Roots of quadratic satisfy equation

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0 replies

jlacosta

Mar 2, 2025

0 replies

Geometry Problem #42

vankhea 2

N 9 minutes ago by kaede_Arcadia

Source: Van Khea

Let $P$ be any point. Let $D, E, F$ be projection point from $P$ to $BC, CA, AB$ . Circumcircle $(ABC)$ cuts circumcircle $(AEF), (BFD), (CDE)$ at $A_1, B_1, C_1$ . Let $A_2, B_2, C_2$ be antipode of $A_1, B_1, C_1$ wrt $(AEF), (BFD), (CDE)$ . Prove that $A_2, B_2, C_2, P$ are cyclic.

2 replies

vankhea

Sep 6, 2023

kaede_Arcadia

9 minutes ago

divisibility

srnjbr 3

N 11 minutes ago by srnjbr

Find all natural numbers n such that there exists a natural number l such that for every m members of the natural numbers the number m+m^2+...m^l is divisible by n.

3 replies

srnjbr

3 hours ago

srnjbr

11 minutes ago

Very easy inequality

pggp 5

N 21 minutes ago by ionbursuc

Source: Polish Junior MO Second Round 2019

Let $x$ , $y$ be real numbers, such that $x^2 + x \leq y$ . Prove that $y^2 + y \geq x$ .

5 replies

pggp

Oct 26, 2020

ionbursuc

21 minutes ago

Solve in gaussian integers

CHESSR1DER 0

30 minutes ago

Solve in gaussian integers.
$\sin\left(\ln\left(x^{x^{x^2}}\right)\right) = x^4$

0 replies

CHESSR1DER

30 minutes ago

0 replies

No more topics!

Roots of quadratic satisfy equation

Marinchoo 3

N Apr 4, 2022 by Moubinool

Source: 2007 Bulgarian Autumn Math Competition, Problem 10.1

Find all integers $b$ and $c$ for which the equation $x^2-bx+c=0$ has two real roots $x_{1}$ and $x_{2}$ satisfying $x_{1}^2+x_{2}^2=5$ .

3 replies

Marinchoo

Mar 17, 2022

Moubinool

Apr 4, 2022

Roots of quadratic satisfy equation

G H J

Reveal topic tags

quadratics

algebra

roots of the equation

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Source: 2007 Bulgarian Autumn Math Competition, Problem 10.1

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Marinchoo

407 posts

Marinchoo

#1 h Mar 17, 2022, 5:43 PM

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Find all integers $b$ and $c$ for which the equation $x^2-bx+c=0$ has two real roots $x_{1}$ and $x_{2}$ satisfying $x_{1}^2+x_{2}^2=5$ .

This post has been edited 1 time. Last edited by Marinchoo, Mar 17, 2022, 7:28 PM

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FIOTN

20 posts

FIOTN

#2 h Mar 17, 2022, 5:53 PM

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$x_1+x_2=b$ and $x_1x_2=c\Longrightarrow{}5=x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=b^2-2c$ . Obviously, $b$ is odd $\Longrightarrow{}b=2k-1\Longrightarrow{}5+2c=b^2=(2k-1)^2=4k^2-4k+1\Longrightarrow{}c=2k^2-2k-2$ . So answer is $(b,c)=(2k-1,2k^2-2k-2)$ $\forall{k}\in{N}$