May 202018

2018 年IMO 国家队队员李一笑——来自江苏天一中学——的大作 “2018 年国家集训队第一阶段选拔试题及解答”. 文档转载自数学新星网.

2018 China IMO team selection test part one

2018 年国家集训队第二阶段选拔试题来自贴吧

2018 China IMO team selection test

2018 China IMO team selection test part two 1

2018 China IMO team selection test

2018 China IMO team selection test part two 2

2018 China IMO team selection test

2018 China IMO team selection test part two 3

2018 China IMO team selection test

2018 China IMO team selection test part two 4

 Posted by at 8:41 am
May 112018

Theorem. There does not exist a group whose commutator subgroup is isomorphic to \(S_4\).

The relevant facts are that \(S_4\) is a complete group(no outer automorphisms, trivial center) which is not perfect(that is, the commutator subgroup of \(S_4\) is not \(S_4\) itself). Any group which has these properties is never a commutator subgroup of anything. Here’s why.

Lemma. If \(K\) is a complete group and \(K\lhd G\), then \(G\) is the direct product \(K\times H\) of \(K\) by its centralizer \(H=C_G(K)\).

In other words, a complete group can be a normal subgroup only in the most trivial fashion: the large group is just a direct product of the normal group by something.

Proof of the lemma. Let \(H=C_G(K)\) be the centralizer of \(K\) in \(G\), namely the set of all elements which commute with all elements of \(K\). \(H\) is a normal subgroup of \(G\), and \(H\cap K=Z(K)=1\) since \(K\) has trivial center. Any element \(g\in G\) induces an automorphisms \(\phi_g\) of \(K\) by conjugation: \(\phi_g(k)=g^{-1}kg\). But \(K\) has no outer automorphisms, so \(\phi_g\) must equal some inner automorphism of \(K\), that is, for some \(k\in K\), \(\phi_g=\phi_k\). Now conjugation by \(gk^{-1}\) does nothing to \(K\), so \(gk^{-1}=h\in H\). In other words \(g=kh\): every element of \(G\) is expressible as product of an element of \(K\) and an element of \(H\). Since \(H\) and \(K\) commute, \(G\) is the direct product of \(H\) and \(K\).            \(\Box\)

Proof. Now suppose that \(G\) is some group such that \(K=G^\prime=[G,G]\), the commutator subgroup, is such that \(K\) is complete and non-perfect. By the lamma, \(G=K\times A\) where \(A\cong G/K\) is an abelian group. So any element of \(G\) can be writeen as \(ka\) with \(k\in K\) and \(a\in A\), and moreover, \(ka=ak\) for any \(k\in K\), \(a\in A\).

Consider a commutator \(c=xyx^{-1}y^{-1}\) in \(G\). Write \(x=ka\) and \(y=lb\). Since \(K\) and \(A\) commute, \(c=aba^{-1}b^{-1}xyx^{-1}y^{-1}\). Since \(A\) is abelian, the first part vanishes and \(c=xyx^{-1}y^{-1}\). So any commutator of \(G\) lies in the commutator sungroup of \(K\), and it follows that \(G^\prime=K^\prime\). Since \(K^\prime\ne K\), \(G^\prime\ne K\), as well.

It remains to show that \(S_4\) is complete and non-perfect.

(i) \(S_4\) has trivial center: this is obvious. No permutation commutes with all other permutations.

(ii) \(S_4\) has no outer automorphisms. This is true for all \(S_n\) except \(n=2\), \(6\). It’s a standard result.

(iii) \(S_4\) is not perfect. This is also ovious: for any two permutations \(\sigma\), \(\tau\in S_n\), the commutator \(\sigma^{-1}\tau^{-1}\sigma\tau\) is an even permutation, so the commutator subgroup is contained in the alternating group \(A_n\). In fact the commutator group eauals \(A_n\), but we don’t need that here.

This completes the proof the theorem.                                  \(\Box\)

Remark. a simple modification: \(K^\prime\subset G^\prime\) is clear, and the other direction follows since \(G/K^\prime=K/{K^\prime\times A}\) is abelian.

Here is a proof not using those well-known facts about \(S_4\). (though it’s easy to derive them with it)

Proof. \(S_4\) has exactly \(4\) Sylow  \(3\)-subgroups

\[P_1=\langle(234)\rangle, P_2=\langle(134)\rangle, P_3=\langle(124)\rangle, P_4=\langle(123)\rangle,\]

where \(P_i\) is the only Sylow \(3\)-subgroup of the stabilizer of \(i\) in \(S_4\) for \(i=1\), \(2\), \(3\), \(4\). So \(\sigma P_i\sigma^{-1}=P_{\sigma(i)}\) for all \(\sigma\in S_4\) and \(i=1\), \(2\), \(3\), \(4\).

Assume \(G^\prime=S_4\) and take \(g\in G\). The conjugation with \(g\) permutes \(P_1\), \(P_2\), \(P_3\), \(P_4\), so we find \(\rho\in G\) with \(g P_ig^{-1}=P_{\rho(i)}=\rho P_i\rho^{-1}\) for \(i=1\), \(2\), \(3\), \(4\). So \(h\colon \rho^{-1}g\in G\) satisfies \(h P_ih^{-1}=P_i\) for \(i=1\), \(2\), \(3\), \(4\).

Then for all \(\sigma\in S_4\) we have also \(h^{-1}\sigma h\in S_4\) and

\begin{equation} \begin{split}P_{h^{-1}\sigma h(i)}&=(h^{-1}\sigma h)P_i(h^{-1}\sigma h)^{-1}\\&=h^{-1}\sigma hP_ih^{-1}\sigma^{-1}h\\&=h^{-1}\sigma P_i\sigma^{-1}h\\&= h^{-1} P_{\sigma(i)}h=P_{\sigma(i)}.\end{split} \end{equation}

and therefore \(h^{-1}\sigma h(i)=\sigma(i)\) for \(i=1\), \(2\), \(3\), \(4\).

This gives \(h\in C_G(S_4)\) and \(g=\rho h\in S_4C_G(S_4)\) for all \(g\in G\). So \(G=S_4C_G(S_4)\) and \(|G\colon A_4C_G(S_4)|\leqslant2\). But then \(A_4C_G(S_4)\trianglelefteq G\) with abelian factor, so \(S_4=G^\prime\leq A_4C_G(S_4)\), and by Dedekind we get

\[S_4=A_4C_G(S_4)\cap S_4=A_4(C_G(S_4)\cap S_4)=A_4Z(S_4)=A_4\]

since \(Z(S_4)=1\), as \(\sigma\in Z(S_4)\) would give \(P_i=\sigma P_i\sigma^{-1}=P_{\sigma(i)}\) and \(i=\sigma(i)\) for \(i=1\), \(2\), \(3\), \(4\). Contradiction!

 Posted by at 11:43 am
Feb 122018

北京时间 2017 年 12 月 24 日上午举行的硕士研究生初试的数学分析

下面的试题, 除了第 3 与第 5 题与实际考卷稍有出入, 其余的题与考场上卷子的用词句子甚至排版都是完全一模一样的!

1. 证明如下极限:

(1)  \(\lim\limits_{n\to\infty}\Big(1+\int_0^1\dfrac{\sin^n x}{x^n}\;dx\Big)^n=+\infty\);
(2)  \(\lim\limits_{n\to\infty}\Big(\int_0^1\dfrac{\sin x^n}{x^n}\;dx\Big)^n=\prod\limits_{k=1}^{+\infty}e^{\frac{(-1)^k}{2k(2k+1)!}}\);
(3) \(\lim\limits_{n\to\infty}\dfrac1n\sum\limits_{k=1}^n\ln\Big(1+\dfrac{k^2-k}{n^2}\Big)=\ln 2-2+\dfrac\pi2\).

2. \(f\in C(0,1)\), \(\dfrac{f(x_2)-f(x_1)}{x_2-x_1}=\alpha\lt\beta=\dfrac{f(x_4)-f(x_3)}{x_4-x_3}\), 这里 \(x_1\), \(x_2\), \(x_3\), \(x_4\in(0, 1)\). 证明: 对任意 \(\lambda\in(\alpha, \beta)\), 存在 \(x_5\), \(x_6\in(0, 1)\), 使得 \(\lambda=\dfrac{f(x_6)-f(x_5)}{x_6-x_5}\).

3. 设 \(\gamma\) 是联结 \(\Bbb R^3\) 中两点 \(A\), \(B\) 的长度为 \(L\) 的光滑曲线, \(U\) 是包含 \(\gamma\) 的 \(\Bbb R^3\) 中的开集, \(f\) 在 \(U\) 中的两个偏导数存在且在 \(\gamma\) 上连续. 梯度 \(\nabla f\) 的长度在 \(\gamma\) 上的界为 \(M\). 证明:

\[|f(A)-f(B)|\leqslant ML.\]

4. \(f\) 在 \((0, 0)\) 点局部三阶连续可微, \(D_R\) 表示圆盘: \(x^2+y^2\leqslant R^2\). 计算:

\[\lim_{R\to0^+}\dfrac1{R^4}\iint_{D_R}\Big(f(x, y)-f(0, 0)\Big) dxdy.\]

5. \(\varphi(x)\) 在 \(0\) 处可导, \(\varphi(0)=0\), \(f(x,y)\) 在 \((0,0)\) 点局部 \(2\) 阶连续可微, \(f\) 的两个偏导数在 \((x,\varphi(x) )\) 上恒为 \(0\). \(\big(\partial_{ij}f(0, 0)\big)_{2\times2}\) 为半正定非 \(0\) 阵. 证明 \(f\) 在 \((0, 0)\) 为取极小值.

6. 证明: \(e^{-x}+\cos2x+x\sin x=0\)
在区间 \(\big((2n-1)\pi, (2n+1)\pi\big)\) 恰有两个根 \(x_{2n-1}\lt x_{2n}\),
\(\forall n=1\), \(2\), \(3\), \(\dotsc\).
证明如下极限存在并求之: \(\lim\limits_{n\to\infty}(-1)^{n-1}n(x_n-n\pi)\).

7. 证明: \(\lim\limits_{x\to0}\sum\limits_{n=1}^\infty\dfrac{\cos nx}n=+\infty\).

8. \(\forall x\in[1,+\infty)\), \(f(x)\gt0\), \(f^{\prime\prime}(x)\leqslant0\), 且 \(\lim\limits_{x\to+\infty}f(x)=+\infty\). 证明如下极限存在并求之:


Nov 162017

不是完整的试题解答, 仅仅在关隘的地点聊一聊.

付云皓的题 3 的解

记 \(a=[nq^{\frac13}]\), \(b=[nq^{\frac23}]\), \(c=nq\). 然后


最后的不等式是因为 \(a^3q^2+b^3q+c^3\gt 3abcq\), 并且 \(c\geqslant aq^{\frac23}\), \(c\geqslant bq^{\frac13}\).

然后, 因为 \(c-aq^{\frac23}\geqslant0\), \(c- bq^{\frac13}\geqslant0\), 以及 \(aq^{\frac23}-bq^{\frac13}=-\Big((c-aq^{\frac23})-(c-bq^{\frac13})\Big)\), 得到



现在, \((1)\) 给出


记得 \(c=nq\), 这也就是

\begin{equation}\Big(q^{\frac13}\cdot \{nq^{\frac23}\}+q^{\frac23}\cdot \{nq^{\frac13}\}\Big)^2\geqslant\frac1{3nq}.\end{equation}


\begin{equation}\Big\{nq^{\frac23}\Big\}+ \Big\{nq^{\frac13}\Big\}\geqslant\frac1{q\sqrt{3qn}}.\end{equation}

然后是邓煜给出的第三题的答案, 从知乎转来.

 Posted by at 8:39 pm  Tagged with:
Nov 162017

第 33 届中国数学奥林匹克

浙江 杭州


(2017 年 11 月 15 日    8:00–12:30)

1. 设 \(A_n\) 是满足以下条件的素数 \(p\) 的集合: \(\exists a\), \(b\in\Bbb N^+\), 使得 \(\dfrac{a+b}p\), \(\dfrac{a^2+b^2}{p^2}\) 都是正整数, 且

\[\Big(\frac{a+b}p, p\Big)=\Big(\frac{a^2+b^2}{p^2}, p\Big)=1.\]

证明: (1) \(A_n\) 为有限集当且仅当 \(n\ne2\);

(2) 记 \(f(n)=|A_n|\). 若 \(k\), \(m\) 为正奇数, \(d=(k, m)\), 则

\[f(d)\leqslant  f(k)+f(m)-f(km)\leqslant 2f(d).\]

2. 设

\[T=\{(x, y, z)|1\leqslant x, y, z\leqslant n\}\]

为空间中 \(n^3\) 个点. 将其中 \((3n^2-3n+1)+k\) 个点染为红色, 且若 \(P\), \(Q\) 为红色, \(PQ\) 平行于任一条坐标轴, 则线段 \(PQ\) 上的所有整点均为红色. 求证: 至少有 \(k\) 个边长为 \(1\) 的立方体的所有顶点均为红色.

3. 设 \(n\), \(q\) 为正整数, \(q\) 不是完全立方数. 求证: 存在正实数 \(c\) 满足
\[\{nq^{\frac13}\}+\{nq^{\frac23}\}\geqslant \frac c{\sqrt n}\]

对所有正整数 \(n\) 成立, 其中 \(\{\cdot\}\) 表示其小数部分.

第 33 届中国数学奥林匹克

浙江 杭州


(2017 年 11 月 16 日    8:00–12:30)

4. 已知圆内接四边形 \(ABCD\), 其对角线 \(AC\) 与 \(BD\) 交于 \(P\) 点, \(\triangle ADP\) 的外接圆交 \(AB\) 于 \(E\), \(\triangle BCP\) 的外接圆交 \(AB\) 于 \(F\). \(\triangle ADE\) 与 \(\triangle BCF\) 的内心分别为 \(I\), \(J\), 直线 \(IJ\) 交 \(AC\) 于 \(K\).
求证: \(A\), \(I\), \(K\) , \(E\) 四点共圆.

cmo 2017

cmo 2017 p4

5. 对 \(n\times n\) 的方格进行黑白染色, 若两个方格 \(a\), \(b\) 有公共顶点且同色, 则称 \(a\), \(b\) 这两个方格相邻. 若 \(a\), \(b\) 能通过一系列的方格 \(c_1\to c_2\to\dotsb\to c_k\), 其中 \(c_1=a\), \(c_2=b\), 且 \(c_i\), \(c_{i+1}\) 相邻, 则称 \(a\), \(b\) 连通. 求最大正整数 \(M\), 使得存在 \(M\) 个方格, 使得其两两不连通.

6. 给定正整数 \(n\), \(k\), \(n\gt k\), 其中 \(a_i\in (k-1, k)\), \(1\leqslant i\leqslant n\). 若正实数 \(x_1\), \(x_2\), \(\dotsc\),\(x_n\) 满足: 对任意集合 \(I\subset\{1,2,\dotsc,n\}\), \(|I|=k\) 有\(\sum\limits_{i\in I}x_i\leqslant \sum\limits_{i\in I}a_i\), 试求 \(\prod\limits_{i=1}^nx_i\) 的最大值.

 Posted by at 8:05 pm  Tagged with:
Oct 122017

齐次多项式(Homogeneous polynomial)在数学中有其特殊的重要性.

在代数几何, Homogeneous polynomial 尤其受到偏爱.

实数域上的的 \(n\) 元多项式环, 以 \(\Bbb R[x_1, x_2,\dotsc, x_n]\) 表之.

Hilbert 限制在齐次多项式.

定义 5.1 设 \(p\in \Bbb R[x_1, x_2,\dotsc, x_n]\), 其次数 \(\leqslant d\). 把 \(n+1\) 元 \(d\) 次齐次多项式

\begin{equation}\overline{p}(x_0, x_1,\dotsc, x_n)=x_0^dp\Big(\frac{x_1}{x_0}, \frac{x_2}{x_0},\dotsc, \frac{x_n}{x_0}\Big)\end{equation}

称为是 \(p\) 的齐次化(Homogenization). 具体来说, 当  \(p=\sum cx_1^{d_1}x_2^{d_2}\dotsm x_n^{d_n}\), 那么

\begin{equation}\begin{split}\overline{p}(x_0, x_1,\dotsc, x_n )&=x_0^d\sum c\Big(\frac{x_1}{x_0}\Big)^{d_1}\Big(\frac{x_2}{x_0}\Big)^{d_2} \dotsm \Big(\frac{x_n}{x_0}\Big)^{d_n}\\&=\sum cx_0^{d-d_1-d_2-\dotsb-d_n}x_1^{d_1}x_2^{d_2}\dotsm x_n^{d_n} \\&=\sum cx_0^{d_0}x_1^{d_1}x_2^{d_2}\dotsm x_n^{d_n},\end{split}\end{equation}

这里 \(d_0=d-d_1-d_2-\dotsb-d_n\).

定理 5.2  设 \(p\in \Bbb R[x_1, x_2,\dotsc, x_n]\), 其次数 \(\leqslant d\). 如果 \(d\) 为偶数, 那么

  • \(p\) 非负当且仅当 \(\overline{p}\)  非负;
  • \(p\) 是多项式的平方和当且仅当 \(\overline{p}\)  能表成 \(\frac d2\) 次齐次多项式的平方和.

引理 5.3  假定 \(p\), \(p_1\), \(p_2\), \(\dotsc\), \(p_k\in \Bbb R[x_1, x_2,\dotsc, x_n]\) 都是多项式, \(p=p_1^2+p_2^2+\dotsm+p_k^2\).  如果 \(p_1\), \(p_2\), \(\dotsc\), \(p_k\) 不全是零多项式, 那么

  1. \(p\ne0\);
  2. \(\deg(p)=2\max\{\deg(p_l)|l=1, 2, \dotsc, k\}\);
  3. 如果 \(p\) 是 \(d\) 次齐次多项式, 则诸 \(p_l\) 皆是 \(\dfrac d2\) 次齐次多项式.

Proof   不妨 \(p_1\ne0\). 于是, 存在 \(x\in\Bbb R^n\), 使得 \(p_1(x)\ne0\). 然后


蕴涵 \(p(x)\ne0\).

写 \(p_l\) 为 \(p_l=p_{l0}+p_{l1}+p_{l2}+\dotsb+p_{ld}\), 这里 \(p_{li}\) 是 \(p_l\) 的 \(i\) 次齐次成分, \(d=\max\{\deg(p_l)|l=1, 2, \dotsc, k\}\). 很明显, \(\deg(p)\leqslant2d\); 1 表明 \(p\) 的 \(2d\) 次齐次成分 \(p^2_{1d}+p^2_{2d}+\dotsb+p^2_{kd}\ne0\), 因为有某 \(l\) 使得 \(p_{ld}\ne0\).

第 3 部分的证明与 2 完全类似, 考虑诸 \(p_l\) 的最低次齐次成分即可.  \(\Box\)

Proof   当 \(p\) 非负之时, 要证明 \(\overline{p}\) 非负, 若 \(x_0\ne0\), 由 \((1)\) 式即可; 若 \(x_0=0\), 由

\begin{equation}\overline{p}(0, x_1,\dotsc, x_n)=\lim_{h\to0}\overline{p}(h, x_1,\dotsc, x_n)\end{equation}


当 \(\overline{p}\) 非负之时, 只要注意

\begin{equation}p(x_1,\dotsc, x_n)=\overline{p}(1, x_1,\dotsc, x_n)\end{equation}

即知 \(p\) 非负.

如果\(p\) 是多项式的平方和, \(p=\sum\limits_{l=1}^kp_l^2\), 那么依据引理 5.3, \(\deg(p_l)\leqslant\dfrac d2\). 然后

\begin{equation}\overline{p}=\sum_{l=1}^k\Bigg(x_0^{\frac d2}p_l\Big(\frac{x_1}{x_0}, \frac{x_2}{x_0},\dotsc, \frac{x_n}{x_0} \Big)\Bigg)^2 \end{equation}

说明  \(\overline{p}\)  能表成 \(\frac d2\) 次齐次多项式的平方和.

如果  \(\overline{p}\)  能表成多项式的平方和, \(\overline{p}=\sum\limits_{l=1}^kh_l^2\),

\begin{equation}p=\overline{p}(1, x_1,\dotsc, x_n)=\sum_{l=1}^k\Big(h_l(1, x_1,\dotsc, x_n)\Big)^2\end{equation}

说明 \(p\)  能表成多项式的平方和.       \(\Box\)

现在, 很容易的, 我们顺便建立二次多项式非负与平方和的联系.

定理 5.4 设二次多项式 \(p\in \Bbb R[x_1, x_2,\dotsc, x_n]\) 非负, 那么 \(p\) 能写成多项式的平方和.

有了定理 5.2, 这个定理就是很显然的了: 只需要考虑与 \(p\) 对应的二次齐次多项式 \(\overline{p}\) 就够了. \(\overline{p}\) 是二次型. 依据高等代数的正定二次型的理论, 我们断言定理 5.4 为真.