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I've a problem whose solution is also stated below. I can't understand the explanation.

There are two disks ,one smaller than the other, are each divided into 20 congruent sectors.

In the larger disk,10 of the sectors are chosen arbitrarily and painted red; the other 10 sectors are painted blue.

In the smaller disk, each sector is painted either red or blue with no stipulations on the number of the red and blue sectors.

The small disk is then placed on the larger disk s.t. their centers coincide.

Show that it is possible to align the two disks s.t. the no. of sectors of the small disk whose color matches the corresponding sector of the large disk is atleast 10.

The steps how to proceed is as follows:

$\texttt1.$ fix the large disk in place, then there are 20 possible positions for the smaller disk such that each sector of the smaller disk is contained in the larger disk.

$\texttt2.$ now count the total no. of color matches over all 20 possible positions of the disk. Since, the large disk has 10 sectors of each of 2 colors,each sector of small disk will match in color the corresponding sector of large disk in exactly 10 of 20 possible positions.

Thus, total no. of color matches over all positions = $20\times10$ . Average no. of color matches per position = $(20\times 10)$$\mathbin{/} 20$. Hence, $\exists$ some position with atleast 10 color matches.

What I can't understand is the main step $2$,in particular why are we multiplying with 10 to get all positions of same matches ?

Please help.

spectraa
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1 Answers1

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suppose there are $k$ red parts in the outer circle. Then as we rotate the little circle if we look at a particular red segment in the little center it will allign which each of the outer red segments exactly once, so it will coincide $k$ times, since there are $10$ inner red segments we get there are $10k$ red "coincidations" as we rotate the inner circle. since there are $k$ red parts in the outer circle there are $20-k$ blue parts. Using the same argument we get there are $(20-k)10$ blue "coincidations" in a complete rotation. Thus there $10k+10(20-k)=200$ coincidations in the full rotation. Then there must be a particular orientation with at least $10$ coinciding segments (since there are $10$ orientations and in total there where 200 "coincidations").

Asinomás
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  • Thanks for the answer I understood your clarification for step 2, but I thing I want to clarify is that why in solution stated by me in question did we easily assume that "Since, the large disk has 10 sectors of each of 2 colors,each sector of small disk will match in color the corresponding sector of large disk in exactly 10 of 20 possible positions." Is there some other reasoning behind this? – spectraa Sep 03 '14 at 05:46
  • the small disk is the one that has 10 sectors of each of the two colors. – Asinomás Sep 03 '14 at 11:47
  • no that is the large disk.You have considered smaller disk as outer circle and large disk as inner circle in your answer. – spectraa Sep 03 '14 at 11:58
  • I can't understand why did they(i.e. book) skip the important step of calculating $10k+10(20−k)=200$ coincidations in the full rotation ,by simply writing $20\times10$. – spectraa Sep 03 '14 at 12:02
  • I can't understand why they wrote as- the large disk has 10 sectors of each of 2 colors,each sector of small disk will match in color the corresponding sector of large disk in exactly 10 of 20 possible positions.Please help... – spectraa Sep 03 '14 at 12:09
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    because there are 10 blue sections in the outer disk, so in 10 out of the 20 rotations it will be paired up with a disk of it's same color, and in the other 10 it will be paired up with a disk of the other color. – Asinomás Sep 04 '14 at 00:03