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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
f(x+f(x)+f(y))=x+f(x+y)
dangerousliri   10
N 19 minutes ago by jasperE3
Source: FEOO, Shortlist A5
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for any positive real numbers $x$ and $y$,
$$f(x+f(x)+f(y))=x+f(x+y)$$Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
10 replies
dangerousliri
May 31, 2020
jasperE3
19 minutes ago
n-variable inequality
ABCDE   66
N 22 minutes ago by ND_
Source: 2015 IMO Shortlist A1, Original 2015 IMO #5
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
66 replies
ABCDE
Jul 7, 2016
ND_
22 minutes ago
Euler Line Madness
raxu   75
N an hour ago by lakshya2009
Source: TSTST 2015 Problem 2
Let ABC be a scalene triangle. Let $K_a$, $L_a$ and $M_a$ be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time at point $X_a$ different from A. Define $X_b$ and $X_c$ analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on the Euler line of ABC.
(The Euler line of ABC is the line passing through the circumcenter, centroid, and orthocenter of ABC.)

Proposed by Ivan Borsenco
75 replies
raxu
Jun 26, 2015
lakshya2009
an hour ago
Own made functional equation
Primeniyazidayi   8
N an hour ago by MathsII-enjoy
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
8 replies
Primeniyazidayi
May 26, 2025
MathsII-enjoy
an hour ago
No more topics!
IMOC 2020 G5 concyclic wanted, parallelogram and concurrent related
parmenides51   5
N Apr 17, 2023 by Mahdi_Mashayekhi
Source: https://artofproblemsolving.com/community/c6h2254883p17398793
Let $O, H$ be the circumcentor and the orthocenter of a scalene triangle $ABC$. Let $P$ be the reflection of $A$ w.r.t. $OH$, and $Q$ is a point on $\odot (ABC)$ such that $AQ, OH, BC$ are concurrent. Let $A'$ be a points such that $ABA'C$ is a parallelogram. Show that $A', H, P, Q$ are concylic.

(ltf0501).
5 replies
parmenides51
Sep 1, 2020
Mahdi_Mashayekhi
Apr 17, 2023
IMOC 2020 G5 concyclic wanted, parallelogram and concurrent related
G H J
G H BBookmark kLocked kLocked NReply
Source: https://artofproblemsolving.com/community/c6h2254883p17398793
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parmenides51
30653 posts
#1
Y by
Let $O, H$ be the circumcentor and the orthocenter of a scalene triangle $ABC$. Let $P$ be the reflection of $A$ w.r.t. $OH$, and $Q$ is a point on $\odot (ABC)$ such that $AQ, OH, BC$ are concurrent. Let $A'$ be a points such that $ABA'C$ is a parallelogram. Show that $A', H, P, Q$ are concylic.

(ltf0501).
This post has been edited 1 time. Last edited by parmenides51, Dec 15, 2022, 7:33 PM
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amar_04
1916 posts
#2 • 1 Y
Y by Gaussian_cyber
Perform a $-\sqrt{HA\cdot HD}$ Inversion around $H$. We get the following Equivalent Problem.

$\textbf{INVERTED PROBLEM:-}$ $ABC$ be a triangle with Circumcenter $O$ and Incenter $I$. Let $\overline{OI}\cap\odot(ABC)=T$ and $\odot(AIT)\cap\odot(ABC)=K$. Let $A'$ be the reflection of $A$ over $\overline{IO}$. and $D$ be the point where the Incircle of $\Delta ABC$ touches $\overline{BC}$. Then $A',D,K$ are collinear.

By Radical Axis Theorem on $\odot(ABC),\odot(BIC),\odot(AIT)$ we get that $\overline{AK},\overline{IO},\overline{BC}$ are concurrent. Now the rest follows from #4 here. $\blacksquare$
This post has been edited 1 time. Last edited by amar_04, Sep 1, 2020, 8:48 PM
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WolfusA
1900 posts
#3 • 1 Y
Y by bin_sherlo
Let $a,b,c$ be complex numbers lying on unit circle such that they are vertices of the triangle. Then
$$h=a+b+c, a'=b+c-a,\ p=\frac{h\overline{a}}{\overline{h}}=bc\cdot\frac{a+b+c}{ab+bc+ca}.$$From $R$ definition
$$\left(\overline{r}=\frac{b+c-r}{bc}\ \wedge\ \frac{\overline{r}}{\overline{h}}=\frac{r}{h}\right)\iff r=\frac{a(b+c)(a+b+c)}{a^2+2a(b+c)+bc}.$$$Q$ definition implies $$\overline{r}=\frac{a+q-r}{aq}\iff q=\frac{a-r}{a\cdot\overline{r}-1}=\frac{(a+b)(a+c)-(b+c)^2}{a^2(b+c)^2-(a+b)(a+c)bc}\cdot abc.$$Moving on...
$$h-p=\frac{a(b+c)(a+b+c)}{ab+bc+ca},\ h-a'=2a,$$$$q-a'=\frac{(a-b)(a-c)(b+c)(ab+bc+ca)}{a^2(b+c)^2-(a+b)(a+c)bc},$$$$q-p=\frac{bc(b+c)(a^2-bc)(a^2+2a(b+c)+bc)}{(ab+bc+ca)(a^2(b+c)^2-(a+b)(a+c)bc)}.$$Therefore $$\frac{h-p}{h-a'}\cdot\frac{q-a'}{q-p}=\frac{(a+b+c)(a-b)(a-c)(b+c)(ab+bc+ca)}{2bc(a^2-bc)(a^2+2a(b+c)+bc)}=\overline{\left(\frac{h-p}{h-a'}\cdot\frac{q-a'}{q-p}\right)}\square.$$#1726
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ltf0501
191 posts
#4 • 1 Y
Y by amar_04
amar_04 wrote:
Perform a $-\sqrt{HA\cdot HD}$ Inversion around $H$. We get the following Equivalent Problem.

$\textbf{INVERTED PROBLEM:-}$ $ABC$ be a triangle with Circumcenter $O$ and Incenter $I$. Let $\overline{OI}\cap\odot(ABC)=T$ and $\odot(AIT)\cap\odot(ABC)=K$. Let $A'$ be the reflection of $A$ over $\overline{IO}$. and $D$ be the point where the Incircle of $\Delta ABC$ touches $\overline{BC}$. Then $A',D,K$ are collinear.

By Radical Axis Theorem on $\odot(ABC),\odot(BIC),\odot(AIT)$ we get that $\overline{AK},\overline{IO},\overline{BC}$ are concurrent. Now the rest follows from #4 here. $\blacksquare$

That's how this problem was produced. :)
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USJL
540 posts
#5 • 2 Y
Y by gnoka, SerdarBozdag
I didn't know how to do the original TST problem, but I somehow manage to solve this one :o
I think my solution is kind of cute so I'm posting it here.

Note that $B,H,C,A'$ are concyclic. Therefore it suffices to show that $BC,A'H,PQ$ are concurrent (by, say, power of a point theorem). Let $A_1$ be the antipedal point of $A$ w.r.t. $\odot(ABC)$. Since the intersections $AA_1\cap BC$ and $A'H\cap BC$ are symmetric w.r.t. the midpoint $M$ of $BC$, by the butterfly theorem (or more precisely, a slightly more generalized statement of it), it suffices to show that $AQ\cap BC$ and $PA_1\cap BC$ are symmetric w.r.t. $M$. Note that $M$ is also a midpoint of $HA_1$ and both $OH, PA_1$ are perpendicular to $AP$. Therefore the lines $OH$ and $PA_1$ are symmetric w.r.t. $M$, which gives the desired result.
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Mahdi_Mashayekhi
697 posts
#6
Y by
Let $OH$ meet $BC$ at $T$. Let $AH$ and $AO$ meet $ABC$ at $S$ and $A''$. Let $M$ be midpoint of $BC$.
Note that $BHCA'$ is cyclic so by Radical Axis Theorem we need to prove $HA'$ and $PQ$ meet at $BC$. Let $PQ$ and $A'H$ meet $BC$ at $X$ and $Y$. Note that $\frac{BX}{XC} = \frac{BQ}{QC}.\frac{BP}{PC}$ and $\frac{BY}{YC} = \frac{BH}{HC}.\frac{BA'}{A'C}$ so we need to prove $\frac{BQ}{QC}.\frac{BP}{PC} = \frac{BS}{SC}.\frac{AC}{AB}$ or $\frac{BQ}{QC}.\frac{AB}{AC} = \frac{BS}{SC}.\frac{PC}{PB}$ or $\frac{TC}{TB} = \frac{CA''}{BA''}.\frac{PC}{PB}$. Let $T'$ be reflection of $T$ across $M$. we need to prove $\frac{T'C}{T'B} = \frac{CA''}{BA''}.\frac{PC}{PB}$ or in fact we need to prove $T',A'',P$ are collinear. Note that $M$ is midpoint of $HA''$ so $THT'A''$ is parallelogram so $TO || T'A''$ and $PA'' \perp AP \perp TO$ so $PA'' || TO$ so $P,A'',T'$ are collinear as wanted.
we're Done.
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