# IMO 2013 解答

Problem 1 (Japan)

$$\left(1+\frac1{2a}\right)\left(1+\frac1{2a+1}\right)=1+\frac1a,$$

$$\left(1+\dfrac1b\right)\left(1+\dfrac1{b+1}\right)\dotsm\left(1+\dfrac1{b+2^t-2}\right)$$

$1+\frac{2^k-1}n=\left(1+\frac1n\right)\left(1+\frac1{n+1}\right)\dotsm\left(1+\frac1{n+2^k-2}\right).$

$1+\frac{2^k-1}n=\left( 1+\frac{2^{k-1}-1}{\frac n2}\right)\left( 1+\frac1{n+2^k-2}\right).$

$1+\frac{2^k-1}n=\left( 1+\frac{2^{k-1}-1}{\frac{n+1}2}\right)\left( 1+\frac1n\right).$

$r=2^{a_1}+2^{a_2}+\dotsc+2^{a_u}, s=2^{b_1}+2^{b_2}+\dotsc+2^{b_v},$

$t_1=2^{a_1},t_2=2^{a_1}+2^{a_2},\dotsc, t_{u-1}=2^{a_1}+2^{a_2}+\dotsc+2^{a_{u-1}},t_u=r,$

$t_{u+1}=r+2^{k-1},t_{u+2}=r+2^{k-1}+2^{b_v}, t_{u+3}= r+2^{k-1}+2^{b_v}+2^{b_{v-1}},\dotsc, t_k = 2^k-1.$

$1+\frac{2^k-1}n=\frac{n+t_1}n\cdot\frac{n+t_2}{n+t_1}\dotsm\frac{n+t_k}{n+t_{k-1}}.$

$$i=u-1$$ 时, $$t_u-t_{u-1}=2^{a_u}$$. 因 $$a_u<k-1$$, 由 $$(t_{u}-t_{u-1})\mid (n+t_u), n+t_u-(t_u-t_{u-1})=n+t_{u-1}$$ 可得 $$(t_{u}-t_{u-1})\mid (n+t_{u-1})$$.

Problem 2 (Ivan Guo, Australia )

Problem 3 (Alexander A. Polyansky, Russia)

IMO 2013 Problem 3 Proof 1

$\angle DBA_1=\angle OAB_1, BA_1=AB_1,$

$\angle OAC_1=\angle ECA_1, AC_1=CA_1$

$\angle DA_1B=\angle OB_1A=180^\circ-\angle OB_1C=180^\circ-\angle OC_1B=\angle OC_1A=\angle EA_1C$

$\angle I_3UB=180^\circ-2\angle I_3=180^\circ-2\cdot\frac{\angle BAC+\angle ABC}2=\angle ACB$

IMO 2013 Problem 3 Proof 2

$$\triangle A_1B_1C_1$$ 的外心在 $$\triangle A_1B_1C_1$$ 外, 因之三角形 $$A_1B_1C_1$$ 是钝角三角形. 不妨 $$\angle B_1A_1C_1\gt90^\circ$$, 于是  $$\triangle A_1B_1C_1$$ 的外心, $$A$$ 在 $$B_1C_1$$ 同侧, 从而 $$U$$ 就是 $$\triangle A_1B_1C_1$$ 的外心, $$UV, UW$$ 分别是 $$A_1C_1, A_1B_1$$ 的中垂线. 注意到四边形 $$UVI_1W$$ 是平行四边形, 由

$\angle BAC=\angle BUC=2\angle VUW=2\angle VI_1W=2\cdot\frac{\angle ABC+\angle ACB}2$

IMO 2013 Problem 3 Proof 3

$$UV$$ 是 $$A_1C_1$$ 的中垂线, 当然 $$A_1C_1\perp UV$$, 于是 $$A_1C_1\perp I_1I_2$$. 记 $$I_3C$$ 与 $$AB$$ 的交点是 $$M$$, 则 $$CM\perp I_1I_2$$. 于是, $$A_1C_1\parallel CM$$, 当然更有

$\frac{BC_1}{BA_1}=\frac{BM}{BC}.$

$\frac{\frac{b+c-a}2}{\frac{a+b-c}2}=\frac{\frac{ac}{a+b}}a,$

Problem 4 (Warut Suksompong and Potcharapol Suteparuk, Thailand )

$$B,C,M$$ 和 $$N$$ 四点共圆, 这圆记为 $$\omega$$. $$\omega$$ 与 $$\omega_1$$ 的公共弦 $$BN$$ 与 $$\omega$$ 与 $$\omega_2$$ 的公共弦 $$CM$$ 的交点是 $$A$$, 于是 $$A$$ 即为 $$\omega_1, \omega_2$$ 和 $$\omega$$ 的根心. $$\omega_1, \omega_2$$ 的另一个交点 $$U$$ 显然在 $$AW$$ 上.

IMO 2013 Problem 4 Proof 1

$$HM,HN$$ 分别与 $$AM,AN$$ 垂直, 因此, $$AH$$ 是 $$\triangle AMN$$ 的外接圆的直径. 此外, 据 Miquel’s theorem, $$U$$ 也在 $$\triangle AMN$$ 的外接圆上, 故而 $$HU\perp AU$$. 于是, $$H$$ 亦在直线 $$XY$$ 上. 换句话说, $$X,Y,U$$ 和 $$H$$ 共线.

IMO 2013 Problem 4 Proof 2

Problem 5 (Nikolai Nikolov, Bulgaria)

1. (i) 中令 $$x=a,y=1$$, 得 $$f(a)f(1)\geqslant f(a)$$. 再利用 (iii), 有 $$f(1)\geqslant1$$;
2. 由 (ii), $$f(nr)\geqslant nf(r), \forall r\in\Bbb Q_{\gt0}, \forall n\in\Bbb N$$. 特别, 令 $$r=1$$, 得 $$f(n)\geqslant n$$;
3. 根据 (i), 注意到 (iii), 可有 $$a^k\geqslant f(a^k),\forall k\in\Bbb N$$;
4. $$\forall \frac pq\in\Bbb Q_{\gt0}$$, 这里 $$p,q\in\Bbb N$$, 因 $$f(\frac pq)f(q)\geqslant f(p)\geqslant p$$, 故 $$f(\frac pq)\geqslant \frac p{f(q)}$$, 当然更有 $$f(\frac pq)\gt0$$;
5. $$\forall x,y\in\Bbb Q_{\gt0}, x\gt y$$, 则 $$r=x-y\in\Bbb Q_{\gt0}$$. 于是 $$f(x)=f(y+r)\geqslant f(y)+f(r)\gt f(y)$$.

$f\left(r\right)\geqslant f\left(\frac{\left[a^kr\right]}{a^k}\right)\geqslant\left[a^kr\right] f\left(\frac1{a^k}\right)\geqslant\frac{\left[a^kr\right]}{a^k}\gt\frac{a^kr-1}{a^k}=r-\frac1{a^k}.$

$a^k\geqslant f\left(a^k\right)\geqslant f\left(a^k-r\right)+f\left(r\right)\geqslant a^k-r+f\left(r\right)$

$f\left(r\right)^k\geqslant f\left(r^k\right)\geqslant f\left(\left[r^k\right]\right)\geqslant \left[r^k\right]\gt r^k-1.$

$f(r)f(\frac1r)\geqslant f\left(r\cdot\dfrac1r\right)= f\left(1\right)\geqslant1$

$f(n)=n, \forall n\in\Bbb N.$

$f(nqm)\geqslant nqf(m)\gt nq(m+\frac1q)=nqm+n,\forall n\in\Bbb N.$

$f(nqm)\leqslant f(a^k).$

$nqm+qm\lt nqm+n\lt f(nqm)\leqslant f(a^k)\leqslant a^k\lt (n+1)qm,$

$$\forall r=\frac pq\in\Bbb Q_{\gt0}$$, 这里 $$p,q\in\Bbb N$$, 据性质 2,

$p=f(p)=f(q\cdot\frac pq)\geqslant qf(\frac pq),$

Problem 6 (Alexander S. Golovanov and Mikhail A.Ivanov, Russia)

$$a_0=0,a_1,a_2,\dotsc,a_n.$$

1. $$a+b\gt n$$;
2. $$\gcd(a,b)=1$$;
3. 若 $$b=n$$, 则 $$a_i\equiv ia\pmod n, i=1,2,\dotsc,n-1$$;
4. 若 $$0\leqslant x<x+a\leqslant n$$, 则 $$i(x+a)=i(x)+1$$, 即 $$x$$ 右边紧挨的那个数就是 $$x+a$$.

• $$1\leqslant a,b\leqslant n$$;
• $$a+b\gt n$$;
• $$\gcd(a,b)=1$$.

$P_a,P_{a-1},\dotsc,P_2,P_1,$

$0,a,\dotsc,b$

$j_1+j_3\equiv j_2+j_4\pmod a.$

$0\leqslant (j_1+j_3)-(j_2+j_4)=(j_1-j_2)+(j_3-j_4)\leqslant j_1-j_4\leqslant a-1$

$\{(x,y)\in\Bbb N^2|\gcd(x,y)=1,x+y\leqslant n\}\to\{(x,y)\in\Bbb N^2|\gcd(x,y)=1, y\lt x\leqslant n\}$

$(x,y)\mapsto (x+y,y)$

$$0,a,2a,\dotsc,(s-1)a\pmod s.$$

1. 当 $$0\leqslant x\leqslant n-a$$ 时, $$i(x+a)=i(a)+1$$;
2. 当 $$n-a<x< b$$ 时, $$i(x+a-b)=i(x)+1$$;
3. 当 $$b<x\leqslant n$$ 时, $$i(x-b)=i(x)+1$$.

$$0,a,2a,\dotsc,(s-1)a\pmod s.$$

$M=\varphi(1)+\varphi(2)+\dotsb+\varphi(n).$

$N=\varphi(2)+\varphi(3)+\dotsb+\varphi(n),$

$$0,a_1,a_2,\dotsc,a_{n+1}$$

(i)  若 $$i^\prime(n+1)=1$$, 则 $$a+b=n+1$$. 若不然, $$a+b>n+1$$, 则$$(n+1)+(a+b-n-1)=a+b$$ 导致矛盾.

$$c_0=0,c_1=n+1,c_2=a,\dotsc,c_{n+b}=b$$

(ii)  若 $$1\lt i^\prime(n+1)\lt n+1$$, 则 $$a+b\gt n+1$$.

$$i(n+1-b)=i(n+1-a)+1$$, 结合 $$i^\prime(n+1)\gt i^\prime(n+1-a), i^\prime(n+1)\lt i^\prime(n+1-b)$$, 定出 $$i^\prime(n+1)-i^\prime(n+1-a)=1, i^\prime(n+1-b)-i^\prime(n+1)=1$$.

(iii)  若 $$i^\prime(n+1)=n+1$$, 则 $$a+b=n+1$$.

$$c_0=0,c_1=a,\dotsc,c_{n+a-1}=b,c_{n+a}=n+1$$

$A_{n+1}=C_n\cup D_n\cup E_n.$

$A_n\setminus\{(1,n),(n,1)\}\to B_n$

$(x,y)\mapsto\begin{cases} (x\bmod y, y), & x>y\$$x, y\bmod x), & x<y\end{cases}$ 从而 \(|A_n|-2=|B_n|$$. 这也就是 $$M-2=N-1$$, 即 $$M=N+1$$.

### Annotations

1. 题 1 的证明 3 是 AoPS Forum 的 crazyfehmy 给出. 此外, 证明 2 最后写出来, 能有这么简洁, 也得益于大家的讨论.
2. 题目 2 的解答 2, 受到了一些网友的启发. 请注意: 对于 $$u$$ 个红点, $$v$$ 个蓝点组成的任意的哥伦比亚式点集, 总存在 $$\left[\dfrac{u+v}2\right]$$ 条直线构成的好直线组, 但 $$\left[\dfrac{u+v}2\right]$$ 不一定是最小值.
3. 题 3 的证明 1 也受益于百度数学竞赛吧 XmlSpinner 的证明; 证明 2 的思路出自贴吧 Richard1018 之手. 这方法可能以不同面目出现, 比如说, $$U,V,W$$ 是$$\triangle ABC$$ 外接圆的弧的中点.
4. 题 4 的证明 2, 使用了一个所谓的 Reim theorem. 这定理是这么说的: $$A,B,C,D$$ 四点共圆, $$M,N$$ 分别是 $$AC,BD$$ 上两点, 则 $$MN\parallel CD$$ 当且仅当 $$A,B,N,M$$ 四点共圆.
5. 题 6 的证明 2 和 3 没有完成, 都是半成品. 这两个证明, 思路是类似的, 还做一点努力, 应该是可以完成. 证明 4 的部分想法来自水木社区数学版的 Cracker. “中等数学” 第九期给出了两种不一般的解法. 建议认真的看一看. “走向IMO2013” 的解答与”中等数学”是一样的.
6. 这 6 个问题, 尤其题目 3 和 4, 还能找到很多精彩的解答. 题 6, 使用归纳法, 有很多大同小异的手段可以指出 $$n$$ 的一部分的漂亮标记有两个位置可以插入 $$n+1$$, 剩下的部分只有一个位置可以插人 $$n+1$$, 进而得到 $$M(n+1)-M(n)=\varphi(n+1)$$, 然后定出 $$M=\varphi(1)+\varphi(2)+\dotsb+\varphi(n)$$. 这办法被很多人采用.
7. 战国时期, 宋玉的作品”登徒子好色赋”中的描写女人美貌的名句“增之一分则太长，减之一分则太短，着粉则太白，施朱则太赤”, 其实也适用来成为解答写的好坏的标准之一.
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# Day $$1$$

Tuesday, July 23, 2013

Problem 1. Prove that for any pair of  positive integers $$k$$ and $$n$$, there exist $$k$$ positive integers $$m_1,m_2,\dotsc, m_k$$(not necessarily different) such that

$1+\frac{2^k-1}n=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\dotsm\left(1+\frac{1}{m_k}\right).$

Problem 2. A configuration of $$4027$$ points in the plane is called Colombian if it consists of $$2013$$ red points and $$2014$$ blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian con guration if the following two conditions are satisfied:

• no line passes through any point of the con guration;
• no region contains points of both colours.

Find the least value of $$k$$ such that for any Colombian con guration of $$4027$$ points, there is a good arrangement of $$k$$ lines.

Problem 3.  Let the excircle of triangle $$ABC$$ opposite the vertex $$A$$ be tangent to the side $$BC$$ at the point $$A_1$$. De fine the points $$B_1$$ on $$CA$$ and $$C_1$$ on $$AB$$ analogously, using the excircles opposite $$B$$ and $$C$$, respectively. Suppose that the circumcentre of triangle $$A_1B_1C_1$$ lies on the circumcircle of triangle $$ABC$$.Prove that triangle $$ABC$$ is right-angled.

The excircle of triangle $$ABC$$ opposite the vertex$$A$$  is the circle that is tangent to the line segment $$BC$$, to the ray $$AB$$ beyond $$B$$, and to the ray $$AC$$ beyond $$C$$. The excircles opposite $$B$$ and $$C$$ are similarly defi ned.

# Day $$2$$

Wednesday, July 24, 2013

Problem 4. Let $$ABC$$ be an acute-triangle with orthocenter $$H$$, and let $$W$$ be a point on the side $$BC$$, lying strictlybetween $$B$$ and $$C$$. The points $$M$$ and $$N$$ are the feet of the altitudes from $$B$$ and $$C$$, respectively. Denote by $$\omega_1$$ the circumcircle of $$BWN$$, and let $$X$$ be the point on $$\omega_1$$ such that $$WX$$ is a diameter of $$\omega_1$$. Similarly, denote by $$\omega_2$$ the circumcircle of triangle $$CWM$$, and let $$Y$$ be the point on $$\omega_2$$ such that $$WY$$ is a diameter of $$\omega_2$$. Prove that the points $$X, Y$$ and $$H$$ are collinear.

Problem 5. Let $$\Bbb Q_{>0}$$ be the set of positive rational numbers. Let $$f\colon\Bbb Q_{>0}\to\Bbb R$$ be a function satisfying the following three conditions:

(i)   for all $$x,y\in\Bbb Q_{>0}$$, we have $$f(x)f(y)\geqslant f(xy)$$;
(ii)   for all $$x,y\in\Bbb Q_{>0}$$, we have $$f(x+y)\geqslant f(x)+f(y)$$;
(iii)  there exists a rational number $$a>1$$ such that $$f(a)=a$$.

Prove  that $$f(x)=x$$ for all $$x\in\Bbb Q_{>0}$$.

Problem 6. Let $$n\geqslant 3$$ be an integer, and consider a circle with $$n+1$$ equally spaced points marked on it. Consider all labellings of these points with the numbers $$0,1,\dotsc, n$$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $$a<b<c<d$$ with $$a+d=b+c$$, the chord joining the points labelled $$a$$ and $$d$$ does not intersect the chord joining the points labelled $$b$$ and $$c$$.

Let $$M$$ be the number of beautiful labellings, and let $$N$$ be the number of ordered pairs $$(x,y)$$ of positive integers such that $$x+y\leqslant n$$ and $$\gcd(x,y)=1$$. Prove that

$M=N+1.$

# Day $$1$$

2013 年 7 月 23 日, 星期二

$1+\frac{2^k-1}n=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\dotsm\left(1+\frac{1}{m_k}\right).$

• 这些直线不经过该哥伦比亚式点集中的任何一个点;
• 每个区域中都不会同时出现两种颜色的点.

# Day $$2$$

2013 年 7 月 24 日, 星期三

(i)  对所有的 $$x,y\in\Bbb Q_{>0}$$, 都有 $$f(x)f(y)\geqslant f(xy)$$;
(ii)  对所有的 $$x,y\in\Bbb Q_{>0}$$, 都有 $$f(x + y) \geqslant f(x) + f(y)$$;
(iii) 存在有理数 $$a > 1$$, 使得 $$f(a) = a$$.

$M=N+1.$

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Problem 1 (Evangelos Psychas, Greece)

It is obvious that

$\angle JFL=\angle JBM-\angle FMB=\frac12(\angle BAC+\angle BCA)-\frac12\angle BCA=\frac12\angle BAC,$

therefore $$J,L,A$$ and $$F$$ belong to a circle which implies that $$\angle JFS=90^{\circ}$$. But $$\angle JMS=90^{\circ}$$, so $$J,M,F$$ and $$S$$ are concyclic, $$\angle JSM=\angle JFM=\frac12\angle BAC$$.

Similarly, $$\angle JTM=\frac12\angle BAC$$, so $$\angle JSM=\angle JTM$$.

Since $$JM\perp ST$$, we obtain $$SM=MT$$.

Problem 2 (Angelo di Pasquale, the current Australian leader)

It is easy to see that

$a_k+1=a_k+\frac 1{k-1}+\dotsb+\frac 1{k-1}\geqslant k\sqrt[k]{\frac{a_k}{(k-1)^{k-1}}},$

so $$(a_k+1)^k\geqslant\frac{k^k}{(k-1)^{k-1}} \,a_k,$$ the inequality is strict unless $$a_k=\frac1{k-1}$$. Multiplying analogous inequalities for $$k=2,3,\dotsc,n$$ yields

$\prod_{k=2}^n(a_k+1)^k\geqslant n^na_2a_3\dotsm a_n=n^n.$

the inequality is strict because $$a_k=\frac1{k-1}$$ holds for all $$k\in\{2,3,\dotsc,n\}$$ is not possible .

The problem was actually written two years ago.

1. We can assume  $$n=2^k, N=n+1$$.

The binary digits of  $$1,2,\dotsc,2^k$$ are $$\overline{a_1a_2\ldots a_{k+1}}$$, where  $$a_i(i=1,2,\dotsc,k+1)$$ is $$0$$ or $$1$$; then, let  $$T$$ is  the set of all the $$2^k$$ binary digits. The binary digit of  $$2^k+1$$ is $$100\ldots01.$$ Define

$S_1=\{100\ldots0\}, S_i=\{\overline{a_1a_2\ldots a_{k+1}}\in T|a_1=0, a_i=1\}, i=2,3,\dotsc,k+1.$

• B chooses  $$S_1$$ repeatedly. If B get an answer of  no $$k+1$$ times, then B can exclude the number $$100\ldots0$$. Otherwise, A will eventually say yes.
• If  A last said yes, B then asks $$S_2,S_3,\dotsc,S_{k+1}$$, and guarantees a win.
2.  Let $$p$$ be a  positive real number such that $$1.99< p<2$$. Let us choose $$k_0$$ such that for every $$k>k_0$$, the following relation holds:

$(2-p)p^{k+1}-1.99^k>2.$

We will prove that if  $$N\in\left(1.99^k+1, (2-p)p^{k+1}\right)$$, then B cannot specify a set $$X(|X|\leqslant N-1)$$ such that $$x\in X$$.

$$T=\{1,2,\dotsc, N\}$$. $$j\in\Bbb N$$, The $$j$$-th question consists of $$B$$ choosing a set $$S_j\subset T$$ and we define $$P_j$$ is $$S_j$$ or $$S_j^C$$ depend on the answer: if A answer it with yes, $$P_j=S_j$$; Otherwise, $$P_j=S_j^C$$. Let $$P_0=T$$. Define $$q_j$$ as follows:

$q_j=q_j(P_j)=a_0+a_1p+a_2p^2+\dotsb+a_jp^j,$

where $$a_0=\left|P_j\right|,a_i=\left|P_{j-i}\setminus \left(P_j\cup P_{j-1}\cup\cdots\cup P_{j-i+1} \right)\right|(i=1,2,\dotsc,j).$$   It is easy to see that $$\sum\limits_{i=0}^ja_i=N.$$   $$q_0=N.$$

The player $$A$$ can keep the following relation holds:

$q_j<\frac N{2-p}, j\geqslant0.$

In fact, $$q_0=N<\frac N{2-p}.$$ We  assume  that $$q_j<\frac N{2-p}$$, then A can choose $$P_{j+1}\in\left\{S_{j+1},S_{j+1}^C\right\}$$ such that $$q_{j+1}<\frac N{2-p}$$. Let

$q_{j+1}(S_{j+1})=b_0+b_1p+b_2p^2+\dotsb+b_jp^j+b_{j+1}p^{j+1},$

$q_{j+1}(S_{j+1}^C)=c_0+c_1p+c_2p^2+\dotsb+c_jp^j+c_{j+1}p^{j+1}.$

Notice that $$b_0+c_0=N,b_i+c_i=a_{i-1}(i=1,2,\dotsc,j+1),$$ so

\begin{equation*}\begin{split}q_{j+1}(S_{j+1})+q_{j+1}(S_{j+1}^C)&=N+p(a_0+a_1p+a_2p^2+\dotsb+a_jp^j)\\
&<N+p\cdot\frac N{2-p},\end{split}\end{equation*}

hence

$\min\left\{q_{j+1}(S_{j+1}),q_{j+1}(S_{j+1}^C)\right\}< \frac{N}2+\frac p2\cdot\frac N{2-p}=\frac N{2-p}.$

so, A can take $$P_{j+1}\in\left\{S_{j+1},S_{j+1}^C\right\}$$.

Since $$q_j<\frac N{2-p}<p^{k+1},$$ we get that for $$i\geqslant k+1, a_i=0,$$ then B cannot exclude a number in $$T$$, and cannot specify a set $$X(|X|\leqslant N-1)$$ such that $$x\in X$$.

1.  可以认为 $$n=2^k, N=n+1$$. 采用二进制.

$S_1=\{100\ldots0\}, S_i=\{\overline{a_1a_2\ldots a_{k+1}}\in T|a_1=0, a_i=1\}, i=2,3,\dotsc,k+1,$

• 连续 $$k+1$$ 次回答 “否”, 则 $$100\ldots0$$ 可以排除;
• 在至多 $$k+1$$ 次回答中, 一旦出现”是”, 乙接下来的 $$k$$ 次提问, 依次选取 $$S_2,S_3,\dotsc,S_{k+1}$$, 就取得胜利. 事实上, 若甲最后的 $$k$$ 次回答都是”是”, 则 $$x\in T$$; 若甲最后的 $$k$$ 次回答有一些是”否”, 则 $$x$$ 绝对不可能是 $$\overline{a_1a_2\ldots a_{k+1}}$$, 这里 $$a_1=0$$, $$a_i=0$$ 还是 $$1$$ 取决于甲对 $$S_i$$ 的答案: 若甲的回答是”是”, $$a_i=0$$, 否则 $$a_i=1( i=2,3,\dotsc,k+1).$$

2. 解答 $$1$$  任取实数 $$p$$ 使得 $$1.99<p<2$$, 再选取正整数 $$k_0$$, 使得当 $$k>k_0$$ 时

$(2-p)p^{k+1}-1.99^k>2.$

$$T=\{1,2,\dotsc, N\}$$. $$j$$ 是正整数, 记 $$S_j\subset T$$ 是乙的第 $$j$$ 个问题展示的集合, 设 $$P_j$$ 为 $$S_j$$ 或者 $$S_j^C$$, 取决于甲对 $$S_j$$ 的答案: 若甲的回答是”是”, $$P_j=S_j$$, 否则 $$P_j=S_j^C$$; $$P_0=T$$. 定义 $$q_j(j\geqslant0)$$ 如下:

$q_j=q_j(P_j)=a_0+a_1p+a_2p^2+\dotsb+a_jp^j,$

$q_j<\frac N{2-p}, j\geqslant0$

$$q_0=N<\frac N{2-p}.$$ 假设已有 $$q_j<\frac N{2-p}$$, 甲可选取 $$P_{j+1}\in\left\{S_{j+1},S_{j+1}^C\right\}$$ 使得 $$q_{j+1}<\frac N{2-p}$$. 事实上,

$q_{j+1}(S_{j+1})=b_0+b_1p+b_2p^2+\dotsb+b_jp^j+b_{j+1}p^{j+1},$

$q_{j+1}(S_{j+1}^C)=c_0+c_1p+c_2p^2+\dotsb+c_jp^j+c_{j+1}p^{j+1}.$

\begin{equation*}\begin{split}q_{j+1}(S_{j+1})+q_{j+1}(S_{j+1}^C)&=N+p(a_0+a_1p+a_2p^2+\dotsb+a_jp^j)\\
&<N+p\cdot\frac N{2-p},\end{split}\end{equation*}

$\min\left\{q_{j+1}(S_{j+1}),q_{j+1}(S_{j+1}^C)\right\}< \frac{N}2+\frac p2\cdot\frac N{2-p}=\frac N{2-p}.$

$\left(\frac pq\right)^{k_0}\leqslant 2\left(1-\frac q2\right),\quad p^{k_0}-1.99^{k_0}>2.$

$P_j\cup P_{j+1}\cup\cdots\cup P_{j+k}=T$

$$x_i^{j+1}=\left\{\begin{array}{rl} 1,& i\in P_{j+1},\\ q\, x_i^j, & i\notin P_{j+1}. \end{array}\right.$$

$T\left(\mathbf y\right)=\sum_{i\in S_{j+1}^C} qx_i^j+\left|S_{j+1}\right|,$

$T\left(\mathbf z\right)=\sum_{i\in S_{j+1}} qx_i^j+\left|S_{j+1}^C\right|.$

$T\left(\mathbf y\right)+T\left(\mathbf z\right)= q\cdot T\left(\mathbf x^j\right)+ N\leqslant q^{k+1}+ p^k,$

$\min\left\{T\left(\mathbf y\right),T\left(\mathbf z\right)\right\}\leqslant \frac{q}2\cdot q^k+\frac{p^k}2\leqslant q^k.$

Problem 4 (Liam Baker, South Africa)

1. 令 $$a=b=c=0$$, 得 $$f(0)=0$$;
2. 令 $$b=-a,c=0$$, 得 $$f(a)=f(-a)$$. 于是只需讨论当 $$a$$ 为正整数时的 $$f(a)$$ 就行了.
3. 若有 $$a\in\Bbb Z$$, 使得 $$f(a)=0$$, 那么, 对于任意整数 $$b$$, 由 $$a+b+(-a-b)=0$$ 得 $$f(a+b)=f(b).$$

$$f(x)=\left\{\begin{array}{rl}0,& 2\mid n,\\ c, & 2\not\mid n.\end{array}\right.$$

$f(1)^2+n^4f(1)^2+f(n+1)^2=2n^2f(1)^2+2(n^2+1)f(n+1)f(1),$

$\left(f(n+1)-(n+1)^2f(1)\right)\left(f(n+1)-(n-1)^2f(1)\right)=0.$

$2(n-1)^4f(1)^2+16f(1)^2=2 (n-1)^4f(1)^2+16(n-1)^2f(1)^2,$

$$f(x)=\left\{\begin{array}{rl}0,& 4\mid n,\\ c, & 2\nmid n,\\ 4c, & n\equiv2\pmod4\end{array}\right.$$

$$f(n)=cn^2,$$

Problem 5 (Josef Tkadlec, Czech Republic)

$$AC$$ 为 $$\odot B$$ 的切线, 所以 $$AL^2=AC^2=AK\cdot AP$$, $$AL$$ 是 $$\odot O$$ 的切线. 同理, $$BK$$ 也是 $$\odot O$$ 的切线, 故而 $$ML=MK$$.

Problem 6 (Dušan Djukić, Serbia)

$1+2+\dotsb+n=\frac{n(n+1)}2\equiv1\pmod2,$

$\mathbf a=\left(2,1,4,3,\dotsc,2k,2k-1,2k+2,2k+1,\dotsc,4k,4k-1,4k+1\right),$

$\mathbf a\prime=\left(2,1,4,3,\dotsc,2k,2k-1,2k+1,\dotsc,4k,4k-1,4k+1,2k+2\right),$

$L(\mathbf a)=\sum_{i=1}^n\frac1{2^{a_i}},R(\mathbf a)=\sum_{i=1}^n\frac i{3^{a_i}}.$

$a_j^\prime=\left\{\begin{array}{rl} a_j,& j\notin \{m+1,2m+2\},\\ a_{m+1}+1,& j\in \{m+1,2m+2\}.\end{array}\right.$

$L\left(\mathbf a^\prime\right)=L(\mathbf a)-\frac1{2^{a_{m+1}}}+2\cdot \frac1{2^{a_{m+1}+1}}=1,$

$R\left(\mathbf a^{\prime}\right)=R(\mathbf a)- \frac{m+1}{3^{a_{m+1}}}+\frac{m+1}{3^{a_{m+1}+1}}+\frac{2m+2}{3^{a_{m+1}+1}}=1.$

$$1$$. 有这么一个趣味的数学游戏, 书上应该不难找到:

$S_i=\{\overline{a_5a_4a_3a_2a_1}\in T| a_i=1\},$

$$2$$. 难度稍大一些, 但我们采取的手段也不新鲜: 考察一个辅助函数, 这个函数有某种性质, 单调性, 有界, 不变量等等, 反映的都是隐藏在游戏中的某个”量”的性质. $$2000$$ IMO 的 problem $$3$$ 的 $$2$$ 是这么解决的, $$1986$$ 年 IMO 的 problem $$3$$ 仍然是这样的思路.

$$3$$. 这里给出的”两种”证法, 实质上没有任何差别.

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# Day $$1$$

Tuesday, July 10, 2012    9:00 am-1:30 pm

Problem 1.  Given triangle $$ABC$$ the point $$J$$ is the centre of the excircle opposite the vertex $$A$$. This excircle is tangent to the side $$BC$$ at $$M$$, and to the lines $$AB$$ and $$AC$$ at $$K$$ and $$L$$, respectively. The lines $$LM$$  and $$BJ$$ meet at $$F$$, and the lines $$KM$$ and $$CJ$$ meet at $$G$$. Let $$S$$ be the point of intersection of the lines $$AF$$ and $$BC$$, and let $$T$$ be the point of intersection of the lines $$AG$$ and $$BC$$.
Prove that $$M$$ is the midpoint of $$ST$$.

(The excircle of $$ABC$$ opposite the vertex $$A$$ is the circle that is tangent to the line segment $$BC$$, to the ray $$AB$$ beyond $$B$$, and to the ray $$AC$$ beyond $$C$$.)

Problem 2.  Let  $$n\geqslant 3$$ be an integer, and let  $$a_2,a_2,\dotsc,a_n$$ be positive real numbers such that $$a_2a_3\dotsm a_n=1.$$ Prove that

$\left(1+a_2\right)^2\left(1+a_3\right)^3\dotsm\left(1+a_n\right)^n>n^n.$
Problem 3.   The liar’s guessing game is a game played between two players $$A$$ and $$B$$. The rules of the game depend on two positive integers $$k$$ and $$n$$ which are known to both players.
At the start of the game $$A$$ chooses integers $$x$$ and $$N$$ with $$1\leqslant x \leqslant N$$. Player $$A$$  keeps $$x$$ secret, and truthfully tells $$N$$ to player $$B$$. Player $$B$$ now tries to obtain information about $$x$$ by asking player $$A$$ questions as follows: each question consists of $$B$$ specifying an arbitrary set $$S$$ of positive integers (possibly one specified in some previous question), and asking $$A$$ whether $$x$$ belongs to $$S$$. Player $$B$$ may ask as many such questions as he wishes. After each question, player $$A$$ must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $$k+1$$ consecutive answers, at least one answer must be truthful.
After $$B$$ has asked as many questions as he wants, he must specify a set $$X$$ of at most $$n$$ positive integers. If $$x$$ belongs to $$X$$, then $$B$$ wins; otherwise, he loses. Prove that:

1. If  $$n\geqslant 2^k$$, then $$B$$ can guarantee a win.
2. For all sufficiently large $$k$$, there exists an integer $$n\geqslant1.99^k$$ such that $$B$$ cannot guarantee a win.

# Day $$2$$

Wednesday, July 11, 2012     9:00 am-1:30 pm

Problem 4.  Find all functions $$f:\Bbb Z \rightarrow\Bbb Z$$ such that, for all integers $$a,b,c$$ that satisfy $$a+b+c=0$$, the following equality holds:

$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$

(Here $$\Bbb Z$$ denotes the set of integers.)

Problem 5.  Let $$ABC$$ be a triangle with $$\angle BCA=90^\circ$$, and let $$D$$ be the foot of the altitude from $$C$$. Let $$X$$ be a point in the interior of the segment $$CD$$. Let $$K$$ be the point on the segment $$AX$$ such that $$BK=BC$$. Similarly, let $$L$$ be the point on the segment $$BX$$ such that $$AL=AC$$. Let $$M$$ be the point of intersection of $$AL$$ and $$BK$$.
Show that $$MK=ML$$.

Problem 6.  Find all positive integers $$n$$ for which there exist non-negative integers $$a_1,a_2,\dotsc,a_n$$ such that

$\frac1{2^{a_1}}+\frac1{2^{a_2}}+\dotsb+\frac1{2^{a_n}}=\frac1{3^{a_1}}+\frac2{3^{a_2}}+\dotsb+\frac n{3^{a_n}}= 1.$

# 第一天

2012年7月10日, 星期二

1.  设 $$J$$ 为三角形 $$ABC$$ 顶点 $$A$$ 所对旁切圆的圆心. 该旁切圆与边 $$BC$$ 相切于点 $$M$$, 与直线 $$AB$$ 和 $$AC$$ 分别相切于点$$K$$ 和 $$L$$. 直线 $$LM$$ 和 $$BJ$$ 相交于点 $$F$$, 直线 $$KM$$ 与 $$CJ$$ 相交于点 $$G$$. 设 $$S$$ 是直线 $$AF$$ 和 $$BC$$ 的交点, $$T$$ 是直线 $$AG$$ 和 $$BC$$ 的交点.

(三角形 $$ABC$$ 的顶点 $$A$$ 所对的旁切圆是指与边 $$BC$$ 相切, 并且与边 $$AB,AC$$ 的延长线相切的圆.)

2. 设整数 $$n\geqslant 3$$, 正实数$$a_2,a_2,\dotsc,a_n$$ 满足 $$a_2a_3\dotsm a_n=1.$$ 证明:

$\left(1+a_2\right)^2\left(1+a_3\right)^3\dotsm\left(1+a_n\right)^n>n^n.$
3.  “欺诈猜数游戏”在两个玩家甲和乙之间进行, 游戏依赖于两个甲和乙都知道的正整数 $$k$$ 和 $$n$$.

1. 若 $$n\geqslant 2^k$$, 则乙可保证获胜;
2. 对所有充分大的整数 $$k$$, 存在整数 $$n\geqslant1.99^k$$, 使得乙无法保证获胜.

# 第二天

2012年7月11日, 星期三

4.  求所有的函数 $$f:\Bbb Z \rightarrow\Bbb Z$$, 使得对所有满足 $$a+b+c=0$$ 的整数 $$a,b,c$$, 都有

$f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).$

(这里 $$\Bbb Z$$ 表示整数集.)

5.  已知三角形 $$ABC$$ 中,  $$\angle BCA=90^\circ$$,  $$D$$ 是过顶点 $$C$$ 的高的垂足. 设 $$X$$ 是线段 $$CD$$ 内部的一点.  $$K$$ 是线段 $$AX$$ 上一点, 使得 $$BK=BC$$.  $$L$$ 是线段 $$BX$$ 上一点, 使得 $$AL=AC$$. 设 $$M$$ 是 $$AL$$ 与 $$BK$$ 的交点.

6.  求所有的正整数 $$n,$$ 使得存在非负整数  $$a_1,a_2,\dotsc,a_n,$$ 满足

$\frac1{2^{a_1}}+\frac1{2^{a_2}}+\dotsb+\frac1{2^{a_n}}=\frac1{3^{a_1}}+\frac2{3^{a_2}}+\dotsb+\frac n{3^{a_n}}= 1.$

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