在泰國清邁舉行的2008 青少年數學國際城市邀請賽有一題所有考生都沒有人得到滿分的隊際賽試題，連大陸隊都沒得分。

在九枚外觀均相同的金幣中，其中有一枚的重量為a，有七枚的重量為b，最後一枚的重量為c，且a小於b小於c。用沒有刻度的兩臂天平，請給出秤四次即可判定a+c小於2b、a+c=2b或a+c大於2b的方案。

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孫文先 敬上

My solution: (it took a lot of time to figure out)
Step 1. Number the weights 123456789. Weigh 12-34, 56-78. There are three cases:
1a. 12=34 and 56=78. Then, if A and C were not in the same group then one of the weighings would not be equal, so A and C are in the same group. Since they were weighed equal to two other B weights we determine that A+C = 2B.

1b. Both are unequal; assume 12>34 and 56>78. The only possibility then that A is in [3478] and C in [1256], and one is in [1234] and one is in [5678].
1b2. Then weigh [1234]-[5678], so you know which one is in [1234] and which is in [5678]. If [1234] is heavier, then C is in [12] and A is in [78]; weigh [1278]=A+2B+C against [3456]=4B for the answer. Or if [5678] is heavier, weigh [3456]=A+2B+C against [1278]=4B.

1c (the hardest case).
There is one weighing that was unequal; call the heavier side weights HH and the lighter side weights LL, and the other equal weighing, call them all XXXX. Call the extra one that hasn't been weighed yet W.
Note that XXXX may not have 4 B's, but their weight is equal to 4 B's because the only other possibility is A+C=2B, and it still adds up.

Weigh HHLL against XXXX. If HHLL is heavier, then:

• 1c.1 CB = HH, and LL = BB.
  
• 1c.2 HH = C and A, and C+A>2B.
  
• 1c.3 Or there is a C in HH and an A in LL.
  

• 1c.1 W must be A, so HHLW = A+2B+C.
  
• 1c.2 LL's weight doesn't change; HHLW = A+2B+C.
  
• 1c.3 (hardest) If the L switched is a B then HHLW = A+2B+C. However if the L switched is the A then HHLW = C+3B and is always larger than 4B. However it still works, because HHLL>XXXX.
  Therefore in this case the result of HHLW-XXXX is the same as A+C to 2B.
  
  If HHLL is lighter, then the arguments are similar. You must switch W with an H, to get HWLL-XXXX.
  
  If HHLL = XXXX, then either XXXX had A and C in it and A+C=2B, or A was in HH and C was in LL, and A+C=2B. Either way it's the same: A+C=2B.
  
  
  

1c (the hardest case).
There is one weighing that was unequal; call the heavier side weights HH and the lighter side weights LL, and the other equal weighing, call them all XXXX. Call the extra one that hasn't been weighed yet W.
Note that XXXX may not have 4 B's, but their weight is equal to 4 B's because the only other possibility is A+C=2B, and it still adds up.


當有一秤平衡有一秤不平衡，平衡那秤上的四個金幣應可知全為真幣。

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孫文先 敬上

So, delete this: "Note that XXXX may not have 4 B's, but their weight is equal to 4 B's because the only other possibility is A+C=2B, and it still adds up." Instead it should be "XXXX must have 4 B's because the only other possibility, A+C=2B, is impossible because the unequal weighing needs at least one different weight."

我認為以下作法不周全，請重新考慮

Weigh HHLL against XXXX. If HHLL is heavier, then:


1c.1 CB = HH, and LL = BB.

1c.2 HH = C and A, and C+A>2B.

1c.3 Or there is a C in HH and an A in LL.


Take W and switch it with either L. Weigh again, which is HHLW-XXXX. As we showed, XXXX = 4B, and therefore, in each case above:


1c.1 W must be A, so HHLW = A+2B+C.

1c.2 LL's weight doesn't change; HHLW = A+2B+C.

1c.3 (hardest) If the L switched is a B then HHLW = A+2B+C. However if the L switched is the A then HHLW = C+3B and is always larger than 4B. However it still works, because HHLL>XXXX.
Therefore in this case the result of HHLW-XXXX is the same as A+C to 2B.

If HHLL is lighter, then the arguments are similar. You must switch W with an H, to get HWLL-XXXX.

If HHLL = XXXX, then either XXXX had A and C in it and A+C=2B, or A was in HH and C was in LL, and A+C=2B. Either way it's the same: A+C=2B.

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孫文先 敬上

(~ is weigh against; == is "gets the same result as")

HHLL>XXXX=BBBB
Clearly there must be C in HHLL, otherwise HHLL could not be larger than 4b because C is the only weight larger than B. Also, C cannot be in LL unless A were there too. But if LL=CA then CA>2B so HHLL>4B=XXXX which contradicts the weighing; therefore C is in HH. A can be the other H, an L, or W, to make 3 possibilities.
1. HHLL=CBBB; W=A; HHLW=CBBA
CBBA~BBBB == c+2b+a~4b == c+a~2b
2. HHLL=CABB; W=B; HHLW=CABB
CABB~BBBB == c+2b+a~4b == c+a~2b
3. HHLL=CBAB; W=B; HHLW=CBAB or CBBB
HHLL>BBBB so CBAB>BBBB and c+a always > 2b
and HHLW>BBBB always, so it is still valid.

HHLL<XXXX=BBBB
Same reasoning, A is in LL.
1. HHLL=BBAB; W=C; HWLL=BCAB
BCAB~BBBB == c+2b+a~4b == c+a~2b
2. HHLL=BBAC; W=B; HWLL=BBAC
BBAC~BBBB == c+2b+a~4b == c+a~2b
3. HHLL=CBAB; W=B; HWLL=CBAB or BBAB
HHLL<BBBB so CBAB<BBBB and c+a always > 2b
and HHLW<BBBB always, so they get the same result, so it is still valid.

您可能有些地方打字錯誤，請修改。切記莫使用小於符號，只可用>符號。

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孫文先 敬上

My mother translated the solution for me:

我的解答如下(花了很久的時間才完成的, 難怪沒有人能在考試時間內作答完成)
首先將9枚硬幣標上1~9號,先秤兩個一組的硬幣12號和34號再秤56號和78號硬幣有三種可能結果
A. 12=34且56=78則我們可以確定AC同在一組因為如果他們不在同一組的話則一定會有任一組硬幣重量不相等的情形出現.既然AC同在一組且重量與其他組真幣重量相等則我們兩次秤重就可以推出A+C=2B
兩組重量都不相等.假設12>34且56>78.則唯一的可能性為A在34,78這兩組硬幣裡, C在12,56這兩組硬幣裡, 接著我們再秤1234和5678來決定A和C在1234裡或是在5678裡, 如果1234比較重, 則C在12這一組裡,A在78這一組裡然後我們就可以將1278放再一起和其他任4個真幣比較重量 則4次就可以得出A+C+2B與4B的重量比
C.(較複雜的一種情形) 其中一組重量不相等, 則令較重的那一組兩枚硬幣為HH較輕那一組為LL另外重量相等的那一組為XXXX,並剩下的第九枚硬幣為W.到此我們可以確定至少有一枚假幣(A或C)在HH和LL兩組硬幣當中, 及另外相等的兩組硬幣XXXX都是真幣(4B)接著我們比較HHLL和XXXX的重量
C1如果HHLL 跟 XXXX 一樣重時，必須 HHLL=CBAB, 則得出結論A+2B+C=4B, A+C=2B.
C2如果HHLL比較重, 則我們可以有以下可能性
1 HH=CB且LL=BB或
2HH=AC則C+A>2B
3 HH當中有C且LL當中有A
然後將W和任何一個L互換, 現在再秤一次比較HHLW和XXXX四個真幣的重量
則以上三種情形分別可以推出下列結果有可能是
1. CB = HH, and LL = BB. 則W為A, 所以HHLW=A+2B+C
2. HH = C and A, and C+A>2B. 則W為B, HHLW= A+2B+C>4B
3. HH當中有C且LL當中有A. 則如果W換掉的是B則HHLW=A+2B+C. 但如果W換掉的是C則HHLW=A+3B>4B是當然的, 但沒關係因為之前我們已比較過出HHLL和XXXX所以之前HHLL=A+2B+C就已經告訴我們A+C>2B
從以上推論情形可以推論在C的模式裡, 第四次秤重HHLW和XXXX的重量比等於A+C+2B和4B的重量比.
C3反之如果第三次秤重時LLHH比XXXX輕則我們要將W和任何一個H互換來檢驗得出結果

我想你的解法是正確的

不過C部分有一點小複雜

這裡有個比較簡單的秤法

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我認為這個部分很重要的一點是

W究竟是真幣還是假幣??

所以我們不妨秤一下W和X

C1.如果W是真幣:

則秤XXXX和HHLL

秤出來的結果即答案

C2.如果W是輕幣:

則秤XXX和HHW

秤出來的結果即答案

C3.W是重幣亦同

提供給大家參考

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我想

這題的關鍵在於

我們沒有辦法明確的指出哪一枚是假幣

我們要做的是

把假幣放在天平一邊

再把真幣放在另一邊

這才是題目要的

所以要搞清楚哪幾枚"一定"是真幣
(想辦法搞清楚)

(至少找出真幣比較容易)

提供給大家參考~~

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Simple

Good thinking. I didn't notice that XXXX had to be all B's when I thought out my solution, so this simplifies it a lot.