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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