It's 6 hours long and consists of 12 questions, broken up into two different 3-hour sessions.

它是一個美加大學部學生的數學競賽

With each question being scored on a 1-10 scale, the highest possible score is 120.

6 個小時的競賽時間中，選手要解 12 道題目

And yet, despite the fact that the only students taking it each year are those who are clearly

而這競賽的過程又分為兩個 3 小時段落

already pretty into math, given that they opt into such a test, the median score tends

每一道題目將被評 0 到 10 分

to be around 1 or 2. So... it's a hard test. And on each section of 6 questions, the problems

全部試題滿分 120 分

tend to get harder as you go from 1 to 6, although of course difficulty is in the eye

然而，儘管每年會參加這一門競賽

of the beholder.

都是對數學特別感興趣的學生

But the thing about the 5's and 6's is that even though they're positioned as the

競賽總分的中位數通常只有 1 到 2 分

hardest problems on a famously hard test, quite often these are the ones with the most

可見這些試題非常難

elegant solutions available. Some subtle shift in perspective that transforms it from challenging

在一個段落中的 6 道試題中

to simple. Here I'll share with you one problem which

往往是一題一題的增加難度

came up as the 6th question on one of these tests a while back.

當然，題目難不難會因人而異的

And those of you who follow the channel know that rather than just jumping straight to

但是說道最難的第五、第六題

the solution, which in this case will be surprisingly short, when possible I prefer to take the

它們雖然是這門超難競賽中的壓軸難題

time to walk through how you might stumble upon the solution yourself.

這些題目卻也往往是有著最精妙的解法的試題

That is, make the video more about the problem-solving process than the particular problem used to

當稍微換個角度切入，這些難題不會是無從入手

exemplify it.

現在，我們就來分享一道

So here's the question: If you choose 4 random points on a sphere, and consider the

在前幾屆 (1992) 競賽中出現的第六題

tetrahedron which has these points as its vertices, what's the probability that the

常光顧這個頻道的朋友們會知道

center of the sphere is inside the tetrahedron? Take a moment to kind of digest the question.

我不會直接開始講答案

You might start thinking about which of these tetrahedra contain the sphere's center,

而且這一題的解法出人意料的簡短

which ones don't, and how you might systematically distinguish the two.

我會多花時間一步一步地討論

And...how do approach a problem like this, where do you even start?

你如何自己探索出這個解法

Well, it's often a good idea to think about simpler cases, so let's bring things down

以及一些關鍵的想法從何而來

into 2 dimensions.

也就是說我製作的影片中

Suppose you choose three random points on a circle. It's always helpful to name things,

重點會放在解決問題的過程

so let's call these guys P1, P2, and P3. What's the probability that the triangle

而問題本身只是一個的例子罷了

formed by these points contains the center of the circle?

總之，問題如下：

It's certainly easier to visualize now, but it's still a hard question.

「在球面上隨機選擇四個點，

So again, you ask yourself if there's a way to simplify what's going on. We still

「考慮該四點為頂點之四面體，

need a foothold, something to build up from. Maybe you imagine fixing P1 and P2 in place,

「球心在四面體內部的機率為何？」

only letting P3 vary. In doing this, you might notice that there's

花一點時間消化一下這個題目

special region, a certain arc, where when P3 is in that arc, the triangle contains the

你可能開始思考：

circle's center. Specifically, if you draw a lines from P1

哪些四面題會包含到球心；哪些不會

and P2 through the center, these lines divide the circle into 4 different arcs. If P3 happens

要怎麼系統化的分辨這兩種四面體

to be in the one opposite P1 and P2, the triangle will contain the center. Otherwise, you're

甚至這道題目到底要怎麼切入？

out of luck.

這題目應該要從何起手？

We're assuming all points of the circle are equally likely, so what's the probability

遇到不會解的題目時 一個好方法事先從簡化的問題開始想

that P3 lands in that arc? It's the length of that arc divided by the

所以我們先把問題減少一個維度

full circumference of the circle; the proportion of the circle that this arc makes up.

考慮圓周上的找三個點

So what is that proportion? This depends on the first two points.

為了方便，我們叫這三個點 P1，P2 和 P3

If they are 90 degrees apart from each other, for example, the relevant arc is ¼ of the

我們考慮的問題變成：

circle. But if those two points are farther apart, the proportion might be closer to ½.

「圓心位在以此隨機三點為頂點的三角形 內的機率為多少？」

If they are really close, that proportion might be closer to 0.

你應該會覺得這道題目變得比較容易想像

Alright, think about this for a moment. If P1 and P2 are chosen randomly, with every

但是還是一道困難的題目

point on the circle being equally likely, what's the average size of the relevant

所以你可以繼續追問

arc? Maybe you imagine fixing P1 in place, and

有沒有方法可以把這個問題繼續簡化

considering all the places that P2 might be. All of the possible angles between these two

好找一個容易構思的立足點？

lines, every angle from 0 degrees up to 180 degrees is equally likely, so every proportion

現在我們先固定 P1，P2

between 0 and 0.5 is equally likely, making the average proportion 0.25.

只讓第三個點可以任意移動

Since the average size of this arc is ¼ this full circle, the average probability that

你這麼做的時候，在腦子裡筆劃筆劃

the third point lands in it is ¼, meaning the overall probability of our triangle containing

你會注意到有一個特別的區間：一段弧長

the center is ¼. Try to extend to 3D

當 P3 落在這段弧長中，三角形會包含到圓心，反之則否

Great! Can we extend this to the 3d case? If you imagine 3 of your 4 points fixed in

更精確的來說，當你從 P1 和 P2 各畫一條通過圓心的線

place, which points of the sphere can that 4th point be on so that our tetrahedron contains

這兩條線會把圓周分成四段弧長

the sphere's center? As before, let's draw some lines from each

而當 P3 落在 P1、P2 對面的弧長上的時候

of our first 3 points through the center of the sphere. And it's also helpful if we

三角形就會包含到圓心

draw the planes determined by any pair of these lines.

如果在其他三段弧長上的話，那就不會包含到圓心

These planes divide the sphere into 8 different sections, each of which is a sort of spherical

我們預設圓周上的每一個點都有相同的機率被選到

triangle. Our tetrahedron will only contain the center of the sphere if the fourth point

那 P3 落在那一段弧長的機率是多少？

is in the section on the opposite side of our three points.

機率會是弧長除以圓周長

Now, unlike the 2d case, it's rather difficult to think about the average size of this section

弧長佔圓周長的比例

as we let our initial 3 points vary. Those of you with some multivariable calculus

那這比例又會是多少？

under your belt might think to try a surface integral. And by all means, pull out some

這當然得看原始的兩個點位置在哪裡

paper and give it a try, but it's not easy. And of course it should be difficult, this

假如他們中間角差距是 90°

is the 6th problem on a Putnam!

那我們在意的那一段弧長就會是圓周長的 1/4

But let's back up to the 2d case, and contemplate if there's a different way of thinking about

但是如果那兩點距離比較遠

it. This answer we got, ¼, is suspiciously clean and raises the question of what that

這個比例會比較接近 1/2

4 represents. One of the main reasons I wanted to make a

而如果這兩點非常靠近

video on this problem is that what's about to happen carries a broader lesson for mathematical

這個比例會接近 0

problem-solving. These lines that we drew from P1 and P2 through

花一點時間思考一下

the origin made the problem easier to think about.

既然 P1 和 P2 都是隨機的被挑選，而圓上的每一個點被選到的機率都一樣

In general, whenever you've added something to your problem setup which makes things conceptually

那我們在意的那一段弧長的平均長度會是多少？

easier, see if you can reframe the entire question in terms of the thing you just added.

先想像你把 P1 固定，只考慮 P2 可能落在的位置

In this case, rather than thinking about choosing 3 points randomly, start by saying choose

這兩點之間角差距從 0° 到 180° 都有相同的機率出現

two random lines that pass through the circle's center.

所以弧長對圓周長的比例從 0 到 0.5 都有相同的機率

For each line, there are two possible points they could correspond to, so flip a coin for

也就是說該比例的平均值是 0.25

each to choose which of those will be P1 and P2.

那既然我們在意的弧長平均是 1/4 圓周長

Choosing a random line then flipping a coin like this is the same as choosing a random

那第三點落在那一段弧長的機率就會是 1/4

point on the circle, with all points being equally likely, and at first it might seem

也就是說，有 1/4 的機率我們的三角形會包含到圓心

needlessly convoluted. But by making those lines the starting point of our random process

不過，我們要怎麼把這個解擴充到三維空的問題？

things actually become easier. We'll still think about P3 as just being

如果我們先把四點中的三點固定

a random point on the circle, but imagine that it was chosen before you do the two coin

那第四點需要落在球面上的哪裡

flips. Because you see, once the two lines and a

四面體才會包含到球心？

random point have been chosen, there are four possibilities for where P1 and P2 end up,

和先前二維的問題一樣 我們畫通過前三點和球心的線

based on the coin flips, each one of which is equally likely. But one and only one of

我們也可以考慮由任意兩線段所定義的平面

those outcomes leaves P1 and P2 on the opposite side of the circle as P3, with the triangle

你可能會注意到 這些平面會把我們的球面分成 8 個區間

they form containing the center. So no matter what those two lines and P3 turned

每個區間有點像是一個球面三角形

out to be, it's always a ¼ chance that the coin flips will leave us with a triangle

而我們的四面體要包含到球心的充要條件

containing the center. That's very subtle. Just by reframing how

是當我們的第四個點剛好落在其餘三點對面的那個區間

we think of the random process for choosing these points, the answer ¼ popped in a different

和我們二維的例子不一樣的是

way from before.

由隨機三點定義的區間的平均面積沒有那麼好想像

And importantly, this style of argument generalizes seamlessly to 3 dimensions.

一些熟知多變數微積分的觀眾可能會想：

Again, instead of starting off by picking 4 random points, imagine choosing 3 random

何不試試面積分？

lines through the center, and then a random point for P4.

這當然無可厚非，你可以拿出紙筆來算算看

That first line passes through the sphere at 2 points, so flip a coin to decide which

但是這方法很不容易

of those two points is P1. Likewise, for each of the other lines flip a coin to decide where

當然，這題目的解應該是難的

P2 and P3 end up. There are 8 equally likely outcomes of these

它畢竟是出在普氏競賽的第六題

coin flips, but one and only one of these outcomes will place P1, P2, and P3 on the

有沒有更好計算的方式呢？

opposite side of the center from P4. So only one of these 8 equally likely outcomes

我們先回到二維的例子

gives a tetrahedron containing the center. Isn't that elegant?

想一想有們有更簡單的方式可以得到答案

This is a valid solution, but admittedly the way I've stated it so far rests on some

這個 1/4 看起來有點過份的簡單

visual intuition. I've left a link in the description to a

不禁讓人想這個 4 是不是有其他潛在的意含

slightly more formal write-up of this same solution in the language of linear algebra

我之所以會特別做一支影片在講解這一道題目

if you're curious. This is common in math, where having the key

正是因為接下來要做的事 在解決數學難題有很廣泛的參考價值

insight and understanding is one thing, but having the relevant background to articulate

想一想我們那兩條通過 P1、P2 和圓心的線段

this understanding more formally is almost a separate muscle entirely, one which undergraduate

那兩條線讓問題好想很多吧？

math students spend much of their time building up.

一般來說，當你在原問題中多加了一些東西

Lesson Now the main takeaway here is not the solution

讓他變得更容易想像

itself, but how you might find the key insight if you were left to solve it. Namely, keep

試試看有沒有辦法用這些新東西重構原本的問題

asking simpler versions of the question until you can get some foothold, and if some added

以這到題目而言 我們試著不考慮隨機的三個點

construct proves to be useful, see if you can reframe the whole question around that

而是把換成考慮：

new construct.

隨機的選通過圓心的兩條線段