MathDB

5

Part of 2012 Iran MO (3rd Round)

Problems(4)

Two perpendicular lines and a locus

Source: Iran 3rd round 2012-Geometry exam-P5

9/20/2012
Two fixed lines l1l_1 and l2l_2 are perpendicular to each other at a point YY. Points XX and OO are on l2l_2 and both are on one side of line l1l_1. We draw the circle ω\omega with center OO and radius OYOY. A variable point ZZ is on line l1l_1. Line OZOZ cuts circle ω\omega in PP. Parallel to XPXP from OO intersects XZXZ in SS. Find the locus of the point SS.
Proposed by Nima Hamidi
conicsparabolageometry proposedgeometry
Infinite base of a prime number!

Source: Iran 3rd round 2011-Number Theory exam-P5

9/19/2012
Let pp be a prime number. We know that each natural number can be written in the form ∑i=0taipi(t,ai∈N∪{0},0≤ai≤p−1)\sum_{i=0}^{t}a_ip^i (t,a_i \in \mathbb N\cup \{0\},0\le a_i\le p-1) Uniquely.
Now let TT be the set of all the sums of the form ∑i=0∞aipi(0≤ai≤p−1).\sum_{i=0}^{\infty}a_ip^i (0\le a_i \le p-1).
(This means to allow numbers with an infinite base pp representation). So numbers that for some N∈NN\in \mathbb N all the coefficients ai,i≥Na_i, i\ge N are zero are natural numbers. (In fact we can consider members of TT as sequences (a0,a1,a2,...)(a_0,a_1,a_2,...) for which ∀i∈N:0≤ai≤p−1\forall_{i\in \mathbb N}: 0\le a_i \le p-1.) Now we generalize addition and multiplication of natural numbers to this set so that it becomes a ring (it's not necessary to prove this fact). For example:
1+(∑i=0∞(p−1)pi)=1+(p−1)+(p−1)p+(p−1)p2+...1+(\sum_{i=0}^{\infty} (p-1)p^i)=1+(p-1)+(p-1)p+(p-1)p^2+... =p+(p−1)p+(p−1)p2+...=p2+(p−1)p2+(p−1)p3+...=p+(p-1)p+(p-1)p^2+...=p^2+(p-1)p^2+(p-1)p^3+... =p3+(p−1)p3+...=...=p^3+(p-1)p^3+...=...
So in this sum, coefficients of all the numbers pk,k∈Np^k, k\in \mathbb N are zero, so this sum is zero and thus we can conclude that ∑i=0∞(p−1)pi\sum_{i=0}^{\infty}(p-1)p^i is playing the role of −1-1 (the additive inverse of 11) in this ring. As an example of multiplication consider (1+p)(1+p+p2+p3+...)=1+2p+2p2+⋯(1+p)(1+p+p^2+p^3+...)=1+2p+2p^2+\cdots
Suppose pp is 11 modulo 44. Prove that there exists x∈Tx\in T such that x2+1=0x^2+1=0.
Proposed by Masoud Shafaei
modular arithmeticnumber theory proposednumber theory
p-th root of rational numbers 2

Source: Iran 3rd round 2012-Algebra exam-P5

9/20/2012
Let pp be an odd prime number and let a1,a2,...,an∈Q+a_1,a_2,...,a_n \in \mathbb Q^+ be rational numbers. Prove that Q(a1p+a2p+...+anp)=Q(a1p,a2p,...,anp).\mathbb Q(\sqrt[p]{a_1}+\sqrt[p]{a_2}+...+\sqrt[p]{a_n})=\mathbb Q(\sqrt[p]{a_1},\sqrt[p]{a_2},...,\sqrt[p]{a_n}).
algebra proposedalgebra
4 three variable polynomials

Source: Iran 3rd rouund 2012-Final exam-P5

9/25/2012
We call the three variable polynomial PP cyclic if P(x,y,z)=P(y,z,x)P(x,y,z)=P(y,z,x). Prove that cyclic three variable polynomials P1,P2,P3P_1,P_2,P_3 and P4P_4 exist such that for each cyclic three variable polynomial PP, there exists a four variable polynomial QQ such that P(x,y,z)=Q(P1(x,y,z),P2(x,y,z),P3(x,y,z),P4(x,y,z))P(x,y,z)=Q(P_1(x,y,z),P_2(x,y,z),P_3(x,y,z),P_4(x,y,z)).
Solution by Mostafa Eynollahzade and Erfan Salavati
algebrapolynomialinvariantalgebra proposed