Equations

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0 replies

jlacosta

May 1, 2025

0 replies

Prove that IMO is isosceles

YLG_123 4

N 29 minutes ago by Blackbeam999

Source: 2024 Brazil Ibero TST P2

Let $ABC$ be an acute-angled scalene triangle with circumcenter $O$ . Denote by $M$ , $N$ , and $P$ the midpoints of sides $BC$ , $CA$ , and $AB$ , respectively. Let $\omega$ be the circle passing through $A$ and tangent to $OM$ at $O$ . The circle $\omega$ intersects $AB$ and $AC$ at points $E$ and $F$ , respectively (where $E$ and $F$ are distinct from $A$ ). Let $I$ be the midpoint of segment $EF$ , and let $K$ be the intersection of lines $EF$ and $NP$ . Prove that $AO = 2IK$ and that triangle $IMO$ is isosceles.

4 replies

YLG_123

Oct 12, 2024

Blackbeam999

29 minutes ago

Geometric mean of squares a knight's move away

Pompombojam 0

33 minutes ago

Source: Problem Solving Tactics Methods of Proof Q27

Each square of an $8 \times 8$ chessboard has a real number written in it in such a way that each number is equal to the geometric mean of all the numbers a knight's move away from it.

Is it true that all of the numbers must be equal?

0 replies

Pompombojam

33 minutes ago

0 replies

Circumcircle of ADM

v_Enhance 71

N 34 minutes ago by judokid

Source: USA TSTST 2012, Problem 7

Triangle $ABC$ is inscribed in circle $\Omega$ . The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$ ), respectively. Let $M$ be the midpoint of side $BC$ . The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$ ), respectively. Let $N$ be the midpoint of segment $PQ$ , and let $H$ be the foot of the perpendicular from $L$ to line $ND$ . Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$ .

71 replies

v_Enhance

Jul 19, 2012

judokid

34 minutes ago

Three operations make any number

awesomeming327. 2

N an hour ago by happymoose666

Source: own

The number $3$ is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.

2 replies

awesomeming327.

4 hours ago

happymoose666

an hour ago

No more topics!

Equations

Jackson0423 2

N Apr 22, 2025 by rchokler

Solve the system of equations
$\begin{cases} x - y z = 1,\\[2pt] y - z x = 2,\\[2pt] z - x y = 4. \end{cases}$

2 replies

Jackson0423

Apr 22, 2025

rchokler

Apr 22, 2025

Equations

G H J

Reveal topic tags

algebra

system of equations

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Jackson0423

104 posts

Jackson0423

#1 h Apr 22, 2025, 4:36 PM

Y by

Solve the system of equations
$\begin{cases} x - y z = 1,\\[2pt] y - z x = 2,\\[2pt] z - x y = 4. \end{cases}$

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Maxklark

6 posts

Maxklark

#2 h Apr 22, 2025, 6:14 PM

Y by

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On what set should the system of equations be solved?

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rchokler

2975 posts

rchokler

#3 h Apr 22, 2025, 8:26 PM

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Plugging in $x=yz+1$ to the second and third equation gives $y-yz^2-z=2$ and $z-y^2z-y=4$ .

Rearrange the former of these equations to $y=\frac{z+2}{1-z^2}$ and plug it in the latter to get $z-\left(\frac{z+2}{1-z^2}\right)^2z-\frac{z+2}{1-z^2}=4$ .

Now simplify:

$z-\left(\frac{z+2}{1-z^2}\right)^2z-\frac{z+2}{1-z^2}=4\implies z(1-z^2)^2-(z+2)^2z+(z+2)(z^2-1)-4(1-z^2)^2=0\implies z^5-4z^4-2z^3+6z^2-4z-6=0$

Starting over and going through similar steps gives $y^5-2y^4-2y^3-16y-6=0$ and $x^5-x^4-2x^3-6x^2-19x-9=0$ .

Each of these quintics is irreducible with three real roots and as a consequence have Galois Group $S_5$ . So they can't be solved closed form by radicals.

The approximate solutions.

Real:
$(-1.50383,3.18296,-0.786634)$
$(-0.563382,-0.371416,4.20925)$
$(3.06006,-1.70257,-1.20997)$

Complex:
$(0.0036+1.8632i,0.4455+1.6681i,0.89368+0.83604i)$
$(0.0036-1.8632i,0.4455-1.6681i,0.89368-0.83604i)$

Z K Y

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