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原题目

如图，在锐角 中， 是 的中点， 是 在 上的投影。设 的外接圆交直线 于另一点 ， 是 的中点，求证：。

绿树教育中心独家解析

作 ，则有 ，则可得到 。再由于 ，则 ，故 ，即证得 。

原题目

求所有的函数 ，使得对任意正实数 ，均有

绿树教育中心独家解析

「结论 1」 是单射

If , then

implies .

「结论 2」

By and , we have

implying by injectivity.

「结论 3」 是线性的

For any , note that summing

gives

The above equation uniquely determines in terms of , , , and it is clear , , satisfies the equation, so are collinear.

由上述结论，不难说明 。

原题目

绿树教育中心独家解析

The answer is . Call a cell blue if its row index and column index are both even, and call a cell red if its row index and column index are both odd. Additionally, color each domino with the color of the colored cell it covers.

Constructions

One possible construction for involves positioning the dominoes covering red cells in a snake-like fashion. An example construction for is shown below. One possible construction for involves positioning the dominoes covering blue cells in a snake-like fashion, blocking the snake's path with a red domino and an empty square, and filling the rest of the grid with red dominoes. An example construction for and is given below.

Proof that no other work

Let be the directed graph whose vertex set is the red and blue cells, and whose directed edges are drawn from to the cell the domino covering points to, if it exists. Let denote the uncovered cell. Observe the following: (i) By a checkerboard coloring argument, must be a vertex of . (ii) Sliding a domino does not change the location of 's edges; it only reverses the direction of an arrow. (iii) The number of cells in any cycle's interior is odd, so if contains a cycle, then is inside the cycle. (iv) The connected component of containing is a tree because it cannot contain any cycles, by (iii). (v) Given the nondirected edges of , specifying the uncovered cell is enough to recover the domino configuration, by (iv). (vi) Hence, is the number of vertices in the connected component of containing . Because arrows only connect cells of the same color, is at most the number of red cells, which is . Therefore, it suffices to prove that . This follows from the following lemma:

Lemma wrote: If the connected component containing has more than vertices, then it contains every red vertex.

Proof. If the connected component containing has more than vertices, then must be red because there are only blue vertices. Additionally, at least one vertex in the connected component containing must border the edge of the grid, so without loss of generality assume borders the edge of the grid by sliding some of the dominoes.

Now, starting from the red vertex and following the arrows must eventually lead to or lead to a cycle. By (iii), it is impossible to get into a cycle because is on the edge. Thus, every red vertex is connected to , as desired.

Therefore, implies that equals the number of red cells, which is . This gives the solution set , as desired.

原题目

给定正整数 以及黑板上的一些正整数，Alice,Bob 在玩如下游戏。轮到 Alice 时，她将黑板上的某个数 换为 ；轮到 Bob 时他将板上的某个偶数 换为 。Alice 先开始，之后两人轮流进行，当 Bob 无法操作时游戏结束。

通过分析黑板上的数，Bob 发现无论 Alice 如何操作，他都能使游戏结束。求证: 事实上，无论 Alice,Bob 如何操作，游戏最终都会结束。

绿树教育中心独家解析

Rewrite each number as its 2-adic valuation; then the problem becomes equivalent to moving tokens on the graph below, where Alice's moves are denoted by magenta edges, and Bob's moves — cyan. (The picture shows an example for .) If Alice has any tokens to the right of or at , then she wins by stalling until Bob moves a token there, and then ''shooting'' that token outwards so that Bob has to move it again. Otherwise it's clear that Alice is basically useless, and Bob is playing with a ticking fuse until the game is over.

原题目

设整数 ，将 填入一个 的方格表中。称一种方法是“行好的”，如果可以将各行中的数重排成一个等差数列；称一种方法是“列好的”，如果可以将各列中的数重排成一个等差数列。求所有的 ，使得可以通过对行中的数重排将任何一种行好的填法变成列好的填法。

绿树教育中心独家解析

If is prime, note that all arithmetic progressions of length either are all the same or contain every residue . Note that if one row has all the same residues, then no row can contain all the residues, since there are only numbers with the same residue less than . Hence, if one row has all the same residues, then all the rows do. In this case, we can permute the rows such that the columns are (in some order)

for all . Otherwise, all rows have all different residues. In this case, sort all the rows, and the columns are (in some order)

for all . Hence, the statement is true for all prime . If is composite, write and consider the construction

Suppose for the sake of contradiction that you can rearrange this table to be column-valid. Consider the arithmetic progression in the column containing . Let it's common difference be . Note that for the th term to not be in the first row, we require

Since the th term is at most , we have

Note that because otherwise the second term would be in the first row. Hence, . However, we now have that the th term is

but then no terms can be from the last row. Hence, no column-valid rearrangement exists.

So, the answer is .

原题目

如图，在 中， 是内心， 分别是 所对的旁心。 是 外接圆上一点，不在直线 上。设 的外接圆交于另一点 ，直线 交于点 。求证：。

绿树教育中心独家解析

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