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Parallelograms and concyclicity
Lukaluce   33
N 9 minutes ago by HamstPan38825
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
33 replies
Lukaluce
Apr 14, 2025
HamstPan38825
9 minutes ago
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Angles in a triangle with integer cotangents
Stear14   1
N 17 minutes ago by Stear14
In a triangle $ABC$, the point $M$ is the midpoint of $BC$ and $N$ is a point on the side $BC$ such that $BN:NC=2:1$. The cotangents of the angles $\angle BAM$, $\angle MAN$, and $\angle NAC$ are positive integers $k,m,n$.
(a) Show that the cotangent of the angle $\angle BAC$ is also an integer and equals $m-k-n$.
(b) Show that there are infinitely many possible triples $(k,m,n)$, some of which consisting of Fibonacci numbers.
1 reply
Stear14
May 21, 2025
Stear14
17 minutes ago
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Two perpendicular lines
jayme   2
N Oct 11, 2023 by ancamagelqueme
Source: Pierre Leteurtre
Dear Mathlinlers,

1. ABC a triangle

2. IJK the median triangle

3. U a point on the segment JK

4. V, W the pointsof 'intersection of UC and KI, AV and IJ

5. H1 the othocenter of the triangle UVW

6. L the De Longchamps's point of de ABC.



Question : LH1 is perpendicular to UA.

Sincerely
Jean-Louis
2 replies
jayme
Oct 9, 2023
ancamagelqueme
Oct 11, 2023
J
Two perpendicular lines
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Source: Pierre Leteurtre
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jayme
9801 posts
jayme
#1 h  Oct 9, 2023, 6:57 AM
Y by
Dear Mathlinlers,

1. ABC a triangle

2. IJK the median triangle

3. U a point on the segment JK

4. V, W the pointsof 'intersection of UC and KI, AV and IJ

5. H1 the othocenter of the triangle UVW

6. L the De Longchamps's point of de ABC.



Question : LH1 is perpendicular to UA.

Sincerely
Jean-Louis
This post has been edited 1 time. Last edited by jayme, Oct 10, 2023, 6:18 AM
Reason: typo
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jayme
9801 posts
jayme
#2 h  Oct 10, 2023, 6:19 AM
Y by
Dear Mathlinkers,

any ideas for this hard problem...

Sincerely
Jean-Louis
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ancamagelqueme
104 posts
ancamagelqueme
#3 h  Oct 11, 2023, 2:49 PM
Y by
I rewrite this topic, rename $UVW$ to $A'B'C'$ and propose an analytical solution.
Quote:
Given a triangle $ABC$, let $M_aM_bM_c$ be the medial triangle, $G$ be the centroid, $O$ be the circumcenter, $H$ be the orthocenter, and $L$ be the De Longchamps point.
Let $A'B'C$' be a variable triangle inscribed in $M_aM_bM_c$ and circumscribed to $ABC$ (the line $B'C'$ passes through $A$ and $A'$ lies in the line $M_bM_c$, etc...). Let $H'$ be the orthocenter of $A'B'C'$.
Question : $LH'$ is perpendicular to $AA$'.

Figure


If $M_bA':A'M_c=t$, then $B'=A'C \cap M_cM_a, C'=A'B\cap M_aM_b$ and $M_cB':B'M_a= -1:(t+1), M_aC':C'M_b = (t+1):-t$. That is, in barycentric coordinates, $A'=(t+1:t:1), B'=(t+1:t:-1), C'=(t+1:-t:1$).
The orthocenter of $A'B'C'$ is

$$ H'=((b^2-c^2) (1+t) (-c^2+b^2 t^2)+a^4 (1+5 t+5 t^2+t^3)-a^2 (c^2 (2+6 t+t^2-t^3)+b^2 (-1+t+6 t^2+2 t^3)):-c^4 t-a^4 t (1+t)^2+b^4 t (2+2 t-t^2)+b^2 c^2 (2+t-4 t^2-t^3)+a^2 (c^2 t (2+2 t+t^2)+b^2 (-2-5 t+2 t^3)):-b^4 t^2-a^4 (1+t)^2+c^4 (-1+2 t+2 t^2)+b^2 c^2 (-1-4 t+t^2+2 t^3)+a^2 (b^2 (1+2 t+2 t^2)+c^2 (2-5 t^2-2 t^3))). $$
The infinity points of the lines $AA'$ and $LH'$ are, respectively, $(-t-1:t:1)$ and $(a^2 (t-1)-(b^2-c^2) (t+1):(-a^2+c^2) t+b^2 (t+2):a^2-b^2-c^2 (2 t+1))$, which determine two perpendicular directions since (see Paul Yiu, Introduction To The Geometry Of The Triangle $\S$4.5 or (12.42) ):
$$ S_A (-t-1)(a^2 (t-1)-(b^2-c^2) (t+1)) + S_B t ((-a^2+c^2) t+b^2 (t+2)) + S_C (a^2-b^2-c^2 (2 t+1))=0. $$

** The locus of $H'$ is a unicursal cubic, with nodal point $L$, asymptotes parallel to the heights through the midpoint of $OL$. It passes through the centers of the index triangle: 1, 5, 20, 1158, 5894, 14135, 30263.



** The locus of the centroid $G'=(-1-4 t-4 t^2-t^3:t (-1-t+t^2):1-t-t^2)$ of $A'B'C'$ is the cubic homothetic of Tucker nodal cubic (K015) under the homothety h($G$, 3/4).



** The lines $GG'$ and $LH'$ are perpendiculars.
This post has been edited 1 time. Last edited by ancamagelqueme, Oct 11, 2023, 8:39 PM
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