MathDB

Phika sextuplets satisfy a strange inequality

Source: 2022 3rd OMpD L2 P4 / L3 P2 - Brazil - Olimpíada Matemáticos por Diversão

7/8/2023

We say that a sextuple of positive real numbers

(a_1, a_2, a_3, b_1, b_2, b_3)

is

<span class='latex-italic'>phika</span>

if

a_1 + a_2 + a_3 = b_1 + b_2 + b_3 = 1

.
(a) Prove that there exists a

<span class='latex-italic'>phika</span>

sextuple

(a_1, a_2, a_3, b_1, b_2, b_3)

such that:

a_1(\sqrt{b_1} + a_2) + a_2(\sqrt{b_2} + a_3) + a_3(\sqrt{b_3} + a_1) > 1 - \frac{1}{2022^{2022}}

(b) Prove that for every

<span class='latex-italic'>phika</span>

sextuple

(a_1, a_2, a_3, b_1, b_2, b_3)

, we have:

a_1(\sqrt{b_1} + a_2) + a_2(\sqrt{b_2} + a_3) + a_3(\sqrt{b_3} + a_1) < 1

algebrainequalitiespositive real

<FCE=? <FEC=< CEB$ , < DFC=< CFE in rectangle ABCD

Source: 2022 2nd OMpD L2 P2 - Brazil - Olimpíada Matemáticos por Diversão

10/13/2022

Let

ABCD

be a rectangle. The point

E

lies on side

\overline{AB}

and the point

F

is lies side

\overline{AD}

, such that

\angle FEC=\angle CEB

and

\angle DFC=\angle CFE

. Determine the measure of the angle

\angle FCE

and the ratio

AD/AB

.

geometryrectangleequal anglesangles

Prove that A^p \neq I_p

Source: 2022 3rd OMpD LU P2 - Brazil - Olimpíada Matemáticos por Diversão

7/8/2023

Let

p \geq 3

be a prime number and let

A

be a matrix of order

p

with complex entries. Assume that

\text{Tr}(A) = 0

and

\det(A - I_p) \neq 0

. Prove that

A^p \neq I_p

.
Note:

\text{Tr}(A)

is the sum of the main diagonal elements of

A

and

I_p

is the identity matrix of order

p

.

linear algebraalgebramatrixMatrix algebratrace

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