28/7/2025: Sixth form maths/Quintic dreams/Frontiermath

The blog is full of my random thoughts. 

The category of random thoughts contributes to the fifth largest categories among tagged entries, after "diary", "notes", "maths" and "works". Most of them were in Chinese though.

And today I come up with something different: three math related random thoughts in a single entry.

This is new in the sense that all "math diary" in the past are more like "math but too casual" that were still focused on a single topic. This is also new because it's 3 math-related entries in a row...

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I recently bought another sixth form "textbook" from an antique store. Titled "Polynomials and equations" and claimed "part of the chapters applicable for HKALE or equivalent examinations". This is nothing special: polynomials is an essential part in the curriculum with questions commonly on binomial sums and Vieta applications.

I bought the book at the price of half a cup of coffee, made myself a proper cup of coffee and started diving in...

- Introduction...starts with the notation $R[x]$ and $C[x]$.

- Factorization...gcd and lcm, then the Euclidean algorithm, unique factorization and FToA.

- Solving equations...then start talking about the meaning of discriminant in these solutions together with the insolvability of quintic equations.

- Integral solutions...some number theory, then Eisenstein without warning.

- Derivatives...Taylor and approximating continuous functions, IVT, Rolle.

- Root bounding and separation...approximates, Strum and Fourier(surprise!)...

Oh yeah, the "new math" style old math textbook.

The context was a bit different though. The book I bought was published in 1992, and indeed part of the chapters fitted the HKALE pure maths syllabus and exam depth adequately. The pure math exam was hard more because HKALE itself was designed to be an ultimate challenge to filter true elite to enter university.

Should it be a true "new math" style book, it has to be much earlier like in the 1970s or 1980s, it would also be covering topics straight at university level abstractness. I have another true "new math" era "sixth form" textbook with the title "group theory" that followed M.Artin's line all the way up to similarity. Can you imagine average sixth form students doing that?

Nonetheless the bridging between high school and university has always been an intriguing topic for discussion especially regions adopting the sixth form/matriculation system. Students at that level are matured enough to take another step above the cert level but not quite ready for university. What would you teach them, and what would you examine?

It always feel weird to me that geometry has been playing the vital role in most A-level or even high school level syllabuses across the globe. Coordinate geometry on conics, vector geometry, 3D vector calculus, parametric equations, geometry on complex plane -- do we really need them? Are they really useful other than serving the purpose of filtering students who can't handle messiness?

These geometry topics are almost never used in universities. You will use vector but they are not for solving Euclidean geometry problems. You will see them in linear algebra or calculus II with completely different intents. Conics are almost never used except for optics which is extremely niche and still look simpler than questions you will encounter in A-level exams. Complex geometry is beautiful but you will only study complex analysis in undergrad, not complex geometry.

There are so many misc topics you can fill into the roster instead. What about functions and relations? Injective and surjective functions? Equivalence relations? Group theory from matrices? Inequalities, epsilon-N and epsilon-delta?

...

Youtube forwarded me a video on quintic insolvability without Galois on the same day. Frankly this is what motivated me to write something, not the sixth form textbook.

The video was about Arnold's proof, who managed to open the field of topological Galois theory from this result. The concept of "multivalued pullback" is somewhat unexpected yet so beautiful to understand. 

When I first read M.Artin's algebra on branched coverings, it was too abstract for me. I simply jotted down and didn't understand a word of it. It wasn't even in the exam so I never came back to the topic until I took algebraic geometry course much, much later. Even so, it was never my focus.

On the other hand, the idea of travelling from a branch to another continuously is something that is much common. Winding is everywhere in geometry, topology and analysis. The moment they showed the loop (the projections of the infinite branched sheets onto complex plane) it's like a spark in my brain and everything started to come together.

The rest is all about bridging between loop commutators and roots permutation, but they looks so complicated even as presented in those modified sources. Here is the logic I would rather present instead:

- Given any solution formula for a given degree of polynomial equation, it should be invariant upon transversal along any loop based at solution point.

- In particular, it must be invariant to the commutators of loops, the commutators of commutators,... and so on because they are all loops.

- The key result is that suppose the minimal (least number of nested roots) formulae that are invariant to two loops are both $k$ and that the two loops do not commute, then the minimal formulae that can be invariant to the two loops and its commutator requires at least $k+1$ nested roots (!!!).

- Thus, the existence of the formula relies on the requirement that commutator of commutators eventually becomes trivial. 

- Quadratic case: the commutators of $S_2$ is trivial, hence there is just 1 square root.

- Cubic case: the commutator group chain is $S_3, A_3, \left\{ e\right\}$, hence two nested roots.

- Quartic case: the commutator group chain is $S_4, A_4, V, \left\{ e \right\}$ so it works too.

- Quintic case? The commutator group is stuck at $A_5$ because it's damn simple!

In the first half the proof may look completely new and creative, but once you saw the subgroup chains it reduces to "oh it's the damn old $A_5$ again...".

The video avoided saying $A_5$ being simple directly. Instead it said commutator (sub)group generates the whole permutation group which means any finite nested root doesn't work. 

Abel's proof (for him proving the result before Galois) also avoided the use of $A_5$. He used the language of algebraic independence to show that no further field extension is possible after the quadratic extension (equivalently from $S_5$ to $A_5$), just without all modern tools.

Both approaches avoided the use of $A_5$ although they are simply equivalent in the algebraic sense. The commutator group is in fact the minimal normal subgroup such that the quotient is Abelian. Since $A_5$ is simple, the only normal subgroup is the trivial group or itself, but the quotient upon trivial group is not Abelian, so the commutator group got stuck at $A_5$. Abel's approach is merely the elementary way of showing that there is no transitive field extension (by prime characteristic each step) to $S_5$ because $A_5$ is simple.

That makes me wondering what counts as "Galois"? The use of Galois theory, or even the fact that $A_5$ is simple? Even in modern algebraic sense, we could have avoided the use of Galois theory simply by using the language of field extension. It could have been much easier than Abel's proof too. Does that count as non-Galois proof?

I have no idea.

But one thing for sure: I would be extremely impressed if someone proved irresolvability without using any algebraic structure related to $A_5$.

Anyway, it's a good math video almost at 3B1B level, a hidden treasure that is definitely worth a look.

Some other great references:

Another video on Arnold's proof

L. GoldMakher, Arnold's elementary proof of the insolvability of the quintic

J. Brown, Abel and the insolvability of the quintic

MSE question explaining Abel's proof

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Speaking of Galois theory I've got one more thing to say: the math LLM benchmarks.

Forget about testing against AMC or AIME. Forget about testing against IMO or Putnam problems. We are now straight into the most advanced problems we can scramble out of our garage.

Previously I have been looking into the HLE (humanity's last exam) benchmarks as they aim to gather top questions from all subjects where math plays a big part of it. However, the mix between math and other subjects means that it is very hard to evaluate how good at math LLMs had became. There are also questions that are pure travia or multiple choice questions that aren't foul proof and hardly represent any meaningful results.

That is, until my friend told me the existence of FrontierMath.

I really like the fact that cover a wide range of breadth while keeping everything hard and foul proof. Top tier questions are only answered using frontier research results and are absolutely into the most niche bit of mathematics at the frontier. I really wonder how did the LLMs managed even to answer any single one of them...we can learn much more from what they managed to answer instead of what they can't for now.

Can I give any prediction on their progress of solving these questions? Not really. I still think they will be stopped by these questions for a long time but I might as well give an opposite answer by the end of 2025...

IMO 2025 and LLM's misleading claim on gold medals

Hello, time for IMO again!

This is one of the IMO hosted that is closest to my base of New Zealand, but I am not a participant for sure and I am not a team staff anyway, so nothing other than the questions themselves. So let's dive into the questions right now, shall we?

As usual, this is not a full attempt with the proper time limit. This is my instinctive brainstorming and observations, together with comments when I check the discussions.

Q1. Do you call that combinatorics? Or algebra? I think it has bits of everything yet so simple that makes it hard to classify.

For $n=3$ that is pretty simple where you come up with the possibility of $k = 0,1,3$. But what about the general $n$? That must be by induction for sure, and it would be nice if we can reduce the $n$.

My instinct is an idea similar to the reciprocity proof: how many lattice points a line would cross given the first crossed point and the rational slope? That is simply the denominator of the slope at simplest form. That says, sunny lines are inefficient in covering the points whose size grows quadratically. However, this approach is so vague and leaves the cover set in a mess. Thus I changed the covering focus to the boundary instead.

To be more precise, we focus on the colinear lattices $(1,b)$. Two cases to consider: if we include the line $x=1$ that passes through all those lattices we can reduce to lower $n$. If not, that is $n$ distinct lines passing through the $n$ lattices. For that case we ask the question: what lines cover the lattices $(b,n+1-b)$? If they are covered by the line $x+y = n+1$ then we reduce to lower $n$, otherwise again they are again covered by $n$ distinct lines.

Now this is a matching between the $(n-1)$ lattices $(1,1),...,(1,n-1)$ and $(2,n-1),...,(n,1)$ so that it covers all lattices not covered by the line connected to $(1,n)$. We focus on the lattices $(2,1),...,(n-1,1)$. If the line connecting $(1,1)$ does not connect $(n,1)$, all such $(n-2)$ lattices are not covered. They can't be covered by any other line connecting $(1,2),...,(1,n-1)$ either because such line won't be able to connect a lattice on $x+y=n$ then. Thus either $n=3$ or we force the line $y=1$. 

From the above we established that we can always reduce $n$ by arguing at least one of the lines would be the edge of the outermost triangle, so we can reduce $n$ until $n=3$ where we already know the answer.

I thought this is a nice argument until I checked AoPS.

oh.

OOOOHHHHHHH.

Point counting around the outermost triangle with inequality $2n \leq 3n-3$ is much better.

Q2. Geometry outside of Q1/4 is hopeless for me as usual. Interestingly if I had 4.5 hours I can actually hardbash this geometry question by co-geom. It gets even more feasible with the fact that I can actually do Q3 timely this year. It is also simple enough in the sense you do not need to use vector calculus or the ratio gimmick stuff.

Q3. Surprisingly easy really. When you see statements in the type of "$f(a)$ satisfies proposition $P(a,b)$ for all $a,b$" you know this is an extremely powerful statement to be exploited by picking the right $b$. It also allows us to approach by thinking about the factorization of $f(a)$.

Believe or not, these two approaches are all you need for this problem.

- For prime $p$ we know that $f(p)\mid p^p$ so $f(p)$ is power of primes. 

- For $f(x)\neq x$ (we know the identity function does not satisfy the functional equation), $f(p) \mid x^p - f(x)^{f(p)}$ but $x^p -f(x)^{p^c} \equiv x-f(x)$ mod $p$ so $p\mid f(x)-x$ which is impossible for large enough $p$. Thus $f(p) = 1$ for all large enough primes.

- Suppose $a$ odd and $p\mid f(a)$ for odd prime $p$. Let $q$ be large enough so that $f(q) = 1$ and is primitive mod $p$. Then $f(a) \mid q^a-1$, but since $q$ is primitive we know $p-1\mid a$ which is impossible since $p-1$ is even and $a$ is odd. Checking $f(a) \mid a^a - f(a)^{f(a)}$ shows that $f(a)$ is odd as well. Thus $f(a) = 1$.

- Suppose $p\mid f(a)$ for prime $p$, then $p\mid f(a)\mid p^a-1$ forces $p=2$, so $f$ maps to powers of 2. (Notice the order, we need the previous claim so that $f(p) = 1$ for all odd primes.

- It is all prime power ($v_2(a)$) chasing and construction of a tight example at the end.

Every step is so natural: divisibility leads to prime behavior, odd behavior then extend to the whole function. The only slightly less natural tool here would be the Dirichlet's theorem but I guess this is well known right? 

Anyway, one of the very few accessible Q3 to me.

Q4. Strangely I find this one causing me more trouble than Q3, due to the troublesome case by case argument. The first part is easy -- the only three distinct unit fractions the sums to 1 is $(2,3,6)$. The rest is all about arguing (1) what would reduce to this case and (2) why the rest do not.

It reminds me of the Aliquot sequence problem! To fully characterize the Aliquot sequence we just need to argue about (1) what would reduce to cycles (primes, amicible numbers or sociable numbers) and (2) what would runaway. Easy right? No. This is an open problem.

It is easy enough to argue $a_1$ must be even or else the prime unit fractions do not sum up to 1 and that keeps going creating infinite descent, but the rest is quite messy and troublesome, like showing that 3 must be a factor with power limits. Bleh.

Q5. Game with inequalities! It immediately becomes one of my favourites in recent years, although that does not tell anything in terms of difficulty. The "critical point" was deemed non-trivial on the AoPS post but I don't really think so.

Alice is maintaining the linear sum $\sum x_i \leq \lambda n$ while Bazza is maintaining the quadratic sum $\sum x_i^2 \leq n$. The way Alice defeats Bazza is by min-maxing: she can keep choosing zero and accumulate (the allowance of $\lambda n - \sum x_i$) until she can pull a big one to break the quadratic sum. The critical point is then $\sqrt{2}/2$ because the best Bazza can do is $\sqrt{2}$ so that the quadratic sum increases by the maximum amount aka 2.

After that it suffices to check the cases $\lambda$ above, below or equal to $\sqrt{2}/2$. And again I would say every step here logical and natural. The choice of $x_{2k} = \sqrt{2-x_{2k-1}^2}$ looks artificial but it's merely "the choice to stuff the quadratic sum". The beautiful ending to the problem is that the equal case is actually the combination of the winning strategies from both cases.

I love setup like this, but again is it too straightforward for a question 5?

Q6. Sometimes I am tired with "general case" combinatorics question. It is always nice to have a question where numbers are concrete and the answer is strongly relying on that number. Last year we had an "algo" problem using the number 2024 but such parameter can be so easily adjusted to any natural numbers. This 2025 is however different because it is a square number! This is probably the once in a lifetime chance for us to have a square number year when the next one is 91 years away. I am glad that they utilized the chance to put up a question like this.

And honestly I have zero idea on the question. The actual tiling is simple and beautiful. Considering how Q4 and 5 are easy I have again 3+ hours for Q6. In that case there will be a slim chance for me to come up with that optimal tiling, although I doubt if it helps in any way other than ending up with 1/7 or so. Erdos-Szekeres sounds way too advanced for IMO but almost every solution seem to use a version of that. It's a beautiful Q6, although I hope to see a real and slightly more elementary solution.

In overall IMO 2025 is special in many ways:

- An actual solvable Q3 and a Q4 that looks messier than Q3.

- Solvable questions can be done quickly allowing time to solve Q2,6

- 2 number theory problems in similar field (divisibility, power chasing)

- Ending with a "square" problem (kind of reminding me the 1988 Q6 which is also a "square" problem although 1988 isn't a square xD)

I really enjoy the problems this year, can't say anything further. I seemed to keep complaining Simon Marais problems but never did the same to IMO or Putnam. Perhaps this is the magic of collaborative efforts, hm?

Oh well, we will see again in 2026.

Update: I read news about OpenAI and google claimed that their ai made good progress with "gold medal" at 35 points. But is it that big of a progress though?

I am not sure if we can solve those question using commercial LLMs, but I would not doubt their ability to compare the score against historical performances:

Q3 is indeed much, much easier than historical mean. Comparing against 2005-2024 data, this is +2.3 in Z score and the highest in 20+ years. This is not only by students getting partial marks (for example, I imagine that knowing $f(p) = 1$ for large enough primes is easy and will get you 1/7 for sure), but also by astonishingly high 7/7 rate at 15%. 

Q5 is also fairly easy as expected with a Z-score of +1 in 2005-2024 data although it seemed to trend easier since 2018 or so:

In overall, Q1-Q5 are all easier than average. Q6 is harder than usual (0.143, Z-score of -0.85), and is among the hardest Q6 historically in terms of average score. But this is fine because Q6 is here to serve as the ultimate challenge anyway. 

The takeaway is, this set of problems is an outlier vs past difficulty, and has a significant gap between mid-hard problems and extremely hard problems. The gap of difficulty is precisely the right difficulty of questions where LLM progressed enough to be able to solve. The lack of such question renders OpenAI and google's "claim in progress" hollow.

Oh by the way the 1988 Q6 average score is 0.6 -- not exactly hard in this regard. Not the legendary level hard at least ;)

16/7/2025: 世冠盃/數學番/AA/A

久違了的閒聊。

嘛，首要的大事當然是世冠盃決賽我車在地球保衛戰中清脆俐落地把大巴黎斬下馬呢。

說起來，我已經不記得上一次一年內熬夜兩次看球是哪年了。今年的兩次也就是兩個決賽，其餘晚場我根本懶得看，又或者看個半場也就剛好對上美股收市，不能算真正熬到天亮那種熬夜。

幸運的是這兩次熬夜都得到對應的回報，還是說這回報過於甜蜜了？全世界都知道我車鋒無力，但大概都沒有誰料到有個正常前鋒後戰鬥力居然直接提升了兩檔。其實浪費機會的可不只有Jackson，還有Nkunku可是上季全歐浪費最多xG的人呢。這兩個中游球隊來的前鋒，一看就知道全身功夫都為禁區肉縛和最後一擊而練，我車缺的就是這種人，讚啦。

至於大家吵得臉紅耳赤的「含金量」問題，我是覺得沒甚麼好吵的。名譽上肯定要叫自己世界冠軍啊。以前的世俱盃贏了都叫自己世界冠軍了，現在這個怎麼不可以？你說歐冠有足夠的歐洲球隊世冠盃沒有？那巴西球隊贏了歐洲的三四場是怎麼回事？世冠盃也有32支球隊，勁旅的數量也夠了啊。不服的話你下屆自己進來試試看？

真正有爭議的競技強度。6、7月歐洲剛好放暑假，南美卻是季中打得火熱的季節。有辦法能證明歐洲球隊輸球是因為競技狀態不足嗎？錢到位球員教練球隊拼命是肯定的，但實際效果可能略有出入。當然歐洲季中也不能保證狀態，不過預期上總比季後再拉來打好不是嗎。這方面我不敢亂說，應該要用上數據科學才能下判斷。

最後一點，說說這比賽的未來。球隊最高達一億美元的分成任誰都會心動，這也是賽事被如此看重的原因之一。漂亮的營收加上川皇加持，這屆比賽算是空前的成功，也吸引了大量國家競投下一輪主辦權。對此我還是懷疑這比賽真的有這麼吸金嗎？如果不是美國的巨型球場、人口基數和消費能力，營收能達到這個高度嗎？我是不太看好，不過如果以後一直放美國舉行的話能有這種成續我也不意外，放其他地方就別想了。

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身為數學家，看到這季新番可是想當興奮，原因正是本季足有兩套「數學」番。

如果只看列表應該會有點疑惑，第二套「數學」番是哪部啊？

Silent Witch沉默魔女。

雖然單憑人家叫了一聲數學博士就把這歸成數學番有點奇怪，不過至少把女主稱為數學腦絕對不為過，那個思考方式真的是做數學的才會有。配上其超級內向的性格，我們可稱之為－－數學小孤獨。

沒錯，這番就是講一個數學小孤獨扮豬吃老虎走向人生巔峰的故事。

扮豬吃虎、魔法世界這種題材每季都有，但這部作畫細膩、劇情合理張弛有度、顏藝到位、聲優陣容豪華，在不涉及劇情下已經很難再給予進一步的讚美。看了兩集目前個人評分為7.2/10，算非常高分了(雖然我已好幾年沒在這邊放上完整評分表)，絕對值得大家去試看。

另一部數學番是費馬的料理，聽名字就就知道是數學x美食的番。

……等等，這不就是我嗎？

進去看不到十分鐘，看到男主用座標硬爆幾何題被白毛嗆怎不用軌跡和不變量解題，說他也就這樣云云。

……欸等等，這真不是我嗎？

本來看到專業等級的數學對談還以為有人真敢把這些搬進動畫裡給廣大觀眾看，沒想到兩集看下來是這樣－－

我、北田岳、上輩子讀數學庸庸碌碌只能當牛馬，最終累到被卡車送走。

重生到高中時期的我只想把舊路再走一次。以為可以靠前世的積累拿下競賽席次，怎料看見天才們的境界後才知道自己真的不是那塊料子，甚至在考試中直接崩掉。

本來被寄予厚望的我被記恨的學園長處處針對，獎學金被要求退還，還被逼去食堂打工。就在此時－－

「叮！至尊美食系統激活中……」

只要嘗過一次的美味就能記住，甚至還能反推菜式的秘密？每日簽到就能拿到菜譜，消費點數就能拿到討好客人的提示？

本來想著低調發育的我本來還想對外還想說是用數學倒推，沒想到食堂打工第一天就被年紀輕輕就拿星的天才廚師識破？

一場未曾設想的剌激美食之旅就此展開－－

……

聽上去很像上架後二百話以內完結的網文，但這的確是我看了兩集的感覺。在同一集內看見專業真實的幾何吐糟和用類似函數關係的草圖就能倒推出料理關鍵步驟這兩種截然不同的數學內容，絕對能引起觀眾帶來尷尬而不失禮貌的微笑。

其實本季的好番和題材真不少，有機會我再來寫吧^.^

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既然要寫日記，那就補兩句DDR好了。

最近不是出了AA的csp譜嗎？我記得我做過AA的太鼓譜，覺得這DDR譜很像我過去會做的太鼓圖。一翻下去發現我做的不是AA而是AAA，風格完全不像我所想像的東西。我那AAA譜面一點都不流暢，虧那還是我代表作Holy Moon半年後才做的圖，那都是甚麼鬼啊……

網絡隨心巡記(2): 三崎港→函館

當我新開這個系列的時候，我完全沒有料到這個系列會有第二集。當時我只是把這當成一種讓有借口寫我這種形式的隨筆而已。以下這個隨心遊記同樣隨心隨機，一切所記之事皆屬偶然。

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1) 三崎港

跟朋友吃飯，電視正播放日本料理巡遊的節目。這一集介紹的是三崎港。這我可熟了，京急的老熟人，還有那用電池漲起的平板點餐的迴轉壽司連鎖店，我怎可能不熟呢？

這次介紹的是某位山田老兄的餐廳。只見他端出他的招牌菜，烤巨型吞拿魚頭。他先給吞拿魚祈福，然後用純熟的手法用純熟的手法把魚頭解體成腦天、魚眼、面頰等部位。每個部位各有不同的口感吃法，讓人看了就流口水。

……沒錯了。

我衝口而出：「我認識這家餐廳和這位老兄。」

2) 愛の貧乏脱出大作戦

如果我是在其他料理節目中看過這家餐廳就算了，偏偏這餐廳出自愛の貧乏脱出大作戦，中文叫拯救貧窮大作戰。

說到料理節目這一塊日本似乎比西方走得更前。當他們在玩鐵人料理和拯救貧窮大作戰的時候拍Master Chef和Kitchen Nightmares的戈登還是個毛頭小子呢。

某一年台灣買了版權回來播，既便距離拍攝當年已經有15年左右，這劇集居然在台灣掀起了不小的熱度。看看中文版維基的詳盡程度就知道了，一般的番組可沒有這種狂粉喔？既然提到了這檔節目，我當然也是粉絲之一。節目裡的貧窮店沒記得多少，達人店倒是很有印象。至少那時候對所謂達人店有多好吃是很有憧憬的。

現在的大多數人自是不會認識這檔節目的，我只好跟朋友聊三崎港坐京急去很方便云云，但我心裡卻一直想著趁我還記得回家搜一下：距離上次檢查又過了好幾年，到底有哪些店還活著呢？

3) Tabelog

二十年前沒有這東西，現在總有了吧？到底達人店有多好吃，一查不就知道了。

原來一些達人的店也不能算最頂級，不過當然這有點過於嚴苛。要找一家肯花幾天幾夜(按照傳聞的話可不只幾天)全心教你的店主也不容易啊。

順帶一提，山田桑的店的分數是3.6，算不錯了。

4) 南醬

拯救貧窮大作戰的超級爭議人物之一。

其實能夠經營不善到被節目挑上，店主多半都有點毛病。比如說不講衛生、味覺跟一般人不同、對無關的事異常執著等等。南醬特別就特別在所有毛病都有一點，偏偏又沒很過份，加上他那不溫不火的性格和辦事調子，除了可以直接DQ他的達人以外大部分人都拿他沒辦法。

偏偏他通過補修，可以賣學來的四五六拉麵了。

偏偏貧乏店的存活率這麼低，這家店卻活下來了。頂著被霸凌的壓力，南醬的拉麵或許不是最好吃的，但他確實找到能讓自己這店活下去的方法。這家在函館開的拉麵小店最後靠俄羅斯遊客活了快二十年，還不是被經營狀況壓倒的，這個傳奇算是有個圓滿的結局吧。

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其實我想到南醬並不單純因為他是這節目的指標性人物，也是因為在愛の貧乏的非官方資料庫最頂看到一條訃聞。才看到第一個み字我想到南醬，而youtube的搜尋建議上「愛の貧乏脱出大作戦 みなみちゃん 死去」這條也的確排很前。考慮到南醬幾年前身體就很差了，現在出事好像也不是很意外。

事實當然不是這樣。嘛，現在南醬的確實狀況也沒人知道，但資料庫上說的卻是另一個同樣重要的番組成員－－

2025/03/01 みのもんたさん逝去。ご冥福をお祈りいたします。