MathDB

5

Part of 2013 Iran MO (3rd Round)

Problems(5)

Maping a circle to a polygon by a polynomial!

Source: Iran 3rd round 2013 - Algebra Exam - Problem 5

9/11/2013
Prove that there is no polynomial P∈C[x]P \in \mathbb C[x] such that set {P(z)  ∣  ∣z∣=1}\left \{ P(z) \; | \; \left | z \right | =1 \right \} in complex plane forms a polygon. In other words, a complex polynomial can't map the unit circle to a polygon. (30 points)
algebrapolynomialtrigonometrycomplex numbersalgebra proposed
Function L

Source: Iran 3rd round 2013 - Number Theory Exam - Problem 5

9/11/2013
p=3k+1p=3k+1 is a prime number. For each m∈Zpm \in \mathbb Z_p, define function LL as follow: L(m)=∑x∈Zp(x(x3+m)p)L(m) = \sum_{x \in \mathbb{Z}_p}^{ } \left ( \frac{x(x^3 + m)}{p} \right ) a) For every m∈Zpm \in \mathbb Z_p and t∈Zp∗t \in {\mathbb Z_p}^{*} prove that L(m)=L(mt3)L(m) = L(mt^3). (5 points)
b) Prove that there is a partition of Zp∗=A∪B∪C{\mathbb Z_p}^{*} = A \cup B \cup C such that ∣A∣=∣B∣=∣C∣=p−13|A| = |B| = |C| = \frac{p-1}{3} and LL on each set is constant. Equivalently there are a,b,ca,b,c for which L(x)={ax∈Abx∈Bcx∈CL(x) = \left\{\begin{matrix} a & & &x \in A \\ b& & &x \in B \\ c& & & x \in C \end{matrix}\right. . (7 points)
c) Prove that a+b+c=−3a+b+c = -3. (4 points)
d) Prove that a2+b2+c2=6p+3a^2 + b^2 + c^2 = 6p+3. (12 points)
e) Let X=2a+b+33,Y=b−a3X= \frac{2a+b+3}{3},Y= \frac{b-a}{3}, show that X,Y∈ZX,Y \in \mathbb Z and also show that :p=X2+XY+Y2p= X^2 + XY +Y^2. (2 points)
(Zp∗=Zp∖{0}{\mathbb Z_p}^{*} = \mathbb Z_p \setminus \{0\})
functionlinear algebramatrixnumber theory proposednumber theoryQuadratic Residues
Collinear Point On Line With Distance R/2 To Circumcenter

Source: Iran Third Round 2013 - Geometry Exam - Problem 5

9/8/2013
Let ABCABC be triangle with circumcircle (O)(O). Let AOAO cut (O)(O) again at A′A'. Perpendicular bisector of OA′OA' cut BCBC at PAP_A. PB,PCP_B,P_C define similarly. Prove that :
I) Point PA,PB,PCP_A,P_B,P_C are collinear.
II ) Prove that the distance of OO from this line is equal to R2\frac {R}{2} where RR is the radius of the circumcircle.
geometrycircumcirclegeometric transformationreflectionperpendicular bisectorgeometry unsolved
Graph with 7n/4 Edges

Source: Iran 3rd round 2013 - Combinatorics exam problem 5

9/18/2014
Consider a graph with nn vertices and 7n4\frac{7n}{4} edges. (a) Prove that there are two cycles of equal length. (25 points) (b) Can you give a smaller function than 7n4\frac{7n}{4} that still fits in part (a)? Prove your claim. We say function a(n)a(n) is smaller than b(n)b(n) if there exists an NN such that for each n>Nn>N ,a(n)<b(n)a(n)<b(n) (At most 5 points) Proposed by Afrooz Jabal'ameli
functioncombinatoricsgraph theory
Recovering Lost Numbers

Source: Iran 3rd round 2013 - final exam problem 5

9/25/2014
A subsum of nn real numbers a1,…,ana_1,\dots,a_n is a sum of elements of a subset of the set {a1,…,an}\{a_1,\dots,a_n\}. In other words a subsum is ϵ1a1+⋯+ϵnan\epsilon_1a_1+\dots+\epsilon_na_n in which for each 1≤i≤n1\leq i \leq n ,ϵi\epsilon_i is either 00 or 11. Years ago, there was a valuable list containing nn real not necessarily distinct numbers and their 2n−12^n-1 subsums. Some mysterious creatures from planet Tarator has stolen the list, but we still have the subsums. (a) Prove that we can recover the numbers uniquely if all of the subsums are positive. (b) Prove that we can recover the numbers uniquely if all of the subsums are non-zero. (c) Prove that there's an example of the subsums for n=1392n=1392 such that we can not recover the numbers uniquely.
Note: If a subsum is sum of element of two different subsets, it appears twice. Time allowed for this question was 75 minutes.
combinatorics unsolvedcombinatorics