MathDB

2

Part of 2013 India IMO Training Camp

Problems(5)

Equidistant circumcenters

Source: Indian IMOTC 2013, Practice Test 1, Problem 2

5/6/2013
Let ABCDABCD by a cyclic quadrilateral with circumcenter OO. Let PP be the point of intersection of the diagonals ACAC and BDBD, and K,L,M,NK, L, M, N the circumcenters of triangles AOP,BOPAOP, BOP, COP,DOPCOP, DOP, respectively. Prove that KL=MNKL = MN.
geometrycircumcircletrigonometrycyclic quadrilateralperpendicular bisectorgeometry proposed
Equal inradii

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

5/10/2013
In a triangle ABCABC with B=90∘B = 90^\circ, DD is a point on the segment BCBC such that the inradii of triangles ABDABD and ADCADC are equal. If ADB^=φ\widehat{ADB} = \varphi then prove that tan⁡2(φ/2)=tan⁡(C/2)\tan^2 (\varphi/2) = \tan (C/2).
trigonometryLaTeXgeometrygeometric transformationreflectiontrig identitiesLaw of Sines
Prove that four points are concyclic

Source: Indian IMOTC 2013, Team Selection Test 3, Problem 2

7/30/2013
In a triangle ABCABC, let II denote its incenter. Points D,E,FD, E, F are chosen on the segments BC,CA,ABBC, CA, AB, respectively, such that BD+BF=ACBD + BF = AC and CD+CE=ABCD + CE = AB. The circumcircles of triangles AEF,BFD,CDEAEF, BFD, CDE intersect lines AI,BI,CIAI, BI, CI, respectively, at points K,L,MK, L, M (different from A,B,CA, B, C), respectively. Prove that K,L,M,IK, L, M, I are concyclic.
geometrycircumcircletrigonometryAsymptotegeometry proposed
Reflection of the orthocenter

Source: Indian IMOTC 2013, Team Selection Test 1, Problem 2

5/15/2013
In a triangle ABCABC, with A^>90∘\widehat{A} > 90^\circ, let OO and HH denote its circumcenter and orthocenter, respectively. Let KK be the reflection of HH with respect to AA. Prove that K,OK, O and CC are collinear if and only if A^−B^=90∘\widehat{A} - \widehat{B} = 90^\circ.
geometrygeometric transformationreflectioncircumcircletrigonometryparallelogramhomothety
Quotients of polynomials

Source: Indian IMOTC 2013, Team Selection Test 4, Problem 2

7/30/2013
Let n≥2n \ge 2 be an integer and f1(x),f2(x),…,fn(x)f_1(x), f_2(x), \ldots, f_{n}(x) a sequence of polynomials with integer coefficients. One is allowed to make moves M1,M2,…M_1, M_2, \ldots as follows: in the kk-th move MkM_k one chooses an element f(x)f(x) of the sequence with degree of ff at least 22 and replaces it with (f(x)−f(k))/(x−k)(f(x) - f(k))/(x-k). The process stops when all the elements of the sequence are of degree 11. If f1(x)=f2(x)=⋯=fn(x)=xn+1f_1(x) = f_2(x) = \cdots = f_n(x) = x^n + 1, determine whether or not it is possible to make appropriate moves such that the process stops with a sequence of nn identical polynomials of degree 1.
algebrapolynomialalgebra proposed