Geometric inequality with Fermat point

by Assassino9931, Apr 27, 2025, 10:21 PM

Let $ABC$ be an acute triangle and let $P$ be an interior point for it such that $\angle APB = \angle BPC = \angle CPA$. Prove that
$$ \frac{PA^2 + PB^2 + PC^2}{2S} + \frac{4}{\sqrt{3}} \leq \frac{1}{\sin \alpha} + \frac{1}{\sin \beta} + \frac{1}{\sin \gamma}. $$When does equality hold?

Inequalities

by hn111009, Apr 27, 2025, 3:29 PM

Let $a,b,c$ be non-negative number. Prove that $$\left(a+bc\right)^2+\left(b+ca\right)^2+\left(c+ab\right)^2\ge \sqrt{2}\left(a+b\right)\left(b+c\right)\left(c+a\right)$$

amazing balkan combi

by egxa, Apr 27, 2025, 1:57 PM

There are $n$ cities in a country, where $n \geq 100$ is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities $A$ and $B$ we define:

$(i)$ A $\emph{path}$ between $A$ and $B$ as a sequence of distinct cities $A = C_0, C_1, \dots, C_k, C_{k+1} = B$, $k \geq 0$, such that there are direct flights between $C_i$ and $C_{i+1}$ for every $0 \leq i \leq k$;
$(ii)$ A $\emph{long path}$ between $A$ and $B$ as a path between $A$ and $B$ such that no other path between $A$ and $B$ has more cities;
$(iii)$ A $\emph{short path}$ between $A$ and $B$ as a path between $A$ and $B$ such that no other path between $A$ and $B$ has fewer cities.
Assume that for any pair of cities $A$ and $B$ in the country, there exist a long path and a short path between them that have no cities in common (except $A$ and $B$). Let $F$ be the total number of pairs of cities in the country that are connected by direct flights. In terms of $n$, find all possible values $F$

Proposed by David-Andrei Anghel, Romania.
This post has been edited 6 times. Last edited by egxa, Yesterday at 10:59 PM

functional equation interesting

by skellyrah, Apr 24, 2025, 8:32 PM

find all functions IR->IR such that $$xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2  + (x+1)f(x)$$
This post has been edited 1 time. Last edited by skellyrah, Apr 25, 2025, 7:42 PM

connected set in grid

by David-Vieta, Sep 8, 2024, 5:06 AM

Given a positive integer $n$. Consider a $3 \times n$ grid, a set $S$ of squares is called connected if for any points $A \neq B$ in $S$, there exists an integer $l \ge 2$ and $l$ squares $A=C_1,C_2,\dots ,C_l=B$ in $S$ such that $C_i$ and $C_{i+1}$ shares a common side ($i=1,2,\dots,l-1$).

Find the largest integer $K$ satisfying that however the squares are colored black or white, there always exists a connected set $S$ for which the absolute value of the difference between the number of black and white squares is at least $K$.
This post has been edited 6 times. Last edited by David-Vieta, Sep 8, 2024, 5:53 AM

JBMO Shortlist 2021 G4

by Lukaluce, Jul 2, 2022, 9:11 PM

Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Let $E$ be the point of intersection of $BC$ with $AD$ and let $M$ be the midpoint of $AE$. On the extension of $CD$, beyond the point $D$, we pick a point $Z$ such that $MZ = \frac{AE}{2}$. Let $U$ and $V$ be the projections of $A$ and $E$ respectively on $BZ$. The circumcircle of the triangle $DUV$ meets again $AE$ at the point $L$. If $I$ is the point of intersection of $BZ$ with $AE$, prove that the lines $BL$ and $CI$ intersect on the line $AZ$.

Pair of multiples

by Jalil_Huseynov, May 17, 2022, 6:44 PM

Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$.

Geometry from Iranian TST 2017

by bgn, Apr 5, 2017, 11:56 AM

In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$.
Prove that $X,Y,Z$ are collinear.

Proposed by Hooman Fattahi
This post has been edited 4 times. Last edited by bgn, Apr 7, 2017, 6:42 PM

A Hard Geometry

by utkarshgupta, Apr 16, 2016, 5:54 AM

It took me so long to solve. Only if I knew a few projective theorems.

Problem : (RMM 2013 Problem 3)
Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\omega$. Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.

Solution :

Let $O$ be the center of $\odot ABCD$.
Let $Q'=\odot PAD \cap PBC$ and $R' = \odot RCD \cap \odot RAB$.
Let $M'$ be the reflection of $P$ in $Q'$

Lemma : $\angle OQ'P = \angle OR'P = 90$
Proof :
Well known

Obviously, $PRR'$ is a straight line (radical axis are concurrent).
Let $OR \cap PQ = Q_1$
Then, since $PQ$ is the polar od $R$, $OQ_1 \perp PQ$
$\implies Q' = Q_1$

Thus $PRR'$ and $ORQ'$ are straight lines..

We know that $OL'Q'K'$ and $OR'Q'P$ are cyclic quadrilaterals.
$\implies RK' \cdot RL' = RO \cdot RQ' = RR' \cdot RP$

Hence $\boxed{R' \in \odot PK'L'}$

It is well known that $Q' \in PQ$
Since $Q'$ lies on the polar of $R$,
$R \in K'L'$
Also since $ORQ'$ is a straight line; $R$ is the midpoint of $K'L'$ and $K'L' || PQ$

Since $M'$ is the reflection of $P$ about $Q'$, $M'$ is the reflection of $P$ about $OQ'$
Hence $L'P = K'M'$
That is $\boxed{M ' \in \odot PK'L'}$


Using the above results,
$PL'R'K'M'$ is a cyclic pentagon.

Now invert with radius $\sqrt{PB \cdot PA}$ and centre $P$.

Obviously $Q' \to Q, M' \to M , R' \to R, \odot(ABCD) \to \odot{ABCD},K' \to K'', L' \to L'' $
The cyclic pentagon implies, $K'',L'' \in MR$ and $K'',L'' \in \odot(ABCD)$

Thus one of $K'',L''$ maps to $K$.
Since $QK',QL'$ are tangent to $\odot (ABCD)$;
$\odot K''PQ$ and $\odot L''PQ$ are tangent to $\odot(ABCD)$.

QED.

Line through orthocenter

by juckter, Jun 22, 2014, 4:09 PM

Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. Let $l$ be the line tangent to $\Gamma$ at $A$. Let $D$ and $E$ be the intersections of the circumference with center $B$ and radius $AB$ with lines $l$ and $AC$, respectively. Prove the orthocenter of $ABC$ lies on line $DE$.

incircle with center I of triangle ABC touches the side BC

by orl, Jun 26, 2005, 12:16 PM

Given a triangle $ABC$. Let $O$ be the circumcenter of this triangle $ABC$. Let $H$, $K$, $L$ be the feet of the altitudes of triangle $ABC$ from the vertices $A$, $B$, $C$, respectively. Denote by $A_{0}$, $B_{0}$, $C_{0}$ the midpoints of these altitudes $AH$, $BK$, $CL$, respectively. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$, respectively. Prove that the four lines $A_{0}D$, $B_{0}E$, $C_{0}F$ and $OI$ are concurrent. (When the point $O$ concides with $I$, we consider the line $OI$ as an arbitrary line passing through $O$.)

Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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utkarshgupta
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  • Here goes first post of 2025! Great blog.

    by math_holmes15, Jan 14, 2025, 8:53 AM

  • First post of 2024

    by Yiyj1, Feb 8, 2024, 5:40 AM

  • First post of 2023

    by HoRI_DA_GRe8, Jul 22, 2023, 7:45 AM

  • Nice blog ! Your isogonality lemma is really powerful !

    by 554183, Oct 14, 2021, 8:55 AM

  • Post plss....

    by samrocksnature, Apr 11, 2021, 10:12 PM

  • alas,this is ded

    by Hamroldt, Mar 18, 2021, 4:13 PM

  • Thanks for the nice blog.

    by Feridimo, Mar 6, 2020, 4:17 PM

  • I think this might be silly but ... when should we expect to have another post ?? I am very keen to see it :D

    by gamerrk1004, Nov 4, 2019, 4:54 PM

  • Let's all echo what's written in the blog description - Stay Insane / 'Cause it's your labor, will and pain/ That takes you to the top of soda fountain :D

    by Kayak, Oct 2, 2017, 7:18 PM

  • hey utkarsh jee is over now ... continue your elementary blog pleaseeeeeee!

    by kk108, Jun 17, 2017, 11:19 AM

  • Congrats on becoming a contest moderator!

    by Ankoganit, Mar 9, 2017, 5:22 AM

  • INTERSTING BLOG

    by kk108, Feb 19, 2017, 2:04 PM

  • I have no plans for this blog right now....
    No time here people !
    But lets see....
    I may try some combinatorics :P

    by utkarshgupta, Feb 15, 2017, 12:47 PM

  • Thanks for the nice blog!

    by Orkhan-Ashraf_2002, Feb 13, 2017, 6:34 PM

  • Revive it!!!
    Best blog out there, for sure!

    by rmtf1111, Jan 12, 2017, 6:02 PM

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