Difference between revisions of "Pigeonhole Principle/Solutions"
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See [http://www.artofproblemsolving.com./Resources/AoPS_R_A_HowWriteName.php this article]. | See [http://www.artofproblemsolving.com./Resources/AoPS_R_A_HowWriteName.php this article]. | ||
===O4=== | ===O4=== | ||
| − | The area of the circle is <math>(\frac{5}{2})^2 \pi = \frac{25}{4} \pi</math>. For the ten points not to be within 2 units from each other, consider circles of 1 diameter each around each of the points. If any 2 of the circles intersect, then the respective centers are within 2 units of each other. <math>\frac{\frac{25}{4} \pi}{1 \pi} = \frac{25}{4}</math>, which is less than 10, which means that less than 10 circles can fit into the larger circle without intersecting (pigeonhole principle). Thus, not all 10 points can be within 2 units of each other and be inside the circle. | + | The area of the circle is <math>\left(\frac{5}{2}\right)^2 \pi = \frac{25}{4} \pi</math>. For the ten points not to be within 2 units from each other, consider circles of 1 diameter each around each of the points. If any 2 of the circles intersect, then the respective centers are within 2 units of each other. <math>\frac{\frac{25}{4} \pi}{1 \pi} = \frac{25}{4}</math>, which is less than 10, which means that less than 10 circles can fit into the larger circle without intersecting (pigeonhole principle). Thus, not all 10 points can be within 2 units of each other and be inside the circle. |
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===O5=== | ===O5=== | ||
The pigeonhole principle is used in [http://usamts.org/Solutions/Solution4_1_18.pdf these solutions (PDF)]. | The pigeonhole principle is used in [http://usamts.org/Solutions/Solution4_1_18.pdf these solutions (PDF)]. | ||
Revision as of 19:45, 9 November 2007
These are the solutions to the problems related to the pigeonhole principle.
Contents
Introductory
This page is in need of some relevant examples or practice problems. Help us out by adding some. Thanks.
Intermediate
M1
The maximum number of friends one person in the group can have is 4, and the minimum is 0. If all of the members have at least one friend, then each individual can have somewhere between 
to 
friends; as there are 
individuals, by pigeonhole there must be at least two with the same number of friends. If one individual has no friends, then we can repeat the above argument for the remaining 4 individuals, and if more than one individual has no friends then we are done.
M2
For the difference to be a multiple of 5, the two integers must have the same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the same remainder. Thus, the answer is 1 (E).
Olympiad
O1
By the Triangle Inequality Theorem, a side of a triangle can be less than the sum of the other two sides. Assume that the first two segments are as short as possible (1 inch). For a triangle to not exist, the next term must be 2, then 3, 5, 8, and 13 (this is the Fibonacci sequence). The 7th term is 13, which is greater than 10 inches.
O2
For the difference to be a multiple of 7, the integers must have equal modulo 7 residues. To avoid having 15 with the same residue, 14 numbers with different modulo 7 residues can be picked (
). Thus, two numbers are left over and have to share a modulo 7 residue with the other numbers under the pigeonhole principle.
O3
See this article.
O4
The area of the circle is 
. For the ten points not to be within 2 units from each other, consider circles of 1 diameter each around each of the points. If any 2 of the circles intersect, then the respective centers are within 2 units of each other. 
, which is less than 10, which means that less than 10 circles can fit into the larger circle without intersecting (pigeonhole principle). Thus, not all 10 points can be within 2 units of each other and be inside the circle.
O5
The pigeonhole principle is used in these solutions (PDF).
O6
This problem needs a solution. If you have a solution for it, please help us out by adding it.
O7
This problem needs a solution. If you have a solution for it, please help us out by adding it.
