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解决一道数学难题背后的思路

The hardest problem on the hardest test

你们听说过普特南数学竞赛吗?
Do you guys know about the Putnam?
这是一个面向美国本科生的数学竞赛
It’s a math competition for undergraduate students.
竞赛时长6小时 一共只有12道题目
It’s a 6-hour long test that just has twelve questions
而且分成上下两场
broken up into two different 3-hour sessions.
每场3小时每道题的得分最低0分
And each one of those questions is scored 1 to 10.
最高10分 所以满分就是120分然而
So the highest possible score would be 120.
尽管
And yet, despite the fact
每年来参加这个竞赛的
that the only students taking this thing each year
很明显都是那些对数学相当感兴趣的学生
are those who clearly are already pretty interested in math,
但是分数的中位数也常常只有一两分所以说
the median score tends to be around 1 or 2. So…
这个竞赛很难
it’s a hard test.
而且对于每场的六个问题来说
And on each one of those sections of six questions,
从第一题到第六题
the problems tend to get harder
难度也在上升当然了
as you go from 1 to 6, although,
难不难是因人而异的
of course, difficulty is in the eye of the beholder.
但是说到第五题和第六题
But the things about those 5s and 6s is that
虽然它们都是超难
even though they are positioned
竞赛中的压轴难题
as the hardest problems on a famously hard test,
但通常来说
quite often, these are the ones
它们也有着最为精妙的解法
with the most elegant solutions available,
稍微转换一下视角 极具挑战性的问题就不再无从下手了
some subtle shift in perspective that transforms it from very challenging to doable. Here,
我想在这里和大家分享一道题
I’m going to share with you one problem
好早以前某届普特南数学
that came up as the sixth question on one
竞赛中的第六题
of these tests a while back.
关注这个频道的朋友们都知道
Those of you who follow the channel,
我并不会直接公布答案
you know that rather than just jumping straight to the solution, which,
而且这道题的答案出人意料地简短
in this case, would be surprisingly short;
如果可能的话
when possible, I like to
我愿意花点时间让你明白
take the time to walk you through
你如何能够独立探索出问题的答案
how you might have stumbled across the solution yourself,
以及重要的想法是怎么来的
where the insight comes from.
也就是说 视频重点关注的是解答问题的过程
That is, make a video more about the problem solving process
而不是用到这个思路的问题本身总之
than about the problem used to exemplify it.
问题是这样的:
So anyway, here’s the question:
在球面上随机选择四个点
if you choose four random points on a sphere,
然后考虑以它们为顶点的四面体
and consider the tetrahedron with these points as its vertices,
那么球心落在四面体内部的概率是多少?
what is the probability that the center of the sphere is inside that tetrahedron?
花点时间 稍微琢磨一下这个问题
Go ahead, take a moment and kind of digest this question.
你可能会想 哪些四面体包含了球心
You might start thinking about which of these tetrahedra contain the sphere’s center,
哪些四面体没有
which ones don’t,
如何系统地区分这两种情况还有
how you might systematically distinguish the two.
你要怎样处理这种问题呢?
And how do you even approach a problem like this, right?
要从何下手呢?
Where do you even start? Well,
通常来说 从问题的简单情形入手是一个好办法
it’s usually a good idea to think about simpler cases.
所以我们把问题简化到二维来此时
So let’s knock things down to two dimensions,
你要在圆上随机选择三个点
where you’ll choose three random points on a circle.
用符号标记更加方便
And it’s always helpful to name things,
所以用P_1 P_2 P_3表示这三个点
so let’s call these guys P1, P2, and P3.
问题就是
The question is
这三个点形成的三角形
what’s the probability that the triangle formed
包含圆心的概率是多少?
by these points contains the center of the circle?
我觉得你也会认为这样更容易想象了
I think you’ll agree it’s way easier to visualize now,
但是问题依然很困难
but it’s still a hard question.
所以你想知道
So again, you ask is there a
能不能简化一下问题
way to simplify what’s going on,
找到一个立足点 方便继续思考
get ourselves to some kind of foothold that we can build up from. Well,
或许你会想 先固定P_1和P_2的位置
maybe you imagine fixing P1 and P2 in place,
只允许第三个点移动
and only letting that third point vary.
你这么做的时候
And when you do this,
在脑子里比划比划
and you play around with it in your mind,
你可能会注意到一个特殊区域
you might notice that there’s a special region, a certain arc,
一段弧P_3落在这条弧上时
where when P3 is in that arc,
三角形就包含圆心 否则就不包含
the triangle contains the center, otherwise not. Specifically,
具体来说 如果分别过P
if you draw lines
_1和P_2以及圆心做直线
from P1 and P2 through the center,
这两条直线会把圆分成四段弧
these lines divide up the circle into four different arcs.
如果P_3恰好落在P_
And if P3 happens to be in the one
1和P_2对面的弧上
on the opposite side as P1 and P2,
那么三角形就包含圆心
the triangle has the center.
如果P_3落在其他弧上
If it’s in any of the other arcs though, no luck.
那就不行我们假定取到圆上任
We are assuming here that all of the points
一点的概率相同
of the circle are equally likely.
那么P_3落在这条弧上的概率是多少?
So what is the probability that P3 lands in that arc?
答案就是这段弧的
It’s the length of that arc divided
长度除以圆的周长
by the full circumference of the circle,
也就是这段弧占整个圆的比例
the proportion of the circle that this arc makes up.
那这个比例是多少呢?
So what is that proportion?
很明显 这和前两个点的位置有关
Obviously that depends on where you put the first two points.
如果它们之间相差90°
I mean, if they’re 90° apart from each other,
那么对应的弧就是1/4个圆
then the relevant arc is 1/4 of the circle.
如果这两个点离得很远
But if those two points were farther apart,
比例就会更接近1/2
that proportion would be something closer to 1/2.
如果这两个点靠得很近 比例就更接近0
And if they were really close together, that proportion gets closer to 0.
花点时间思考一下
So think about this for a moment.
P_1和P_2都是随机选出的
P1 and P2 are chosen randomly with every point
而任一点被取到的概率相同
on the circle being equally likely.
那这段弧的平均长度是多少呢?
So what is the average size of this relevant arc?
你或许会固定P_1
Maybe you imagine fixing P1 in place,
只考虑P_2所有可能的位置
and just considering all the places that P2 might be.
这两条直线形成的所有可能的夹角
All of the possible angles between these two lines,
也就是从0°到180°之间的所有角度 出现的概率都一样所以
every angle from 0° up to 180° is equally likely.
从0到0.5之间的每个比例出现的概率也一样
So every proportion between 0 and 0.5 is equally likely.
这就意味着 平均比例就是0.25所以
And that means that the average proportion is 0.25. So,
如果这段弧的平均长度是圆周
if the average size of this arc is 1/4
长的1/4
of the full circle,
第三个点落在这段弧上的平均概率就是1/4
the average probability that the third point lands in it is 1/4.
也就是说
And that means that the overall probability
三角形包含圆心的概率是1/4
that our triangle contains the center is 1/4.
但是我们能把它推广到三维情况吗?
But can we extend this into the three-dimensional case?
一共四个点
If you imagine three
如果其中三个点固定不动
out of those fourth points just being fixed in place,
第四个点要落在球面的哪些位置
which points of the sphere can the fourth one be on
才能让形成的四面体包含球心?
so that the tetrahedron that they form contain the center of the sphere?
和之前一样
Just like before, let’s go ahead and
我们分别过这三个点以及
draw some lines from each of those fixed 3
球心做直线
points through the center of the sphere.
这些直线两两确定
And here, it’s also helpful
一个平面 画出这些平面也很有用
if we draw some planes that are determined by any pair of these lines.
你可能注意到了
Now what these planes do, you might notice,
这些平面把球面分成了八个区域
is divide the sphere into eight different sections,
每个区域都有点像球面三角形
each of which is a sort of spherical triangle.
要想让形成的四面体
And our tetrahedron is only going
包含球心的话
to contain the center of the sphere
第四个点就必须落在前三个
if the fourth point is in the spherical triangle
点相对的球面三角形上
on the opposite side as the first three.
和二维情况不同
Now unlike the 2D case,
让最初的三个点变化 然后求这个区域的
it’s pretty difficult to think about the average size
平均面积 这就很困难了
of the section as we let the initial three points vary.
掌握多元微积分的朋友
Those of you with some multivariable calculus
可能会想:
under your belt might think:
用曲面积分试试看
let’s just try a surface integral.
不妨拿出纸笔算一算吧
And by all means, pull out some paper and give it a try.
但是这并不简单当然了
But it’s not easy.
这道题本来就应该很难
And of course, it should be difficult.
这是普特南数学竞赛的第六题啊
I mean, this is the sixth problem on a Putnam,
你指望它能有多简单?
what do you expect? And…
更何况 算出了三角面积又有什么用呢?
what do you even do with that? Well,
你可以回头
one thing you can
看看二维的情况
do is back up to the two-dimensional case,
仔细想想 是否可以通过
and contemplate if there is a different way to think
别的方法解出问题
about the same answer that we got.
“1/4”这个答案看上去异常简洁
That answer”1/4″ look suspiciously clean,
这也提出了一个问题:“4”代表着什么?
and it raises the question of what that”4″ represents.
我之所以要特别为这个问题做一期视频
One of the main reasons I wanted to make a video
主要的原因之一是
about this particular problem is that
接下来的事情 对求解数学问题
what’s about to happen carries with it a broader lesson
有广泛的参考价值
for mathematical problem solving.
回想我们之前画出的以P_1与P_2为端点
Think about those two lines that we drew
通过圆心的两条线
for P1 and P2 through the origin.
这两条线让问题更容易思考了
They made the problem a lot easier to think about.
一般而言
And in general,
只要你往原问题中添加了新东西
whenever you’ve added something to the problem set up
使得问题的概念更简洁的话
that makes it conceptually easier,
那就看看 你能否将整个问题
see if you can reframed the entire question
只用加进来的新东西来重新叙述在这里
in terms of those things that you just added.
我们不再考虑随机选择三个点
In this case, rather than thinking about choosing three points randomly,
而是考虑
start by saying:
随机选择两条过圆心的直线
choose two random lines that pass through the circle center;
每条直线
for each line,
都对应着圆上的两个点
there are two possible points that could correspond to,
所以就二选一
so just flip a coin for each one to choose which
确定哪个端点是P_1类似地
of the endpoints is going to be P1;
在另一条直线上确定哪个端点是P_2
and likewise, for the other which endpoint is going to be P2.
随机选择一条直线 然后二选一确定端点
Choosing a random line and flipping a coin like this
这种做法就等于随机在圆上选取一点乍一看
is the same thing as choosing a random point on the circle.
这让人感觉有点绕弯子
It just feels a little bit convoluted at first.
但之所以
But the reason
用这种方式来考虑随机过程
for thinking about the random process this way is that
是因为这能让事情变得简单许多
things are actually about to become easier.
我们依然认为点P_3
We’ll still think about that third point P3
不过是圆上的任意一点
as just being a random point on the circle.
但我们是在扔两次硬币前就确定好P_3了
But imagine that it was chosen before you do the two coin flips.
因为你看嘛
Because you see,
一旦这两条线和第三个点定死了以后
once the two lines and that third point are set in stone,
P_1和P_2落在哪里
there are only four possibilities
依扔硬币的结果而定就只有四种情况了
for where P1 and P2 might end up based on those coin flips,
每种情况都是等可能的但是
each one being equally likely.
有且仅有一种情况
But one and only one of those four outcomes
能使得P_1和P_2在圆上落
leaves P1 and P2 on the opposite side
在P_3的对面
of the circle as P3
因此三点围成的三角形能包含圆心所以说
with the triangle that they form containing the center.
不管这两条线或者
So no matter where those two lines end
P_3点最后落在哪里
up and where that P3 ends up,
扔硬币总是会给
it’s always a 1/4 chance
我们1/4的机率使得 三角形包含圆心
that the coin flips leave us with a triangle containing the center.
这就很巧妙了
Now that’s very subtle.
重新修改一下
Just by reframing how we think
我们随机选点的顺序
about the random process for choosing points,
1/4这个答案就以一种很不一样
the answer 1/4 popped out in a very different way
的方式蹦了出来
from how it did before.
重要的是 这种推导过程可以毫无痕迹地被推广至三维
And importantly, this style of argument generalizes seamlessly up into three dimensions. Again,
再来一次 这次我们从选四个随机点开始
instead of starting off by picking four random points,
想象一下 任选三条穿过球心的线
imagine choosing three random lines through the center of the sphere,
再随便来一个点P_4
and then some random point for P4.
第一条线与球面相交于两点
That first line passes through the sphere at two points,
扔个硬币
so flip a coin to decide which
决定一下哪个点是P_1类似地
of those two points is going to be P1. Likewise,
扔硬币决定
for each of the other lines,
P_2和P_3的落点
flip a coin to decide where P2 and P3 end up.
扔硬币就有8种等可能的结果
Now there’s eight equally likely outcomes of those coin flips,
但有且仅有一种结果
but one and only one of them is going
能使P_1 P_2和P_3
to place P1, P2 and P3 on the
处于与P_4相对的位置上
opposite side of the center as P4.
所以这8个概率
So one and only one
一样的结果中 有且仅有一个 能让我们得到包含球心的四面体
of these eight equally likely outcomes gives us a tetrahedron the contains the center. Again,
这个结果再一次以巧妙的方式出现在了我们的面前但是
it’s kind of subtle how that pops up to us; but,
这的确很简洁啊
isn’t that elegant?
这就是此问题的一个确实的解了
This is a valid solution to the problem,
但是必须承认
but admittedly, the way
到目前为止我的讲解都是基于几何直观的
that I’ve stated it so far rests on some visual intuition.
如果你有点好奇 应该如何不依靠
If you’re curious about how you might write it up
几何直观来写出这个解的话
in a way that doesn’t rely on visual intuition,
我在简介里留了个解答的链接
I’ve left a link in the description
它用线性代数
to one such write up in the language
的语言解答了这个问题
of linear algebra if you’re curious.
这在数学中是挺常见的:
And this is pretty common in math,
理解问题并知道关键是一回事
where having the key insight and understanding is one thing,
但有相关背景知识 能更正式更清晰地阐述这个理解
but having the relevant background to articulate that understanding more formally
基本就完全是另一码事了
is almost a separate muscle entirely,
这种能力也是数学系本科生要花大时间来培养的
one that undergraduate math students kind of spend most of their time building up.
你应该从本题中学到的并不是这个解本身
But the main takeaway here is not the solution itself,
而是如果是你来做这道题的时候
but how you might find that key insight if
应该怎么
it was put in front of you and you
找到关键的想法
were just left to solve it. Namely,
也就是说 不断去找这个问题的简化版本
just keep asking simpler versions of the question
直到你能找到个落脚点为止
until you can get some kind of foothold.
在这么做的过程中
And then when you do,
如果你发现有什么新添的结构能为自己所用
if there’s any kind of added construct that proves to be useful,
就试试看能不能根据这些新构造来重述整个问题
see if you can reframe the whole question around that new construct.
视频结束之前 我还有一个关于概率的问题
To close things off here, I’ve got another probability puzzle,
这个问题取自这期视频的赞助商Brilliant.org
one that comes from this video sponsor – Brilliant.org.
假设八个参加普特南数学竞赛的学生环坐在一起
Suppose that you have eight students sitting in a circle taking the Putnam.
竞赛很难
It’s a hard test,
所以每个学生都试图抄邻座的答案
so each student tries to cheat off of his neighbor
而且是随机选择一个邻座来抄
choosing randomly which neighbor to cheat from.
现在圈出所有
Now circle all of the students that do
没被抄到的学生
n’t have somebody cheating off of their test.
被圈出的学生数目的期望值是多少?
What is the expected number of such circled students?
这个问题很有趣
It’s an interesting question, right?
对吧~Brilliant.org就是个能让你用许许多多类似的问题
Brilliant.org is a site where you can practice your problem solving abilities
来锻炼自己解题能力的地方
with questions like this and many many more,
这也的确是最好的学习方式
and that really is the best way to learn.
你能找到数不胜数
You are going to find countless
的有趣问题 而它们都被组织得很有深度
interesting questions curated in a pretty thoughtful way
所以你一定会在解题方面有所提高的
so that you really do come away better at problem solving.
如果你还想探索更多的可能性 他们也有相当棒的概率课程当然
If you want more probability, they have a really good course on probability.
他们也有
But they’ve got all sorts
各种其他的数理课程
of other math and science as well,
所以你怎么着都能找到点你感兴趣的东西的我嘛
so you’re almost certainly going to find something that interests you. Me?
我是已经用这个好久了
I’ve been a fan for a while.
如果你访问brilliant.org/3b1b
And if you go to brilliant.org/3b1b,
他们就知道你是从这个频道来的了
it lets them know that you came from here.
前256位通过此链接访问
And the first 256 of you to visit
的同学可以得到他们高级会员的八折优惠
that link can get 20 % off of their Premium Membership,
如果你考虑升级一下的话 我就是用的这个还有
which is the one I use if you want to upgrade. Also,
如果你心痒
if you’re just
难耐想看这个问题的解答的话插一句
itching to see a solution to this puzzle, which,
这个解
by the way,
使用了概率论中的一个策略 该策略在其他情况下也很有用
uses a certain tactic in probability that’s useful in a lot of other circumstances,
我也在简介中留了链接
I also left a link
点一下你就能直接看到答案了
in the description that just jumps you straight to the solution.

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视频来源

https://www.youtube.com/watch?v=OkmNXy7er84

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