Bulgaria team selection test 2015

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0 replies

jlacosta

Yesterday at 3:18 PM

0 replies

Inequality from China

sqing 1

N 24 minutes ago by SunnyEvan

Source: lemondian(https://kuing.cjhb.site/thread-13667-1-1.html)

Let $x\in (0,\frac{\pi}{2}) .$ Prove that $tanx\ge x^k$ Where $k=1,2,3,4.$

1 reply

+1 w

sqing

3 hours ago

SunnyEvan

24 minutes ago

Burak0609

Burak0609 0

35 minutes ago

So $b=x+y+z$ $x^3+a=-3y-3z$ $P(t)=t^3-3t+a-3b\implies x+y+z=0$ from vieta $x^3+a=-3y-3z,y^3+a=-3x-3z\implies x^2+xy+y^2=3\implies y^2+xy+(x^2-3)=0\implies \Delta\ge 0 \implies 12\ge 3x^2 \implies x\in [-2,2]$ . Notice in similar fashion we get $y\in [-2,2] , z\in [-2,2]$ , also its easy to observe that none of $\{x,y,z\} \in \{-2,2,0\}$ . So we have $\boxed{a \in (-2,2)-\{0\}}$

0 replies

Burak0609

35 minutes ago

0 replies

Problem 1 IMO 2005 (Day 1)

Valentin Vornicu 92

N an hour ago by Baimukh

Six points are chosen on the sides of an equilateral triangle $ABC$ : $A_1$ , $A_2$ on $BC$ , $B_1$ , $B_2$ on $CA$ and $C_1$ , $C_2$ on $AB$ , such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.

Prove that the lines $A_1B_2$ , $B_1C_2$ and $C_1A_2$ are concurrent.

Bogdan Enescu, Romania

92 replies

Valentin Vornicu

Jul 13, 2005

Baimukh

an hour ago

Inspired by old results

sqing 2

N an hour ago by sqing

Source: Own

Let $a,b,c \ge \frac{1}{21}$ and $a+b+c=1.$ . Prove that
$(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)\geq\frac{1}{729}$ Let $a,b,c \ge \frac{1}{10}$ and $a+b+c=2.$ . Prove that
$(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)\geq\frac{64}{729}$ Let $a,b,c \ge \frac{1}{11}$ and $a+b+c=2.$ . Prove that
$(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)\geq \frac{145161}{1771561}$

2 replies

sqing

2 hours ago

sqing

an hour ago

No more topics!

Bulgaria team selection test 2015

luutrongphuc 2

N Apr 1, 2025 by internationalnick123456

Find all real-coefficient polynomials of the form
$f(x) = x^{2n} + a_1 x^{2n-1} + \dots + a_{n-1} x^{n+1} + a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + 1$ that have $2n$ real roots and satisfy $|a_i| \leq 2$ for all $i = 1, 2, \dots, n$ .

2 replies

luutrongphuc

Mar 30, 2025

internationalnick123456

Apr 1, 2025

Bulgaria team selection test 2015

G H J

Reveal topic tags

polynomial

G H BBookmark kLocked kLocked NReply

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luutrongphuc

27 posts

luutrongphuc

#1 h Mar 30, 2025, 4:45 PM

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luutrongphuc

27 posts

luutrongphuc

#2 h Apr 1, 2025, 1:07 PM

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hmm any idea?

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internationalnick123456

113 posts

internationalnick123456

#3 h Apr 1, 2025, 3:17 PM • 2 Y

Y by luutrongphuc, kiyoras_2001

Let $2n$ real roots of $f$ be $z_1,z_2,\ldots,z_{2n}$ . Since $z_1\ldots z_{2n} = 1$ , we have $f(i)f(-i)=\left(1+z_1^2\right)\ldots\left(1+z_{2n}^2\right) \geq 2^{2n}$ Hence, $|f(i)f(-i)|\geq 2^{2n}$ . Moreover, $|f(i)| = |f(-i)|$ . By constructing the upperbound of $|f(i)|=|f(-i)|$ , we derive $n\in\left\{1,2\right\}$ and if $n=2$ , the inequalities must become equalities, i.e, $|z_1|=\cdots = |z_{2n}| = 1$ .
Finally, we find out that only the polynomials: $x^2\pm 2x + 1, x^4 \pm 2x^2 + 1$ work.

This post has been edited 3 times. Last edited by internationalnick123456, Apr 1, 2025, 3:19 PM

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