MathDB

5

Part of 2017 China Team Selection Test

Problems(5)

China 2017 TSTST1 Day 2 Geometry Problem

Source: China 2017 TSTST1 Day 2 Problem 5

3/7/2017
In the non-isosceles triangle ABCABC,DD is the midpoint of side BCBC,EE is the midpoint of side CACA,FF is the midpoint of side ABAB.The line(different from line BCBC) that is tangent to the inscribed circle of triangle ABCABC and passing through point DD intersect line EFEF at XX.Define Y,ZY,Z similarly.Prove that X,Y,ZX,Y,Z are collinear.
geometry
China Team Selection Test 2017 TST 2 Day 2 Q5

Source: China Shanghai ,Mar 12, 2017

3/12/2017
Let φ(x) \varphi(x) be a cubic polynomial with integer coefficients. Given that φ(x) \varphi(x) has have 3 distinct real roots u,v,wu,v,w and u,v,wu,v,w are not rational number. there are integers a,b,c a, b,c such that u=av2+bv+cu=av^2+bv+c. Prove that b2−2b−4ac−7b^2 -2b -4ac - 7 is a square number .
number theoryalgebrapolynomialChina TST
Pairwise gcd is equal to gcd of all elements

Source: China TSTST 3 Day 2 Problem 2

3/18/2017
Show that there exists a positive real CC such that for any naturals H,NH,N satisfying H≥3,N≥eCHH \geq 3, N \geq e^{CH}, for any subset of {1,2,…,N}\{1,2,\ldots,N\} with size ⌈CHNln⁡N⌉\lceil \frac{CHN}{\ln N} \rceil, one can find HH naturals in it such that the greatest common divisor of any two elements is the greatest common divisor of all HH elements.
greatest common divisornumber theory
An inequality problem from China TST

Source: 2017 China TST 4 Problem 5

3/22/2017
Given integer m≥2m\geq2,x1,...,xmx_1,...,x_m are non-negative real numbers,prove that:(m−1)m−1(x1m+...+xmm)≥(x1+...+xm)m−mmx1...xm(m-1)^{m-1}(x_1^m+...+x_m^m)\geq(x_1+...+x_m)^m-m^mx_1...x_mand please find out when the equality holds.
inequalitiesTST
sum of square like a discriminent

Source: 2017 China TST 5 P5

4/14/2017
A(x,y), B(x,y), and C(x,y) are three homogeneous real-coefficient polynomials of x and y with degree 2, 3, and 4 respectively. we know that there is a real-coefficient polinimial R(x,y) such that B(x,y)2−4A(x,y)C(x,y)=−R(x,y)2B(x,y)^2-4A(x,y)C(x,y)=-R(x,y)^2. Proof that there exist 2 polynomials F(x,y,z) and G(x,y,z) such that F(x,y,z)2+G(x,y,z)2=A(x,y)z2+B(x,y)z+C(x,y)F(x,y,z)^2+G(x,y,z)^2=A(x,y)z^2+B(x,y)z+C(x,y) if for any x, y, z real numbers A(x,y)z2+B(x,y)z+C(x,y)≥0A(x,y)z^2+B(x,y)z+C(x,y)\ge 0
algebrapolynomialChina TST