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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Help my diagram has too many points
MarkBcc168   27
N 9 minutes ago by Om245
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
27 replies
+1 w
MarkBcc168
Jul 17, 2024
Om245
9 minutes ago
Geometry, SMO 2016, not easy
Zoom   18
N an hour ago by SimplisticFormulas
Source: Serbia National Olympiad 2016, day 1, P3
Let $ABC$ be a triangle and $O$ its circumcentre. A line tangent to the circumcircle of the triangle $BOC$ intersects sides $AB$ at $D$ and $AC$ at $E$. Let $A'$ be the image of $A$ under $DE$. Prove that the circumcircle of the triangle $A'DE$ is tangent to the circumcircle of triangle $ABC$.
18 replies
Zoom
Apr 1, 2016
SimplisticFormulas
an hour ago
A touching question on perpendicular lines
Tintarn   2
N an hour ago by pi_quadrat_sechstel
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
2 replies
Tintarn
Mar 17, 2025
pi_quadrat_sechstel
an hour ago
Woaah a lot of external tangents
egxa   2
N 2 hours ago by soryn
Source: All Russian 2025 11.7
A quadrilateral \( ABCD \) with no parallel sides is inscribed in a circle \( \Omega \). Circles \( \omega_a, \omega_b, \omega_c, \omega_d \) are inscribed in triangles \( DAB, ABC, BCD, CDA \), respectively. Common external tangents are drawn between \( \omega_a \) and \( \omega_b \), \( \omega_b \) and \( \omega_c \), \( \omega_c \) and \( \omega_d \), and \( \omega_d \) and \( \omega_a \), not containing any sides of quadrilateral \( ABCD \). A quadrilateral whose consecutive sides lie on these four lines is inscribed in a circle \( \Gamma \). Prove that the lines joining the centers of \( \omega_a \) and \( \omega_c \), \( \omega_b \) and \( \omega_d \), and the centers of \( \Omega \) and \( \Gamma \) all intersect at one point.
2 replies
egxa
Apr 18, 2025
soryn
2 hours ago
No more topics!
Locus of centers of rectangles
Simo_the_Wolf   10
N Mar 25, 2008 by Ilthigore
Source: Italian TST day 1, n°1
Let $ABC$ an acute triangle.

(a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$;

(b) Determine if exist some points that are centers of $3$ distinct rectangles whose vertices lie on the sides of $ABC$.
10 replies
Simo_the_Wolf
Jun 1, 2007
Ilthigore
Mar 25, 2008
Locus of centers of rectangles
G H J
Source: Italian TST day 1, n°1
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Simo_the_Wolf
106 posts
#1 • 1 Y
Y by Adventure10
Let $ABC$ an acute triangle.

(a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$;

(b) Determine if exist some points that are centers of $3$ distinct rectangles whose vertices lie on the sides of $ABC$.
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N.T.TUAN
3595 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Simo_the_Wolf wrote:
(a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$;
Assume that $AH\perp BC=H$ and $A_{1}$ is the midpoint of $BC$ , $A_{2}$ is the midpoint of $AH$. Similary we'll have $B_{1},B_{2},C_{1},C_{2}$. I think , that locus is $A_{1}A_{2}\cup B_{1}B_{2}\cup C_{1}C_{2}$.
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vittasko
1327 posts
#3 • 2 Y
Y by Adventure10, Mango247
For a generalization of this problem, see at http://www.mathlinks.ro/Forum/viewtopic.php?t=80163

Kostas Vittas.
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pavel kozlov
615 posts
#4 • 1 Y
Y by Adventure10
Really, in part (b) answer is $I$ - incentre of triangle $ABC$
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vittasko
1327 posts
#5 • 2 Y
Y by Adventure10, Mango247
pavel kozlov wrote:
Really, in part (b) answer is $I$ - incentre of triangle $ABC$
I think that this is not correct.

In my opinion, if we denote as $D,$ $E,$ $F,$ the midpoints of the side segments $BC,$ $AC,$ $AB$ respectively and as $D',$ $E',$ $F',$ the midpoints of the altitudes of $\bigtriangleup ABC,$ through vertices $A,$ $B,$ $C$ respectively, then the answer in part $(b)$ is the concurrency point of the segment lines $DD',$ $EE',$ $FF',$ not necessary coincided with the incenter of $\bigtriangleup ABC.$

Kostas Vittas.

PS. Please, see also the N.T.TUAN's message.
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¬[ƒ(Gabriel)³²¹º]¼
347 posts
#6 • 2 Y
Y by Adventure10, Mango247
pavel kozlov wrote:
Really, in part (b) answer is $I$ - incentre of triangle $ABC$
really it is the Lemoine point. :wink:
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vittasko
1327 posts
#7 • 2 Y
Y by Adventure10, Mango247
[ƒ(Gabriel)³²¹º]¼ wrote:
pavel kozlov wrote:
Really, in part (b) answer is $I$ - incentre of triangle $ABC$
really it is the Lemoine point. :wink:

Thank you dear ƒ(Gabriel)³²¹º]¼ for your remark. It is a new result to me and I see that it is true by the drawing.

Have you in mind a synthetic proof about it?

Kostas vittas.
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EvaristeG
102 posts
#8 • 3 Y
Y by Adventure10 and 2 other users
Well ... it's enough to show that the Lemoine point is the center of three inscribed rectangles at the same time.
In order to achieve this, you only have to notice that the antiparallels of the sides through Lemoine point are bisected by it and, considering two of them, you obtain the diagonals of an inscribed rectangle.
Obviously, this is not a proof, but merely a hint.
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Ilthigore
315 posts
#9 • 4 Y
Y by Adventure10, Mango247, Mango247, Mango247
I'm not convinced it's the Lemoine point. I make it the isotomic conjugate of the orthocentre of the medial triangle (which I don't think is the Lemoine point is it? :S).

Method: By inspection and thinking about homothety, the loci are the three lines between the midpoints of BC and AD (altitude from A) and the other two symmetric situations. it is clear that the midpoint P (say) of AD lies on the side MN of the medial triangle (if we take it to be LMN) and furthermore that NP/PM=BD/DC by homothety factor 2. Looking at this from the point of view of the medial triangle, which is oppositely oriented, we see that the reflection of P in the midpoint of MN will be the foot of the altitude from L to MN, whence the point of concurrency is the isotomic conjugate of the orthocentre of LMN.
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¬[ƒ(Gabriel)³²¹º]¼
347 posts
#10 • 2 Y
Y by Adventure10, Mango247
Ilthigore wrote:
(which I don't think is the Lemoine point is it? :S).
Yes, the symmedian point of the anticomplementary triangle is the isotomic conjugate of the orthocenter.
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Ilthigore
315 posts
#11 • 2 Y
Y by Adventure10, Mango247
Ah, that's cool, and putting our two arguments together we've proved it. Is there a nice reason for this to be true? I don't know much about isotomic conjugation beyond the definition and existence properties.
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