MathDB

3

Part of 2013 India IMO Training Camp

Problems(5)

Translations of a cubic polynomial with integer roots

Source: Indian IMOTC 2013, Team Selection Test 1, Problem 3

5/15/2013
For a positive integer nn, a cubic polynomial p(x)p(x) is said to be nn-good if there exist nn distinct integers a1,a2,…,ana_1, a_2, \ldots, a_n such that all the roots of the polynomial p(x)+ai=0p(x) + a_i = 0 are integers for 1≤i≤n1 \le i \le n. Given a positive integer nn prove that there exists an nn-good cubic polynomial.
geometrygeometric transformationalgebrapolynomialalgebra proposed
Bijection of natural numbers

Source: Indian IMOTC 2013, Practice Test 1, Problem 3

5/6/2013
We define an operation ⊕\oplus on the set {0,1}\{0, 1\} by 0⊕0=0 ,0⊕1=1 ,1⊕0=1 ,1⊕1=0 . 0 \oplus 0 = 0 \,, 0 \oplus 1 = 1 \,, 1 \oplus 0 = 1 \,, 1 \oplus 1 = 0 \,. For two natural numbers aa and bb, which are written in base 22 as a=(a1a2…ak)2a = (a_1a_2 \ldots a_k)_2 and b=(b1b2…bk)2b = (b_1b_2 \ldots b_k)_2 (possibly with leading 0's), we define a⊕b=ca \oplus b = c where cc written in base 22 is (c1c2…ck)2(c_1c_2 \ldots c_k)_2 with ci=ai⊕bic_i = a_i \oplus b_i, for 1≤i≤k1 \le i \le k. For example, we have 7⊕3=47 \oplus 3 = 4 since 7=(111)2 7 = (111)_2 and 3=(011)23 = (011)_2.
For a natural number nn, let f(n)=n⊕[n/2]f(n) = n \oplus \left[ n/2 \right], where [x]\left[ x \right] denotes the largest integer less than or equal to xx. Prove that ff is a bijection on the set of natural numbers.
floor functioninductionvectorGaussnumber theoryrelatively primenumber theory proposed
Back to the origin

Source: Indian IMOTC 2013, Practice Test 2, Problem 3

5/10/2013
A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers a,ba, b, neither of which was chosen earlier by any player and move the marker by aa units in the horizontal direction and bb units in the vertical direction. Hobbes wins if the marker is back at the origin any time after the first turn. Prove or disprove that Calvin can prevent Hobbes from winning.
Note: A move in the horizontal direction by a positive quantity will be towards the right, and by a negative quantity will be towards the left (and similar directions in the vertical case as well).
vectoranalytic geometrycombinatorics proposedcombinatorics
Sums and products

Source: Indian IMOTC 2013, Team Selection Test 3, Problem 3

7/30/2013
Let h≥3h \ge 3 be an integer and XX the set of all positive integers that are greater than or equal to 2h2h. Let SS be a nonempty subset of XX such that the following two conditions hold:
[*]if a+b∈Sa + b \in S with a≥h,b≥ha \ge h, b \ge h, then ab∈Sab \in S;
[*]if ab∈Sab \in S with a≥h,b≥ha \ge h, b \ge h, then a+b∈Sa + b \in S. Prove that S=XS = X.
inductionalgebra proposedalgebra
A circle passing through a vertex and touching an altitude

Source: Indian IMOTC 2013, Team Selection Test 4, Problem 3

7/30/2013
In a triangle ABCABC, with AB≠BCAB \ne BC, EE is a point on the line ACAC such that BEBE is perpendicular to ACAC. A circle passing through AA and touching the line BEBE at a point P≠BP \ne B intersects the line ABAB for the second time at XX. Let QQ be a point on the line PBPB different from PP such that BQ=BPBQ = BP. Let YY be the point of intersection of the lines CPCP and AQAQ. Prove that the points C,X,Y,AC, X, Y, A are concyclic if and only if CXCX is perpendicular to ABAB.
geometrycircumcircleangle bisectorperpendicular bisectorgeometry proposed