Transforming letters into words

Question

^{I found this weird-looking puzzle in the 2007 Tournament of the Towns.}

For each letter A to Z, I allot a specific English word containing (but not necessarily starting with) that letter. Then I write pages as follows:

• The first page consists only of the word corresponding to the letter A.
• On each subsequent page, I replace each letter of the preceding page by its corresponding word.

The opening of the 40th page begins with:

Till whatsoever star that guides my moving.

Prove that this sentence is written again later on the same page.

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Bonus question: what properties of the above-quoted sentence are essential to the proof? Or in other words, what other sentences could we replace this by so that the result still holds?

Answer 1

First, note that

every time a page begins with a certain letter, the next page will always begin with the same word. The first letter of that word will determine the first word on the next page.

There are 26 letters in the alphabet. Therefore,

If we look at the first words of pages, they must form a cycle of length 26 at most. This is the 40th page; therefore, we are well into at least the second such cycle.

The first word of the present page is "Till".

Because we are in the second (or subsequent) cycle, there must be a prior page that began with "Till". We know that the first word of the preceding page was not "Till" or the third and fourth word on this page would be the same. Therefore, we know that "till" is not the word corresponding to T. It must then be the word corresponding to I or L.

Consider the first word of the 39th page. Call it 39th.

Since we are so far into the cycle, we know that 39th must also be appearing for at least the second time. 39th must begin with either I or L. None of the seven letters on this page begins with I or L, therefore none of them is 39th. Because each word must contain the letter it corresponds to, that word must appear in its own expansion. Therefore, 39th must be more than seven letters long. Since 39th appears in its own expansion and it has occurred as the first word at least once before, we know that it occurs in some position after the seventh on every subsequent page including the 39th page. Therefore, the expansion of its first seven letters occurs later on this page.

For the bonus,

I got nothing, other than the above. Definitely, it is sufficient that none of the words in the sentence could be the first word on the preceding page but but I feel certain that there is something more intelligent to be said.

Answer 2

Random observations:

Each page contains a copy of the previous page. The word assigned to A cannot start with A - otherwise the words on all the pages would start with A, and the 40th page starts with "Till". This means that there exist two pages at most 26 pages apart that start with the same letter. Let's call those pages a and b. Then pages a+1 and b+1 would start with the same word, etc, and pages a+n and b+n would start with the same n words. This may have nothing to do with what we are trying to prove, who knows.

EDIT: Just ate some ice cream, my mind is clear now.

So the part I was missing was that the copy of page i that page (i+1) has does not start with the content of page i. Therefore page 40 contains some page (14-40) that starts with "Till whatsoever star that guides my moving" not from the beginning, so the sentence must appear somewhere else.

As for the bonus:

Since the pages are at most 26 pages apart, the maximum length of the sentence would be 13 words, so any sentence from 1-13 words that does not start with the letter A(so each page can't be the same) satisfies this property.