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2009 USAMO 第 3 题

几何 · P3/P6 · 压轴题

题面据 USAMO 可核档案整理;中文题意为本站自译,公式请以原始来源为准。

USAMO 2009 P3 geometry

We define a chessboard polygon to be a polygon whose sides are situated along lines of the form x=ax = a or y=by = b, where aa and bb are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping 1×21 \times 2 rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a 3×43 \times 4 rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.

[asy] size(400); pathpen = linewidth(2.5); void chessboard(int a, int b, pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle); } chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12)); /* draw lines */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3)); [/asy]

a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.

b) Prove that such a tasteful tiling is unique.

我们将棋盘多边形定义为边沿 x=ax = ay=by = b 形式的线定位的多边形,其中 aabb 是整数。这些线将内部划分为单位方块,这些方块的阴影交替为灰色和白色,以便相邻的方块具有不同的颜色。用多米诺骨牌平铺棋盘多边形就是用不重叠的 1×21 \times 2 矩形精确地覆盖多边形。最后,一种有品味的瓷砖可以避免左下图所示的两种多米诺骨牌配置。显示了 3×43 \times 4 矩形的两个平铺;第一个很有品味,而第二个则不然,因为右上角有垂直的多米诺骨牌。

$$

[asy] 大小(400);路径笔=线宽(2.5); void chessboard(int a, int b,pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--循环); } 棋盘(2,2,(2.5,0));填充(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6));棋盘(4,3,(6,0));棋盘(4,3,(11,0)); MP("\mathrm{令人厌恶的\瓷砖}",(2.25,3),fontsize(12)); /* 画线 */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3)); [/asy]

$$

a) 证明如果棋盘多边形可以用多米诺骨牌平铺,那么它可以做得如此有品味。

b) 证明这种有品位的瓷砖是独一无二的。

提示 1

先标出固定点、动点、角、圆和长度关系。

提示 2

尝试角追、相似、圆幂、面积比、反演或坐标化中的一种。

提示 3

把关键等式还原成标准定理,或补出一个让结构闭合的辅助点。

完整解答

这页先给题面、题型和提示阶梯,完整证明留给读者逐步展开。2009 年 USAMO P3 可先归入几何:第一步把题设翻成对象、条件、目标三行;第二步沿提示寻找不变量、标准构型或关键变形;第三步补齐边界情形,并回到题目原要求核对。

这题适合先独立想一轮再打开提示。不要急着搜索完整解答,先问自己:题面里最硬的限制是哪一句?