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Nice problem about the Lemoine point of triangle JaBC and OI line
Ktoan07   0
an hour ago
Source: Own
Let \(\triangle ABC\) be an acute-angled, non-isosceles triangle with circumcenter \(O\) and incenter \(I\), such that

\[
\prod_{\text{cyc}} \left( \frac{1}{a+b-c} + \frac{1}{a+c-b} - \frac{2}{b+c-a} \right) \neq 0,
\]
where \(a = BC\), \(b = CA\), and \(c = AB\).

Let \(J_a\), \(J_b\), and \(J_c\) be the excenters opposite to vertices \(A\), \(B\), and \(C\), respectively, and let \(L_a\), \(L_b\), and \(L_c\) be the Lemoine points of triangles \(J_aBC\), \(J_bCA\), and \(J_cAB\), respectively.

Prove that the circles \((L_aBC)\), \((L_bCA)\), and \((L_cAB)\) all pass through a common point \(P\). Moreover, the isogonal conjugate of \(P\) with respect to \(\triangle ABC\) lies on the line \(OI\).

Note (Hint)
0 replies
Ktoan07
an hour ago
0 replies
Rectangular line segments in russia
egxa   0
an hour ago
Source: All Russian 2025 9.1
Several line segments parallel to the sides of a rectangular sheet of paper were drawn on it. These segments divided the sheet into several rectangles, inside of which there are no drawn lines. Petya wants to draw one diagonal in each of the rectangles, dividing it into two triangles, and color each triangle either black or white. Is it always possible to do this in such a way that no two triangles of the same color share a segment of their boundary?
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egxa
an hour ago
0 replies
external tangents of circumcircles
egxa   0
an hour ago
Source: All Russian 2025 9.2
The diagonals of a convex quadrilateral \(ABCD\) intersect at point \(E\). The points of tangency of the circumcircles of triangles \(ABE\) and \(CDE\) with their common external tangents lie on a circle \(\omega\). The points of tangency of the circumcircles of triangles \(ADE\) and \(BCE\) with their common external tangents lie on a circle \(\gamma\). Prove that the centers of circles \(\omega\) and \(\gamma\) coincide.
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egxa
an hour ago
0 replies
Writing letters in grid
egxa   0
an hour ago
Source: All russian 2025 10.1
Petya and Vasya are playing a game on an initially empty \(100 \times 100\) grid, taking turns. Petya goes first. On his turn, a player writes an uppercase Russian letter in an empty cell (each cell can contain only one letter). When all cells are filled, Petya is declared the winner if there are four consecutive cells horizontally spelling the word ``ПЕТЯ'' (PETYA) from left to right, or four consecutive cells vertically spelling ``ПЕТЯ'' from top to bottom. Can Petya guarantee a win regardless of Vasya's moves?
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egxa
an hour ago
0 replies
Existence of a circle tangent to four lines
egxa   0
an hour ago
Source: All Russian 2025 10.2
Inside triangle \(ABC\), point \(P\) is marked. Point \(Q\) is on segment \(AB\), and point \(R\) is on segment \(AC\) such that the circumcircles of triangles \(BPQ\) and \(CPR\) are tangent to line \(AP\). Lines are drawn through points \(B\) and \(C\) passing through the center of the circumcircle of triangle \(BPC\), and through points \(Q\) and \(R\) passing through the center of the circumcircle of triangle \(PQR\). Prove that there exists a circle tangent to all four drawn lines.
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egxa
an hour ago
0 replies
Existence of perfect squares
egxa   0
2 hours ago
Source: All Russian 2025 10.3
Find all natural numbers \(n\) for which there exists an even natural number \(a\) such that the number
\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]is a perfect square.
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egxa
2 hours ago
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Weighted graph problem
egxa   0
2 hours ago
Source: All Russian 2025 10.4
In the plane, $106$ points are marked, no three of which are collinear. All possible segments between them are drawn. Grisha assigned to each drawn segment a real number with absolute value no greater than $1$. For every group of $6$ marked points, he calculated the sum of the numbers on all $15$ connecting segments. It turned out that the absolute value of each such sum is at least \(C\), and there are both positive and negative such sums. What is the maximum possible value of \(C\)?
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egxa
2 hours ago
0 replies
Board problem with complex numbers
egxa   0
2 hours ago
Source: All Russian 2025 11.1
$777$ pairwise distinct complex numbers are written on a board. It turns out that there are exactly 760 ways to choose two numbers \(a\) and \(b\) from the board such that:
\[
a^2 + b^2 + 1 = 2ab
\]Ways that differ by the order of selection are considered the same. Prove that there exist two numbers \(c\) and \(d\) from the board such that:
\[
c^2 + d^2 + 2025 = 2cd
\]
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egxa
2 hours ago
0 replies
two 3D problems in one day
egxa   0
2 hours ago
Source: All Russian 2025 11.2
A right prism \(ABCA_1B_1C_1\) is given. It is known that triangles \(A_1BC\), \(AB_1C\), \(ABC_1\), and \(ABC\) are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle \(ABC\), lie on the same sphere.
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egxa
2 hours ago
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Important pairs of polynomials
egxa   0
2 hours ago
Source: All Russian 2025 11.3
A pair of polynomials \(F(x, y)\) and \(G(x, y)\) with integer coefficients is called $\emph{important}$ if from the divisibility of both differences \(F(a, b) - F(c, d)\) and \(G(a, b) - G(c, d)\) by $100$, it follows that both \(a - c\) and \(b - d\) are divisible by 100. Does there exist such an important pair of polynomials \(P(x, y)\), \(Q(x, y)\), such that the pair \(P(x, y) - xy\) and \(Q(x, y) + xy\) is also important?
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egxa
2 hours ago
0 replies
Interesting Factoring Result
paganiniana   10
N Mar 1, 2025 by theclockhasstruck
Factor $b^5+b+1$.

Hint
10 replies
paganiniana
Jul 5, 2024
theclockhasstruck
Mar 1, 2025
Interesting Factoring Result
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G H BBookmark kLocked kLocked NReply
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paganiniana
211 posts
#1 • 1 Y
Y by MC_ADe
Factor $b^5+b+1$.

Hint
This post has been edited 1 time. Last edited by paganiniana, Jul 5, 2024, 5:04 PM
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fruitmonster97
2476 posts
#2
Y by
Solution
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anduran
476 posts
#3 • 1 Y
Y by mathMagicOPS
Alternatively, we can completely ditch the hint and let $x=e^\frac{2\pi i}{3}$ and get that the expression simplifies to
$$\omega^2+\omega+1=0,$$Hence by the factor theorem $x^2+x+1$ is such a factor. Then the other factor is $x^3-x^2+1$ by polynomial division.
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anduran
476 posts
#4
Y by
If you want to follow the hint, note that
$$x^4+x^2+1=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1)$$Then we do
$$x^5+x+1-(x^4+x^2+1)=x^5-x^4-x^2+x$$$$=(x^4-x)(x-1)$$$$=x(x-1)(x-1)(x^2+x+1)$$Then
$$(x^5+x+1)-(x^2+x+1)(x^2-x+1)=(x^3-2x^2+x)(x^2+x+1)$$Giving
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$
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hypertension
101 posts
#5 • 1 Y
Y by anduran
Amazing solution congratulations anduran I wish you eternal happiness bro!
Hugs
Best Wishes
George
Greece
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lord_of_the_rook
137 posts
#6
Y by
Note that
\[ x^5 + x + 1 = (x^5 + x^4 + x^3 + x^2 + x + 1) - (x^4 + x^3  + x^2) \]Using geometric series,
\[ x^5 + x + 1 = \frac{x^6 - 1}{x-1} - x^2 \cdot \frac{x^3 - 1}{x-1} \]Factoring out an $\frac{x^3 - 1}{x-1} = x^2 + x + 1$,
\[ x^5 + x + 1 = (x^2 + x + 1) \left ( x^3 + 1 - x^2 \right ) = (x^2 + x + 1)(x^3 -x^2 + 1) \]
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sadas123
1229 posts
#7
Y by
anduran wrote:
Alternatively, we can completely ditch the hint and let $x=e^\frac{2\pi i}{3}$ and get that the expression simplifies to
$$\omega^2+\omega+1=0,$$Hence by the factor theorem $x^2+x+1$ is such a factor. Then the other factor is $x^3-x^2+1$ by polynomial division.

I got the same thing except I am to lazy to write it in latex...
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maxamc
551 posts
#8
Y by
Could be simplified to $x^5+x+1=(x^2+x+1)(x^3-x^2+1)=(x^2+x+1)(x+\frac{1}{3}(-1+a+\frac{1}{a}))(x+\frac{1}{3}(-1+\omega a+\omega^2 \frac{1}{a}))(x+\frac{1}{3}(-1+\omega^2 a+\omega \frac{1}{a}))\text{ where }\omega=\frac{-1+\sqrt{3}i}{2}\text{ and }a=(\frac{1}{2}(25-3\sqrt{69}))^{\frac{1}{3}}.$ This is the final answer.
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BS2012
990 posts
#9
Y by
maxamc wrote:
Could be simplified to $x^5+x+1=(x^2+x+1)(x^3-x^2+1)=(x^2+x+1)(x+\frac{1}{3}(-1+a+\frac{1}{a}))(x+\frac{1}{3}(-1+\omega a+\omega^2 \frac{1}{a}))(x+\frac{1}{3}(-1+\omega^2 a+\omega \frac{1}{a}))\text{ where }\omega=\frac{-1+\sqrt{3}i}{2}\text{ and }a=(\frac{1}{2}(25-3\sqrt{69}))^{\frac{1}{3}}.$ This is the final answer.

except you forgot to factor $x^2+x+1=(x-w)(x-w^2).$
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maxamc
551 posts
#10
Y by
BS2012 wrote:
maxamc wrote:
Could be simplified to $x^5+x+1=(x^2+x+1)(x^3-x^2+1)=(x^2+x+1)(x+\frac{1}{3}(-1+a+\frac{1}{a}))(x+\frac{1}{3}(-1+\omega a+\omega^2 \frac{1}{a}))(x+\frac{1}{3}(-1+\omega^2 a+\omega \frac{1}{a}))\text{ where }\omega=\frac{-1+\sqrt{3}i}{2}\text{ and }a=(\frac{1}{2}(25-3\sqrt{69}))^{\frac{1}{3}}.$ This is the final answer.

except you forgot to factor $x^2+x+1=(x-w)(x-w^2).$

:oops_sign:
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theclockhasstruck
4 posts
#11
Y by
The #6 it is a natural solution!
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