Question

请翻译大神帮忙翻译一段文献，不要翻译工具的。翻译通顺就好！！分数可以追加的！！

Our last example is also a geometric one, but here the “divisions”, i. e. the boxes, are not
whole parts of the given geometric figure, indeed, they are points in it. The difficulty is
realizing that it is these precise points that will help (or having a hunch of what those
points are by making a few drawings).
We are given a square and 9 lines cutting through it. Each line cuts the square
into two quadrilaterals, and the ration of the areas of those quadrilaterals is 2:3.
Prove that at least three of these 9 lines pass through the same point.
Since each line divides the square in two quadrilaterals, each line cuts the square
at two non consecutive edges. The figures obtained are trapezoids with two right angles,
and their area is equal to their base (here the side of the square)
multiplied by the height of their “midline”, the line perpendicular to
their two parallel edges and at equal distance from those two edges.
Since for any of those trapezoids their base is the same, the ration of
their midlines must be 2:3. But there are only two possible midlines
for the trapezoids, so we get four different points, call them
midpoints, two on each midline, where the midline of two
trapezoids intersects with one of the 9 given lines. Every one of those lines must pass
through one of these midpoints. We have 4 points and 9 lines, so there is at least one
point through which pass three lines.
We have shown that, given 9 lines, each cutting a square into two trapezoids
whose areas are in the ratio 2:3, at least three of the lines pass through the same point.
We now move on to more difficult problems, which generally necessitate a little
more twist, or work, before applying the pigeonhole principle

Answer 1

Our last example is also a geometric one, but here the “divisions”, i. e. the boxes, are not whole parts of the given geometric figure, indeed, they are points in it.
我们最后的一个例子仍然是关于几何的，但是这里用的是“分割”，例如，这些方框，不是这个几何图形的全貌，事实上，这几个方框只是这个几何图形中的几个点。
The difficulty is realizing that it is these precise points that will help (or having a hunch of what those points are by making a few drawings).
理解这里的困难之处在于这几个精确的点是否有用（或者说可以通过做一些辅助来预测到这几个点是干什么用的）。
We are given a square and 9 lines cutting through it. Each line cuts the square into two quadrilaterals, and the ration of the areas of those quadrilaterals is 2:3.
我们有一个有9条线穿过的正方形，每条线都将这个正方形切成两个面积比为2:3的四边形。
Prove that at least three of these 9 lines pass through the same point.
试证明这9条线里面至少有3条相交于同一点。
Since each line divides the square in two quadrilaterals, each line cuts the square at two non consecutive edges.
因为每条线都将这个正方形分成两个四边形，每条线都切割在了正方形的两条不相交的边上。
The figures obtained are trapezoids with two right angles, and their area is equal to their base (here the side of the square) multiplied by the height of their “midline”, the line perpendicular to their two parallel edges and at equal distance from those two edges.
最后得出来的图形是一些有着两个直角的梯形，并且这些梯形的面积都等于它们的底边（这里指正方形的边）乘以它们高的“中线”，这条中线垂直于梯形的两条平行的边上，并且这条中线到两边的距离相等。
Since for any of those trapezoids their base is the same, the ration of their midlines must be 2:3.
因为所有梯形的底边都相同，那么他们的中线和中线的比必然是2:3。
But there are only two possible midlines for the trapezoids, so we get four different points, call them midpoints, two on each midline, where the midline of two trapezoids intersects with one of the 9 given lines.
但是这些梯形有两条可行的中线，所以我们取4个点，叫它们中点，每条中线上都有两个中点，并且每两个梯形的中线和已知的9条线相交。
Every one of those lines must pass through one of these midpoints.
每条线都必须通过一个中点。
We have 4 points and 9 lines, so there is at least one point through which pass three lines.
现在我们有4个点和9条线，所以这里至少有3条线相交于一点。
We have shown that, given 9 lines, each cutting a square into two trapezoids whose areas are in the ratio 2:3, at least three of the lines pass through the same point.
我们已经知道，已知的9条线，每条线将正方形切割成两个面积比为2:3的梯形，至少有3条线会相交于一点。
We now move on to more difficult problems, which generally necessitate a little more twist, or work, before applying the pigeonhole principle
我们现在再看一看更难的问题，在应用鸽巢原理之前需要一点更多的思考和工作。

*****************************
本人是英语系大三学生，长时间不接触数学，有不恰当的地方还请见谅！：）

Answer 2

Our last example is also a geometric one, but here the “divisions”, i. e. the boxes, are not
whole parts of the given geometric figure, indeed, they are points in it. The difficulty is
realizing that it is these precise points that will help (or having a hunch of what those
points are by making a few drawings).
我们最后一个例子也是几何的，但是这里的分格 也就是小方格 不是给出的几何图形的整体，实际上它们是里面的点。难点是理会到这些精确的点是否有作用（或通过画一些草图来推测那些点是什么）
We are given a square and 9 lines cutting through it. Each line cuts the square
into two quadrilaterals, and the ration of the areas of those quadrilaterals is 2:3.
Prove that at least three of these 9 lines pass through the same point.
我们现在给出一个正方形，用九条线分割它。每条线都将这个正方形分割成两个矩形，且这两个矩形的面积之比为2：3。证明：九条线中至少有三条线经过同一点。
Since each line divides the square in two quadrilaterals, each line cuts the square
at two non consecutive edges. The figures obtained are trapezoids with two right angles,
and their area is equal to their base (here the side of the square)
multiplied by the height of their “midline”, the line perpendicular to
their two parallel edges and at equal distance from those two edges.
因为每根线都将这个正方形分成两个矩形，且每根线切割的都是经过两个不连续的边缘。这样就会得到有两个直角的不规则四边形，且它们的面积等于它们的底（这里是这个正方形的边）乘以它们高的“中线”，这根线垂直与它们两平行边，且两边到这条线的距离相等。
Since for any of those trapezoids their base is the same, the ration of
their midlines must be 2:3. But there are only two possible midlines
for the trapezoids, so we get four different points, call them
midpoints, two on each midline, where the midline of two
trapezoids intersects with one of the 9 given lines. Every one of those lines must pass
through one of these midpoints. We have 4 points and 9 lines, so there is at least one
point through which pass three lines.
因为任何这些不规则四边形的底都是一样的，它们中线的比必定是2:3。但是这些不规则四边形只有两种可能的中线，所以我们得到四个不同的点，我们叫它中点，每条中线上两个，这样每两个不规则四边形的中线就与给出的九条线中的一条相交。这些线中的每一条都必定经过这些中点中的某一点。我们有四个点九条线，所以至少有一点经过三条线。
We have shown that, given 9 lines, each cutting a square into two trapezoids
whose areas are in the ratio 2:3, at least three of the lines pass through the same point.
我们已经证明 给出九条线每条线都将一正方形分割成两个不规则的四边形且四边形的面积比为2:3 这些线中至少会有三条经过同一点。
We now move on to more difficult problems, which generally necessitate a little
more twist, or work, before applying the pigeonhole principle
现在我们来探讨更难的问题，这些在引用鸽巢原理之前一般来说都需要更多的思考或工作。
PS：本人数学一般所以有些翻译难免偏颇，译文仅供参考。还有那个 “鸽巢原理”你可在百度上查查，见谅！

Answer 3

不是数学系的，但是大概好像明白了...以下正文：
我们最后一个例子还是几何方面的，但是是从“分解”的角度来看。比如，一个方框，并不是一块一块图形构成的，实际上，它由许多“点”构成。问题在于如何认识到这些点（画一个图会帮助你明白）
假设有一个正方形，用9条线段将它分割，每条线都会把这个正方形按面积2：3的比例分割，请证明9条线中至少有3条会相交在同一个点。 既然每条线都会把这个正方形分成2个四边形，那么每条线一定会与该正方形2条不连续的边相交，那么我们得到的一定是一个至少有2个直角的梯形，它的面积等于它的底边长度乘以他中线的长度（梯形面积=边长*高）。显然所有梯形的底边都是一样长度的，那么他们高的比例一定也是2：3. 但是一个梯形肯定只能有2条中线（纵横各1条），那么我们就能得到4个中点，每条中线上有2个。每个中点至少会成为2个梯形的共同中点，且该中点一定包含于已知的9条线。（我这样跟你解释吧：我们每个梯形有4个中点，每个中点至少要经过2条线，但是我们已知有9条这样的线，那么就必须有1个中点要经过3条线---原句翻译说不到这么透彻）
现在我们已经证明了上面这个了，下面我要来点更难的，需要把脑瓜子拧成麻花那样思考才行。接下来就是鸽巢原理。（它的简单形式是 ： 把n＋1个物体放入n个盒子里，则至少有一个盒子里含有两个或两个以上的物体 。）

Answer 4

我们的最后一个例子是几何，但这里的“师”，即盒，不

整个零件的几何图形，事实上，他们是在它。困难的是

实现它是这些精确的点，将帮助（或有预感那些什么

点是通过做一些图纸）。

我们有一个广场和9线贯穿它。每个线切割方

为四边形，和定量的地区的四边形是2 : 3。

证明至少三这些9线通过相同的一点。

由于每个分广场2四边形，每线切割方

在非连续的边缘。人物获得双直角梯形，

其面积等于其基地（这里广场侧）

乘以高度的“中线”，本线垂直

他们的两根平行的边缘和等距离从边。

因为任何一种梯形基础是相同的，比

中线必须3。但还有一个可能的中线

为梯形，所以我们得到四个不同的点，称他们为

中点，2每中线，在中线2

梯形相交与9个给定线。每一个这些行必须通过

通过这些中点。我们有4分和9线，那么至少有一个

点通过通过三线。

我们已经表明，给定9行，每一平方成梯形切割

该地区是在比为2 : 3，至少三的线路通过相同的一点。

现在我们进入更困难的问题，这通常需要一点

更多的扭曲，或工作，申请前的鸽巢原理

Answer 5

我们的最后一个例子是几何，但这里的“师”，即盒，不

整个零件的几何图形，事实上，他们是在它。困难的是

实现它是这些精确的点，将帮助（或有预感那些什么

点是通过做一些图纸）。

我们有一个广场和9线贯穿它。每个线切割方

为四边形，和定量的地区的四边形是2 : 3。

证明至少三这些9线通过相同的一点。

由于每个分广场2四边形，每线切割方

在非连续的边缘。人物获得双直角梯形，

其面积等于其基地（这里广场侧）

乘以高度的“中线”，本线垂直

他们的两根平行的边缘和等距离从边。

因为任何一种梯形基础是相同的，比

中线必须3。但还有一个可能的中线

为梯形，所以我们得到四个不同的点，称他们为

中点，2每中线，在中线2

梯形相交与9个给定线。每一个这些行必须通过

通过这些中点。我们有4分和9线，那么至少有一个

点通过通过三线。

我们已经表明，给定9行，每一平方成梯形切割

该地区是在比为2 : 3，至少三的线路通过相同的一点。

现在我们进入更困难的问题，这通常需要一点。

更多的扭曲，或工作，申请前的鸽巢原理

应该就是这样子~