MathDB

Problem 4

Part of 2001 Moldova National Olympiad

Problems(6)

permutation of [9] subject to constraint (2001 Moldova MO Grade 7 P4)

Source:

4/12/2021
Find all permutations of the numbers 1,2,…,91,2,\ldots,9 in which no two adjacent numbers have a sum divisible by 77 or 1313.
combinatorics
Im((a+b)(b+c)(c+a)/abc)

Source: 2001 Moldova MO Grade 8 P4

4/12/2021
Find all integers that can be written as (a+b)(b+c)(c+a)abc\frac{(a+b)(b+c)(c+a)}{abc}, where a,b,ca,b,c are pairwise coprime positive integers.
number theory
prove concurrency of lines (altitude and lines such that angles are equal)

Source: 2001 Moldova MO Grade 9 P4

4/12/2021
In a triangle ABCABC the altitude ADAD is drawn. Points MM on side ACAC and NN on side ABAB are taken so that ∠MDA=∠NDA\angle MDA=\angle NDA. Prove that the lines AD,BMAD,BM and CNCN are concurrent.
geometry
prove that quadrilateral is cyclic and circle is tangent to lines

Source: 2001 Moldova MO Grade 10 P4

4/13/2021
In a triangle ABCABC, the angle bisector at AA intersects BCBC at DD. The tangents at DD to the circumcircles of the triangles ABDABD and ACDACD meet ACAC and ABAB at NN and MM, respectively. Prove that the quadrilateral AMDNAMDN is inscribed in a circle tangent to BCBC.
geometry
bisector equal to altitude

Source: 2001 Moldova MO Grade 12 P4

4/13/2021
In a triangle ABCABC, BC=aBC=a, AC=bAC=b, ∠B=β\angle B=\beta and ∠C=γ\angle C=\gamma. Prove that the bisector of the angle at AA is equal to the altitude from BB if and only if b=acos⁡β−γ2b=a\cos\frac{\beta-\gamma}2.
geometryTriangles
inequality for polynomial in Z[x]

Source: 2001 Moldova MO Grade 11 P4

4/13/2021
Let P(x)=xn+a1xn−1+…+anP(x)=x^n+a_1x^{n-1}+\ldots+a_n (n≥2n\ge2) be a polynomial with integer coefficients having nn real roots b1,…,bnb_1,\ldots,b_n. Prove that for x0≥max⁡{b1,…,bn}x_0\ge\max\{b_1,\ldots,b_n\}, P(x0+1)(1x0−b1+…+1x0−bn)≥2n2.P(x_0+1)\left(\frac1{x_0-b_1}+\ldots+\frac1{x_0-b_n}\right)\ge2n^2.
inequalitiesalgebrapolynomial