g(x)) d x . $$The line and the hyperbola intersect at the points $\\left(\\frac{1}{3}, \\frac{2}{3}\\right)$ and $\\left(\\frac{2}{3}, \\frac{1}{3}\\right)$. Therefore,$$ \\operatorname{Area}\\left(D_{f}\\right)=\\int_{0}^{1 \/ 3}(1-x) d x+\\int_{1 \/ 3}^{2 \/ 3} \\frac{2}{9 x} d x+\\int_{2 \/ 3}^{1}(1-x) d x=\\frac{1}{3}+\\frac{2}{9} \\ln 2 . $$We conclude that $P=\\frac{1}{3}+\\frac{2}{9} \\ln 2 \\approx 0.487$."
+ },
+ {
+  "instruction":"Evaluate\n\n$$\n\\int_{0}^{2 \\pi} e^{\\left(e^{i \\theta}-i \\theta\\right)} d \\theta .\n$$\n\n",
+  "response":" By Cauchy's Integral Formula for derivatives, we have\n\ntherefore,\n\n$$\n\\left.\\frac{d}{d z} e^{z}\\right|_{z=0}=\\frac{1}{2 \\pi i} \\int_{|z|=1} \\frac{e^{z}}{z^{2}} d z=\\frac{1}{2 \\pi} \\int_{0}^{2 \\pi} e^{e^{i \\theta}-i \\theta} d \\theta\n$$\n\n$$\n\\int_{0}^{2 \\pi} e^{e^{i \\theta}-i \\theta} d \\theta=2 \\pi .\n$$\n\n"
+ },
+ {
+  "instruction":"Compute the product$$ \\left(1-\\frac{4}{1}\\right)\\left(1-\\frac{4}{9}\\right)\\left(1-\\frac{4}{25}\\right) \\cdots $$",
+  "response":"For $N \\geq 2$, define$$ a_{N}=\\left(1-\\frac{4}{1}\\right)\\left(1-\\frac{4}{9}\\right)\\left(1-\\frac{4}{25}\\right) \\cdots\\left(1-\\frac{4}{(2 N-1)^{2}}\\right) . $$The problem asks us to find $\\lim _{N \\rightarrow \\infty} a_{N}$. The defining product for $a_{N}$ telescopes as follows:$$ \\begin{aligned} a_{N} &=\\left[\\left(1-\\frac{2}{1}\\right)\\left(1+\\frac{2}{1}\\right)\\right]\\left[\\left(1-\\frac{2}{3}\\right)\\left(1+\\frac{2}{3}\\right)\\right] \\cdots\\left[\\left(1-\\frac{2}{2 N-1}\\right)\\left(1+\\frac{2}{2 N-1}\\right)\\right] \\\\ &=(-1 \\cdot 3)\\left(\\frac{1}{3} \\cdot \\frac{5}{3}\\right)\\left(\\frac{3}{5} \\cdot \\frac{7}{5}\\right) \\cdots\\left(\\frac{2 N-3}{2 N-1} \\cdot \\frac{2 N+1}{2 N-1}\\right)=-\\frac{2 N+1}{2 N-1} . \\end{aligned} $$Hence the infinite product is equal to$$ \\lim _{N \\rightarrow \\infty} a_{N}=-\\lim _{N \\rightarrow \\infty} \\frac{2 N+1}{2 N-1}=-1 . $$"
+ },
+ {
+  "instruction":"Suppose the coefficients of the power series\n\n$$\n\\sum_{n=0}^{\\infty} a_{n} z^{n}\n$$\n\nare given by the recurrence relation\n\n$$\na_{0}=1, a_{1}=-1,3 a_{n}+4 a_{n-1}-a_{n-2}=0, n=2,3, \\ldots .\n$$\n\nFind the radius of convergence $r$.\n\n",
+  "response":" From the recurrence relation, we see that the coefficients $a_{n}$ grow, at most, at an exponential rate, so the series has a positive radius of convergence. Let $f$ be the function it represents in its disc of convergence, and consider the polynomial $p(z)=3+4 z-z^{2}$. We have\n\n$$\n\\begin{aligned}\np(z) f(z) &=\\left(3+4 z-z^{2}\\right) \\sum_{n=0}^{\\infty} a_{n} z^{n} \\\\n&=3 a_{0}+\\left(3 a_{1}+4 a_{0}\\right) z+\\sum_{n=0}^{\\infty}\\left(3 a_{n}+4 a_{n-1}-a_{n-2}\\right) z^{n} \\\\n&=3+z\n\\end{aligned}\n$$\n\nSo\n\n$$\nf(z)=\\frac{3+z}{3+4 z-z^{2}} .\n$$\n\nThe radius of convergence of the series is the distance from 0 to the closest singularity of $f$, which is the closest root of $p$. The roots of $p$ are $2 \\pm \\sqrt{7}$. Hence, the radius of convergence is $\\sqrt{7}-2$.\n\n"
+ },
+ {
+  "instruction":"For integers $n \\geq 2$ and $0 \\leq k \\leq n-2$, compute the determinant$$ \\left|\\begin{array}{ccccc} 1^{k} & 2^{k} & 3^{k} & \\cdots & n^{k} \\\\ 2^{k} & 3^{k} & 4^{k} & \\cdots & (n+1)^{k} \\\\ 3^{k} & 4^{k} & 5^{k} & \\cdots & (n+2)^{k} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ n^{k} & (n+1)^{k} & (n+2)^{k} & \\cdots & (2 n-1)^{k} \\end{array}\\right| . $$",
+  "response":"The polynomials $P_{j}(x)=(x+j)^{k}, j=0,1, \\ldots, n-1$, lie in the $(k+1)$-dimensional real vector space of polynomials of degree at most $k$. Because $k+1<n$, they are linearly dependent. The columns consist of the evaluations of these polynomials at $1,2, \\ldots, n$, so the columns are linearly dependent. It follows that the determinant is zero."
+ },
+ {
+  "instruction":"Compute$$ \\lim _{n \\rightarrow \\infty}\\left[\\frac{1}{\\sqrt{4 n^{2}-1^{2}}}+\\frac{1}{\\sqrt{4 n^{2}-2^{2}}}+\\cdots+\\frac{1}{\\sqrt{4 n^{2}-n^{2}}}\\right] . $$",
+  "response":"We have$$ \\begin{aligned} s_{n} &=\\frac{1}{\\sqrt{4 n^{2}-1^{2}}}+\\frac{1}{\\sqrt{4 n^{2}-2^{2}}}+\\cdots+\\frac{1}{\\sqrt{4 n^{2}-n^{2}}} \\\\ &=\\frac{1}{n}\\left[\\frac{1}{\\sqrt{4-\\left(\\frac{1}{n}\\right)^{2}}}+\\frac{1}{\\sqrt{4-\\left(\\frac{2}{n}\\right)^{2}}}+\\cdots+\\frac{1}{\\sqrt{4-\\left(\\frac{n}{n}\\right)^{2}}}\\right] . \\end{aligned} $$Hence $s_{n}$ is the Riemann sum of the function $f:[0,1] \\rightarrow \\mathbb{R}, f(x)=\\frac{1}{\\sqrt{4-x^{2}}}$ associated to the subdivision $x_{0}=0<x_{1}=\\frac{1}{n}<x_{2}=\\frac{2}{n}<\\cdots<x_{n}=\\frac{n}{n}=1$, with the intermediate points $\\xi_{i}=\\frac{i}{n} \\in\\left[x_{i}, x_{i+1}\\right]$. The answer to the problem is therefore$$ \\lim _{n \\rightarrow \\infty} s_{n}=\\int_{0}^{1} \\frac{1}{\\sqrt{4-x^{2}}} d x=\\left.\\arcsin \\frac{x}{2}\\right|_{0} ^{1}=\\frac{\\pi}{6} . $$"
+ },
+ {
+  "instruction":"Consider the sequences $\\left(a_{n}\\right)_{n}$ and $\\left(b_{n}\\right)_{n}$ defined by $a_{1}=3, b_{1}=100, a_{n+1}=3^{a_{n}}$, $b_{n+1}=100^{b_{n}}$. Find the smallest number $m$ for which $b_{m}>a_{100}$. ",
+  "response":"We need to determine $m$ such that $b_{m}>a_{n}>b_{m-1}$. It seems that the difficult part is to prove an inequality of the form $a_{n}>b_{m}$, which reduces to $3^{a_{n-1}}>100^{b_{m-1}}$, or $a_{n-1}>\\left(\\log _{3} 100\\right) b_{m-1}$. Iterating, we obtain $3^{a_{n-2}}>\\left(\\log _{3} 100\\right) 100^{b_{m-2}}$, that is,$$ a_{n-2}>\\log _{3}\\left(\\log _{3} 100\\right)+\\left(\\left(\\log _{3} 100\\right) b_{m-2} .\\right. $$Seeing this we might suspect that an inequality of the form $a_{n}>u+v b_{n}$, holding for all $n$ with some fixed $u$ and $v$, might be useful in the solution. From such an inequality we would derive $a_{n+1}=3^{a_{n}}>3^{u}\\left(3^{v}\\right)^{b_{m}}$. If $3^{v}>100$, then $a_{n+1}>3^{u} b_{m+1}$, and if $3^{u}>u+v$, then we would obtain $a_{n+1}>u+v b_{m+1}$, the same inequality as the one we started with, but with $m+1$ and $n+1$ instead of $m$ and $n$.The inequality $3^{v}>100$ holds for $v=5$, and $3^{u}>u+5$ holds for $u=2$. Thus $a_{n}>2+5 b_{m}$ implies $a_{n+1}>2+5 b_{m+1}$. We have $b_{1}=100, a_{1}=3, a_{2}=27, a_{3}=3^{27}$, and $2+5 b_{1}=502<729=3^{6}$, so $a_{3}>2+5 b_{1}$. We find that $a_{n}>2+5 b_{n-2}$ for all $n \\geq 3$. In particular, $a_{n} \\geq b_{n-2}$.On the other hand, $a_{n}<b_{m}$ implies $a_{n+1}=3^{a_{n}}<100^{b_{m}}<b_{m+1}$, which combined with $a_{2}<b_{1}$ yields $a_{n}<b_{n-1}$ for all $n \\geq 2$. Hence $b_{n-2}<a_{n}<b_{n-1}$, which implies that $m=n-1$, and for $n=100, m=99$."
+ },
+ {
+  "instruction":"Let the sequence $a_{0}, a_{1}, \\ldots$ be defined by the equation\n\n$$\n1-x^{2}+x^{4}-x^{6}+\\cdots=\\sum_{n=0}^{\\infty} a_{n}(x-3)^{n} \\quad(0<x<1) .\n$$\n\nFind\n\n$$\n\\limsup _{n \\rightarrow \\infty}\\left(\\left|a_{n}\\right|^{\\frac{1}{n}}\\right) .\n$$\n\n",
+  "response":" As\n\n$$\n1-x^{2}+x^{4}-x^{6}+\\cdots=\\frac{1}{1+x^{2}}\n$$\n\nwhich has singularities at $\\pm i$, the radius of convergence of\n\n$$\n\\sum_{n=0}^{\\infty} a_{n}(x-3)^{n}\n$$\n\nis the distance from $3$ to $\\pm i, and |3 \\mp i|=\\sqrt{10}$. We then have\n\n$$\n\\limsup _{n \\rightarrow \\infty}\\left(\\left|a_{n}\\right|^{\\frac{1}{n}}\\right)=\\frac{1}{\\sqrt{10}} \\cdot\n$$\n\n"
+ },
+ {
+  "instruction":"Find the radius of convergence $R$ of the Taylor series about $z=1$ of the function $f(z)=1 \/\\left(1+z^{2}+z^{4}+z^{6}+z^{8}+z^{10}\\right)$. \n\n",
+  "response":" The rational function\n\n$$\nf(z)=\\frac{1-z^{2}}{1-z^{12}}\n$$\n\nhas poles at all nonreal twelfth roots of unity (the singularities at $z^{2}=1$ are removable). Thus, the radius of convergence is the distance from 1 to the nearest singularity:\n\n$$\nR=|\\exp (\\pi i \/ 6)-1|=\\sqrt{(\\cos (\\pi \/ 6)-1)^{2}+\\sin ^{2}(\\pi \/ 6)}=\\sqrt{2-\\sqrt{3}} .\n$$\n\n"
+ },
+ {
+  "instruction":"Find the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ of the numbers $1,2,3,4,5,6$ that can be transformed into $1,2,3,4,5,6$ through exactly four transpositions (and not fewer).",
+  "response":"The condition from the statement implies that any such permutation has exactly two disjoint cycles, say $\\left(a_{i_{1}}, \\ldots, a_{i_{r}}\\right)$ and $\\left(a_{i_{r+1}}, \\ldots, a_{i_{6}}\\right)$. This follows from the fact that in order to transform a cycle of length $r$ into the identity $r-1$, transpositions are needed. Moreover, we can only have $r=5,4$, or 3 .When $r=5$, there are $\\left(\\begin{array}{l}6 \\\\ 1\\end{array}\\right)$ choices for the number that stays unpermuted. There are $(5-1) !$ possible cycles, so in this case we have $6 \\times 4 !=144$ possibilities. When $r=4$, there are $\\left(\\begin{array}{l}6 \\\\ 4\\end{array}\\right)$ ways to split the numbers into the two cycles (two cycles are needed and not just one). One cycle is a transposition. There are $(4-1) !=6$ choices for the other. Hence in this case the number is 90 . Note that here exactly four transpositions are needed.Finally, when $r=3$, then there are $\\left(\\begin{array}{l}6 \\\\ 3\\end{array}\\right) \\times(3-1) ! \\times(3-1) !=40$ cases. Therefore, the answer to the problem is $144+90+40=274$."
+ },
+ {
+  "instruction":"Given the fact that $\\int_{-\\infty}^{\\infty} e^{-x^{2}} d x=\\sqrt{\\pi}$, evaluate the integral\n\n$$\nI=\\int_{-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} e^{-\\left(x^{2}+(y-x)^{2}+y^{2}\\right)} d x d y .\n$$",
+  "response":" We write\n\n$$\n\\begin{aligned}\nI &=\\int_{-\\infty}^{\\infty} e^{-3 y^{2} \/ 2}\\left(\\int_{-\\infty}^{\\infty} e^{-2 x^{2}+2 x y-y^{2} \/ 2} d x\\right) d y \\\\n&=\\int_{-\\infty}^{\\infty} e^{-3 y^{2} \/ 2}\\left(\\int_{-\\infty}^{\\infty} e^{-2\\left(x-\\frac{y}{2}\\right)^{2}} d x\\right) d y\n\\end{aligned}\n$$\n\nmaking the substitution $t=\\sqrt{2}\\left(x-\\frac{y}{2}\\right), d t=\\sqrt{2} d x$ on the inner integral, we get\n\n$$\n\\begin{aligned}\nI &=\\frac{1}{\\sqrt{2}} \\int_{-\\infty}^{\\infty} e^{-3 y^{2} \/ 2}\\left(\\int_{-\\infty}^{\\infty} e^{-t^{2}} d t\\right) d y \\\\n&=\\sqrt{\\frac{\\pi}{2}} \\int_{-\\infty}^{\\infty} e^{-3 y^{2} \/ 2} d y\n\\end{aligned}\n$$\n\nnow making the substitution: $s=\\sqrt{\\frac{3}{2}} y, d s=\\sqrt{\\frac{3}{2}} d y$ we obtain\n\n$$\n\\begin{aligned}\nI &=\\sqrt{\\frac{\\pi}{3}} \\int_{-\\infty}^{\\infty} e^{-s^{2}} d s \\\\n&=\\frac{\\pi}{\\sqrt{3}} .\n\\end{aligned}\n$$\n\n"
+ },
+ {
+  "instruction":"The zeros of the polynomial $P(x)=x^{3}-10 x+11$ are $u$, $v$, and $w$. Determine the value of $\\arctan u+\\arctan v+\\arctan w$.",
+  "response":"First solution: Let $\\alpha=\\arctan u, \\beta=\\arctan v$, and $\\arctan w$. We are required to determine the sum $\\alpha+\\beta+\\gamma$. The addition formula for the tangent of three angles, $$ \\tan (\\alpha+\\beta+\\gamma)=\\frac{\\tan \\alpha+\\tan \\beta+\\tan \\gamma-\\tan \\alpha \\tan \\beta \\tan \\gamma}{1-(\\tan \\alpha \\tan \\beta+\\tan \\beta \\tan \\gamma+\\tan \\alpha \\tan \\gamma)}, $$implies$$ \\tan (\\alpha+\\beta+\\gamma)=\\frac{u+v+w-u v w}{1-(u v+v w+u v)} . $$Using Vi\u00e8te's relations,$$ u+v+w=0, \\quad u v+v w+u w=-10, \\quad u v w=-11, $$we further transform this into $\\tan (\\alpha+\\beta+\\gamma)=\\frac{11}{1+10}=1$. Therefore, $\\alpha+\\beta+\\gamma=\\frac{\\pi}{4}+k \\pi$, where $k$ is an integer that remains to be determined.From Vi\u00e8te's relations we can see the product of the zeros of the polynomial is negative, so the number of negative zeros is odd. And since the sum of the zeros is 0 , two of them are positive and one is negative. Therefore, one of $\\alpha, \\beta, \\gamma$ lies in the interval $\\left(-\\frac{\\pi}{2}, 0\\right)$ and two of them lie in $\\left(0, \\frac{\\pi}{2}\\right)$. Hence $k$ must be equal to 0 , and $\\arctan u+$ $\\arctan v+\\arctan w=\\frac{\\pi}{4}$.Second solution: Because$$ \\operatorname{Im} \\ln (1+i x)=\\arctan x, $$we see that$$ \\begin{aligned} \\arctan u+\\arctan v+\\arctan w &=\\operatorname{Im} \\ln (i P(i))=\\operatorname{Im} \\ln (11+11 i) \\\\ &=\\arctan 1=\\frac{\\pi}{4} . \\end{aligned} $$"
+ },
+ {
+  "instruction":"Find the maximum of $x_{1}^{3}+\\ldots+x_{10}^{3}$ for $x_{1}, \\ldots, x_{10} \\in[-1,2]$ such that  $$ x_{1}+\\ldots+x_{10}=10.$$",
+  "response":"Look at the behavior of expression $x_{1}^{3}+x_{2}^{3}$ when we move the variables $x_{1} \\leq x_{2}$ together, that is, replace them with $x_{1}+\\varepsilon$ and $x_{2}-\\varepsilon, 0<\\varepsilon \\leq \\frac{x_{2}-x_{1}}{2}$, and when we move them apart, that is, replace them with $x_{1}-\\varepsilon$ and $x_{2}+\\varepsilon, \\varepsilon>0$. After moving together, the product $x_{1} x_{2}$ is not decreasing, and after moving apart, it is not increasing. Hence the sum of cubes of two variables   $$ x_{1}^{3}+x_{2}^{3}=\\left(x_{1}+x_{2}\\right)\\left(\\left(x_{1}+x_{2}\\right)^{2}-3 x_{1} x_{2}\\right) $$  is not decreasing as two variables move apart, if $x_{1}+x_{2} \\geq 0$, and as they move together, if $x_{1}+x_{2} \\leq 0$.  We will move apart couples of variables $x_{i}, x_{j}$, for which $x_{i}+x_{j} \\geq 0$, until one of the variables coincides with an end of the segment $[-1 ; 2]$. That can be done while at least one couple of variables with nonnegative sum remains in the interval $(-1 ; 2)$. After that we get one of the next two situations:  (1) some variables are equal to 2 or $-1$, and the rest of the variables have negative pairwise sums (and moreover each of the variables in the interval $(-1 ; 2)$ has negative sum with -1), or  (2) some variables are equal to 2 or $-1$, and a single variable in the interval $(-1 ; 2)$ has nonnegative sum with $-1$.  In the first case, let $x$ be the mean value of all the variables in $[-1 ; 2)$. Then $x<0$ because all pairwise sums are negative. We will move together couples of variables from the interval $[-1 ; 2)$ as described above, until one of the variables in the couple equals $x$.  Thus, we can start with an arbitrary collection of points $x_{1}, \\ldots, x_{10}$ and use the described above moving together and apart (which preserve the sum of elements and not decrease the sum of cubes) to reach one of the following collections:  (1) $k$ variables are at a point $x \\in[-1 ; 0]$ and $10-k$ variables are at the point 2 $(k=0, \\ldots, 10)$  (2) $k$ variables are at the point $-1$, a single variable equals $x \\in[1 ; 2]$, and $9-k$ variables are at the point $2(k=0, \\ldots, 9)$.  From the conditions $x_{1}+\\ldots+x_{10}=10$ and $x \\in[-1 ; 0]$ (or $x \\in[1 ; 2]$ ), we obtain that either $k=4$ or $k=5$ for collections of the first type and $k=3$ for collections of the second type. It remains to examine the following collections:  $$ \\begin{gathered} \\left(-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2}, 2,2,2,2,2,2\\right), \\quad(0,0,0,0,0,2,2,2,2,2), \\\\ (-1,-1,-1,1,2,2,2,2,2,2) . \\end{gathered} $$  The maximal value of the sum of cubes is equal to $47.5$ and attained at the first collection.  Answer: $47"
+ },
+ {
+  "instruction":"Two airplanes are supposed to park at the same gate of a concourse. The arrival times of the airplanes are independent and randomly distributed throughout the 24 hours of the day. What is the probability that both can park at the gate, provided that the first to arrive will stay for a period of two hours, while the second can wait behind it for a period of one hour?",
+  "response":"The set of possible events is modeled by the square $[0, 24] \\times[0,24]$. It is, however, better to identify the 0th and the 24th hours, thus obtaining a square with opposite sides identified, an object that in mathematics is called a torus (which is, in fact, the Cartesian product of two circles. The favorable region is outside a band of fixed thickness along the curve $x=y$ on the torus as depicted in Figure 110. On the square model this region is obtained by removing the points $(x, y)$ with $|x-y| \\leq 1$ together with those for which $|x-y-1| \\leq 1$ and $|x-y+1| \\leq 1$. The required probability is the ratio of the area of the favorable region to the area of the square, and is$$ P=\\frac{24^{2}-2 \\cdot 24}{24^{2}}=\\frac{11}{12} \\approx 0.917 . $$"
+ },
+ {
+  "instruction":"Find a limit  $$ \\lim _{n \\rightarrow \\infty}\\left(\\int_{0}^{1} e^{x^{2} \/ n} d x\\right)^{n}. $$",
+  "response":"Fix an arbitrary $\\varepsilon>0$. Since the function $y=e^{t}$ is convex, for $t \\in[0, \\varepsilon]$ its graph lies above the tangent $y=1+t$ but under the secant $y=1+a_{\\varepsilon} t$, where $a_{\\varepsilon}=\\frac{1}{\\varepsilon}\\left(e^{\\varepsilon}-1\\right)$. Hence for $\\frac{1}{n} \\leq \\varepsilon$ it holds  $$ \\left(\\int_{0}^{1} e^{x^{2} \/ n} d x\\right)^{n} \\geq\\left(\\int_{0}^{1}\\left(1+\\frac{x^{2}}{n}\\right) d x\\right)^{n}=\\left(1+\\frac{1}{3 n}\\right)^{n} \\rightarrow e^{1 \/ 3}, \\text { as } n \\rightarrow \\infty, $$  and  $$ \\left(\\int_{0}^{1} e^{x^{2} \/ n} d x\\right)^{n} \\leq\\left(\\int_{0}^{1}\\left(1+\\frac{a_{\\varepsilon} x^{2}}{n}\\right) d x\\right)^{n}=\\left(1+\\frac{a_{\\varepsilon}}{3 n}\\right)^{n} \\rightarrow e^{a_{\\varepsilon} \/ 3}, \\text { as } n \\rightarrow \\infty $$  Since $a_{\\varepsilon}=\\frac{1}{\\varepsilon}\\left(e^{\\varepsilon}-1\\right) \\rightarrow 1$, as $\\varepsilon \\rightarrow 0+$, we get $\\lim _{n \\rightarrow \\infty}\\left(\\int_{0}^{1} e^{x^{2} \/ n} d x\\right)^{n}=e^{1 \/ 3}$.  Answer: $e^{1 \/ 3}$"
+ },
+ {
+  "instruction":"Let $x_{0}=1$ and\n\n$$\nx_{n+1}=\\frac{3+2 x_{n}}{3+x_{n}}, \\quad n \\geqslant 0 .\n$$\n\nFind the limit $x_{\\infty}=\\lim _{n \\rightarrow \\infty} x_{n}$.\n\n",
+  "response":" Obviously, $x_{n} \\geqslant 1$ for all $n$; so, if the limit exists, it is $\\geqslant 1$, and we can pass to the limit in the recurrence relation to get\n\n$$\nx_{\\infty}=\\frac{3+2 x_{\\infty}}{3+x_{\\infty}} \\text {; }\n$$\n\nin other words, $x_{\\infty}^{2}+x_{\\infty}-3=0$. So $x_{\\infty}$ is the positive solution of this quadratic equation, that is, $x_{\\infty}=\\frac{1}{2}(-1+\\sqrt{13})$.\n\nTo prove that the limit exists, we use the recurrence relation to get\n\n$$\n\\begin{aligned}\nx_{n+1}-x_{n} &=\\frac{3+2 x_{n}}{3+x_{n}}-\\frac{3+2 x_{n-1}}{3+x_{n-1}} \\\\n&=\\frac{3\\left(x_{n}-x_{n-1}\\right)}{\\left(3+x_{n}\\right)\\left(3+x_{n+1}\\right)}\n\\end{aligned}\n$$\n\nHence, $\\left|x_{n+1}-x_{n}\\right| \\leqslant \\frac{1}{3}\\left|x_{n}-x_{n-1}\\right|$. Iteration gives\n\n$$\n\\left|x_{n+1}-x_{n}\\right| \\leqslant 3^{-n}\\left|x_{1}-x_{0}\\right|=\\frac{1}{3^{n} \\cdot 4} .\n$$\n\nThe series $\\sum_{n=1}^{\\infty}\\left(x_{n+1}-x_{n}\\right)$, of positive terms, is dominated by the convergent series $\\frac{1}{4} \\sum_{n=1}^{\\infty} 3^{-n}$ and so converges. We have $\\sum_{n=1}^{\\infty}\\left(x_{n+1}-x_{n}\\right)=\\lim x_{n}-x_{1}$ and we are done."
+ },
+ {
+  "instruction":"Consider a sequence $x_{n}=x_{n-1}-x_{n-1}^{2}, n \\geq 2, x_{1} \\in(0,1)$. Calculate  $$ \\lim _{n \\rightarrow \\infty} \\frac{n^{2} x_{n}-n}{\\ln n} . $$",
+  "response":"By the monotone convergence theorem, it is easy to show that $x_{n} \\rightarrow 0$, as $n \\rightarrow \\infty$. Next, by the Stolz-Cesaro theorem it holds  $$ \\begin{aligned} n x_{n}=\\frac{n}{\\frac{1}{x_{n}}} \\sim \\frac{1}{\\frac{1}{x_{n}}-\\frac{1}{x_{n-1}}}=\\frac{x_{n-1} x_{n}}{x_{n-1}-x_{n}}=\\\\ \\quad=\\frac{x_{n-1}\\left(x_{n-1}-x_{n-1}^{2}\\right)}{x_{n-1}^{2}}=1-x_{n-1} \\rightarrow 1, \\text { as } n \\rightarrow \\infty . \\end{aligned} $$  Transform the expression in question and use the Stolz-Cesaro theorem once again:   $$ \\begin{aligned} \\frac{n^{2} x_{n}-n}{\\ln n}=\\frac{n x_{n} \\cdot\\left(n-\\frac{1}{x_{n}}\\right)}{\\ln n} \\sim \\frac{n-\\frac{1}{x_{n}}}{\\ln n} \\sim \\frac{1-\\frac{1}{x_{n}}+\\frac{1}{x_{n-1}}}{\\ln n-\\ln (n-1)}=\\frac{1-\\frac{1}{x_{n}}+\\frac{1}{x_{n-1}}}{-\\ln \\left(1-\\frac{1}{n}\\right)} \\sim \\\\ \\sim n\\left(1-\\frac{1}{x_{n-1}-x_{n-1}^{2}}+\\frac{1}{x_{n-1}}\\right)=-\\frac{n x_{n-1}}{1-x_{n-1}} \\rightarrow-1, \\text { as } n \\rightarrow \\infty \\end{aligned} $$  Answer: $-1"
+ },
+ {
+  "instruction":"Let $\\varphi(x, y)$ be a function with continuous second order partial derivatives such that\n\n1. $\\varphi_{x x}+\\varphi_{y y}+\\varphi_{x}=0$ in the punctured plane $\\mathbb{R}^{2} \\backslash\\{0\\}$,\n\n2. $r \\varphi_{x} \\rightarrow \\frac{x}{2 \\pi r}$ and $r \\varphi_{y} \\rightarrow \\frac{y}{2 \\pi r}$ as $r=\\sqrt{x^{2}+y^{2}} \\rightarrow 0$.\n\nLet $C_{R}$ be the circle $x^{2}+y^{2}=R^{2}$. Evaluate the line integral\n\n$$\n\\int_{C_{R}} e^{x}\\left(-\\varphi_{y} d x+\\varphi_{x} d y\\right)\n$$\n\n",
+  "response":"Consider the annular region $\\mathcal{A}$ between the circles of radius $r$ and $R$, then by Green's Theorem\n\n$$\n\\begin{aligned}\n&\\int_{R} e^{x}\\left(-\\varphi_{y} d x+\\varphi_{x} d y\\right)-\\int_{r} e^{y}\\left(-\\varphi_{y} d x+\\varphi_{x} d y\\right)= \\\\n&=\\int_{\\partial \\mathcal{A}} e^{x}\\left(-\\varphi_{x} d x+\\varphi_{x} d y\\right) \\\\n&=\\iint_{\\mathcal{A}} d\\left(e^{x}\\left(-\\varphi_{y} d x+\\varphi_{x} d y\\right)\\right) \\\\n&=\\iint_{\\mathcal{A}}-e^{x} \\varphi_{y y} d y \\wedge d x+\\left(e^{x} \\varphi_{x}+e^{x} \\varphi_{x x}\\right) d x \\wedge d y \\\\n&=\\iint_{\\mathcal{A}} e^{x}\\left(\\varphi_{x x}+e^{x} \\varphi_{x x}+\\varphi_{x}\\right) d x \\wedge d y=0\n\\end{aligned}\n$$\n\nshowing that the integral does not depend on the radius $r$. Now, parametrizing the circle of radius $r$\n\n$$\n\\begin{aligned}\n&\\int_{C_{r}} e^{x}\\left(-\\varphi_{y} d x+\\varphi_{x} d y\\right)= \\\\n&=\\int_{0}^{2 \\pi} e^{r \\cos \\theta}\\left(\\varphi_{y}(r \\cos \\theta, r \\sin \\theta) \\sin \\theta+\\varphi_{x}(r \\cos \\theta, r \\sin \\theta) \\cos \\theta\\right) r d \\theta\n\\end{aligned}\n$$\n\nbut when $r \\rightarrow 0$ the integrand converges uniformly to\n\n$$\n\\frac{\\sin \\theta}{2 \\pi} \\cdot \\sin \\theta+\\frac{\\cos \\theta}{2 \\pi} \\cdot \\cos \\theta=\\frac{1}{2 \\pi}\n$$\n\nso the integral approaches $1$ when $r \\rightarrow 0$ and that is the value of the integral.\n\n"
+ },
+ {
+  "instruction":"Compute$$ \\lim _{n \\rightarrow \\infty} \\sum_{k=1}^{n}\\left(\\frac{k}{n^{2}}\\right)^{\\frac{k}{n^{2}}+1} . $$",
+  "response":"We use the fact that$$ \\lim _{x \\rightarrow 0^{+}} x^{x}=1 . $$As a consequence, we have$$ \\lim _{x \\rightarrow 0^{+}} \\frac{x^{x+1}}{x}=1 . $$For our problem, let $\\epsilon>0$ be a fixed small positive number. There exists $n(\\epsilon)$ such that for any integer $n \\geq n(\\epsilon)$,$$ 1-\\epsilon<\\frac{\\left(\\frac{k}{n^{2}}\\right)^{\\frac{k}{n^{2}}+1}}{\\frac{k}{n^{2}}}<1+\\epsilon, \\quad k=1,2, \\ldots, n . $$From this, using properties of ratios, we obtain$$ 1-\\epsilon<\\frac{\\sum_{k=1}^{n}\\left(\\frac{k}{n^{2}}\\right)^{\\frac{k}{n^{2}}+1}}{\\sum_{k=1}^{n} \\frac{k}{n^{2}}}<1+\\epsilon, \\quad \\text { for } n \\geq n(\\epsilon) . $$Knowing that $\\sum_{k=1}^{n} k=\\frac{n(n+1)}{2}$, this implies$$ (1-\\epsilon) \\frac{n+1}{2 n}<\\sum_{k=1}^{n}\\left(\\frac{k}{n^{2}}\\right)^{\\frac{k}{n^{2}}+1}<(1+\\epsilon) \\frac{n+1}{2 n}, \\quad \\text { for } n \\geq n(\\epsilon) . $$It follows that$$ \\lim _{n \\rightarrow \\infty} \\sum_{k-1}^{n}\\left(\\frac{k}{n^{2}}\\right)^{\\frac{k}{n^{2}}+1}=\\frac{1}{2} . $$"
+ },
+ {
+  "instruction":"A sequence $\\left\\{x_{n}, n \\geq 1\\right\\}$ satisfies $x_{n+1}=x_{n}+e^{-x_{n}}, n \\geq 1$, and $x_{1}=1$. Find $\\lim _{n \\rightarrow \\infty} \\frac{x_{n}}{\\ln n}$.",
+  "response":"The sequence $\\left\\{x_{n}\\right\\}$ is increasing, hence it has a finite or infinite limit. If $\\lim _{n \\rightarrow \\infty} x_{n}=$ $x<+\\infty$, then $x=x+e^{-x}$, a contradiction. Therefore, $\\lim _{n \\rightarrow \\infty} x_{n}=+\\infty$. We use twice the Stolz-Cesaro theorem and obtain  $$ \\lim _{n \\rightarrow \\infty} \\frac{x_{n}}{\\ln n}=\\lim _{n \\rightarrow \\infty} \\frac{x_{n+1}-x_{n}}{\\ln (n+1)-\\ln n}=\\lim _{n \\rightarrow \\infty} \\frac{e^{-x_{n}}}{\\ln \\left(1+\\frac{1}{n}\\right)}=\\lim _{n \\rightarrow \\infty} \\frac{n}{e^{x_{n}}}=\\lim _{n \\rightarrow \\infty} \\frac{1}{e^{x_{n+1}}-e^{x_{n}}} . $$  It remains to notice that  $$ \\lim _{n \\rightarrow \\infty}\\left(e^{x_{n+1}}-e^{x_{n}}\\right)=\\lim _{n \\rightarrow \\infty}\\left(e^{x_{n}+e^{-x_{n}}}-e^{x_{n}}\\right)=\\lim _{n \\rightarrow \\infty} \\frac{e^{e^{-x_{n}}}-1}{e^{-x_{n}}}=\\lim _{t \\rightarrow 0} \\frac{e^{t}-1}{t}=1 . $$  Answer: 1"
+ },
+ {
+  "instruction":"Find the limit  $$ \\lim _{N \\rightarrow \\infty} \\sqrt{N}\\left(1-\\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\}\\right), $$  where $\\{x\\}$ denotes the fractional part of $x$.",
+  "response":"Estimate $\\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\}$. For $k^{2} \\leq n<(k+1)^{2}$ it holds  $$ \\{\\sqrt{n}\\}=\\sqrt{n}-k \\leq \\sqrt{(k+1)^{2}-1}-k, $$  where equality is attained for $n=(k+1)^{2}-1$. If $k<l$ then  $$ \\begin{aligned} 1 &-\\left\\{\\sqrt{(k+1)^{2}-1}\\right\\}=k+1-\\sqrt{(k+1)^{2}-1}=\\\\ &=\\frac{1}{k+1+\\sqrt{(k+1)^{2}-1}}>\\frac{1}{l+1+\\sqrt{(l+1)^{2}-1}}=1-\\left\\{\\sqrt{(l+1)^{2}-1}\\right\\}, \\end{aligned} $$  hence $\\left\\{\\sqrt{(k+1)^{2}-1}\\right\\}$ is increasing in $k$. Thus, for $N=m^{2}-1$ we get  $$ \\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\}=\\left\\{\\sqrt{m^{2}-1}\\right\\}=\\sqrt{m^{2}-1}-m+1 . $$  Now consider $m^{2} \\leq N \\leq(m+1)^{2}-1$. Since $\\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\}$ is increasing in $N$, it holds  $$ \\sqrt{m^{2}-1}-m+1 \\leq \\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\} \\leq \\sqrt{(m+1)^{2}-1}-m . $$  Therefore,  $$ 1+m-\\sqrt{(m+1)^{2}-1} \\leq 1-\\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\} \\leq m-\\sqrt{m^{2}-1} $$  Also we have $m \\leq \\sqrt{N}<m+1$, so  $$ \\begin{aligned} &\\frac{m}{m+1} \\cdot(m+1)\\left(m+1-\\sqrt{(m+1)^{2}-1}\\right) \\leq \\\\ &\\quad \\leq \\sqrt{N}\\left(1-\\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\}\\right)<\\frac{m+1}{m} \\cdot m\\left(m-\\sqrt{m^{2}-1}\\right) . \\end{aligned} $$  Since $m\\left(m-\\sqrt{m^{2}-1}\\right)=\\frac{m}{m+\\sqrt{m^{2}-1}} \\rightarrow \\frac{1}{2}$ and $\\frac{m}{m+1} \\rightarrow 1$, as $m \\rightarrow \\infty$, we have $\\lim _{N \\rightarrow \\infty} \\sqrt{N}\\left(1-\\max _{1 \\leq n \\leq N}\\{\\sqrt{n}\\}\\right)=\\frac{1}{2}$  Answer: $\\frac{1}{2}"
+ },
+ {
+  "instruction":"Compute\n\n$$\nL=\\lim _{n \\rightarrow \\infty}\\left(\\frac{n^{n}}{n !}\\right)^{1 \/ n} .\n$$\n\n",
+  "response":" Let $p_{1}=1, p_{2}=(2 \/ 1)^{2}, p_{3}=(3 \/ 2)^{3}, \\ldots, p_{n}=(n \/(n-1))^{n}$. Then\n\n$$\n\\frac{p_{1} p_{2} \\cdots p_{n}}{n}=\\frac{n^{n}}{n !},\n$$\n\nand since $p_{n} \\rightarrow e$, we have $\\lim \\left(n^{n} \/ n !\\right)^{1 \/ n}=e$ as well (using the fact that $\\left.\\lim n^{1 \/ n}=1\\right)$.\n\n"
+ },
+ {
+  "instruction":"Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be the function of period $2 \\pi$ such that $f(x)=x^{3}$ for $-\\pi \\leqslant x<\\pi$.\n\nThe Fourier series for $f$ has the form $\\sum_{1}^{\\infty} b_{n} \\sin n x$. Find\n$$\n\\sum_{n=1}^{\\infty} b_{n}^{2}.\n$$\n\n",
+  "response":"Using Parseval's Identity \n\n$$\n\\frac{1}{2} a_{0}^{2}+\\sum_{n=1}^{\\infty}\\left(a_{n}^{2}+b_{n}^{2}\\right)=\\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} f^{2}(x) d x\n$$\n\nand the fact that all $a_{n}=0$,\n\n$$\n\\sum_{n=1}^{\\infty} b^{2}=\\frac{1}{\\pi} \\int_{-\\pi}^{\\pi} x^{6} d x=\\frac{2}{7} \\pi^{6} .\n$$\n\n"
+ },
+ {
+  "instruction":"For a real $2 \\times 2$ matrix\n\n$$\nX=\\left(\\begin{array}{ll}\n x & y \\\\\n z & t\n\\end{array}\\right),\n$$\n\nlet $\\|X\\|=x^{2}+y^{2}+z^{2}+t^{2}$, and define a metric by $d(X, Y)=\\|X-Y\\|$. Let $\\Sigma=\\{X \\mid \\operatorname{det}(X)=0\\}$. Let\n\n$$\nA=\\left(\\begin{array}{ll}\n 1 & 0 \\\\\n 0 & 2\n\\end{array}\\right) \\text {. }\n$$\n\nFind the minimum distance from $A$ to $\\Sigma$.",
+  "response":"For $X \\in \\Sigma$, we have\n\n$$\n\\begin{aligned}\n\\|A-X\\|^{2} &=(1-x)^{2}+y^{2}+z^{2}+(2-t)^{2} \\\\n&=y^{2}+z^{2}+1-2 x+x^{2}+(2-t)^{2} \\\\n& \\geqslant \\pm 2 y z+1-2 x+2|x|(2-t) \\\\n&=4|x|-2 x+2(\\pm y z-|x| t)+1\n\\end{aligned}\n$$\n\nWe can choose the sign, so $\\pm y z-|x| t=0$ because $\\operatorname{det} X=0$. As $4|x|-2 x \\geqslant 0$, we have $\\|A-X\\| \\geqslant 1$ with equality when $4|x|-2 x=0,|x|=2-t, y=\\pm z$,\n\n"
+ },
+ {
+  "instruction":"Find the probability that in the process of repeatedly flipping an unbiased coin, one will encounter a run of 5 heads before one encounters a run of 2 tails.",
+  "response":"We call a successful string a sequence of $H$ 's and $T$ 's in which $H H H H H$ appears before $T T$ does. Each successful string must belong to one of the following three types:(i) those that begin with $T$, followed by a successful string that begins with $H$;(ii) those that begin with $H, H H, H H H$, or $H H H H$, followed by a successful string that begins with $T$;(iii) the string $H H H H H$.Let $P_{H}$ denote the probability of obtaining a successful string that begins with $H$, and let $P_{T}$ denote the probability of obtaining a successful string that begins with $T$. Then$$ \\begin{aligned} P_{T} &=\\frac{1}{2} P_{H}, \\\\ P_{H} &=\\left(\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}\\right) P_{T}+\\frac{1}{32} . \\end{aligned} $$Solving these equations simultaneously, we find that$$ P_{H}=\\frac{1}{17} \\quad \\text { and } \\quad P_{T}=\\frac{1}{34} . $$Hence the probability of obtaining five heads before obtaining two tails is $\\frac{3}{34}$."
+ },
+ {
+  "instruction":"Compute the integral $\\iint_{D} x d x d y$, where$$ D=\\left\\{(x, y) \\in \\mathbb{R}^{2} \\mid x \\geq 0,1 \\leq x y \\leq 2,1 \\leq \\frac{y}{x} \\leq 2\\right\\} . $$",
+  "response":"The domain is bounded by the hyperbolas $x y=1, x y=2$ and the lines $y=x$ and $y=2 x$. This domain can mapped into a rectangle by the transformation$$ T: \\quad u=x y, \\quad v=\\frac{y}{x} . $$Thus it is natural to consider the change of coordinates$$ T^{-1}: \\quad x=\\sqrt{\\frac{u}{v}}, \\quad y=\\sqrt{u v} . $$The domain becomes the rectangle $D^{*}=\\left\\{(u, v) \\in \\mathbb{R}^{2} \\mid 1 \\leq u \\leq 2,1 \\leq v \\leq 2\\right\\}$. The Jacobian of $T^{-1}$ is $\\frac{1}{2 v} \\neq 0$. The integral becomes$$ \\int_{1}^{2} \\int_{1}^{2} \\sqrt{\\frac{u}{v}} \\frac{1}{2 v} d u d v=\\frac{1}{2} \\int_{1}^{2} u^{1 \/ 2} d u \\int_{1}^{2} v^{-3 \/ 2} d v=\\frac{1}{3}(5 \\sqrt{2}-6) . $$"
+ },
+ {
+  "instruction":"A husband and wife agree to meet at a street corner between 4 and 5 o'clock to go shopping together. The one who arrives first will await the other for 15 minutes, and then leave. What is the probability that the two meet within the given time interval, assuming that they can arrive at any time with the same probability? ",
+  "response":"Denote by $x$, respectively, $y$, the fraction of the hour when the husband, respectively, wife, arrive. The configuration space is the square $$ D=\\{(x, y) \\mid 0 \\leq x \\leq 1,0 \\leq y \\leq 1\\} . $$In order for the two people to meet, their arrival time must lie inside the region$$ D_{f}=\\left\\{(x, y)|| x-y \\mid \\leq \\frac{1}{4}\\right\\} . $$The desired probability is the ratio of the area of this region to the area of the square.The complement of the region consists of two isosceles right triangles with legs equal to $\\frac{3}{4}$, and hence of areas $\\frac{1}{2}\\left(\\frac{3}{4}\\right)^{2}$. We obtain for the desired probability$$ 1-2 \\cdot \\frac{1}{2} \\cdot\\left(\\frac{3}{4}\\right)^{2}=\\frac{7}{16} \\approx 0.44 . $$"
+ },
+ {
+  "instruction":"Evaluate the product$$ \\left(1-\\cot 1^{\\circ}\\right)\\left(1-\\cot 2^{\\circ}\\right) \\cdots\\left(1-\\cot 44^{\\circ}\\right) \\text {. } $$",
+  "response":"We have$$ \\begin{aligned} \\left(1-\\cot 1^{\\circ}\\right)\\left(1-\\cot 2^{\\circ}\\right) \\cdots\\left(1-\\cot 44^{\\circ}\\right) \\\\ &=\\left(1-\\frac{\\cos 1^{\\circ}}{\\sin 1^{\\circ}}\\right)\\left(1-\\frac{\\cos 2^{\\circ}}{\\sin 2^{\\circ}}\\right) \\cdots\\left(1-\\frac{\\cos 44^{\\circ}}{\\sin 44^{\\circ}}\\right) \\\\ &=\\frac{\\left(\\sin 1^{\\circ}-\\cos 1^{\\circ}\\right)\\left(\\sin 2^{\\circ}-\\cos 2^{\\circ}\\right) \\cdots\\left(\\sin 44^{\\circ}-\\cos 44^{\\circ}\\right)}{\\sin 1^{\\circ} \\sin 2^{\\circ} \\cdots \\sin 44^{\\circ}} \\end{aligned} $$Using the identity $\\sin a-\\cos a=\\sqrt{2} \\sin \\left(a-45^{\\circ}\\right)$ in the numerators, we transform this further into$$ \\begin{gathered} \\frac{\\sqrt{2} \\sin \\left(1^{\\circ}-45^{\\circ}\\right) \\cdot \\sqrt{2} \\sin \\left(2^{\\circ}-45^{\\circ}\\right) \\cdots \\sqrt{2} \\sin \\left(44^{\\circ}-45^{\\circ}\\right)}{\\sin 1^{\\circ} \\sin 2^{\\circ} \\cdots \\sin 44^{\\circ}} \\\\ =\\frac{(\\sqrt{2})^{44}(-1)^{44} \\sin 44^{\\circ} \\sin 43^{\\circ} \\cdots \\sin 1^{\\circ}}{\\sin 44^{\\circ} \\sin 43^{\\circ} \\cdots \\sin 1^{\\circ}} . \\end{gathered} $$After cancellations, we obtain $2^{22}$."
+ },
+ {
+  "instruction":"Find the number of roots of\n\n$$\nz^{7}-4 z^{3}-11=0\n$$\n\nwhich lie between the two circles $|z|=1$ and $|z|=2$.\n\n",
+  "response":" Let $p(z)=z^{7}-4 z^{3}-11$. For $z$ in the unit circle, we have\n\n$$\n|p(z)-11|=\\left|z^{7}-4 z^{3}\\right| \\leqslant 5<11\n$$\n\nso, by Rouch\\u00e9's Theorem, the given polynomial has no zeros in the unit disc. For $|z|=2$,\n\n$$\n\\left|p(z)-z^{7}\\right|=\\left|4 z^{3}+11\\right| \\leqslant 43<128=\\left|z^{7}\\right|\n$$\n\nso there are seven zeros inside the disc $\\{z|| z \\mid<2\\}$ and they are all between the two given circles.\n\n"
+ },
+ {
+  "instruction":"Let $\\mathcal{S}=\\left\\{(x, y, z) \\in \\mathbb{R}^{3} \\mid x^{2}+y^{2}+z^{2}=1\\right\\}$ denote the unit sphere in $\\mathbb{R}^{3}$. Evaluate the surface integral over $\\mathcal{S}$ :\n\n$$\n\\iint_{\\mathcal{S}}\\left(x^{2}+y+z\\right) d A .\n$$\n\n",
+  "response":" Using the change of variables\n\n$$\n\\begin{cases}x=\\sin \\varphi \\cos \\theta & 0<\\theta<2 \\pi \\\\ y=\\sin \\varphi \\sin \\theta & 0<\\varphi<\\pi \\\\ z=\\cos \\varphi & \\end{cases}\n$$\n\nwe have\n\n$$\nd A=\\sin \\varphi d \\theta d \\varphi\n$$\n\nand\n\n$$\n\\iint_{\\mathcal{S}}\\left(x^{2}+y+z\\right) d A=\\int_{0}^{\\pi} \\int_{0}^{2 \\pi}\\left(\\sin ^{2} \\varphi \\cos ^{2} \\theta+\\sin \\varphi \\sin \\theta+\\cos \\varphi\\right) \\sin \\varphi d \\theta d \\varphi .\n$$\n\nBreaking the integral in three terms, we get\n\n$$\n\\begin{aligned}\n\\int_{0}^{\\pi} \\int_{0}^{2 \\pi} \\sin \\varphi \\cos \\varphi d \\theta d \\varphi=2 \\pi \\cdot \\frac{1}{2} \\int_{0}^{\\pi} \\sin 2 \\varphi d \\varphi=0 \\\\n\\int_{0}^{\\pi} \\int_{0}^{2 \\pi} \\sin ^{2} \\varphi \\sin \\theta d \\theta d \\varphi=\\left(\\int_{0}^{\\pi} \\sin ^{2} \\varphi d \\varphi\\right) \\int_{0}^{2 \\pi} \\sin \\theta d \\theta=0 \\\\n\\int_{0}^{\\pi} \\int_{0}^{2 \\pi} \\sin ^{3} \\varphi \\cos ^{2} \\theta d \\theta d \\varphi &=\\left(\\int_{0}^{\\pi} \\sin ^{3} \\varphi d \\varphi\\right)\\left(\\int_{0}^{2 \\pi} \\cos ^{2} \\theta d \\theta\\right) \\\\n&=\\int_{0}^{\\pi} \\frac{1}{4}\\left(3 \\sin \\varphi-\\sin ^{3} \\varphi\\right) d \\varphi \\int_{0}^{2 \\pi} \\cos ^{2} \\theta d \\theta \\\\n&=\\frac{1}{4}\\left(-3 \\cos \\varphi+\\left.\\frac{1}{3} \\cos ^{3} \\varphi\\right|_{0} ^{\\pi}\\right)\\left(\\int_{0}^{2 \\pi} \\frac{1+\\cos 2 \\theta}{2} d \\theta\\right) \\\\n&=\\frac{1}{4}\\left(-3(-2)+\\frac{1}{3}(-2)\\right) \\pi \\\\n&=\\frac{1}{4}\\left(6-\\frac{2}{3}\\right) \\pi=\\frac{4}{3} \\pi .\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n\\iint_{\\mathcal{S}}\\left(x^{2}+y+z\\right) d A=\\frac{4}{3} \\pi\n$$\n\n"
+ },
+ {
+  "instruction":"Find the sum of the series  $$ \\sum_{n=0}^{\\infty} \\frac{1}{n ! 2^{n}} \\cos \\frac{\\pi n-1}{2} . $$",
+  "response":"For all $x \\in \\mathbb{R}$, it holds  $$ \\begin{aligned} &\\sum_{n=0}^{\\infty} \\frac{\\cos \\left(\\frac{\\pi n}{2}-x\\right)}{n !} x^{n}=\\cos x \\sum_{n=0}^{\\infty} \\frac{\\cos \\frac{\\pi n}{2}}{n !} x^{n}+\\sin x \\sum_{n=0}^{\\infty} \\frac{\\sin \\frac{\\pi n}{2}}{n !} x^{n}= \\\\ &=\\cos x \\sum_{k=0}^{\\infty} \\frac{(-1)^{k}}{(2 k) !} x^{2 k}+\\sin x \\sum_{k=0}^{\\infty} \\frac{(-1)^{k}}{(2 k+1) !} x^{2 k+1}=\\cos ^{2} x+\\sin ^{2} x=1 . \\end{aligned} $$  In particular for $x=\\frac{1}{2}$ we obtain the sum of the series in question.  Answer: 1"
+ },
+ {
+  "instruction":"What is the probability that a uniformly random permutation of the first $n$ positive integers has the numbers 1 and 2 within the same cycle?",
+  "response":"The total number of permutations is of course $n$ !. We will count instead the number of permutations for which 1 and 2 lie in different cycles.If the cycle that contains 1 has length $k$, we can choose the other $k-1$ elements in $\\left(\\begin{array}{c}n-2 \\\\ k-1\\end{array}\\right)$ ways from the set $\\{3,4, \\ldots, n\\}$. There exist $(k-1)$ ! circular permutations of these elements, and $(n-k)$ ! permutations of the remaining $n-k$ elements. Hence the total number of permutations for which 1 and 2 belong to different cycles is equal to$$ \\sum_{k=1}^{n-1}\\left(\\begin{array}{l} n-2 \\\\ k-1 \\end{array}\\right)(k-1) !(n-k) !=(n-2) ! \\sum_{k=1}^{n-1}(n-k)=(n-2) ! \\frac{n(n-1)}{2}=\\frac{n !}{2} . $$It follows that exactly half of all permutations contain 1 and 2 in different cycles, and so half contain 1 and 2 in the same cycle. The probability is $\\frac{1}{2}$."
+ },
+ {
+  "instruction":"Find$$ \\int_{0}^{1} \\frac{\\ln (1+x)}{1+x^{2}} d x .$$",
+  "response":"With the substitution $\\arctan x=t$ the integral takes the form$$ I=\\int_{0}^{\\frac{\\pi}{4}} \\ln (1+\\tan t) d t . $$This we already computed in the previous problem. (\"Happiness is longing for repetition,\" says M. Kundera.) So the answer to the problem is $\\frac{\\pi}{8} \\ln 2$."
+ },
+ {
+  "instruction":"Let $C$ be the unit circle $x^{2}+y^{2}=1$. A point $p$ is chosen randomly on the circumference of $C$ and another point $q$ is chosen randomly from the interior of $C$ (these points are chosen independently and uniformly over their domains). Let $R$ be the rectangle with sides parallel to the $x$-and $y$-axes with diagonal $p q$. What is the probability that no point of $R$ lies outside of $C$?",
+  "response":"The pair $(p, q)$ is chosen randomly from the three-dimensional domain $C \\times$ int $C$, which has a total volume of $2 \\pi^{2}$ (here int $C$ denotes the interior of $C$ ). For a fixed $p$, the locus of points $q$ for which $R$ does not have points outside of $C$ is the rectangle whose diagonal is the diameter through $p$ and whose sides are parallel to the coordinate axes (Figure 112). If the coordinates of $p$ are $(\\cos \\theta, \\sin \\theta)$, then the area of the rectangle is $2|\\sin 2 \\theta|$Figure 112The volume of the favorable region is therefore$$ V=\\int_{0}^{2 \\pi} 2|\\sin 2 \\theta| d \\theta=4 \\int_{0}^{\\pi \/ 2} 2 \\sin 2 \\theta d \\theta=8 . $$Hence the probability is$$ P=\\frac{8}{2 \\pi^{2}}=\\frac{4}{\\pi^{2}} \\approx 0.405 . $$"
+ },
+ {
+  "instruction":"Let $M$ be a subset of $\\{1,2,3, \\ldots, 15\\}$ such that the product of any three distinct elements of $M$ is not a square. Determine the maximum number of elements in $M$.",
+  "response":"Note that the product of the three elements in each of the sets $\\{1,4,9\\},\\{2,6,12\\}$, $\\{3,5,15\\}$, and $\\{7,8,14\\}$ is a square. Hence none of these sets is a subset of $M$. Because they are disjoint, it follows that $M$ has at most $15-4=11$ elements.Since 10 is not an element of the aforementioned sets, if $10 \\notin M$, then $M$ has at most 10 elements. Suppose $10 \\in M$. Then none of $\\{2,5\\},\\{6,15\\},\\{1,4,9\\}$, and $\\{7,8,14\\}$ is a subset of $M$. If $\\{3,12\\} \\not \\subset M$, it follows again that $M$ has at most 10 elements. If $\\{3,12\\} \\subset M$, then none of $\\{1\\},\\{4\\},\\{9\\},\\{2,6\\},\\{5,15\\}$, and $\\{7,8,14\\}$ is a subset of $M$, and then $M$ has at most 9 elements. We conclude that $M$ has at most 10 elements in any case.Finally, it is easy to verify that the subset$$ M=\\{1,4,5,6,7,10,11,12,13,14\\} $$has the desired property. Hence the maximum number of elements in $M$ is 10 ."
+ },
+ {
+  "instruction":"Compute the limit \n\n$$\n\\lim _{z \\rightarrow 0}\\left((\\tan z)^{-2}-z^{-2}\\right)\n$$\n\n where $z$ is a complex variable.\n\n",
+  "response":" As\n\n$$\n(\\tan z)^{-2}-z^{-2}=\\frac{z^{2}-(\\tan z)^{2}}{z^{2}(\\tan z)^{2}}\n$$\n\nthe Maclaurin expansion of the numerator has no terms of degree up to 3 , whereas the expansion of the denominator starts with $z^{4}$, therefore, the limit is finite.\n\nAs\n\n$$\n\\tan z=z+\\frac{1}{3} z^{3}+o\\left(z^{4}\\right) \\quad(z \\rightarrow 0)\n$$\n\nwe have\n\n$$\n(\\tan z)^{-2}-z^{-2}=\\frac{z^{2}-z^{2}-\\frac{2}{3} z^{4}+o\\left(z^{4}\\right)}{z^{4}+o\\left(z^{4}\\right)} \\quad(z \\rightarrow 0)\n$$\n\nso the limit at 0 is $-2 \/ 3$.\n\n"
+ },
+ {
+  "instruction":"Let $f(x)=\\frac{1}{4}+x-x^{2}$. For any real number $x$, define a sequence $\\left(x_{n}\\right)$ by $x_{0}=x$ and $x_{n+1}=f\\left(x_{n}\\right)$. If the sequence converges, let $x_{\\infty}$ denote the limit. For $x=0$, find $x_{\\infty}=\\lambda$.\n\n",
+  "response":"We have\n\n$$\nf(x)=\\frac{1}{2}-\\left(x-\\frac{1}{2}\\right)^{2}\n$$\n\nso $x_{n}$ is bounded by $1 \/ 2$ and, by the Induction Principle, nondecreasing. Let $\\lambda$ be its limit. Then\n\n$$\n\\lambda=\\frac{1}{2}-\\left(\\lambda-\\frac{1}{2}\\right)^{2}\n$$\n\nand, as the sequence takes only positive values,\n\n$$\n\\lambda=\\frac{1}{2} \\text {. }\n$$\n\n"
+ },
+ {
+  "instruction":"The clock-face is a disk of radius 1 . The hour-hand is a disk of radius $1 \/ 2$ is internally tangent to the circle of the clock-face, and the minute-hand is a line segment of length 1 . Find the area of the figure formed by all intersections of the hands in 12 hours (i.e., in one full turn of the hour-hand).",
+  "response":"Let the minute hand makes an angle $\\varphi \\in[0 ; 2 \\pi)$ with the vertical position. At that moment, the hour hand makes one of the angles  $$ \\psi_{k}=\\frac{\\varphi+2 \\pi k}{12}(\\bmod 2 \\pi), k \\in \\mathbb{Z}, $$  with the vertical position. The hands intersect along a segment of length  $$ \\max \\left(0, \\cos \\left(\\psi_{k}-\\varphi\\right)\\right), $$  hence the area in question equals $\\frac{1}{2} \\int_{0}^{2 \\pi} \\rho^{2}(\\varphi) d \\varphi$, where  $$ \\rho(\\varphi)=\\max \\left(0, \\max _{k} \\cos \\left(\\psi_{k}-\\varphi\\right)\\right), \\varphi \\in \\mathbb{R} . $$  Notice that  $$ \\rho(\\varphi)=\\cos \\left(\\min _{k}\\left|\\frac{\\varphi+2 \\pi k}{12}-\\varphi\\right|\\right)=\\cos \\left(\\min _{k}\\left|\\frac{11 \\varphi}{12}-\\frac{\\pi k}{6}\\right|\\right) . $$  This function is periodic with period of $\\frac{2 \\pi}{11}$. Since for $\\varphi \\in\\left[-\\frac{\\pi}{11}, \\frac{\\pi}{11}\\right]$ it holds $\\rho(\\varphi)=$ $\\cos \\left(\\frac{11 \\varphi}{12}\\right)$, the area of the figure formed by intersections of hands equals  $$ \\begin{gathered} \\frac{1}{2} \\int_{0}^{2 \\pi} \\rho^{2}(\\varphi) d \\varphi=\\frac{11}{2} \\int_{-\\frac{\\pi}{11}}^{\\frac{\\pi}{11}} \\rho^{2}(\\varphi) d \\varphi=\\frac{11}{2} \\int_{-\\frac{\\pi}{11}}^{\\frac{\\pi}{11}} \\cos ^{2}\\left(\\frac{11 \\varphi}{12}\\right) d \\varphi= \\\\ =\\frac{11}{4} \\int_{-\\frac{\\pi}{11}}^{\\frac{\\pi}{11}}\\left(1+\\cos \\left(\\frac{11 \\varphi}{6}\\right)\\right) d \\varphi=\\left.\\left(\\frac{11 \\varphi}{4}+\\frac{3}{2} \\sin \\left(\\frac{11 \\varphi}{6}\\right)\\right)\\right|_{-\\frac{\\pi}{11}} ^{\\frac{\\pi}{11}}=\\frac{\\pi+3}{2} . \\end{gathered} $$  Answer: $\\frac{\\pi+3}{2}"
+ },
+ {
+  "instruction":"Evaluate\n\n$$\n\\iint_{\\mathcal{A}} e^{-x^{2}-y^{2}} d x d y,\n$$\n\nwhere $\\mathcal{A}=\\left\\{(x, y) \\in \\mathbb{R}^{2} \\mid x^{2}+y^{2} \\leqslant 1\\right\\}$.\n\n",
+  "response":" Using polar coordinates, we have\n\n$$\n\\begin{aligned}\n\\iint_{\\mathcal{A}} e^{-x^{2}-y^{2}} d x d y &=\\int_{0}^{2 \\pi} \\int_{0}^{1} \\rho e^{-\\rho^{2}} d \\rho d \\theta \\\\n&=-\\frac{1}{2} \\int_{0}^{2 \\pi} \\int_{0}^{1}-2 \\rho e^{-\\rho^{2}} d \\rho d \\theta \\\\n&=-\\frac{1}{2} \\int_{0}^{2 \\pi}\\left(e^{-1}-1\\right) \\\\n&=\\pi\\left(e^{-1}-1\\right) .\n\\end{aligned}\n$$\n\n"
+ },
+ {
+  "instruction":"Let $S_{9}$ denote the group of permutations of nine objects and let $A_{9}$ be the subgroup consisting of all even permutations. Denote by $1 \\in S_{9}$ the identity permutation. Determine the minimum of all positive integers $m$ such that every $\\sigma \\in A_{9}$ satisfies $\\sigma^{m}=1$.\n\n",
+  "response":" The order of a $k$-cycle is $k$, so the smallest $m$ which simultaneously annihilates all 9-cycles, 8-cycles, 7-cycles, and 5-cycles is $2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7= 2520$. Any $n$-cycle, $n \\leqslant 9$, raised to this power is annihilated, so $n=2520$.\n\nTo compute $n$ for $A 9$, note that an 8-cycle is an odd permutation, so no 8-cycles are in $A_{9}$. Therefore, $n$ need only annihilate 4-cycles (since a 4-cycle times a transposition is in A9), 9-cycles, 7-cycles, and 5-cycles. Thus, $n=2520 \/ 2= 1260$.\n\n"
+ },
+ {
+  "instruction":"A bag contains 1993 red balls and 1993 black balls. We remove two uniformly random balls at a time repeatedly and\n\n(i) discard them if they are of the same color,\n\n(ii) discard the black ball and return to the bag the red ball if they are of different colors.\n\nThe process ends when there are less than two balls in the bag. What is the probability that this process will terminate with one red ball in the bag?",
+  "response":"First, observe that since at least one ball is removed during each stage, the process will eventually terminate, leaving no ball or one ball in the bag. Because red balls are removed 2 at a time and since we start with an odd number of red balls, the number of red balls in the bag at any time is odd. Hence the process will always leave red balls in the bag, and so it must terminate with exactly one red ball. The probability we are computing is therefore 1 ."
+ },
+ {
+  "instruction":"Let $T_{1}, T_{2}, T_{3}$ be points on a parabola, and $t_{1}, t_{2}, t_{3}$ the tangents to the parabola at these points. Compute the ratio of the area of triangle $T_{1} T_{2} T_{3}$ to the area of the triangle determined by the tangents.",
+  "response":"Choose a Cartesian system of coordinates such that the equation of the parabola is $y^{2}=4 p x$. The coordinates of the three points are $T_{i}\\left(4 p \\alpha_{i}^{2}, 4 p \\alpha_{i}\\right)$, for appropriately chosen $\\alpha_{i}, i=1,2,3$. Recall that the equation of the tangent to the parabola at a point $\\left(x_{0}, y_{0}\\right)$ is $y y_{0}=2 p\\left(x+x_{0}\\right)$. In our situation the three tangents are given by$$ 2 \\alpha_{i} y=x+4 p \\alpha_{i}^{2}, \\quad i=1,2,3 . $$If $P_{i j}$ is the intersection of $t_{i}$ and $t_{j}$, then its coordinates are $\\left(4 p \\alpha_{i} \\alpha_{j}, 2 p\\left(\\alpha_{i}+\\alpha_{j}\\right)\\right)$. The area of triangle $T_{1} T_{2} T_{3}$ is given by a Vandermonde determinant: $$ \\pm \\frac{1}{2}\\left|\\begin{array}{lll} 4 p \\alpha_{1}^{2} & 4 p \\alpha_{1} & 1 \\\\ 4 p \\alpha_{2}^{2} & 4 p \\alpha_{2} & 1 \\\\ 4 p \\alpha_{3}^{2} & 4 p \\alpha_{3} & 1 \\end{array}\\right|=\\pm 8 p^{2}\\left|\\begin{array}{lll} \\alpha_{1}^{2} & \\alpha_{1} & 1 \\\\ \\alpha_{2}^{2} & \\alpha_{2} & 1 \\\\ \\alpha_{3}^{2} & \\alpha_{3} & 1 \\end{array}\\right|=8 p^{2}\\left|\\left(\\alpha_{1}-\\alpha_{2}\\right)\\left(\\alpha_{1}-\\alpha_{3}\\right)\\left(\\alpha_{2}-\\alpha_{3}\\right)\\right| . $$The area of the triangle $P_{12} P_{23} P_{31}$ is given by$$ \\begin{aligned} & \\pm \\frac{1}{2}\\left|\\begin{array}{lll}4 p \\alpha_{1} \\alpha_{2} & 2 p\\left(\\alpha_{1}+\\alpha_{2}\\right) & 1 \\\\4 p \\alpha_{2} \\alpha_{3} & 2 p\\left(\\alpha_{2}+\\alpha_{3}\\right) & 1 \\\\4 p \\alpha_{3} \\alpha_{1} & 2 p\\left(\\alpha_{3}+\\alpha_{1}\\right) & 1\\end{array}\\right| \\\\ & =\\pm 4 p^{2}\\left|\\begin{array}{ll}\\alpha_{1} \\alpha_{2}\\left(\\alpha_{1}+\\alpha_{2}\\right) & 1 \\\\\\alpha_{2} \\alpha_{3}\\left(\\alpha_{2}+\\alpha_{3}\\right) & 1 \\\\\\alpha_{3} \\alpha_{1}\\left(\\alpha_{3}+\\alpha_{1}\\right) & 1\\end{array}\\right|=\\pm 4 p^{2}\\left|\\begin{array}{rr}\\left(\\alpha_{1}-\\alpha_{3}\\right) \\alpha_{2}\\left(\\alpha_{1}-\\alpha_{3}\\right) & 0 \\\\\\left(\\alpha_{2}-\\alpha_{1}\\right) \\alpha_{3}\\left(\\alpha_{2}-\\alpha_{1}\\right) & 0 \\\\\\alpha_{3} \\alpha_{1}\\left(\\alpha_{3}+\\alpha_{1}\\right) & 1\\end{array}\\right| \\\\ & =4 p^{2}\\left|\\left(\\alpha_{1}-\\alpha_{3}\\right)\\left(\\alpha_{1}-\\alpha_{2}\\right)\\left(\\alpha_{2}-\\alpha_{3}\\right)\\right| \\text {. } \\end{aligned} $$We conclude that the ratio of the two areas is 2 , regardless of the location of the three points or the shape of the parabola."
+ },
+ {
+  "instruction":"For each continuous function $f:[0,1] \\rightarrow \\mathbb{R}$, we define $I(f)=\\int_{0}^{1} x^{2} f(x) d x$ and $J(f)=\\int_{0}^{1} x(f(x))^{2} d x$. Find the maximum value of $I(f)-J(f)$ over all such functions $f$.",
+  "response":"We change this into a minimum problem, and then relate the latter to an inequality of the form $x \\geq 0$. Completing the square, we see that$$ \\left.x(f(x))^{2}-x^{2} f(x)=\\sqrt{x} f(x)\\right)^{2}-2 \\sqrt{x} f(x) \\frac{x^{\\frac{3}{2}}}{2}=\\left(\\sqrt{x} f(x)-\\frac{x^{\\frac{3}{2}}}{2}\\right)^{2}-\\frac{x^{3}}{4} $$Hence, indeed,$$ J(f)-I(f)=\\int_{0}^{1}\\left(\\sqrt{x} f(x)-\\frac{x^{\\frac{3}{2}}}{2}\\right)^{2} d x-\\int_{0}^{1} \\frac{x^{3}}{4} d x \\geq-\\frac{1}{16} $$It follows that $I(f)-J(f) \\leq \\frac{1}{16}$ for all $f$. The equality holds, for example, for $f:[0,1] \\rightarrow \\mathbb{R}, f(x)=\\frac{x}{2}$. We conclude that$$ \\max _{f \\in \\mathcal{C}^{0}([0,1])}(I(f)-J(f))=\\frac{1}{16} . $$"
+ },
+ {
+  "instruction":"Compute the integral$$ \\iint_{D} \\frac{d x d y}{\\left(x^{2}+y^{2}\\right)^{2}}, $$where $D$ is the domain bounded by the circles$$ \\begin{array}{ll} x^{2}+y^{2}-2 x=0, & x^{2}+y^{2}-4 x=0, \\\\ x^{2}+y^{2}-2 y=0, & x^{2}+y^{2}-6 y=0 . \\end{array} $$",
+  "response":"The domain $D$ is depicted in Figure 71 . We transform it into the rectangle $D_{1}=$ $\\left[\\frac{1}{4}, \\frac{1}{2}\\right] \\times\\left[\\frac{1}{6}, \\frac{1}{2}\\right]$ by the change of coordinates$$ x=\\frac{u}{u^{2}+v^{2}}, \\quad y=\\frac{v}{u^{2}+v^{2}} . $$The Jacobian is Figure 71$$ J=-\\frac{1}{\\left(u^{2}+v^{2}\\right)^{2}} . $$Therefore,$$ \\iint_{D} \\frac{d x d y}{\\left(x^{2}+y^{2}\\right)^{2}}=\\iint_{D_{1}} d u d v=\\frac{1}{12} . $$"
+ },
+ {
+  "instruction":"Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$. Find the probability that $\\sqrt{2+\\sqrt{3}} \\leq|v+w|$.",
+  "response":"Because the 1997 roots of the equation are symmetrically distributed in the complex plane, there is no loss of generality to assume that $v=1$. We are required to find the probability that$$ |1+w|^{2}=|(1+\\cos \\theta)+i \\sin \\theta|^{2}=2+2 \\cos \\theta \\geq 2+\\sqrt{3} . $$This is equivalent to $\\cos \\theta \\geq \\frac{1}{2} \\sqrt{3}$, or $|\\theta| \\leq \\frac{\\pi}{6}$. Because $w \\neq 1, \\theta$ is of the form $\\pm \\frac{2 k \\pi}{1997} k$, $1 \\leq k \\leq\\left\\lfloor\\frac{1997}{12}\\right\\rfloor$. There are $2 \\cdot 166=332$ such angles, and hence the probability is $\\frac{332}{1996}=\\frac{83}{499} \\approx 0.166$"
+ },
+ {
+  "instruction":"The numbers $1,2,3,4,5,6,7$, and 8 are written on the faces of a regular octahedron so that each face contains a different number. Find the probability that no two consecutive numbers are written on faces that share an edge, where 8 and 1 are considered consecutive.",
+  "response":"Consider the dual cube to the octahedron. The vertices $A, B, C, D, E, F, G$, $H$ of this cube are the centers of the faces of the octahedron (here $A B C D$ is a face of the cube and $(A, G),(B, H),(C, E),(D, F)$ are pairs of diagonally opposite vertices). Each assignment of the numbers $1,2,3,4,5,6,7$, and 8 to the faces of the octahedron corresponds to a permutation of $A B C D E F G H$, and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. The problem requires us to count octagonal circuits that can be formed by eight of these diagonals.Six of these diagonals are edges of the tetrahedron $A C F H$, six are edges of the tetrahedron $D B E G$, and four are long diagonals, joining opposite vertices of the cube. Notice that each vertex belongs to exactly one long diagonal. It follows that an octagonal circuit must contain either 2 long diagonals separated by 3 tetrahedron edges (Figure 108a), or 4 long diagonals (Figure 108b) alternating with tetrahedron edges.\nFigure 108When forming a (skew) octagon with 4 long diagonals, the four tetrahedron edges need to be disjoint; hence two are opposite edges of $A C F H$ and two are opposite edges of $D B E G$. For each of the three ways to choose a pair of opposite edges from the tetrahedron $A C F H$, there are two possible ways to choose a pair of opposite edges from tetrahedron $D B E G$. There are $3 \\cdot 2=6$ octagons of this type, and for each of them, a circuit can start at 8 possible vertices and can be traced in two different ways, making a total of $6 \\cdot 8 \\cdot 2=96$ permutations.An octagon that contains exactly two long diagonals must also contain a three-edge path along the tetrahedron $A C F H$ and a three-edge path along tetrahedron the $D B E G$. A three-edge path along the tetrahedron the $A C F H$ can be chosen in $4 !=24$ ways. The corresponding three-edge path along the tetrahedron $D B E G$ has predetermined initial and terminal vertices; it thus can be chosen in only 2 ways. Since this counting method treats each path as different from its reverse, there are $8 \\cdot 24 \\cdot 2=384$ permutations of this type.In all, there are $96+384=480$ permutations that correspond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is $\\frac{480}{8 !}=\\frac{1}{84}$."
+ },
+ {
+  "instruction":"Let $V$ be the vector space of all real $3 \\times 3$ matrices and let $A$ be the diagonal matrix\n\n$$\n\\left(\\begin{array}{lll}\n1 & 0 & 0 \\\\\n0 & 2 & 0 \\\\\n0 & 0 & 1\n\\end{array}\\right) .\n$$\n\nCalculate the determinant of the linear transformation $T$ on $V$ defined by $T(X)=\\frac{1}{2}(A X+X A)$.\n\n",
+  "response":" Let $X=\\left(x_{i j}\\right)$ be any element of $M_{3}(\\mathbb{R})$. A calculation gives\n\n$$\nT(X)=\\left(\\begin{array}{ccc}\nx_{11} & 3 x_{12} \/ 2 & x_{13} \\\\n3 x_{21} \/ 2 & 2 x_{22} & 3 x_{23} \/ 2 \\\\nx_{31} & 3 x_{32} \/ 2 & x_{33}\n\\end{array}\\right) .\n$$\n\nIt follows that the basis matrices $M_{i j}$ are eigenvectors of $T$. Taking the product of their associated eigenvalues, we get $\\operatorname{det} T=2(3 \/ 2)^{4}=81 \/ 8$.\n\n"
+ },
+ {
+  "instruction":"An old woman went to the market and a horse stepped on her basket and smashed her eggs. The rider offered to pay for the eggs and asked her how many there were. She did not remember the exact number, but when she had taken them two at a time there was one egg left, and the same happened when she took three, four, five, and six at a time. But when she took them seven at a time, they came out even. What is the smallest number of eggs she could have had? ",
+  "response":"We are to find the smallest positive solution to the system of congruences$$ \\begin{aligned} &x \\equiv 1(\\bmod 60), \\\\ &x \\equiv 0(\\bmod 7) . \\end{aligned} $$The general solution is $7 b_{1}+420 t$, where $b_{1}$ is the inverse of 7 modulo 60 and $t$ is an integer. Since $b_{1}$ is a solution to the Diophantine equation $7 b_{1}+60 y=1$, we find it using Euclid's algorithm. Here is how to do it: $60=8 \\cdot 7+4,7=1 \\cdot 4+3,4=1 \\cdot 3+1$. Then$$ \\begin{aligned} 1 &=4-1 \\cdot 3=4-1 \\cdot(7-1 \\cdot 4)=2 \\cdot 4-7=2 \\cdot(60-8 \\cdot 7)-7 \\\\ &=2 \\cdot 60-17 \\cdot 7 \\end{aligned} $$Hence $b_{1}=-17$, and the smallest positive number of the form $7 b_{1}+420 t$ is $-7 \\cdot 17+$ $420 \\cdot 1=301$."
+ }
+]