MathDB

2021 239 Open Mathematical Olympiad

Part of 239 Open Math Olympiad

Subcontests

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integer root of x^n+a_1x^{n -1}+..+a_n=0 where a_i=product of first i primes

You are given nn different primes p1,p2,...,pnp_1, p_2,..., p_n. Consider the polynomial xn+a1xn−1+a2xn−2+...+an−1x+anx^n + a_1x^{n -1} + a_2x^{n - 2} + ...+ a_{n - 1}x + a_n, where aia_i is the product of the first ii given prime numbers. For what nn can it have an integer root?

angle wanted, arc midpoints of circumcircle, angle bisectors

Points XX and YY are the midpoints of arcs ABAB and BCBC of the circumscribed circle of triangle ABCABC. Point TT lies on side ACAC. It turned out that the bisectors of the angles ATBATB and BTCBTC pass through points XX and YY respectively. What angle BB can be in triangle ABCABC?
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concurrency lines _|_ plane ABC at Lemoine point, by equilateral triangles

Symedians of an acute-angled non-isosceles triangle ABCABC intersect at a point at point LL, and AA1AA_1, BB1BB_1 and CC1CC_1 are its altitudes. Prove that you can construct equilateral triangles A1B1C′A_1B_1C', B1C1A′B_1C_1A' and C1A1B′C_1A_1B' not lying in the plane ABCABC, so that lines AA′,BB′AA' , BB' and CC′CC' and also perpendicular to the plane ABCABC at point LL intersected at one point.

238^2 (ab)^{239} <1 if a^{239} = ac- 1 , b^{239} = bc- 1 for different a,b,c>0

Different positive a,b,ca, b, c are such that a239=ac−1a^{239} = ac- 1 and b239=bc−1b^{239} = bc- 1.Prove that 2382(ab)239<1238^2 (ab)^{239} <1.
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Hard inequality involving complex numbers from Saint Petersburg

Let a,b,ca,b,c be some complex numbers. Prove that ∣a2ab+ac−bc∣+∣b2ba+bc−ac∣+∣c2ca+cb−ab∣≥32|\dfrac{a^2}{ab+ac-bc}| + |\dfrac{b^2}{ba+bc-ac}| + |\dfrac{c^2}{ca+cb-ab}| \ge \dfrac{3}{2} if the denominators are not 0

exists triangle of sides BD, DF,AF with area &gt;1/2 ABC if&lt;ABC =&lt;AED, AF // BC

The median ADAD is drawn in triangle ABCABC. Point EE is selected on segment ACAC, and on the ray DEDE there is a point FF, and ∠ABC=∠AED\angle ABC = \angle AED and AF//BCAF // BC. Prove that from segments BD,DFBD, DF and AFAF, you can make a triangle, the area of ​​which is not less half the area of ​​triangle ABCABC.