x******g
发帖数: 318 | 1 将一个圆形的饼掰成两个，恰好两部分面积相等，求证：裂痕的长不小于饼的直径 | B****n
发帖数: 11290 | 2 I believe we can assume it's a differential curve that divides a sphere into
two equal areas. But maybe someone can give a more rigous proof for this
approximation.
In the following, I denote the curve as (t,f(t)) t in [0,1] and assume it's a
unit sphere with center at 1/2
1. We can only consider bounded variation curve otherwise its length can not
be minimal.
2. For bounded variation curve, its root is countable. Here I define root as a
real number a such that f(a)=0 and there exists a epison s
【在 x******g 的大作中提到】
: 将一个圆形的饼掰成两个，恰好两部分面积相等，求证：裂痕的长不小于饼的直径
| I***e
发帖数: 1136 | 3 太复杂了.
简单答案:
假设这两块饼在圆周上端点为A与B. 过圆心作与AB平行的直线. 注意到, 如果饼的
裂痕的长度小于直径, 则裂痕不可能与这条直线相交. 显然一块饼比半圆大, 一块
比半圆小.
其实, 过AB两点,而且长度不长于直径的线形成一个椭圆. 这个椭圆在圆心与那条
直线相切.由此得出, 就算裂痕长度等于直径,假如AB两点不是恰好某条直径的两
个端点, 那两块饼还是一大一小.
-iCare-
a
a
【在 B****n 的大作中提到】
: I believe we can assume it's a differential curve that divides a sphere into
: two equal areas. But maybe someone can give a more rigous proof for this
: approximation.
: In the following, I denote the curve as (t,f(t)) t in [0,1] and assume it's a
: unit sphere with center at 1/2
: 1. We can only consider bounded variation curve otherwise its length can not
: be minimal.
: 2. For bounded variation curve, its root is countable. Here I define root as a
: real number a such that f(a)=0 and there exists a epison s
| B****n
发帖数: 11290 | 4 Thanks for your simple answer.
Can you explain why it can not cross the straight line.
【在 I***e 的大作中提到】
: 太复杂了.
: 简单答案:
: 假设这两块饼在圆周上端点为A与B. 过圆心作与AB平行的直线. 注意到, 如果饼的
: 裂痕的长度小于直径, 则裂痕不可能与这条直线相交. 显然一块饼比半圆大, 一块
: 比半圆小.
: 其实, 过AB两点,而且长度不长于直径的线形成一个椭圆. 这个椭圆在圆心与那条
: 直线相切.由此得出, 就算裂痕长度等于直径,假如AB两点不是恰好某条直径的两
: 个端点, 那两块饼还是一大一小.
: -iCare-
:
| I***e
发帖数: 1136 | 5 对称性呀. 设B'为B点关于那条直线的对称点. 如果裂痕与那条直线相交于C,
可以构造从A到B'的裂痕与原裂痕等长. 可是, AB'实际上是一条直径.
-iCare-
【在 B****n 的大作中提到】
: Thanks for your simple answer.
: Can you explain why it can not cross the straight line.
| B****n
发帖数: 11290 | 6 very good.
thanks
【在 I***e 的大作中提到】
: 对称性呀. 设B'为B点关于那条直线的对称点. 如果裂痕与那条直线相交于C,
: 可以构造从A到B'的裂痕与原裂痕等长. 可是, AB'实际上是一条直径.
: -iCare-
| n******t
发帖数: 4406 | 7 设那个裂痕得端点是A, B.
则作过A得直径交于C.
如果B和C重合,显然成立.
否则曲线AB显然和直径AC相交于AC间得一点P.
线段AP显然小于曲线AP.而且线段PC显然小于曲线PB.(因为线段PC小于线段PB)
所以结论成立.
【在 x******g 的大作中提到】
: 将一个圆形的饼掰成两个，恰好两部分面积相等，求证：裂痕的长不小于饼的直径
| I***e
发帖数: 1136 | 8 Wrong.
^^^^^^^^^^^^^^^^^^^^
why is PC
can easily construct an example where this is not true.
【在 n******t 的大作中提到】
: 设那个裂痕得端点是A, B.
: 则作过A得直径交于C.
: 如果B和C重合,显然成立.
: 否则曲线AB显然和直径AC相交于AC间得一点P.
: 线段AP显然小于曲线AP.而且线段PC显然小于曲线PB.(因为线段PC小于线段PB)
: 所以结论成立.
| n******t
发帖数: 4406 | 9 Yes, I had taken PC<= r for granted. 3x.
anyway, I think it can be corrected.
Because we can investigate the diameter BD through B too.
if BD intersects curve AB at Q and QD <=r then job is done.
If not there must be a diameter between AC and BD which intersects
Curve AB at 2 points Q', P', and both AP' and BQ' not bigger than r.
【在 I***e 的大作中提到】
: Wrong.
:
: ^^^^^^^^^^^^^^^^^^^^
: why is PC: can easily construct an example where this is not true.
| B****n
发帖数: 11290 | 10 我對於Icare的證明非常佩服
我的證明確實比較囉唆 不過它應該可以推廣到任意有對稱軸的二維圖形都是對稱軸
為最短的等分曲線 我稍微修改了一點之前的證明 讓它可以推廣
I denote the graph as (t,f(t)), t in [0,1], with x-axis as its symmetric axis.
1. We can only consider bounded variation curve otherwise its length can not
be minimal.
2. For bounded variation curve, its root is countable. Here I define root as a
real number a such that f(a)=0 and there exists a epison such that f(a-c)<=0<
f
(a+c) or f(a-c)<0<=f(a+c) for any 0
3.Since the root is countable, we can deno | | | I***e
发帖数: 1136 | 11 All you have proven is that for all curves that start from 0
and end at 1, the line segment [0, 1] is the shortest...
isn't this an axiom? :)
The problem is, what if the begining and ending points does
not pass through the symmetric axis?
An easy example: consider triangle ABC where A=(-1, 0),
B=(1, 0) and C=(0, 10000). The symmetric axis is the Y axis,
however that'll give you a length of 1000. It can easily
be shown that you can be also draw a horizontal line at
height 1000-500 * Sqrt(2) to cu
【在 B****n 的大作中提到】
: 我對於Icare的證明非常佩服
: 我的證明確實比較囉唆 不過它應該可以推廣到任意有對稱軸的二維圖形都是對稱軸
: 為最短的等分曲線 我稍微修改了一點之前的證明 讓它可以推廣
: I denote the graph as (t,f(t)), t in [0,1], with x-axis as its symmetric axis.
:
: 1. We can only consider bounded variation curve otherwise its length can not
: be minimal.
: 2. For bounded variation curve, its root is countable. Here I define root as a
: real number a such that f(a)=0 and there exists a epison such that f(a-c)<=0<
: f
| B****n
发帖数: 11290 | 12 Ya, I made a big mistake. sorry for waisting your time.
It does not even prove the case for sphere.
thanks anyway
【在 I***e 的大作中提到】
: All you have proven is that for all curves that start from 0
: and end at 1, the line segment [0, 1] is the shortest...
: isn't this an axiom? :)
: The problem is, what if the begining and ending points does
: not pass through the symmetric axis?
: An easy example: consider triangle ABC where A=(-1, 0),
: B=(1, 0) and C=(0, 10000). The symmetric axis is the Y axis,
: however that'll give you a length of 1000. It can easily
: be shown that you can be also draw a horizontal line at
: height 1000-500 * Sqrt(2) to cu
| x******g
发帖数: 318 | 13 漂亮的证明.
第二个问题(高维的推广)
曲面将一个球分成两个体积相等的部分，
证明或否定
该曲面的面积大于等于大圆的面积
【在 I***e 的大作中提到】
: 太复杂了.
: 简单答案:
: 假设这两块饼在圆周上端点为A与B. 过圆心作与AB平行的直线. 注意到, 如果饼的
: 裂痕的长度小于直径, 则裂痕不可能与这条直线相交. 显然一块饼比半圆大, 一块
: 比半圆小.
: 其实, 过AB两点,而且长度不长于直径的线形成一个椭圆. 这个椭圆在圆心与那条
: 直线相切.由此得出, 就算裂痕长度等于直径,假如AB两点不是恰好某条直径的两
: 个端点, 那两块饼还是一大一小.
: -iCare-
:
| x******g
发帖数: 318 | 14 太长了……
a
a
0
【在 B****n 的大作中提到】
: I believe we can assume it's a differential curve that divides a sphere into
: two equal areas. But maybe someone can give a more rigous proof for this
: approximation.
: In the following, I denote the curve as (t,f(t)) t in [0,1] and assume it's a
: unit sphere with center at 1/2
: 1. We can only consider bounded variation curve otherwise its length can not
: be minimal.
: 2. For bounded variation curve, its root is countable. Here I define root as a
: real number a such that f(a)=0 and there exists a epison s
| x******g
发帖数: 318 | 15 我开始也这样想的，但这并不对.
因为|AC|可能小于半径,这样下面的结论就不对了
【在 n******t 的大作中提到】
: 设那个裂痕得端点是A, B.
: 则作过A得直径交于C.
: 如果B和C重合,显然成立.
: 否则曲线AB显然和直径AC相交于AC间得一点P.
: 线段AP显然小于曲线AP.而且线段PC显然小于曲线PB.(因为线段PC小于线段PB)
: 所以结论成立.
| x******g
发帖数: 318 | 16
^^^^^^^^^^^^这个不对，实际上正三角
形就是反例.
不过我想推广到中心对称图形倒是可能的
axis.
a
0<
【在 B****n 的大作中提到】
: 我對於Icare的證明非常佩服
: 我的證明確實比較囉唆 不過它應該可以推廣到任意有對稱軸的二維圖形都是對稱軸
: 為最短的等分曲線 我稍微修改了一點之前的證明 讓它可以推廣
: I denote the graph as (t,f(t)), t in [0,1], with x-axis as its symmetric axis.
:
: 1. We can only consider bounded variation curve otherwise its length can not
: be minimal.
: 2. For bounded variation curve, its root is countable. Here I define root as a
: real number a such that f(a)=0 and there exists a epison such that f(a-c)<=0<
: f
| x******g
发帖数: 318 | 17 对于正三角形问同样的问题，也是很有趣的.
角
【在 x******g 的大作中提到】
:
: ^^^^^^^^^^^^这个不对，实际上正三角
: 形就是反例.
: 不过我想推广到中心对称图形倒是可能的
: axis.
: a
: 0<
|
|
