Minimum number of points

by Ecrin_eren, May 15, 2025, 4:09 PM

There are 18 teams in a football league. Each team plays against every other team twice in a season—once at home and once away. A win gives 3 points, a draw gives 1 point, and a loss gives 0 points. One team became the champion by earning more points than every other team. What is the minimum number of points this team could have?

Pells equation

by Entrepreneur, May 15, 2025, 3:56 PM

A Pells Equation is defined as follows $$x^2-1=ky^2.$$Where $x,y$ are positive integers and $k$ is a non-square positive integer. If $(x_n,y_n)$ denotes the n-th set of solution to the equation with $(x_0,y_0)=(1,0).$ Then, prove that $$x_{n+1}x_n-ky_{n+1}y_n=x_1,$$$$x_n\pm y_n\sqrt k=(x_1\pm y_1\sqrt k)^n.$$

Inequalities

by sqing, May 14, 2025, 3:46 AM

Inequalities

by sqing, May 13, 2025, 11:31 AM

Let $a,b,c >2 $ and $ ab+bc+ca \leq 75.$ Show that
$$\frac{1}{a-2}+\frac{1}{b-2}+\frac{1}{c-2}\geq 1$$Let $a,b,c >2 $ and $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{6}{7}.$ Show that
$$\frac{1}{a-2}+\frac{1}{b-2}+\frac{1}{c-2}\geq 2$$

Incircle concurrency

by niwobin, May 11, 2025, 4:28 PM

Triangle ABC with incenter I, incircle is tangent to BC, AC, and AB at D, E and F respectively.
DT is a diameter for the incircle, and AT meets the incircle again at point H.
Let DH and EF intersect at point J. Prove: AJ//BC.
Attachments:

Weird locus problem

by Sedro, May 11, 2025, 3:12 AM

Points $A$ and $B$ are in the coordinate plane such that $AB=2$. Let $\mathcal{H}$ denote the locus of all points $P$ in the coordinate plane satisfying $PA\cdot PB=2$, and let $M$ be the midpoint of $AB$. Points $X$ and $Y$ are on $\mathcal{H}$ such that $\angle XMY = 45^\circ$ and $MX\cdot MY=\sqrt{2}$. The value of $MX^4 + MY^4$ can be expressed in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Inequalities

by sqing, May 9, 2025, 8:50 AM

Let $ 0\leq x,y,z\leq 2. $ Prove that
$$-48\leq (x-yz)( 3y-zx)(z-xy)\leq 9$$$$-144\leq (3x-yz)(y-zx)(3z-xy)\leq\frac{81}{64}$$$$-144\leq (3x-yz)(2y-zx)(3z-xy)\leq\frac{81}{16}$$
This post has been edited 1 time. Last edited by sqing, May 9, 2025, 8:58 AM

trigonometric functions

by VivaanKam, Apr 29, 2025, 8:29 PM

Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!

IOQM P23 2024

by SomeonecoolLovesMaths, Sep 8, 2024, 10:20 AM

Consider the fourteen numbers, $1^4,2^4,...,14^4$. The smallest natural numebr $n$ such that they leave distinct remainders when divided by $n$ is:
This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Sep 26, 2024, 12:05 PM
L

The centroid of ABC lies on ME [2023 Abel, Problem 1b]

by Amir Hossein, Mar 12, 2024, 4:49 PM

In the triangle $ABC$, points $D$ and $E$ lie on the side $BC$, with $CE = BD$. Also, $M$ is the midpoint of $AD$. Show that the centroid of $ABC$ lies on $ME$.

Problem 63: Concurrent lines

by henderson, Mar 26, 2017, 12:23 PM

$$\color{red}\bf{Problem \ 63} $$Let $O$ be the circumcenter of an acute $\triangle ABC.$ The points $E$ and $F$ are chosen on the lines $AB$ and $AC,$ respectively, such that $O$ is the midpoint of $EF.$ Let $D$ be the intersection point of $AO$ $($ other than $A$ $)$ with the circumcircle of $\triangle ABC.$ Prove that the lines $EF, BC$ and the tangent to the circumcircle of $\triangle ABC$ at $D$ are concurrent.
$$\color{red}\bf{My \ solution} $$Denote by $l$ the tangent to the circumcircle of $\triangle ABC$ at $D$ and let $\{ S \}=BC \cap l.$ Applying the butterfly theorem to the degenerated quadrilateral $BDCD,$ we obtain that, if $m$ is the line passing through the point $S$ such that $m \perp OS,$ then the intersection points of $BD$ and $CD$ with the line $m$ have equal distances from the point $S.$ In other words, if $\{ E_2 \}=BD\cap m$ and $\{ F_2 \}=CD\cap m,$ then $SE_2=SF_2.$

$\color{blue}\textbf{Claim.}$ $\quad$ $\triangle DE_2F_2 \sim \triangle AEF.$

$\color{blue}\textbf{Proof.}$ $\quad$ Since $l$ is tangent to the circumcircle of $\triangle ABC,$ $\angle EAO=\angle E_2DS$ and $\angle FAO=\angle F_2DS$ hold. On the other hand, sines' law imply
\[ \frac{E_2D}{\sin \angle E_2SD}=\frac{E_2S}{\sin \angle E_2DS} \ \text{and} \ \frac{F_2D}{\sin \angle F_2SD}=\frac{F_2S}{\sin \angle F_2DS} \]$\Longrightarrow$

\[ \frac{E_2D}{F_2D}=\frac{\sin \angle F_2DS}{\sin \angle E_2DS}=\frac{\sin \angle FAO}{\sin \angle EAO} .\]Moreover, \[ \frac{EA}{FA}=\frac{\sin \angle FAO}{\sin \angle EAO} \]in $\triangle AEF.$
Thus, we obtain \[\frac{E_2D}{F_2D}=\frac{EA}{FA}.\]Taking into account the fact that
\[\angle E_2DF_2=\angle E_2DS+\angle F_2DS= \angle EAO+\angle FAO=\angle EAF,\]we conclude
\[\triangle DE_2F_2 \sim \triangle AEF ,\]as desired. $\quad$ $\square$

Let $\{ E_1 \}=OS\cap AB.$ Since $\angle E_2SE_1=\angle E_2BE_1,$ the quadrilateral $E_2SBE_1$ is cyclic $\Longrightarrow$ $\angle SE_2B=\angle SE_1B.$ As $\angle SE_2B=\angle F_2E_2D=\angle FEA=\angle OEA,$ we obtain $\angle SE_1B=\angle OEA,$ which means that $E$ and $E_1$ coincide.
Therefore, the lines $l, BC$ and $EO$ are concurrent at the point $S$ $\Longrightarrow$ $l, BC$ and $EF$ are concurrent at the point $S.$ $\quad$ $\square$
This post has been edited 4 times. Last edited by henderson, Mar 30, 2017, 4:49 PM

"Do not worry too much about your difficulties in mathematics, I can assure you that mine are still greater." - Albert Einstein

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