2022 USAMO 第 2 题

Let $b\geq2$ and $w\geq2$ be fixed integers, and $n=b+w$. Given are $2b$ identical black rods and $2w$ identical white rods, each of side length $1$.

We assemble a regular $2n$-gon using these rods so that parallel sides are the same color. Then, a convex $2b$-gon $B$ is formed by translating the black rods, and a convex $2w$-gon $W$ is formed by translating the white rods. An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below, as well as the resulting polygons $B$ and $W$.

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[asy] size(10cm); real w = 2*Sin(18); real h = 0.10 * w; real d = 0.33 * h; picture wht; picture blk; draw(wht, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle); fill(blk, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle, black); // draw(unitcircle, blue+dotted); // Original polygon add(shift(dir(108))*blk); add(shift(dir(72))*rotate(324)*blk); add(shift(dir(36))*rotate(288)*wht); add(shift(dir(0))*rotate(252)*blk); add(shift(dir(324))*rotate(216)*wht); add(shift(dir(288))*rotate(180)*blk); add(shift(dir(252))*rotate(144)*blk); add(shift(dir(216))*rotate(108)*wht); add(shift(dir(180))*rotate(72)*blk); add(shift(dir(144))*rotate(36)*wht); // White shifted real Wk = 1.2; pair W1 = (1.8,0.1); pair W2 = W1 + w*dir(36); pair W3 = W2 + w*dir(108); pair W4 = W3 + w*dir(216); path Wgon = W1--W2--W3--W4--cycle; draw(Wgon); pair WO = (W1+W3)/2; transform Wt = shift(WO)*scale(Wk)*shift(-WO); draw(Wt * Wgon); label("$W$", WO); /* draw(W1--Wt*W1); draw(W2--Wt*W2); draw(W3--Wt*W3); draw(W4--Wt*W4); */ // Black shifted real Bk = 1.10; pair B1 = (1.5,-0.1); pair B2 = B1 + w*dir(0); pair B3 = B2 + w*dir(324); pair B4 = B3 + w*dir(252); pair B5 = B4 + w*dir(180); pair B6 = B5 + w*dir(144); path Bgon = B1--B2--B3--B4--B5--B6--cycle; pair BO = (B1+B4)/2; transform Bt = shift(BO)*scale(Bk)*shift(-BO); fill(Bt * Bgon, black); fill(Bgon, white); label("$B$", BO); [/asy]
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Prove that the difference of the areas of $B$ and $W$ depends only on the numbers $b$ and $w$, and not on how the $2n$-gon was assembled.

设$b\geq2$和$w\geq2$为固定整数，且$n=b+w$。给出 $2b$ 相同的黑棒和 $2w$ 相同的白棒，每根边长 $1$。

我们使用这些杆组装一个常规的 $2n$ 边形，以便平行边具有相同的颜色。然后，通过平移黑棒形成凸$2b$-gon $B$，通过平移白棒形成凸$2w$-gon $W$。下面显示了当 $b=3$ 和 $w=2$ 时进行组装的一种方法的示例，以及生成的多边形 $B$ 和 $W$。

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[asy] 尺寸(10cm)；实数 w = 2*Sin(18);实际 h = 0.10 * w；实际 d = 0.33 * h；图片内容是什么；图片块；绘制(wht,(0,0)--(w,0)--(w+d,h)--(-d,h)--循环); fill(blk, (0,0)--(w,0)--(w+d,h)--(-d,h)--循环，黑色)； // 绘制(单位圆，蓝色+点)； // 原始多边形 add(shift(dir(108))*blk);添加(移位(目录(72))*旋转(324)*块)；添加(移位(目录(36))*旋转(288)*什么)；添加(移位(目录(0))*旋转(252)*块)；添加(移位(目录(324))*旋转(216)*什么)；添加(移位(目录(288))*旋转(180)*块)；添加(移位(目录(252))*旋转(144)*块)；添加(移位(目录(216))*旋转(108)*什么)；添加(移位(目录(180))*旋转(72)*块)；添加(移位(目录(144))*旋转(36)*什么)； // 白移实数 Wk = 1.2;对 W1 = (1.8,0.1)；对 W2 = W1 + w*dir(36);对 W3 = W2 + w*dir(108);对 W4 = W3 + w*dir(216);路径Wgon=W1--W2--W3--W4--循环；绘制(Wgon)；对 WO = (W1+W3)/2;变换 Wt = 移位(WO)*缩放(Wk)*移位(-WO)；绘制(Wt * Wgon)；标签(“$W$”，WO)； /* 绘制(W1--Wt*W1);绘制(W2--Wt*W2);绘制(W3--Wt*W3);绘制(W4--Wt*W4); */ // 黑移实数 Bk = 1.10;对 B1 = (1.5,-0.1);对 B2 = B1 + w*dir(0);对 B3 = B2 + w*dir(324);对 B4 = B3 + w*dir(252);对 B5 = B4 + w*dir(180);对 B6 = B5 + w*dir(144);路径Bgon = B1--B2--B3--B4--B5--B6--循环；对 BO = (B1+B4)/2;变换 Bt = 移位(BO)*缩放(Bk)*移位(-BO)；填充(Bt * Bgon，黑色)；填充(Bgon，白色)；标签(“$B$”，BO)； [/asy]

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证明 $B$ 和 $W$ 的面积之差仅取决于数字 $b$ 和 $w$，而不取决于 $2n$-gon 的组装方式。