# 2019 第 60 届 IMO 解答

## Problem 1 ()

$f(0) + 2f(b) = f(f(b)).$

$f(2a) + 2f(b) = f(0) + 2f(a+b)$

$f(2) + 2f(b) = f(0) + 2f(1+b).$

$B+2(An+B)=A(An+B)+B.$

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# 2017 第 58 届 IMO 解答

## Problem 1 (South Africa)

Lemma 1 当实数 $$x\geqslant6$$, 则 $$x\lt(x-3)^2$$.

Lemma 2 若整数 $$m$$ 符合 $$m \equiv 2\pmod3$$, 则 $$m$$ 不能是完全平方.

$$a_s, a_{s+1}, a_{s+2}, \dotsc.$$

$a_s\lt(a_s-3)^2,\qquad a_s \equiv (a_s-3)^2\pmod3$

$3, 6, 9, 3, 6, 9, 3, 6, 9, \dotsc.$

## Problem 2 (Dorlir Ahmeti, Albania)

$f(0)+f(x)=f(0),$

$f(f^2(0))+f(0)=f(0).$

$$f(f(\frac b{b-1})f(b))+f(\dfrac b{b-1}+b)=f(\frac {b^2}{b-1}),$$

$f(0)+f(x+1)=f(x),$

$$f(x+1)=f(x)-1.$$

$$\big(a^2-4(b-1)\big)+\big(b^2-4(a-1)\big)=(a-2)^2+(b-2)^2\geqslant0$$

$f(f(r)f(s))+f(r+s) = f(rs).$

$f(f(r)f(s)+1)+f(r+s) = f(rs+1).$

$$f(f(r)f(s)+1) = 0$$

$f(f(x)f(-x))+f(0) = f(-x^2).$

$f(f(x)f(-x)) = f(-x^2)-1=f(1-x^2).$

$$f(x)$$ 为单射导出

$$f(x)f(-x) =1-x^2.$$

$f(f(x)f(1-x))+f(1) = f(x-x^2).$

$f(f(x)f(1-x))= f(x-x^2).$

$$f(x)$$ 为单射导出 $$f(x)f(1-x)=x-x^2$$, 即

$$f(x)(f(-x)-1) =x-x^2.$$

$f(x)=f(x)f(-x)-(x-x^2)=(1-x^2)-(x-x^2)=1-x.$

## Problem 4 (Charles Leytem, Luxembourg)

IMO 2017 p4

$$K$$, $$R$$, $$J$$, $$S$$ 四点共圆, 以及$$S$$, $$J$$, $$A$$, $$B$$ 亦是四点共圆, 定出 $$\angle KRS=\angle KJS=\angle RBS$$.

$$RA$$ 为 $$\Omega$$ 的切线蕴涵 $$\angle RKS=\angle BRS$$. 于是, $$\triangle RKS\sim\triangle BRS$$, 故此

$$\angle RSK=\angle BSR$$, 然后  $$\angle RSK=\angle BSR$$;

$$\frac{KS}{RS}=\frac{RS}{BS}$$,  结合 $$RS=TS$$, 然后

$\frac{KS}{TS}=\frac{TS}{BS}.$

## Problem 5 (Russia)

$$N=2$$ 时, 这排球员左边 $$3$$ 人必有 $$2$$ 人属同一组, 选出这 $$2$$ 人; 右边 $$3$$ 人必有 $$2$$ 人亦属同一组, 选出这 $$2$$ 人. 如此, 教练移走了 $$2$$ 人, 剩下 $$4$$ 人的左边  $$2$$ 人与右边 $$2$$ 人分属不同的组.

## Problem 6 (John Berman, USA)

$ap_0+bq_0=1.$

$$up_{m+1}+vq_{m+1}=1.$$

$$F(x, y) = \big (G(x, y) \big )^h – w\Big(\prod_{k=1}^m\left ( q_kx-p_ky \right ) \Big) \Big(ux+vy\Big)^{gh-m},$$

$$q_{m+1}^g = q_{m+1}^g G(p_k, q_k) = G(p_kq_{m+1}, q_kq_{m+1}) \equiv G(q_kp_{m+1}, q_kq_{m+1}) = q_k^g G(p_{m+1}, q_{m+1}) \equiv 0 \pmod p$$

$$A^h – wB = 1.$$

$$\begin{split}F(p_{m+1}, q_{m+1}) &= \big (G(p_{m+1}, q_{m+1}) \big )^h – w\Big(\prod_{i=1}^m\left ( q_ip_{m+1}-p_i q_{m+1} \right ) \Big)\Big(up_{m+1}+v q_{m+1}\Big)^{gh-m}\\&=A^h – wB=1.\end{split}$$

### Annotations

1. 第 2 题的函数方程, 没有什么新奇的. 得分那么低, 倒是有点出乎意料.
2. 第二天的题其实没啥特别, 难度也不大. 题 6 可以推广为更普遍的形式.
3. 第三题确实有独到之处, 是这个试卷仅有的好题.  可以对 $$m$$ 个猎人考虑同样的问题.
4. 这六个题何以成为史上得分最低的试卷呢! 除了第三题, 别的题为啥得分也不高?
5. 大陆居然在函数方程载了跟头: 函数方程应该是必须掌握的基本功. 最难的题最可能来自组合, 这没有疑问; 数论的难题虽多, 但能成为竞赛卷子的妙题却不容易; 考场上的几何再难, 也没太大意思.
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# Day $$1$$

Tuesday, July 18, 2017

Problem 1. For each integer  $$a_0\gt1$$, define the sequence $$a_0$$, $$a_1$$, $$a_2$$, $$\dotsc$$ by:

$a_{n+1} = \begin{cases}\sqrt{a_n} & \text{if } \sqrt{a_n} \text{ is an integer,} \\a_n + 3 & \text{otherwise.}\end{cases}\quad \text{for each}\; n\geqslant 0.$

Determine all values of $$a_0$$ so that there exists a number $$A$$ such that $$a_n = A$$ for infinitely many values of $$n$$.

Problem 2. Let $$\Bbb R$$ be the set of real numbers. Determine all functions $$f\colon \Bbb R \rightarrow \Bbb R$$ such that, for all real numbers $$x$$ and $$y$$,

$f\big(f(x)f(y)\big) + f(x+y) = f(xy).$

Problem 3. A hunter and an invisible rabbit play a game in the Euclidean plane. The rabbit’s starting point, $$A_0$$ , and the hunter’s starting point, $$B_0$$ are the same. After $$n-1$$ rounds of the game, the rabbit is at point $$A_{n-1}$$ and the hunter is at point $$B_{n-1}$$ . in the $$n^{\text{th}}$$ round of the game, three things occur in order:

i) The rabbit moves invisibly to a point $$A_n$$ such that the distance between $$A_{n-1}$$ and $$A_n$$ is exactly $$1$$ .

ii) A tracking device reports a point $$P_n$$ to the hunter. The only guarantee provided by the tracking device to the hunter is that the distance between $$P_n$$ and $$A_n$$ is at most $$1$$ .

iii) The hunter moves visibly to a point $$B_n$$ such that the distance between $$B_{n-1}$$ and $$B_n$$ is exactly $$1$$ .

Is it always possible, no matter how the rabbit moves, and no matter what points are reported by the tracking device, for the hunter to choose her moves so that after $$10^9$$ rounds, she can ensure that the distance between her and the rabbit is at most $$100$$ ?

# Day $$2$$

Wednesday, July 19, 2017

Problem 4. Let $$R$$ and $$S$$ be different points on a circle $$\Omega$$ such that $$RS$$ is not a diameter. Let $$\ell$$ be the tangent line to at $$R$$. Point $$T$$ is such that $$S$$ is the midpoint of the line segment $$RT$$. Point $$J$$ is chosen on the shorter arc $$RS$$ of $$\Omega$$ so that the circumcircle  $$\Gamma$$ of triangle $$JST$$ intersects $$\ell$$ at two distinct points. Let $$A$$ be the common point of $$\Gamma$$ and $$\ell$$ that is closer to $$R$$. Line $$AJ$$ meets $$\Omega$$ again at $$K$$. Prove that the line $$KT$$ is tangent to $$\Gamma$$.

Problem 5. An integer $$N\geqslant2$$ is given. A collection of $$N(N + 1)$$ soccer players, no two of whom are of the same height, stand in a row. Sir Alex wants to remove $$N(N-1)$$ players from this row leaving a new row of $$2N$$ players in which the following $$N$$ conditions hold:

(1) no one stands between the two tallest players,

(2) no one stands between the third and fourth tallest players,

$$\vdots$$

(N) no one stands between the two shortest players.

Show that this is always possible.

Problem 6. An ordered pair $$(x, y)$$ of integers is a primitive point if the greatest common divisor of $$x$$ and $$y$$ is $$1$$. Given a finite set $$S$$ of primitive points, prove that there exist a positive integer $$n$$ and integers $$a_0$$, $$a_1$$ , $$\dotsc$$, $$a_n$$ such that, for each $$(x, y)$$ in $$S$$, we have:

$a_0x^n + a_1x^{n-1}y + a_2x^{n-2}y^2 + \dotsb + a_{n-1}xy^{n-1} + a_ny^n = 1.$

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# 2016 第 57 届 IMO 解答

## Problem 1 (The Kingdom Of Belgium)

IMO 2016 Problem 1

$\angle FBA=\angle FAB=\angle DAC=\angle DCA= \angle EAD=\angle EDA.$

$$\begin{split}\angle FDC&=180^\circ-\angle ADF-\angle DAC-\angle DCA\\&=180^\circ-\angle ACB-\angle FAB-\angle FBA\\&=\angle FBC=90^\circ,\end{split}$$

$$D$$ 落在以 $$M$$ 为心, $$MB$$ 为半径的圆上, 即 $$D$$, $$F$$, $$B$$, $$C$$ 四点共圆, 并且 $$FC$$ 即为此圆的直径. 记这个圆为 $$\Gamma_1$$. 然后, $$\angle FBD=\angle FCD=\angle FBA$$ 表明 $$FB$$ 平分 $$\angle DBA$$. 结合 $$AF$$ 是 $$\angle DAB$$ 的平分线, 我们知道 $$F$$ 就是 $$\triangle DAB$$ 的内心, 并且 $$DA=DB$$, 这是因为

$\angle DBA=2\angle FBA=\angle DAB.$

$$M$$ 是直角三角形 $$FBC$$ 的斜边 $$FC$$ 的中点, 因此 $$MF=MB$$. 由

$\angle MFB= \angle FBA+\angle FBA=\angle DAB$

$\angle DEA+\angle EAC= \angle DEA+(\angle EAD+\angle DAC)=\angle DEA+(\angle EAD+\angle EDA)=180^\circ$

## Problem 2 ( Australia)

• 第一类: 第 $$2$$, $$5$$, $$8$$, $$\dotsc$$, $$3k-1$$ 行的所有格子;
• 第二类: 第 $$2$$, $$5$$, $$8$$, $$\dotsc$$, $$3k-1$$ 列的所有格子; 以及
• 第三类: 小方格个数是三的倍数的所有对角线上的全部格子.

IMO 2016 Problem 2 Proof 1

## Problem 3 ( Russia)

$16S^2=2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$

$A_iA_j^2=p^{\alpha_{ij}}z_{ij},\;\alpha_{ij}\in\Bbb N,\; z_{ij}\in\Bbb N, \;\big(z_{ij}, p\big)=1, \; 1\leqslant i\lt j\leqslant m,$

$u=\max\{\alpha_{ij}, \; 1\leqslant i\lt j\leqslant m,\; j-i\gt1\}.$

IMO 2016 Problem 3 Proof 1

Ptolemy 定理给出

$ab+cd=ef.$

$a^2b^2+c^2d^2+2abcd=e^2f^2.$

$2abcd=2\sqrt{p^{\alpha_{(i-1)i}+\alpha_{(i+1)j}+\alpha_{i(i+1)}+\alpha_{(i-1)j}}z_{(i-1)i}z_{(i+1)j}z_{i(i+1)}z_{(i-1)j}}=2p^{\alpha+v}\sqrt z,$

## Problem 4 (Luxembourg)

1. $$\big(P(n),P(n+1)\big)=1$$;
2. $$\big(P(n),P(n+2)\big)\mid7;\;\big(P(n),P(n+2)\big)=7 \iff n\equiv 2\pmod 7$$;
3. $$\big(P(n),P(n+3)\big)\mid3;\;\big(P(n),P(n+3)\big)=3 \iff n \equiv 1 \pmod3$$;
4. $$\big(P(n),P(n+4)\big)\mid19;\;\big(P(n),P(n+4)\big) = 19 \iff n \equiv 7 \pmod{19}$$.

$a \equiv 7\pmod{19},\; a+1 \equiv 2\pmod7,\; a+2 \equiv 1\pmod 3,$

$\big(P(a),P(a+4)\big)=19,\; \big(P(a+1),P(a+3)\big)=7,\; \big(P(a+2),P(a+5)\big)=3.$

$$P(a+1)$$, $$P(a+2)$$, $$P(a+3)$$, $$P(a+4)$$, 因为 $$\big(P(a+1),P(a+3)\big)=7$$ 与 $$\big(P(a+2),P(a+4)\big)=7$$ 不能同时成立, 故 $$b=4$$ 不可能存在非负整数 $$a$$ 满足要求.

$\big(P(a+2),P(a+5)\big)=3,\; \big(P(a+1),P(a+4)\big)=3$

Lemma 1   当 $$n$$ 为正整数, $$9\not\mid P(n)$$.

$4(n^2+n+1)=(2n+1)^2+3,$

Lemma 2   当 $$n$$, $$m$$ 都是正整数,  $$\big(P(n),P(n+m)\big)\mid (m^3+3m)$$.

$$\begin{split}n\big(P(n+m)-P(n)\big)-2mP(n)&=\big(2mn^2+(m^2+m)n\big)-\big(2mn^2+2mn+2m\big)\\&=\big(m^2-m\big)n-2m.\end{split}$$

$\big(m-1\big)X-2Y=\big(m-1\big)(m^2+2nm+m)-2\Big(\big(m^2-m\big)n-2m\Big)=m^3+3m.$

Lemma 3  命 $$p$$ 为素数. 同余方程

$x^2+a_1x+a_0\equiv 0\pmod p$

Lemma 4  当整数 $$t \equiv n, n^2\pmod{P(n)}$$, 必定 $$P(t) \equiv 0 \pmod{P(n)}$$.

$P(t)\equiv n^4 + n^2 + 1 = (n^2-n+1) (n^2+n+1) \equiv 0 \pmod {P(n)}.$

$$n \equiv 1 \pmod 3$$, 则 $$P(n) \equiv 0 \pmod 3$$;
$$n \equiv 2,4 \pmod 7$$, 则 $$P(n)\equiv 0 \pmod 7$$;
$$n\equiv7, 49\pmod{57}$$, 则 $$P(n)\equiv 0\pmod {57}$$. 这导致当 $$n\equiv 7, 11\pmod {19}$$ 时, 有 $$P(n)\equiv0\pmod{19}$$

## Problem 5 ( Russia)

$$\begin{split}&\hspace3.25ex(x-1)(x-4)(x-5)(x-8)\dotsm(x-2013)(x-2016)\\&=(x-2)(x-3)(x-6)(x-7)\dotsm(x-2014)(x-2015)\end{split}$$

\begin{split}
(x-1)(x-4)&\lt(x-2)(x-3);\\
(x-5)(x-8)&\lt(x-6)(x-7);\\
&\vdots\\
(x-2013)(x-2016)&\lt(x-2014)(x-2015).\end{split}

\begin{split}
(x-4)(x-5)&\gt(x-3)(x-6);\\
(x-8)(x-9)&\gt(x-7)(x-10);\\
&\vdots\\
(x-2012)(x-2013)&\gt(x-2011)(x-2014).\end{split}

$x-1\gt x-2\gt0,$

$-(x-2016)\gt-(x-2015)\gt0.$

\begin{equation*}-(x-1)(x-4)(x-5)(x-8)\dotsm(x-2013)(x-2016)\gt-(x-2)(x-3)(x-6)(x-7)\dotsm(x-2014)(x-2015)\end{equation*}

## Problem 6 (The Czech Republic)

(a) 在 $$n$$ 为奇数, 把青蛙放在 $$P_1$$, $$P_3$$, $$\dotsc$$, $$P_{2n-1}$$, 可以实现他的愿望.

$$P_i$$, $$P_j$$ 之间有奇数个点(不包括这两点本身). 这奇数个点组成集合 $$S$$. 不以 $$S$$ 中的点为端点的弦如果与线段 $$P_iA$$, $$P_jA$$ 中的一个相交, 则必定也与另一个相交, 因为此弦的端点不属于 $$S$$, 即不能在$$P_i$$, $$P_j$$ 之间; 以 $$S$$ 中的点为端点的弦必定与线段 $$P_iA$$, $$P_jA$$ 中的恰好一个相交. $$S$$ 有奇数个点, 这表明线段 $$P_iA$$, $$P_jA$$ 与所有的 $$n$$ 条弦的交点个数的奇偶性不同. 进而, 从 $$P_i$$, $$P_j$$ 出发的青蛙, 任何时刻都不会落在同一个交点.

(b) 杰夫想实现他的理想的话, 青蛙不能放在圆上相邻的端点.

### Annotations

1. 今年的题, 如果时间充裕一点, 应该都能做出来
2. 又一次的证明, 出现精彩万分的数论题是多么不容易.
3. 最好的题是第 3 题, 毫无疑问. 本题需要一点智慧. 如果知道一点代数数论, 本题是有好几种突破口, 请参看续集 IMO 2016 solutions II.
4. 题 6 不适合作为 Q3 或 Q6, 难度不够, 似乎比 Q2 容易.
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# Day $$1$$

Monday, July 11, 2016

Problem 1. Triangle $$BCF$$ has a right angle at $$B$$. Let $$A$$ be the point on line $$CF$$ such that $$FA=FB$$ and $$F$$ lies between $$A$$ and $$C$$. Point $$D$$ is chosen so that $$DA=DC$$ and $$AC$$ is the bisector of $$\angle DAB$$. Point $$E$$ is chosen so that $$EA=ED$$ and $$AD$$ is the bisector of $$\angle EAC$$. Let $$M$$ be the midpoint of $$CF$$. Let $$X$$ be the point such that $$AMXE$$ is a parallelogram(where $$AM\parallel EX$$ and $$AE\parallel MX$$). Prove that $$BD$$, $$FX$$ and $$ME$$ are concurrent.

Problem 2. Find all positive integers $$n$$ for which each cell of $$n \times n$$ table can be filled with one of the letters $$I$$, $$M$$ and $$O$$ in such a way that:

• in each row and each column, one third of the entries are $$I$$, one third are $$M$$ and one third are $$O$$; and
• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are $$I$$, one third are $$M$$ and one third are $$O$$.

Note. The rows and columns of an $$n\times n$$ table are each labelled $$1$$ to $$n$$ in a natural order. Thus each cell corresponds to a pair of positive integer $$(i$$, $$j)$$ with $$1 \leqslant i$$, $$j\leqslant n$$. For $$n\gt 1$$, the table has $$4n-2$$ diagonals of two types. A diagonal of first type consists all cells $$(i$$, $$j)$$ for which $$i+j$$ is a constant, and the diagonal of this second type consists all cells $$(i$$, $$j)$$ for which $$i-j$$ is constant.

Problem 3. Let $$P=A_1A_2\dotsm A_k$$ be a convex polygon in the plane. The vertices $$A_1$$, $$A_2$$, $$\dotsc$$, $$A_k$$ have integral coordinates and lie on a circle. Let $$S$$ be the area of $$P$$. An odd positive integer $$n$$ is given such that the squares of the side lengths of $$P$$ are integers divisible by $$n$$. Prove that $$2S$$ is an integer divisible by $$n$$.

# Day $$2$$

Tuesday, July 12, 2016

Problem 4. A set of postive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let $$P(n)=n^2+n+1$$. What is the least possible positive integer value of $$b$$ such that there exists a non-negative integer $$a$$ for which the set

$\{P(a+1),P(a+2),\dotsc,P(a+b)\}$

is fragrant?

Problem 5. The equation

$(x-1)(x-2)\dotsm(x-2016)=(x-1)(x-2)\dotsm (x-2016)$

is written on the board, with $$2016$$ linear factors on each side. What is the least possible value of $$k$$ for which it is possible to erase exactly $$k$$ of these $$4032$$ linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?

Problem 6. There are $$n\geqslant 2$$ line segments in the plane such that every two segments cross, and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it, facing the other endpoint. Then he will clap his hands $$n-1$$ times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will every occupy the same intersection point at the same time.

(a) Prove that Geoff can always fulfill his wish if $$n$$ is odd.

(b) Prove that Geoff can never fulfill his wish if $$n$$ is even.

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$\angle MKH=\angle YKF.$

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# 2015 第 56 届 IMO 解答

## Problem 1 (Netherlands 荷兰)

(a)  $$n$$ 是奇数.

$$\mathcal S$$ 亦是无中心的. $$\mathcal S$$ 中任意三个不同的点 $$A$$, $$B$$, $$C$$ 形成的三角形的外心即是 $$\omega$$ 的中心. $$\omega$$ 的中心显然不在 $$\omega$$ 上, 当然也就不在 $$\mathcal S$$ 中.

$$n$$ 是偶数, 记 $$n=2k+2$$, 这里 $$k$$ 是正整数.

$\mathcal S=\{O, A_1, A_2, \dotsc, A_k, B_1, B_2, \dotsc, B_k, C\}.$

(b) 先指出, 若 $$n\gt3$$ 为偶数, 不存在由 $$n$$ 个点构成的平衡且无中心的点集.

$\frac1n\dbinom n2=\frac{n-1}2$

$X_1Y_1, X_2Y_2, \dotsc , X_{\frac n2}Y_{\frac n2}.$

(a) 已经说明对所有不小于 $$3$$ 的奇数 $$n$$, 存在由 $$n$$ 个点构成的平衡且无中心的点集.

## Problem 2 (Serbia 塞尔维亚)

$$a^2-c=2^u,$$

$$a=2^v,$$

$$c=2^w+1,$$

$$2^{2v}-2=2^u.$$

$$2^u\equiv2\pmod4$$ 定出 $$u=1$$, 进而 $$v=1$$. 于是 $$a=2$$, $$b=a=2$$.

$$2^{2v}-c=1.$$

$$2^{2v}=2^w+2.$$

$$2^w\equiv2\pmod4$$ 定出 $$w=1$$, $$c=2^1+1=3$$, 进而 $$v=1$$.  于是 $$a=2$$, $$b=a=2$$.

$$ab-c=2^x,\;ca-b=2^y,\;bc-a=2^z$$

$2^x-2^y=(ab-c)-(ca-b)=(b-c)(a+1)$

$$2^y\mid(b-c)(a+1).$$

$2^x+2^y=(ab-c)+(ca-b)=(b+c)(a-1)$

$$2^y\mid(b+c)(a-1).$$

$$a-1$$, $$a+1$$ 是两个相差 $$2$$ 的正整数, 至少有一个不是 $$4$$ 的倍数.

$c(b+1)\leqslant ca=2^y+b\leqslant 2(b+c)+b$

$bc\lt3b+c\lt4b,$

$$c=2$$.

$9(2b^2-b-2)=(3b-2)(6b+1)-16$

$$c=3$$.

$$2^x=ab-3\gt3^2-3\gt4$$ 定出 $$x\gt2$$, 并且 $$ab-3$$ 是偶数, 从而 $$ab$$ 是, 进而 $$a$$, $$b$$ 都是, 奇数.  $$2^x$$, $$2^y$$, $$2^z$$ 都是偶数, 从而 $$x$$, $$y$$, $$z$$ 都是正整数, $$x\gt y\gt z\geqslant1$$.

$$(3a-b)+(3b-a)=2^y+2^z$$

$$2^y=3a-b\gt2a$$ 说明 $$a\lt2^{y-1}$$.

$$(8)$$, $$(9)$$ 表明

$2^y\mid(b-3)(a+1),\;2^y\mid(b+3)(a-1).$

$$b$$ 为奇数, 所以 $$b-3$$, $$b+3$$ 恰有一个不被 $$4$$ 整除.

$2^y-2^z=8.$

$2^z(2^{y-z}-1)=8.$

$$z$$, $$y-z$$ 都是正整数, 因此 $$2^z=8$$, $$2^{y-z}-1=1$$. 所以 $$z=3$$, $$y-z=1$$. 后一式说明 $$y=4$$. 于是

$a=2^{4-1}-1=7,\;b=2^{4-1}-3=5.$

##### 解答二

$(2^v)^2-(2^w+1)=2^u.$

$2^{2v}=2^w+1+2^u.$

$$v\geqslant1$$ 说明

$2^w+1+2^u\equiv0\pmod4.$

$$2^w+2^u$$ 为奇数表明 $$w$$, $$u$$ 一个是 $$0$$, 另一个大于 $$0$$.

$$w=0$$ 导出 $$4\mid (1+1+2^u)$$, 进而 $$2^u$$ 是偶数但不被 $$4$$ 整除, 于是 $$u=1$$. 此时 $$v=1$$,

$a=2,\; b=a=2,\; c=2^0+1=2.$

$a=2,\; b=a=2,\; c=2^1+1=3.$

## Problem 3 (Ukraine 乌克兰)

$$AX$$ 是 $$\Gamma$$ 的直径蕴涵 $$\angle AQX = 90^{\circ}$$. $$\angle AQH = 90^{\circ}$$ 表明 $$\angle AQX = \angle AQH$$, 进而 $$Q$$, $$H$$, $$X$$ 三点共线.

$$AX$$ 是 $$\Gamma$$ 的直径蕴涵 $$XB$$, $$XC$$ 分别与 $$AB$$, $$AC$$ 垂直. $$H$$ 是 $$\triangle ABC$$ 的垂心宣示 $$HC$$, $$HB$$ 分别与 $$AB$$, $$AC$$ 垂直. 因此, 四边形 $$BXCH$$ 是平行四边形. $$M$$ 是 $$BC$$, 进而也是 $$XH$$, 的中点. 从而, $$X$$, $$M$$, $$H$$ 三点共线.

IMO 2015 Problem 3 Proof 1

$\angle HKQ=\angle HYX=\angle HFM= 90^{\circ},$

$$QK$$ 与 $$XY$$ 的延长线交于点 $$V$$. 于是

$VQ\cdot VK=VX\cdot VY.$

$$AX$$ 为 $$\Gamma$$ 的直径蕴涵 $$XY\perp AY$$. 于是 $$BC\perp AF$$ 表明 $$BC\parallel XY$$. 设 $$U$$ 是 $$VH$$ 与 $$BC$$ 的交点. $$H$$ 是 $$\triangle ABC$$ 的垂心揭示 $$F$$ 是 $$HY$$ 的中点. 至此, $$U$$ 是 $$HV$$ 的中点. $$\angle HKV= 90^{\circ}$$ 蕴涵 $$UK=UH$$. $$UH$$ 与三角形 $$HKQ$$ 的外接圆相切, 故而 $$UK$$ 也与三角形 $$HKQ$$ 的外接圆相切.

$$U$$ 在三角形 $$HMF$$ 的外接圆与三角形 $$FKM$$ 的外接圆的根轴 $$MF$$ 上. $$UH$$ 与三角形 $$HMF$$ 的外接圆相切. 然后, $$UK=UH$$ 蕴涵 $$UK$$ 是三角形 $$FKM$$ 的外接圆的切线. 于是 $$HKQ$$ 的外接圆与三角形 $$FKM$$ 的外接圆相切, 因为这两个圆都与 $$UK$$ 切于点 $$K$$.

##### 解答二

IMO 2015 Problem 3 Proof 2

$\angle KHQ = \angle KXA =\angle KYA=\angle KYH.$

$$UH\perp HM$$, $$HF\perp MU$$, 根据射影定理

$UK^2=UH^2=UF\cdot UM.$

##### 解答三

IMO 2015 Problem 3 Proof 3

$$H$$ 是 $$\triangle ABC$$ 的垂心揭示 $$\angle QMC = \angle YMC$$, 于是,  四边形 $$BYCQ$$ 是调和四边形, 由此 $$\angle QBY = \angle QMC$$.

$$ZQ$$ 为 $$\Gamma$$ 的直径, 于是  $$\angle ZKY+\angle QBY= 90^{\circ}$$. 故此

$\angle HKY=\angle ZKY= 90^{\circ}-\angle QBY=90^{\circ}-\angle QMC=\angle MHY,$

##### 解答四

IMO 2015 Problem 3 Proof 4

$$QK$$ 与 $$BC$$ 的延长线交于点 $$W$$. 既然 $$HK\perp KW$$, $$HF\perp FW$$, 于是 $$H$$, $$F$$, $$W$$, $$K$$ 四点共圆. 故而

$\angle KFW=\angle KHW.$

$\angle QHS=180^{\circ}-\angle MHS=180^{\circ}-\angle MXS^\prime=180^{\circ}-\angle QK^\prime S.$

$\angle KFM=180^{\circ}-\angle KFW=180^{\circ}-\angle KHW=\angle KTS.$

##### 解答五

IMO 2015 Problem 3 Proof 5

$\angle KHQ=\angle HQZ+\angle HZQ=\angle AYQ+\angle QYK=\angle AYK.$

$\angle XQK+\angle AYK=90^{\circ}.$

$\angle KHQ=90^{\circ}-\angle KQH=\angle HYK.$

$\angle MKH=\angle YKF.$

$\angle QHK+\angle MKH=\angle HYK+\angle YKF=\angle HFK.$

$$QK\perp HK$$, $$HF\perp FC$$ 给出 $$\angle QHK=90^{\circ}-\angle KQH$$,  $$\angle HFK=90^{\circ}-\angle KFC$$,  我们有

$(90^{\circ}-\angle KQH)+\angle MKH=90^{\circ}-\angle KFC.$

$\angle KFC+\angle MKH=\angle KQH.$

## Problem 4 (Vaggelis Psychas&Silouanos Brazitikos , Greece 希腊)

IMO 2015 Problem 4 Proof 1

$$FG$$ 是 $$\Gamma$$ 和 $$\Omega$$ 的公共弦蕴涵 $$AO$$ 是 $$FG$$ 的中垂线. 于是, $$\widehat{AF}=\widehat{AG}$$, 进而

$\angle AGR=\angle AGF=\angle ACF=\angle ACG.$

$\angle GRC=180^{\circ}-\angle ARG=180^{\circ}-\angle AGC=\angle ABC=\angle KFD.$

$\angle GFD=\angle GEC=\angle GLC=\angle GLR.$

$\angle XGF=\angle GRC-\angle GLR=\angle KFD-\angle GFD=\angle XFG.$

## Problem 5 (Dorlir Ahmeti，Albania 阿尔巴尼亚)

$$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x).$$

$f(0+a)+a=0+a.$

$f(f(a))+f(0)=0+f(a)+af(0).$

$$f(x+f(x))=x+f(x).$$

$f(x+f(x+1))+f(x)=x+f(x+1)+f(x).$

$$f(x+f(x+1))=x+f(x+1).$$

$$f(x)+f(-x^2)=x-xf(x).$$

$$f(-1)=-1.$$

$$f(1)=1.$$

$f(1+f(x+f(x)))+f(x-1+f(x))=1+f(x+f(x))+(x-1+f(x)).$

$f(1+x+f(x))+(x-1+f(x))=1+(x+f(x))+(x-1+f(x)).$

$$f(1+x+f(x))=1+x+f(x).$$

$f(x+f(x-1))+f(-x)=x+f(x-1)-f(x).$

$x+f(x-1)+f(-x)=x+f(x-1)-f(x).$

$$f(-x)=-f(x).$$

$$f(-x)+f(-x^2)=-x+xf(-x).$$

$$(14)$$, $$(19)$$ 两式相减, 得

$(f(x)+f(-x^2))-(f(-x)+f(-x^2))=(x-xf(x))-(-x+xf(-x)).$

$f(x)-f(-x)=2x-x(f(x)+f(-x)).$

$$(18)$$ 说明这就是 $$2f(x)=2x$$. 所以, $$f(x)=x$$.

$$f(x+f(x+1))=x+f(x+1).$$

$f(f(x-1+f(x)))+2=f(x-1+f(x))+2(x-1+f(x)).$

$x-1+f(x)+2=x-1+f(x)+2(x-1+f(x)).$

## Problem 6 (Ross Atkins&Ivan Guo, Australia 澳大利亚)

$N_j=\{1, 2, 3, \dotsc, j\},\;D_j=\{d_1, d_2, d_3, \dotsc, d_j\}.$

$d_j\leqslant j+2015\leqslant(L-2015)+2015=L,$

$$Y$$ 中的元素小于 $$N$$, 当然更小于 $$m+2015$$; $$D_m$$ 中的元素 $$d_k$$($$k\leqslant m$$) 必定 $$d_k\leqslant k+2015\leqslant m+2015$$, 因此 $$Y\cup D_m\subset N_{m+2015}$$. 此外, 当 $$k\leqslant m+1$$ 时, $$k$$ 要么属于 $$Y$$, 要么存在正整数 $$t\lt k$$, 使得 $$d_t=k$$. 因此, $$N_{m+1}\subset Y\cup D_m$$, $$Y\cup D_m$$ 恰有 $$b-1$$ 个元素属于 $$N_{m+2015}\setminus N_{m+1}$$. 于是

$$\sum_{k=1}^{m+b}k\leqslant\sum_{k=1}^md_k+\sum_{k=1}^bz_k\leqslant\sum_{k=1}^{m+1}k+\sum_{k=m+2017-b}^{m+2015}k.$$

$$\sum_{k=1}^md_k+\sum_{k=1}^bz_k=\sum_{k=1}^{m+1}k,$$

$$\sum_{k=1}^nd_k+\sum_{k=1}^bz_k=\sum_{k=1}^{n+1}k.$$

$$\sum_{k=m+1}^nd_k=\sum_{k=m+2}^{n+1}k.$$

$$\sum_{k=m+1}^nd_k=\sum_{k=m+1}^n\Big(k+1\Big).$$

$$\sum_{k=m+1}^n\Big(a_k-b\Big)=0.$$

$$\left(\sum_{k=1}^nd_k+\sum_{k=1}^bz_k\right)-\left(\sum_{k=1}^md_k+\sum_{k=1}^bz_k\right)\geqslant\sum_{k=1}^{n+b}k-\left(\sum_{k=1}^{m+1}k+\sum_{k=m+2017-b}^{m+2015}k\right),$$

$$\left(\sum_{k=1}^nd_k+\sum_{k=1}^bz_k\right)-\left(\sum_{k=1}^md_k+\sum_{k=1}^bz_k\right)\leqslant\left(\sum_{k=1}^{n+1}k+\sum_{k=n+2017-b}^{n+2015}k\right)-\sum_{k=1}^{m+b}k.$$

$$(27)$$ 即

$$\sum_{k=m+1}^nd_k\geqslant\sum_{k=m+2}^{n+b}k-\sum_{k=m+2017-b}^{m+2015}k.$$

$$\sum_{k=m+2}^{n+b}k=\sum_{k=m+2}^{m+b}k+\sum_{k=m+b+1}^{n+b}k=\sum_{k=2}^b\Big(k+m\Big)+\sum_{k=m+1}^n\Big(k+b\Big).$$

$$\sum_{k=m+2017-b}^{m+2015}k=\sum_{k=2}^b\Big(k+m+2015-b\Big)=\sum_{k=2}^b\Big(k+m\Big)+\sum_{k=2}^b\Big(2015-b\Big).$$

$$(29)$$ 就是

$$\sum_{k=m+1}^nd_k\geqslant\sum_{k=m+1}^n\Big(k+b\Big)-\sum_{k=2}^b\Big(2015-b\Big).$$

$$\sum_{k=m+1}^n\Big(d_k-k-b\Big)\geqslant-\sum_{k=2}^b\Big(2015-b\Big).$$

$$\sum_{k=m+1}^n\Big(a_k-b\Big)\geqslant-\Big(b-1\Big)\Big(2015-b\Big).$$

$$\sum_{k=m+1}^nd_k\leqslant\sum_{k=n+2017-b}^{n+2015}k+\sum_{k=m+b+1}^{n+1}k;$$

$$\sum_{k=m+1}^nd_k\leqslant\sum_{k=n+2017-b}^{n+2015}k-\sum_{k=n+2}^{m+b}k.$$

$$\begin{split}\sum_{k=n+2017-b}^{n+2015}k&=\sum_{k=2}^b\Big(k+n+2015-b\Big)\\&=\sum_{k=2}^b\Big(k+n\Big)+\sum_{k=2}^b\Big(2015-b\Big)\\&=\sum_{k=n+2}^{n+b}k+\sum_{k=2}^b\Big(2015-b\Big),\end{split}$$

$$\begin{split}\sum_{k=m+1}^nd_k&\leqslant\sum_{k=m+b+1}^{n+1}k+\sum_{k=n+2}^{n+b}k+\sum_{k=2}^b\Big(2015-b\Big)\\&=\sum_{k=m+b+1}^{n+b}k+\sum_{k=2}^b\Big(2015-b\Big)\\&=\sum_{k=m+1}^n\Big(k+b\Big)+\sum_{k=2}^b\Big(2015-b\Big);\end{split}$$

$$\begin{split}\sum_{k=m+1}^nd_k&\leqslant\sum_{k=n+2}^{n+b}k+\sum_{k=2}^b\Big(2015-b\Big)-\sum_{k=n+2}^{m+b}k\\&=\sum_{k=m+b+1}^{n+b}k+\sum_{k=2}^b\Big(2015-b\Big)\\&=\sum_{k=m+1}^n\Big(k+b\Big)+\sum_{k=2}^b\Big(2015-b\Big).\end{split}$$

$$\sum_{k=m+1}^n\Big(a_k-b\Big)\leqslant\Big(b-1\Big)\Big(2015-b\Big).$$

$\left|\sum_{k=m+1}^n\Big(a_k-b\Big)\right|\leqslant\Big(b-1\Big)\Big(2015-b\Big)\leqslant1007^2.$

### Annotations

1. 本届 IMO 其实不是最难, 恰恰相反, 应该是史上最简单, 大概与 1989 年相当. 分数低与训练有关, 不完全真实反映试题的难易.
2. 题 1(a) 是陈题, 已经多次出现; 把 (a) 中对偶数 $$2k+2$$ 构造的平衡点集的点 $$C$$ 去掉, 得到的 $$2k+1$$ 个点形成的点集亦是平衡的; 找出 (a) 所有符合要求的构造可能是困难的. 也就是说, 对整数 $$n\geq3$$, 希望找出所有的由 $$n$$ 个点构成的平衡点集.
3. 题 3 的几何, 其实比最近几年的以 3 或 6 出现的几何简单很多. 通常, 如果考察的是三角形一些常见的点及其外接圆的性质, 不应当有太多人做不出的.
4. 题 5 是函数方程. 一般来说, 一个函数方程可能是真正有难度的问题, 如果任何解法都避不了某关键的性质, 诸如特定集合的一个非常特殊的性质, 或者必须使用不等式.
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# Day $$1$$

Friday, July 10, 2015

Problem 1. We say that a finite set $$\mathcal S$$ of points in the plane is balanced if, for any two different points $$A$$ and $$B$$ in $$\mathcal{S}$$, there is a point $$C$$ in $$\mathcal{S}$$ such that $$AC=BC$$. We say that $$\mathcal{S}$$ is centre-free if for any three different points $$A$$, $$B$$ and $$C$$ in $$\mathcal{S}$$, there is no points $$P$$ in $$\mathcal{S}$$ such that $$PA=PB=PC$$.

(a) Show that for all integers $$n\ge 3$$, there exists a balanced set consisting of $$n$$ points.

(b) Determine all integers $$n\ge 3$$ for which there exists a balanced centre-free set consisting of $$n$$ points.

Problem 2.  Determine all triples $$(a, b, c)$$ of positive integers such that each of the numbers

$ab-c,\;bc-a, \;ca-b$

is a power of $$2$$.

(A power of $$2$$ is an integer of the form $$2^n$$, where $$n$$ is a non-negative integer. )

Problem 3. Let $$ABC$$ be an acute triangle with $$AB \gt AC$$. Let $$\Gamma$$ be its cirumcircle, $$H$$ its orthocenter, and $$F$$ the foot of the altitude from $$A$$. Let $$M$$ be the midpoint of $$BC$$. Let $$Q$$ be the point on $$\Gamma$$ such that $$\angle HQA = 90^{\circ}$$ and let $$K$$ be the point on $$\Gamma$$ such that $$\angle HKQ = 90^{\circ}$$. Assume that the points $$A$$, $$B$$, $$C$$, $$K$$ and $$Q$$ are all different, and lie on $$\Gamma$$ in this order.

Prove that the circumcircles of triangles $$KQH$$ and $$FKM$$ are tangent to each other.

# Day $$2$$

Saturday, July 11, 2015

Problem 4. Triangle $$ABC$$ has circumcircle $$\Omega$$ and circumcenter $$O$$. A circle $$\Gamma$$ with center $$A$$ intersects the segment $$BC$$ at points $$D$$ and $$E$$, such that $$B$$, $$D$$, $$E$$, and $$C$$ are all different and lie on line $$BC$$ in this order. Let $$F$$ and $$G$$ be the points of intersection of $$\Gamma$$ and $$\Omega$$, such that $$A$$, $$F$$, $$B$$, $$C$$, and $$G$$ lie on $$\Omega$$ in this order. Let $$K$$ be the second point of intersection of the circumcircle of triangle $$BDF$$ and the segment $$AB$$. Let $$L$$ be the second point of intersection of the circumcircle of triangle $$CGE$$ and the segment $$CA$$.

Suppose that the lines $$FK$$ and $$GL$$ are different and intersect at the point $$X$$. Prove that $$X$$ lies on the line $$AO$$.

Problem 5. Let $$\Bbb R$$ be the set of real numbers. Determine all functions $$f\colon\Bbb R\to\Bbb R$$ satisfying the equation

$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$

for all real numbers $$x$$ and $$y$$.

Problem 6. The sequence $$a_1,a_2,\dotsc$$ of integers satisfies the following conditions:

(i) $$1\leqslant a_j\leqslant2015$$ for all $$j\geqslant1$$;

(ii) $$k+a_k\neq \ell+a_\ell$$ for all $$1\leqslant k\lt \ell$$.

Prove that there exist two positive integers $$b$$ and $$N$$ such that

$\left\vert\sum_{j=m+1}^n(a_j-b)\right\vert\leqslant1007^2$

for all integers $$m$$ and $$n$$ satisfying $$n\gt m\geqslant N$$.

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