Difference between revisions of "1990 IMO Problems/Problem 6"
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− | + | ==Problem== | |
+ | Prove that there exists a convex <math>1990\text{-gon}</math> with the following two properties: | ||
(a) All angles are equal. | (a) All angles are equal. | ||
(b) The lengths of the <math>1990</math> sides are the numbers <math>1^2, 2^2,3^2,\dots, 1990^2</math> in some order. | (b) The lengths of the <math>1990</math> sides are the numbers <math>1^2, 2^2,3^2,\dots, 1990^2</math> in some order. | ||
+ | |||
+ | ==Solution 1== | ||
+ | Let <math>\{a_1,a_2,\cdots,a_{1990}\}=\{1,2,\cdots,1990\}</math> and <math>\theta=\frac{\pi}{995}</math>. | ||
+ | Then the problem is equivalent to that there exists a way to assign <math>a_1,a_2,\cdots,a_{1990}</math> such that <math>\sum_{i=1}^{1990}a_i^2e^{i\theta}=0</math>. | ||
+ | Note that <math>e^{i\theta}+e^{i\theta+\pi}=0</math>, then if we can find a sequence <math>\{b_n\}</math> such that <math>b_i=a_p^2-a_q^2, (p\not=q)</math> and <math>\sum_{i=1}^{995}b_ie^{i\theta}=0</math>, the problem will be solved. | ||
+ | Note that <math>2^2-1^2=3,4^2-3^2=7,\cdots,1990^2-1989^2=3979</math>, then <math>3,7,\cdots,3979</math> can be written as a sum from two elements in sets <math>A=\{3,3+4\times199,3+4\times398,3+4\times597,3+4\times796\}</math> and <math>B=\{0,4\times1,4\times2,\cdots,4\times 198\}</math>. | ||
+ | If we assign the elements in <math>A</math> in the way that <math>l_{a1}=3,l_{a2}=3+4\times199,l_{a3}=3+4\times398,l_{a4}=3+4\times597,l_{a5}=3+4\times796</math>, then clearly <math>\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}=0, (j\in\{1,2,3,4,5\})</math> | ||
+ | Similarly, we could assign elements in <math>B</math> in that way (<math>l_{b1}=0,l_{b2}=4,\cdots</math>) to <math>e^{i\frac{2\pi}{5}}</math>. | ||
+ | Then we make <math>b_i</math> according to the previous steps. Let <math>i\equiv j</math> (mod 5), <math>i\equiv k</math> (mod 199), and <math>b_i=l_{aj}+l_{bk}</math>, then each <math>b_i</math> will be some <math>a_ p^2-a_q^2</math> and <math>\sum_{i=1}^{995}b_ie^{i\theta}=\sum_{j=1}^5\sum_{i=1}^{199}l_{aj}e^{i\frac{2\pi}{199}}+\sum_{k=1}^{199}\sum_{i=1}^{5}l_{bk}e^{i\frac{2\pi}{5}}=0 </math> | ||
+ | And we are done. | ||
+ | |||
+ | This solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [https://aops.com/community/p2114824] | ||
+ | |||
+ | ==Solution 2== | ||
+ | Throughout this solution, <math>\omega</math> denotes a primitive <math>995</math>th root of unity. | ||
+ | |||
+ | We first commit to placing <math>1^2</math> and <math>2^2</math> on opposite sides, <math>3^2</math> and <math>4^2</math> on opposite sides, etc. Since <math>2^2-1^2=3</math>, <math>4^2-3^2=7</math>, <math>6^2-5^2=11</math>, etc., this means the desired conclusion is equivalent to<cmath> 0 = \sum_{n=0}^{994} c_n \omega^n </cmath>being true for some permutation <math>(c_0, \dots, c_{994})</math> of <math>(3, 7, 11, \dots, 3981)</math>. | ||
+ | |||
+ | Define <math>z = 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796}</math>. Then notice that\begin{align*} z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ \omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ \omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdotswithin{=} \end{align*}and so summing yields the desired conclusion, as the left-hand side becomes<cmath> (1+\omega^5+\omega^{10}+\cdots+\omega^{990})z = 0 </cmath>and the right-hand side is the desired expression. | ||
+ | |||
+ | This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [https://aops.com/community/p17584021] | ||
+ | |||
+ | == See Also == {{IMO box|year=1990|num-b=5|after=Last Question}} |
Revision as of 12:54, 30 January 2021
Contents
Problem
Prove that there exists a convex with the following two properties: (a) All angles are equal. (b) The lengths of the sides are the numbers in some order.
Solution 1
Let and . Then the problem is equivalent to that there exists a way to assign such that . Note that , then if we can find a sequence such that and , the problem will be solved. Note that , then can be written as a sum from two elements in sets and . If we assign the elements in in the way that , then clearly Similarly, we could assign elements in in that way () to . Then we make according to the previous steps. Let (mod 5), (mod 199), and , then each will be some and And we are done.
This solution was posted and copyrighted by YCHU. The original thread for this problem can be found here: [1]
Solution 2
Throughout this solution, denotes a primitive th root of unity.
We first commit to placing and on opposite sides, and on opposite sides, etc. Since , , , etc., this means the desired conclusion is equivalent tobeing true for some permutation of .
Define . Then notice that\begin{align*} z &= 3 \omega^0 + 7 \omega^{199} + 11\omega^{398} + 15 \omega^{597} + 19 \omega^{796} \\ \omega^5 z &= 23 \omega^5 + 27 \omega^{204} + 31\omega^{403} + 35 \omega^{602} + 39 \omega^{801} \\ \omega^{10} z &= 43 \omega^{10} + 47 \omega^{209} + 51\omega^{408} + 55 \omega^{607} + 59 \omega^{806} \\ &\vdotswithin{=} \end{align*}and so summing yields the desired conclusion, as the left-hand side becomesand the right-hand side is the desired expression.
This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]
See Also
1990 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |