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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
H not needed
dchenmathcounts   44
N 16 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
16 minutes ago
IZHO 2017 Functional equations
user01   51
N 37 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
37 minutes ago
chat gpt
fuv870   2
N 39 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
39 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 41 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
41 minutes ago
No more topics!
bisected segment
zhaoli   10
N Jul 11, 2019 by AlastorMoody
Source: CSEMO 2005-2
Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.
10 replies
zhaoli
Jul 17, 2005
AlastorMoody
Jul 11, 2019
bisected segment
G H J
Source: CSEMO 2005-2
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zhaoli
417 posts
#1 • 1 Y
Y by Adventure10
Circle $C$ (with center $O$) does not have common point with line $l$. Draw $OP$ perpendicular to $l$, $P \in l$. Let $Q$ be a point on $l$ ($Q$ is different from $P$), $QA$ and $QB$ are tangent to circle $C$, and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$. $PM$, $PN$ are perpendicular to $QB$, $QA$, respectively, $M \in QB$, $N \in QA$. Prove that segment $KP$ is bisected by line $MN$.
Z K Y
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shobber
3498 posts
#2 • 2 Y
Y by Adventure10, Mango247
After making a sketch, I have a strong feeling that this problem can be solved by using analytic geometry. But I am having troubles finding the equations of the tangants......
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nttu
486 posts
#3 • 2 Y
Y by Adventure10, Mango247
Grobber gave a solution long time ago . Try to search with the keyword : " circle_and_tangent "

[Moderator edit: In fact, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5198 .]
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mecrazywong
606 posts
#4 • 2 Y
Y by Adventure10, Mango247
Let QA and QB intersect OP at X and Y respectively. Then the problem is equivalent to the fact P,X,K,Y are harmonic conjugate(why ;) ), which is easy to justify.
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Ashegh
858 posts
#5 • 1 Y
Y by Adventure10
dear shober i love solving problem by using analytic geometry.but i think :
solving th e problems with out using analytic geometry is more powerfull and nice.
dont u think so?
:D
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shobber
3498 posts
#6 • 2 Y
Y by Adventure10, Mango247
Yes, I love to solve a problem with pure geometry. But the perpendicular lines in this problem really reminded of the coordinate system.
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trainer14
2 posts
#7 • 1 Y
Y by Adventure10
draw PX perpendicular to AB ( X is a point on AB)
We know that O, A, P, Q, B are cyclic.
Then apply Simson's theorem from point P to the triangle ABQ.
K, N, M are on the same line.
angle QOP = angle QAP = angle NXP = angle KPX (because OQ are parallel to XP)
So KP is bisected by line NM
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Ashegh
858 posts
#8 • 2 Y
Y by Adventure10, Mango247
dear zaoli ithink u question is wrang :!: .
be cause u know by moving Q on line L the locus of point X (inter section ofAQ andKP)is only a point .
and it is stable.but by by changing Qthe locus of Kwont be stable .and it change on the line OP.
then how can line QA bisect KP :?: :!: :?: :!:
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darij grinberg
6555 posts
#9 • 1 Y
Y by Adventure10
ashegh wrote:
dear zaoli ithink u question is wrang :!: .
be cause u know by moving Q on line L the locus of point X (inter section ofAQ andKP)is only a point .
and it is stable.but by by changing Qthe locus of Kwont be stable .and it change on the line OP.
then how can line QA bisect KP :?: :!: :?: :!:

Why on earth is X stable?

And yes, K is stable. This is something most solvers showed in their solutions as an auxiliary fact. Please read the solutions posted by others before posting such comments, since they could clear up your doubts.

And, by the way, it's MN and not QA that bisects KP.

Darij
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Ashegh
858 posts
#10 • 2 Y
Y by Adventure10, Mango247
ya ya i finally found where was the mistake.
but itsreallyanicequestion .isnt it?
because as u know the locus of point X is fixed(X is the inter section ofKP and MN).
and again the lucas of point K,will be stable.
because K is the pole of line L.and for each point like Q on L we have:
the polar of Q will go through the pole of L wich is K.and it means k is stable :D .
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AlastorMoody
2125 posts
#11 • 2 Y
Y by Adventure10, Mango247
Lemma: Let $\Delta ABC$ be an isosceles triangle with $AB=AC$, $H$ as its orthocenter and $D$ as $A-$antipode. Take a point $F$ on $\odot(ABC)$. Let $FD\cap BC=G$. Then the simson line of $F$ is parallel to $GH$
Proof: Assume, $F$ on minor arc $AB$. Let $N$ be foot from $F$ to $AB$ and $DF$ intersects $F-$simson line at $O$. $$\angle FON=\angle 180^{\circ} -\angle FAN-\angle FAC=180^{\circ}-2\angle FAD=\angle FGH \qquad \blacksquare$$
Using the fact, of orthocentric bisection by simson line, the problem is proved
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