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interesting geo config (2/3)
Royal_mhyasd   1
N 2 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
1 reply
Royal_mhyasd
2 hours ago
Royal_mhyasd
2 hours ago
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interesting geo config (1\3)
Royal_mhyasd   0
2 hours ago
Source: own
Let $\triangle ABC$ be an acute triangle with $AC > AB$, $H$ its orthocenter and $O$ it's circumcenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = \angle ABC - \angle ACB$ and $P$ and $C$ are on different sides of $AB$. Denote by $S$ the intersection of the circumcircle of $\triangle ABC$ and $PA'$, where $A'$ is the reflection of $H$ over $BC$, $M$ the midpoint of $PH$, $Q$ the intersection of $OA$ and the parallel through $M$ to $AS$, $R$ the intersection of $MS$ and the perpendicular through $O$ to $PS$ and $N$ a point on $AS$ such that $NT \parallel PS$, where $T$ is the midpoint of $HS$. Prove that $Q, N, R$ lie on a line.

fiy it's 2am and i'm bored so i decided to look further into this interesting config that i had already made some observations on, maybe this problem is trivial from some theorem so if that's the case then i'm sorry lol :P i'll probably post 2 more problems related to it soon, i'd say they're easier than this though
0 replies
Royal_mhyasd
2 hours ago
0 replies
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Parallel lines..
ts0_9   9
N 2 hours ago by OutKast
Source: Kazakhstan National Olympiad 2014 P3 D1 10 grade
The triangle $ABC$ is inscribed in a circle $w_1$. Inscribed in a triangle circle touchs the sides $BC$ in a point $N$. $w_2$ — the circle inscribed in a segment $BAC$ circle of $w_1$, and passing through a point $N$. Let points $O$ and $J$ — the centers of circles $w_2$ and an extra inscribed circle (touching side $BC$) respectively. Prove, that lines $AO$ and $JN$ are parallel.
9 replies
ts0_9
Mar 26, 2014
OutKast
2 hours ago
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KMN and PQR are tangent at a fixed point
hal9v4ik   4
N 2 hours ago by OutKast
Let $ABCD$ be cyclic quadrilateral. Let $AC$ and $BD$ intersect at $R$, and let $AB$ and $CD$ intersect at $K$. Let $M$ and $N$ are points on $AB$ and $CD$ such that $\frac{AM}{MB}=\frac{CN}{ND}$. Let $P$ and $Q$ be the intersections of $MN$ with the diagonals of $ABCD$. Prove that circumcircles of triangles $KMN$ and $PQR$ are tangent at a fixed point.
4 replies
hal9v4ik
Mar 19, 2013
OutKast
2 hours ago
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one cyclic formed by two cyclic
CrazyInMath   40
N 3 hours ago by HamstPan38825
Source: EGMO 2025/3
Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.
40 replies
CrazyInMath
Apr 13, 2025
HamstPan38825
3 hours ago
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geometry problem
Medjl   5
N 3 hours ago by LeYohan
Source: Netherlands TST for IMO 2017 day 3 problem 1
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
5 replies
Medjl
Feb 1, 2018
LeYohan
3 hours ago
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Connected, not n-colourable graph
mavropnevma   7
N 3 hours ago by OutKast
Source: Tuymaada 2013, Day 1, Problem 4 Juniors and 3 Seniors
The vertices of a connected graph cannot be coloured with less than $n+1$ colours (so that adjacent vertices have different colours).
Prove that $\dfrac{n(n-1)}{2}$ edges can be removed from the graph so that it remains connected.

V. Dolnikov

EDIT. It is confirmed by the official solution that the graph is tacitly assumed to be finite.
7 replies
mavropnevma
Jul 20, 2013
OutKast
3 hours ago
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Homothety with incenter and circumcenters
Ikeronalio   8
N 3 hours ago by LeYohan
Source: Korea National Olympiad 2009 Problem 1
Let $I, O$ be the incenter and the circumcenter of triangle $ABC$, and $D,E,F$ be the circumcenters of triangle $ BIC, CIA, AIB$. Let $ P, Q, R$ be the midpoints of segments $ DI, EI, FI $. Prove that the circumcenter of triangle $PQR $, $M$, is the midpoint of segment $IO$.
8 replies
Ikeronalio
Sep 9, 2012
LeYohan
3 hours ago
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2-var inequality
sqing   11
N 3 hours ago by ytChen
Source: Own
Let $ a,b>0 , a^2+b^2-ab\leq 1 . $ Prove that
$$a^3+b^3 -\frac{a^4}{b+1} -\frac{b^4}{a+1} \leq 1 $$
11 replies
sqing
May 27, 2025
ytChen
3 hours ago
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Sums of products of entries in a matrix
Stear14   0
3 hours ago
(a) $\ $Each entry of an $\ 8\times 8\ $ matrix equals either $\ 1\ $ or $\ 2.\ $ Let $\ A\ $ denote the sum of eight products of entries in each row. Also, let $\ B\ $ denote the sum of eight products of entries in each column. Find the maximum possible value of $\ A-B.\ $ In other words, find
$$ {\rm max}\ \left[ \sum_{i=1}^8\ \prod_{j=1}^8\ a_{ij} - \sum_{j=1}^8\ \prod_{i=1}^8\ a_{ij} \right] $$
(b) $\ $Same question, but for a $\ 2025\times 2025\ $ matrix.
0 replies
Stear14
3 hours ago
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a father and his son are skating around a circular skating rink
parmenides51   2
N 4 hours ago by thespacebar1729
Source: Tournament Of Towns Spring 1999 Junior 0 Level p1
A father and his son are skating around a circular skating rink. From time to time, the father overtakes the son. After the son starts skating in the opposite direction, they begin to meet five times more often. What is the ratio of the skating speeds of the father and the son?

(Tairova)
2 replies
parmenides51
May 7, 2020
thespacebar1729
4 hours ago
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Sums of n mod k
EthanWYX2009   1
N 4 hours ago by Martin.s
Source: 2025 May 谜之竞赛-3
Given $0<\varepsilon <1.$ Show that there exists a constant $c>0,$ such that for all positive integer $n,$
\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]Proposed by Cheng Jiang
1 reply
EthanWYX2009
May 26, 2025
Martin.s
4 hours ago
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Incircle and circumcircle
stergiu   6
N Apr 10, 2025 by Sadigly
Source: tst- Greece 2019
Let a triangle $ABC$ inscribed in a circle $\Gamma$ with center $O$. Let $I$ the incenter of triangle $ABC$ and $D, E, F$ the contact points of the incircle with sides $BC, AC, AB$ of triangle $ABC$ respectively . Let also $S$ the foot of the perpendicular line from $D$ to the line $EF$.Prove that line $SI$ passes from the antidiametric point $N$ of $A$ in the circle $\Gamma$.( $AN$ is a diametre of the circle $\Gamma$).
6 replies
stergiu
Sep 23, 2019
Sadigly
Apr 10, 2025
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Incircle and circumcircle
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Source: tst- Greece 2019
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stergiu
1648 posts
stergiu
#1 h  Sep 23, 2019, 12:06 PM • 2 Y
Y by mathematicsy, Adventure10
Let a triangle $ABC$ inscribed in a circle $\Gamma$ with center $O$. Let $I$ the incenter of triangle $ABC$ and $D, E, F$ the contact points of the incircle with sides $BC, AC, AB$ of triangle $ABC$ respectively . Let also $S$ the foot of the perpendicular line from $D$ to the line $EF$.Prove that line $SI$ passes from the antidiametric point $N$ of $A$ in the circle $\Gamma$.( $AN$ is a diametre of the circle $\Gamma$).
This post has been edited 1 time. Last edited by stergiu, Sep 23, 2019, 12:08 PM
Reason: explanation
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TheDarkPrince
3042 posts
TheDarkPrince
#2 h  Sep 23, 2019, 12:09 PM • 3 Y
Y by Pluto1708, Mathasocean, Adventure10
Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$. Inverting about the incircle gives that $S$ maps to $R$, so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.
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stergiu
1648 posts
stergiu
#3 h  Sep 23, 2019, 3:46 PM • 1 Y
Y by Adventure10
TheDarkPrince wrote:
Let $R$ be the intersection of $\odot(AI)$ and $\odot(ABC)$. Inverting about the incircle gives that $S$ maps to $R$, so $\angle ARI = 90^{\circ}$ and $R,I,S$ are collinear which completes the problem.

Just a simple question to this nice solution:

Why $S$ goes to $R$ and not to another point $Q$ ,of the circle $(AI)$ ? Thank you !

( Ok ! The circle $(A,B,C)$ has invers the Euler circle of triangle $DEF$ and $S$ belongs to this circle.Allmost obvious, but ....)
This post has been edited 3 times. Last edited by stergiu, Sep 23, 2019, 5:41 PM
Reason: correction
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jayme
9801 posts
jayme
#5 h  Sep 24, 2019, 8:15 AM • 2 Y
Y by Adventure10, Mango247
Dear Matlinkers,

http://www.artofproblemsolving.com/community/c6h614584

Sincerely
Jean-Louis
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Euler365
143 posts
Euler365
#6 h  Sep 24, 2019, 5:04 PM • 2 Y
Y by Adventure10, Mango247
Lets complex bash
Set the incircle as the unit circle with incentre as the origin and let $d = 1$.

Then $a = \frac{2ef}{e + f} , b = \frac{2f}{1 + f} , c = \frac{2e}{1 + e}$
$\therefore \frac{a - b}{\overline{a - b}} = -f^2$.

$ nb \perp ba \implies -\frac{a - b}{\overline{a - b}} = \frac{c - n}{\overline{c - n}} = f^2$

$\therefore \overline{n} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$

Similarly $\overline{n} = \frac{(1+e)n-2e+2e^2}{e^2(1+e)}$

$\therefore \frac{(1+e)n-2e+2e^2}{e^2(1+e)} = \frac{(1+f)n-2f+2f^2}{f^2(1+f)}$
After simplifying we obtain that
$n = (1 + e + f - ef)k$

where $k = \frac{2ef}{(e + 1)(f + 1)(e + f)}$
Now we also obtain that $\overline{k} = \frac{2ef}{(e + 1)(f + 1)(e + f)}$
So $\overline{k} = k \implies k \in\mathbb{R}$
Now also note that $s = \frac{1}{2}(1 + e + f - ef)$
So $\frac{n}{s} = 2k \in \mathbb{R}$
So $N$, $I$ and $S$ are collinear.
This post has been edited 4 times. Last edited by Euler365, Sep 25, 2019, 10:53 AM
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gambi
82 posts
gambi
#8 h  Oct 17, 2021, 2:05 PM
Y by
Let $L$ be the midpoint of arc $BAC$ in $\Gamma$, and let $M$ be the antipode of $L$ in $\Gamma$.
Let $K$ be the second intersection point of $MD$ with $\Gamma$.

Claim 1. $K$ is the Miquel point of $BFEC$.
Let $\Psi$ be the inversion around $(BIC)$. By the Incenter Lemma, we get that such inversion is centered in $M$. Clearly,
$$ \left\{\begin{array}{lll} \Psi ((ABC))=BC \\ \Psi (KM)=KM \\ \end{array} \right. \Longrightarrow \Psi (K)=D $$Therefore,
$$ MD\cdot MK=MI^2 \Longrightarrow (KID) \thickspace \text{tangent to} \thickspace IM $$Hence
$$ \measuredangle DKI=\measuredangle DIM=\measuredangle LMA=\measuredangle LKA $$But then
$$ 90^o=\measuredangle MKL=\measuredangle DKL=\measuredangle DKI+\measuredangle IKL=\measuredangle LKA+\measuredangle IKL=\measuredangle IKA $$This way $K\in (AEIF)$ and so Claim 1 is proved.

Claim 2. $K,S,I$ are collinear.
Right triangles $\triangle FSD$ and $\triangle IDC$ are similar because
$$ \measuredangle DFS=\measuredangle DFI+\measuredangle IFS=\measuredangle DBI+\measuredangle IAE=90^o-\measuredangle ICD $$Analogously, right triangles $\triangle ESD$ and $\triangle IDB$ are similar.
From these two similarities, we get
$$ \left\{\begin{array}{lll} \frac{FS}{SD}=\frac{ID}{DC} \\ \\ \frac{SE}{SD}=\frac{ID}{BD} \\ \end{array} \right. \Longrightarrow \frac{FS}{SE}=\frac{BD}{DC} \qquad (\star) $$From Claim 1, we get that $K$ is the center of the spiral similarity sending $BC$ to $FE$. But from $(\star)$, we get that such spiral similarity also sends $D$ to $S$.
Hence $\triangle KFS$ and $\triangle KBD$ are similar. Consequently,
$$ \measuredangle FKS=\measuredangle BKD=\measuredangle BKM=\measuredangle BAM=\measuredangle FAI=\measuredangle FKI $$Therefore, $K,S,I$ are collinear and so Claim 2 is proved.

Finally, from Claim 1 we have $K\in (AEFI) \Longrightarrow IK\perp KA$.
But, in light of Claim 2, this implies that lines $ISK$ and $AK$, are perpendicular, so ray $SI$ will intersect $\Gamma$ at $N$, the antipode of $A$.
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Sadigly
229 posts
Sadigly
#9 h  Apr 10, 2025, 1:46 AM
Y by
Easiest bash exercise

$(DEF)\in\mathbb{S}^1~D=d~E=e~F=f$
$A=\frac{2ef}{e+f}$
$B=\frac{2fd}{f+d}$
$C=\frac{2de}{d+e}$

$O=\frac{2def(d+e+f)}{(d+e)(e+f)(f+d)}$

$N=2O-A=\frac{2ef(d^2+de+df-ef)}{(d+e)(e+f)(f+d)}$

$S=\frac12(d+e+f-\frac{ef}{d})$

$\frac{s-i}{n-i}=\frac{(d+e+f-\frac{ef}{d})(d+e)(e+f)(f+d)}{4ef(d^2+de+df-ef)}=\frac{(d+e)(e+f)(f+d)}{4def}\in\mathbb{R}$
This post has been edited 2 times. Last edited by Sadigly, Apr 10, 2025, 1:56 AM
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