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Circumcircle excircle chaos
CyclicISLscelesTrapezoid   25
N 2 hours ago by bin_sherlo
Source: ISL 2021 G8
Let $ABC$ be a triangle with circumcircle $\omega$ and let $\Omega_A$ be the $A$-excircle. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_A$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_A$ at $X$ and $Y$ respectively. The tangent line at $P$ to the circumcircle of the triangle $APX$ intersects the tangent line at $Q$ to the circumcircle of the triangle $AQY$ at a point $R$. Prove that $\overline{AR} \perp \overline{BC}$.
25 replies
CyclicISLscelesTrapezoid
Jul 12, 2022
bin_sherlo
2 hours ago
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hard problem
Cobedangiu   7
N 3 hours ago by arqady
Let $x,y,z>0$ and $xy+yz+zx=3$ : Prove that :
$\sum \ \frac{x}{y+z}\ge\sum \frac{1}{\sqrt{x+3}}$
7 replies
Cobedangiu
Apr 2, 2025
arqady
3 hours ago
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Combo problem
soryn   2
N 3 hours ago by Anulick
The school A has m1 boys and m2 girls, and ,the school B has n1 boys and n2 girls. Each school is represented by one team formed by p students,boys and girls. If f(k) is the number of cases for which,the twice schools has,togheter k girls, fund f(k) and the valute of k, for which f(k) is maximum.
2 replies
soryn
Today at 6:33 AM
Anulick
3 hours ago
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Calculate the distance of chess king!!
egxa   4
N 3 hours ago by Primeniyazidayi
Source: All Russian 2025 9.4
A chess king was placed on a square of an \(8 \times 8\) board and made $64$ moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called $\emph{pleasant}$ if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
4 replies
egxa
Apr 18, 2025
Primeniyazidayi
3 hours ago
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As some nations like to say "Heavy theorems mostly do not help"
Assassino9931   9
N 3 hours ago by EVKV
Source: European Mathematical Cup 2022, Senior Division, Problem 2
We say that a positive integer $n$ is lovely if there exist a positive integer $k$ and (not necessarily distinct) positive integers $d_1$, $d_2$, $\ldots$, $d_k$ such that $n = d_1d_2\cdots d_k$ and $d_i^2 \mid n + d_i$ for $i=1,2,\ldots,k$.

a) Are there infinitely many lovely numbers?

b) Is there a lovely number, greater than $1$, which is a perfect square of an integer?
9 replies
Assassino9931
Dec 20, 2022
EVKV
3 hours ago
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congruence
moldovan   5
N 4 hours ago by EVKV
Source: Canada 2004
Let $p$ be an odd prime. Prove that:
\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]
5 replies
moldovan
Jun 26, 2009
EVKV
4 hours ago
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Checking a summand property for integers sufficiently large.
DinDean   1
N 4 hours ago by Double07
For any fixed integer $m\geqslant 2$, prove that there exists a positive integer $f(m)$, such that for any integer $n\geqslant f(m)$, $n$ can be expressed by a sum of positive integers $a_i$'s as
\[n=a_1+a_2+\dots+a_m,\]where $a_1\mid a_2$, $a_2\mid a_3$, $\dots$, $a_{m-1}\mid a_m$.
1 reply
DinDean
5 hours ago
Double07
4 hours ago
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real+ FE
pomodor_ap   4
N 4 hours ago by jasperE3
Source: Own, PDC001-P7
Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$$for all $x, y \in \mathbb{R}^+$. Determine all such functions $f$.
4 replies
pomodor_ap
Yesterday at 11:24 AM
jasperE3
4 hours ago
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FE solution too simple?
Yiyj1   8
N 4 hours ago by lksb
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
8 replies
Yiyj1
Apr 9, 2025
lksb
4 hours ago
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Polynomials in Z[x]
BartSimpsons   16
N 5 hours ago by bin_sherlo
Source: European Mathematical Cup 2017 Problem 4
Find all polynomials $P$ with integer coefficients such that $P (0)\ne 0$ and $$P^n(m)\cdot P^m(n)$$is a square of an integer for all nonnegative integers $n, m$.

Remark: For a nonnegative integer $k$ and an integer $n$, $P^k(n)$ is defined as follows: $P^k(n) = n$ if $k = 0$ and $P^k(n)=P(P(^{k-1}(n))$ if $k >0$.

Proposed by Adrian Beker.
16 replies
BartSimpsons
Dec 27, 2017
bin_sherlo
5 hours ago
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Two Orthocenters and an Invariant Point
Mathdreams   2
N Apr 6, 2025 by hukilau17
Source: 2025 Nepal Mock TST Day 1 Problem 3
Let $\triangle{ABC}$ be a triangle, and let $P$ be an arbitrary point on line $AO$, where $O$ is the circumcenter of $\triangle{ABC}$. Define $H_1$ and $H_2$ as the orthocenters of triangles $\triangle{APB}$ and $\triangle{APC}$. Prove that $H_1H_2$ passes through a fixed point which is independent of the choice of $P$.

(Kritesh Dhakal, Nepal)
2 replies
Mathdreams
Apr 6, 2025
hukilau17
Apr 6, 2025
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Two Orthocenters and an Invariant Point
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G H BBookmark kLocked kLocked NReply
Source: 2025 Nepal Mock TST Day 1 Problem 3
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Mathdreams
1465 posts
Mathdreams
#1 h  Apr 6, 2025, 1:30 PM • 3 Y
Y by RANDOM__USER, PikaPika999, khan.academy
Let $\triangle{ABC}$ be a triangle, and let $P$ be an arbitrary point on line $AO$, where $O$ is the circumcenter of $\triangle{ABC}$. Define $H_1$ and $H_2$ as the orthocenters of triangles $\triangle{APB}$ and $\triangle{APC}$. Prove that $H_1H_2$ passes through a fixed point which is independent of the choice of $P$.

(Kritesh Dhakal, Nepal)
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RANDOM__USER
7 posts
RANDOM__USER
#2 h  Apr 6, 2025, 2:14 PM • 1 Y
Y by PikaPika999
Let us linearly move \(P\) on the line \(AO\), then \(H_1\) and \(H_2\) move linearly on the perpendicular lines from \(B\) and \(C\) onto \(AO\) respectively. Additionally the lines on which \(H_1\) and \(H_2\) move are parallel to each other, because the perpendicular lines from \(B\) and \(C\) onto \(AO\) are parallel.

Claim: If \(D\) is the base of the altitude from \(A\) onto \(BC\), then \(D \in H_1H_2\).
Proof: Because of the properties stated above, we only need to verify this statement for two positions of \(P\) to prove it holds for a general position of \(P\). When \(P\) is the antipode of \(A\) in \((ABC)\) the statement is more than trivial, and when \(P \in BC\), \(H_1\) and \(H_2\) lie on \(AD\).

Thus \(H_1H_2\) always passes through \(D\). \(\blacksquare\)
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hukilau17
283 posts
hukilau17
#3 h  Apr 6, 2025, 7:05 PM • 1 Y
Y by PikaPika999
Complex bash with $\triangle ABC$ inscribed in the unit circle, so that
$$|a|=|b|=|c|=1$$$$o=0$$$$p = \lambda a$$for some real number $\lambda$. Then the circumcenter of $\triangle APB$ has coordinate
$$o_1 = \frac{ab(p\overline{p}-1)}{p-a-b+ab\overline{p}} = \frac{ab(1+\lambda)}{a+b}$$and so its orthocenter has coordinate
$$h_1 = a + b + p - 2o_1 = \frac{a^2+b^2+\lambda a^2-\lambda ab}{a+b}$$Similarly,
$$h_2 = \frac{a^2+c^2+\lambda a^2-\lambda ac}{a+c}$$To guess the fixed point, we pick some convenient values of $\lambda$. When $\lambda = -1$, we have $h_1 = b$ and $h_2 = c$. When $\lambda = 0$, we have $h_1 = \frac{a^2+b^2}{a+b}$ and $h_2 = \frac{a^2+c^2}{a+c}$. Thus, we let $Q$ be the intersection of these two lines. Since $Q$ is on line $BC$, we have
$$\overline{q} = \frac{b+c-q}{bc}$$Since $Q$ lies on the line through $\frac{a^2+b^2}{a+b}$ and $\frac{a^2+c^2}{a+c}$, we have
$$\frac{\frac{a^2+b^2}{a+b} - q}{\frac{a^2+b^2}{a+b} - \frac{a^2+c^2}{a+c}} \in \mathbb{R}$$We set this equal to its conjugate, and use the fact that $\overline{q} = \frac{b+c-q}{bc}$, to get
$$\frac{(a+c)(a^2+b^2-aq-bq)}{(b-c)(-a^2+ab+ac+bc)} = \frac{(a+c)(b^2c-a^2b-ab^2-abc+a^2q+abq)}{a(c-b)(a^2+ab+ac-bc)}$$$$a(a^2+b^2-aq-bq)(a^2+ab+ac-bc) = (-a^2+ab+ac+bc)(a^2b+ab^2+abc-b^2c-a^2q-abq)$$$$q = \frac{a(a^2+b^2)(a^2+ab+ac-bc) - b(a^2+ab+ac-bc)(-a^2+ab+ac+bc)}{a(a+b)(a^2+ab+ac-bc) - a(a+b)(-a^2+ab+ac+bc)}$$We simplify this to
$$q = \frac{(a^2+ab+ac-bc)(a^3+a^2b-abc-b^2c)}{a(a+b)(2a^2-2bc)} = \frac{a^2+ab+ac-bc}{2a}$$So $Q$ is the foot of the altitude from $A$ onto line $BC$. It remains to show that $Q$ always lies on line $H_1H_2$ as $\lambda$ varies. Indeed, we find
$$h_1-q = \frac{2a(a^2+b^2+\lambda a^2-\lambda ab) - (a+b)(a^2+ab+ac-bc)}{2a(a+b)} = \frac{(a-b)(2\lambda a^2+a^2-ab-ac-bc)}{2a(a+b)}$$and similarly
$$h_2-q = \frac{(a-c)(2\lambda a^2+a^2-ab-ac-bc)}{2a(a+c)}$$So the factor involving $\lambda$ cancels out, and
$$\frac{h_1-q}{h_2-q} = \frac{(a-b)(a+c)}{(a-c)(a+b)}$$which is real. $\blacksquare$
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