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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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0 replies
jlacosta
May 1, 2025
0 replies
Asymmetric FE
sman96   15
N a minute ago by youochange
Source: BdMO 2025 Higher Secondary P8
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that$$f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$$for all $x, y \in \mathbb{R}$.
15 replies
sman96
Feb 8, 2025
youochange
a minute ago
Mount Inequality erupts in all directions!
BR1F1SZ   3
N 3 minutes ago by sqing
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
3 replies
+1 w
BR1F1SZ
May 5, 2025
sqing
3 minutes ago
Looks like power mean, but it is not
Nuran2010   4
N 36 minutes ago by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
4 replies
+1 w
Nuran2010
2 hours ago
sqing
36 minutes ago
ISI UGB 2025 P6
SomeonecoolLovesMaths   1
N 40 minutes ago by ronitdeb
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
1 reply
SomeonecoolLovesMaths
3 hours ago
ronitdeb
40 minutes ago
Real parameter equation
L.Lawliet03   1
N an hour ago by Mathzeus1024
For which values of the real parameter $a$ does only one solution to the equation $(a+1)x^{2} -(a^{2} + a + 6)x +6a = 0$ belong to the interval (0,1)?
1 reply
L.Lawliet03
Nov 3, 2019
Mathzeus1024
an hour ago
a inequality problem
Polus425   1
N 2 hours ago by Mathzeus1024
$x_1,x_2\; are\; such\; two\; different\; real\; numbers:\; $
$(x_1 ^2 -2x_1 +4ln\, x_1)+(x_2 ^2 -2x_2 +4ln\, x_2)- x_1 ^2 x_2 ^2=0$
$prove\; that:\; x_1+x_2\ge 3$
1 reply
Polus425
Dec 19, 2019
Mathzeus1024
2 hours ago
Find the range of 'f'
agirlhasnoname   1
N 2 hours ago by Mathzeus1024
Consider the triangle with vertices (1,2), (-5,-1) and (3,-2). Let Δ denote the region enclosed by the above triangle. Consider the function f:Δ-->R defined by f(x,y)= |10x - 3y|. Then the range of f is in the interval:
A)[0,36]
B)[0,47]
C)[4,47]
D)36,47]
1 reply
agirlhasnoname
May 14, 2021
Mathzeus1024
2 hours ago
Function of Common Area [China HS Mathematics League 2021]
HamstPan38825   1
N 3 hours ago by Mathzeus1024
Define the regions $M, N$ in the Cartesian Plane as follows:
\begin{align*}
M &= \{(x, y) \in \mathbb R^2 \mid 0 \leq y \leq \text{min}(2x, 3-x)\} \\
N &= \{(x, y) \in \mathbb R^2 \mid t \leq x \leq t+2 \}
\end{align*}for some real number $t$. Denote the common area of $M$ and $N$ for some $t$ be $f(t)$. Compute the algebraic form of the function $f(t)$ for $0 \leq t \leq 1$.

(Source: China National High School Mathematics League 2021, Zhejiang Province, Problem 5)
1 reply
HamstPan38825
Jun 29, 2021
Mathzeus1024
3 hours ago
Functions
Entrepreneur   2
N 3 hours ago by alexheinis
Let $f(x)$ be a polynomial with integer coefficients such that $f(0)=2020$ and $f(a)=2021$ for some integer $a$. Prove that there exists no integer $b$ such that $f(b) = 2022$.
2 replies
Entrepreneur
Aug 18, 2023
alexheinis
3 hours ago
Inequalities
sqing   3
N 4 hours ago by sqing
Let $ a,b,c\geq 0 , (a+8)(b+c)=9.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{38}{23}$$Let $ a,b,c\geq 0 , (a+2)(b+c)=3.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{3}+1)}{5}$$
3 replies
sqing
Yesterday at 12:50 PM
sqing
4 hours ago
Plz help
Bet667   1
N 4 hours ago by Mathzeus1024
f:R-->R for any integer x,y
f(yf(x)+f(xy))=(x+f(x))f(y)
find all function f
(im not good at english)
1 reply
Bet667
Jan 28, 2024
Mathzeus1024
4 hours ago
Minimum value of 2 variable function
girishpimoli   6
N 4 hours ago by Mathzeus1024
Minimum value of $x^2+y^2-xy+3x-3y+4$ , Where $x,y\in\mathbb{R}$
6 replies
girishpimoli
Apr 1, 2024
Mathzeus1024
4 hours ago
Function prob
steven_zhang123   4
N 5 hours ago by Mathzeus1024
If the function $f(x)=x^2+ax+b$ has a maximum value of $M$ and a minimum value of $m$ in the interval $[0,1]$. Confirm whether the value of $M-m$ depends on $a$ or $b$.
4 replies
steven_zhang123
Sep 22, 2024
Mathzeus1024
5 hours ago
Angle Formed by Points on the Sides of a Triangle
xeroxia   4
N Today at 8:03 AM by jainam_luniya

In triangle $ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that
$BD = 20$, $DC = 15$, $CE = 13$, $EA = 8$, $AF = 6$, $FB = 22$.

What is the measure of $\angle EDF$?


4 replies
xeroxia
Yesterday at 10:28 AM
jainam_luniya
Today at 8:03 AM
4 lines concurrent
Zavyk09   7
N May 2, 2025 by bin_sherlo
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
7 replies
Zavyk09
Apr 9, 2025
bin_sherlo
May 2, 2025
4 lines concurrent
G H J
Source: Homework
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Zavyk09
13 posts
#1 • 1 Y
Y by PikaPika999
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
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aidenkim119
33 posts
#2 • 1 Y
Y by PikaPika999
............
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aidenkim119
33 posts
#3 • 1 Y
Y by PikaPika999
First three are trivial by pascal, but AD looks a bit hard / '
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ItzsleepyXD
141 posts
#4 • 1 Y
Y by PikaPika999
Redefine
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $A'$ is antipode of $A$ . $O'$ is circumcenter of $(BOC)$ . Point $E,F$ satisfied $OE // A'F // AB , OF // A'E // AC$ then prove $OH,O'A',BE,CF$ concurrent .

Let $B',C'$ be antipode of $B,C$ respectively.
MMP: Fix $(O),B,C$ . Move $A$ on $(O)$ deg 2.
Since $A',E,C'$ collinear and $A',F,B'$ collinear
By $\angle C'EO = \angle BAC = \angle BC'C = \angle C'BO$ so $C',B,O,E$ concyclic.
implies that $E$ deg 2. Also $F$ deg 2.
So line $BE,CF$ deg 1.
$H=$ reflection of $A \infty_{\perp BC} \cap (O)$ across $BC$
Since $H,A'$ deg 2. implies that line $O'A'.OH$ deg 2.
We want to prove $BE,CF,OH$ concurrent first and $BE,CF,O'A'$ concurrent.
but both have deg 1+1+2+1 = 5 .

choose $A= B,C,B',C'$ and midpoint of arc $BC$
the rest of problem is easy. $\square$
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pingupignu
49 posts
#5 • 1 Y
Y by PikaPika999
My solution may not be elegant but here's some DDIT spam for you guys to enjoy :blush:.
Let $X = KE \cap LF$. I will show that $X, A, D$ and $X, O, H$ are collinear.

Part 1:
I first claim that $\angle \infty_{CH}XL = \angle \infty_{BH}XK$. This follows from
$$\angle \infty_{CH}XL = \angle FLH = \angle FHL = 90^\circ - \angle BHL = 90^\circ - \angle CHK = \angle EHK = \angle XKH = \angle \infty_{BH}XK.$$Then, applying DDIT on $X \cup LDKH$ we see that $$(XK, XL), (XH, XD), (X\infty_{BH}, X\infty_{CH})$$are reciprocal pairs under some involution on $\mathcal{P}_X$. This involution must be a reflection in the angle bisector of $\angle KXL$. Hence $XH, XD$ are isogonal in $\angle KXL$.

Next, since $AK=AL$ (well-known), $LF=FH=AE$, $AF=EH=EK$, we yield $\triangle LAF \cong \triangle AKE$.
I claim that $X\infty_{AB}, X\infty_{AC}$ are isogonal in $\angle XKL$. This is because $$\angle \infty_{AB}XF = 180^\circ - \angle AFL = 180^\circ - \angle AEK = \angle AEX = \angle \infty_{AC}XE.$$Applying DDIT on $X \cup AEHF$ would then give $XA, XH$ are isogonal in $\angle EXF$. Since $XH, XD$, $XH, XA$ are isogonal in $\angle KXL = \angle EXF$ we conclude that $XA \equiv XD$, or $X \in AD$.

Part 2:
I first prove that $X\infty_{CH}, OL, AK$ concur at a point $S$ on $(XLK)$. For this, let $S = X\infty_{CH} \cap OL$, where from a short angle chase we get $$\angle XSL = \angle OLH = B-A = (90^\circ - A) - (90^\circ - B) = \angle AKL - \angle LCB$$$$= \angle AKL - \angle LAF = \angle AKL - \angle  AKE = \angle EKL = \angle XKL$$Hence $XSKL$ are cyclic, and from
$$\angle SKX = \angle SLX = \angle FLO = \angle ALO - \angle ALF = A - \angle EAK = A - (90^\circ - C)$$which equals
$$= A+C-90^\circ = 90^\circ - B = \angle LAB = \angle AKE = \angle AKX$$$\implies S \in AK$. The claim is proven.

From DDIT in $X \cup AKOL$ we have the reciprocal pairs $$(XA, XO), (XK, XL), (XS, XT)$$where $T = AL \cap KO \cap X \infty_{BH}$ (similarly).
since $(XS, XT) = (X\infty_{CH}, X\infty_{BH})$ and we have established $(XA, XH), (X\infty_{CH}, X\infty_{BH})$ are isogonal in $\angle KXL$ we get $XH \equiv XO \implies X \in OH$. The problem is solved. $\blacksquare$
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tomsuhapbia
6 posts
#6 • 1 Y
Y by Amkan2022
We need two well-known lemmas about the isogonal line:
1. Given a $\triangle ABC$ and a point $P$ satisfied $\angle ABP=\angle ACP$. Let $Q$ be the reflection of $P$ in the midpoint of $BC$. Then $AP$ and $AQ$ are isogonals wrt $\angle BAC$.
2. Given a trapezoid $ABCD$ $(AB\parallel CD)$ is inscribed $(O)$. Let $E,F$ be the intersections of $BC$ and $AD$; $AC$ and $BD$. Let $S$ be a abitary point on $(O)$. Then $SE,SG$ are isogonals wrt $\angle ASB$.

Back to the problem: Let $X,Y$ be the intersections of $OL$ and $AK$; $AL$ and $OK$. By symmetric, we have $AK=AH=AL$ so $AO$ is the perpendicular bisector of $KL$ then $XKLY$ is a trapezoid. Let $LK$ intersects $KE$ at $Z$.

We have
$$\angle FZK=\angle ZEH-\angle ZFC=180^\circ-\angle KEH-\angle HFC=180^\circ-3\angle BAC$$and
$$\angle LYK=\angle AOK-\angle OAL=2(90^\circ-\angle OAL)-\angle OAL=180^\circ-3\angle OAL=180^\circ-3\angle BAC=\angle FZK\,(2)$$since
$$\angle OAL=\dfrac{1}{2}\angle KAL=\dfrac{1}{2}(\angle KAB+\angle BAH+\angle HAC+\angle CAL)=\angle BAC$$$(2)$ leads to $Z$ lies on the circumcircle of $XKLY$. From the lemma 2, we obtain that $ZA,ZO$ are isogonals wrt $\angle LZK$ $(1)$. We also have $\angle ZFE=\angle LFE=\angle KEH=180^\circ-\angle ZEH$ and $\angle HEZ=\angle HKE=\angle FLH=180^\circ-\angle HLZ$, using lemma 1 and get $(XD,XH)$ and $(XH,XA)$ are two isogonal pairs wrt $\angle LZK\equiv\angle FZE$, so $A,D,Z$ are collinear. Combine with $(1)$ and we conclude $Z,O,H$ are collinear or $AD,LF,KE,OH$ are concurrent at $Z$.

https://i.postimg.cc/hJQdTNdW/image.png
This post has been edited 2 times. Last edited by tomsuhapbia, May 1, 2025, 5:25 PM
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hectorleo123
344 posts
#7
Y by
I apologize for the complex bash, but I couldn't find another way.

Let \( B' \) and \( C' \) be the antipodes of \( B \) and \( C \).
Since \( L \) is the reflection of \( H \) over \( AB \), we have \( \angle FLB = \angle FHB = 90^\circ \) (since \( FH \parallel AC \perp BH \)).
Analogously, \( \angle EKC = 90^\circ \).
\(\Rightarrow B', F, L \) are collinear and \( C', E, K \) are collinear.
By Pascal's Theorem on
\[ \binom{B, L, C'}{C, K, B'} \]we get that \( KE, LF \), and \( OH \) are concurrent.
Now it suffices to prove that \( AD, KE \), and \( LF \) are concurrent.

We use complex numbers, where \( (ABC) \) is the unit circle and \( a = 1 \).
Let
\[
k = -\frac{c}{b}, \quad l = -\frac{b}{c}, \quad c' = -c, \quad b' = -b, \quad h = b + c + 1, \quad o = 0
\]\[
d + b + c + 1 = d + h = l + k = -\frac{b}{c} - \frac{c}{b}
\]\[
\Rightarrow d = -\frac{b^2 + c^2 + b^2c + bc^2 + bc}{bc}
\]
Let \( X = KC' \cap LB' \).
We have:
\[
\frac{x + c}{\overline{x + c}} = \frac{c - \frac{c}{b}}{\overline{c - \frac{c}{b}}} = -\frac{c^2}{b}
\Rightarrow \overline{x} = -\frac{xb + bc + c}{c^2}
\]Analogously,
\[
\overline{x} = -\frac{xc + bc + b}{b^2}
\]Equating both expressions:
\[
(b - c)(b^2c + bc^2 + bc - x(b^2 + bc + c^2)) = 0
\Rightarrow x = -\frac{bc(b + c + 1)}{b^2 + bc + c^2}
\]
Points \( A, D, X \) are collinear if and only if
\[
\frac{d - 1}{x - 1} \in \mathbb{R}
\]Substituting:
\[
\frac{(b^2 + c^2 + b^2c + bc^2 + 2bc)/bc}{(b^2c + bc^2 + 2bc + b^2 + c^2)/(b^2 + bc + c^2)} = \frac{b^2 + bc + c^2}{bc}\in \mathbb{R}_\blacksquare
\]
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bin_sherlo
722 posts
#8
Y by
Let $B',C'$ be the antipodes of $B,C$ on $(ABC)$. Also let $LC'\cap KB'=W,KC'\cap LB'=P$.
Claim: $B',F,L$ and $C',E,K$ are collinear.
Proof: Pascal at $KB'LCAB$ yields $AC_{\infty},B'L\cap AB,H$ are collinear thus, $B'L\cap AB=F$. Similarily $C',E,K$ are collinear.
Claim: $P$ lies on $OH$.
Proof: Pascal at $BKC'CLB'$ gives $H,P,O$ are collinear.
Claim: $A,D,P$ are collinear.
Proof: Notice that $W,L,H,K$ lie on the circle with diameter $WH$ and since $AH=AK=AL$, $A$ must be the circumcenter of $(KLHW)$. Hence $W,A,H$ are collinear.
DDIT at $DLHK$ implies $(\overline{AD},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. DDIT at $B'KC'L$ gives $(\overline{AP},\overline{AH}),(\overline{AK},\overline{AL}),(\overline{AC'},\overline{AB'})$ is an involution. Combining these implies $AD\equiv AP$ hence $A,D,P$ are collinear as desired.$\blacksquare$
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