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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
Inspired by Kazakhstan 2017
sqing   1
N a few seconds ago by sqing
Source: Own
Let $a,b,c\ge \frac{1}{2}$ and $a^2+b^2+c^2=2. $ Prove that
$$\left(\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)\ge \frac{8}{3}$$$$  \left( 2a^2+\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left( 2a^2+\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right)  \ge 9\sqrt[3]{4}$$
1 reply
1 viewing
sqing
a minute ago
sqing
a few seconds ago
Number theory
Ecrin_eren   0
12 minutes ago
Show that there are no prime numbers satisfying the equation

(p + r)^q + (q + r)^p = (p + q)^r.

0 replies
Ecrin_eren
12 minutes ago
0 replies
Wait wasn't it the reciprocal in the paper?
Supercali   7
N 13 minutes ago by kes0716
Source: India TST 2023 Day 2 P1
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$, and $$f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)}$$for all integers $m, n \ge 1$. Prove that $f(99, 99)<1.99$.

Proposed by Navilarekallu Tejaswi
7 replies
Supercali
Jul 9, 2023
kes0716
13 minutes ago
Functional equation
Ecrin_eren   0
18 minutes ago
Find all functions f:R R satisfying the equation

f(f(x)y) + f(x+ f(y)) = x f(y) + f(x+y)

for all x,y real numbers

0 replies
Ecrin_eren
18 minutes ago
0 replies
Spheres and a point source of light
mofidy   3
N Yesterday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Yesterday at 9:22 PM
Real Analysis
rljmano   4
N Yesterday at 4:38 PM by alexheinis
In [0 1], {f(t)}*{f’(t)-1} =0 and f is continuously differentiable. How do we conclude that either f is identically zero or f’(t) is identically 1 in [0 1]?
4 replies
rljmano
Yesterday at 6:32 AM
alexheinis
Yesterday at 4:38 PM
find the isomorphism
nguyenalex   14
N Yesterday at 1:49 PM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Yesterday at 1:49 PM
find the convex sets satisfied
nguyenalex   2
N Yesterday at 9:54 AM by ILOVEMYFAMILY
In $\mathbb{R}^2$, let $B = \{(x, y) \mid x \geq 0\}$. Find all convex sets $C$ such that

\[\mathcal{E}(B \cup C) = B \cup C.\]
2 replies
nguyenalex
Yesterday at 8:57 AM
ILOVEMYFAMILY
Yesterday at 9:54 AM
A great result
steven_zhang123   10
N Yesterday at 9:35 AM by teomihai
Show that $\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4} $.
10 replies
steven_zhang123
Oct 31, 2024
teomihai
Yesterday at 9:35 AM
Weird family of sequences
AndreiVila   9
N Yesterday at 8:40 AM by Fibonacci_math
Source: Romanian District Olympiad 2025 12.3
[list=a]
[*] Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$. Show that $f(a)\cdot f(b)<0$.
[*] Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$$
9 replies
AndreiVila
Mar 8, 2025
Fibonacci_math
Yesterday at 8:40 AM
Monster Integral
Entrepreneur   1
N Yesterday at 7:41 AM by RezerdPrime
$$\color{blue}{\int_{0}^{\pi} \frac{\tan^{-1}\left(\frac{\ln(\sin(x))}{x}\right)dx}{\ln^2\left(x^2 + \ln^2(\sin(x))\right) + 4\arctan^2\left(\frac{\ln(\sin(x))}{x}\right)}= -\frac{\pi \tan^{-1}\left(\frac{2\ln(2)}{\pi}\right)}{\ln^2\left(\frac{\pi^2}{4} + \ln^2(2)\right) + 4\arctan^2\left(\frac{2\ln(2)}{\pi}\right)}.}$$
1 reply
Entrepreneur
Jan 10, 2025
RezerdPrime
Yesterday at 7:41 AM
convex closed set with a nonempty interior
ILOVEMYFAMILY   0
Yesterday at 6:11 AM
a) When $n = 2$, prove that a convex closed set with a nonempty interior that contains exactly one extreme point must contain a ray. (SOLVED)

b) When $n = 2$, find all convex closed sets with a nonempty interior that contain exactly one extreme point.
0 replies
ILOVEMYFAMILY
Yesterday at 6:11 AM
0 replies
Differentiation Marathon!
LawofCosine   178
N Yesterday at 5:26 AM by awzhang10
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
178 replies
LawofCosine
Feb 1, 2025
awzhang10
Yesterday at 5:26 AM
Integrals problems and inequality
tkd23112006   1
N Yesterday at 3:55 AM by removablesingularity
Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
1 reply
tkd23112006
Feb 16, 2025
removablesingularity
Yesterday at 3:55 AM
Parallelogram and a simple cyclic quadrilateral
Noob_at_math_69_level   5
N Yesterday at 9:29 PM by awesomeming327.
Source: DGO 2023 Individual P1
Let $ABC$ be an acute triangle with point $D$ lie on the plane such that $ABDC$ is a parallelogram. $H$ is the orthocenter of $\triangle{ABC}.$ $BH$ intersects $CD$ at $Y$ and $CH$ intersects $BD$ at $X.$ The circle with diameter $AH$ intersects the circumcircle of $\triangle{ABC}$ again at $Q.$ Prove that: The circumcircle of $\triangle{XQY}$ passes through the reflection point of $D$ over $BC.$

Proposed by MathLuis
5 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
Yesterday at 9:29 PM
Parallelogram and a simple cyclic quadrilateral
G H J
Source: DGO 2023 Individual P1
The post below has been deleted. Click to close.
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Noob_at_math_69_level
191 posts
#1 • 2 Y
Y by MathLuis, Rounak_iitr
Let $ABC$ be an acute triangle with point $D$ lie on the plane such that $ABDC$ is a parallelogram. $H$ is the orthocenter of $\triangle{ABC}.$ $BH$ intersects $CD$ at $Y$ and $CH$ intersects $BD$ at $X.$ The circle with diameter $AH$ intersects the circumcircle of $\triangle{ABC}$ again at $Q.$ Prove that: The circumcircle of $\triangle{XQY}$ passes through the reflection point of $D$ over $BC.$

Proposed by MathLuis
This post has been edited 1 time. Last edited by Noob_at_math_69_level, Dec 18, 2023, 7:40 PM
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v4913
1650 posts
#2 • 1 Y
Y by MathLuis
If $Z = BC \cap XY$ then $Z$ has equal power with respect to $(XBCY), (BQAD'C)$, so it suffices to show that $Z, Q, D'$ are collinear, where $D'$ is the reflection of $D$ over $BC$. If $J = DH \cap XY$ then by Ceva-Menelaus, $(Z, J; B, C) = -1$. $QH$ passes through the midpoint of $BC$, and the $A$-antipode wrt $(ABC)$, so if $H'$ is the reflection of $H$ over $BC$ then $QBH'C$ is harmonic, and $(Z, J; B, C) \overset D'= (D'Z \cap (ABC), H'; B, C) \implies D', Q, Z$ are collinear as desired. $\square$
Z K Y
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SerdarBozdag
892 posts
#3 • 1 Y
Y by Infinityfun
Definitions: $AD\cap ABC=E$, $A'$ is the antipode of $A$. $D'$ is the reflection of $D$ over $BC$.

$Q,H,M,A'$ are collinear. $\angle XBY= \angle XCY= 90^{\circ} \implies XBCY$ is cyclic. If we prove that $XY, BC, QD'$ is concurrent at $Z$, then $ZX \cdot ZY = ZB \cdot ZC= ZQ \cdot ZD'$ gives the desired result.

To prove it, note that by butterfly theorem, midpoint of segment $(QE\cap BC)(AA'\cap BC)$ is $M$. In addition, $HDA'A$ is a parallelogram so $M$ is the midpoint of segment $(HD \cap BC)(AA'\cap BC)$. Thus $HD, QE,BC$ are concurrent.

We have $(XY \cap BC,DH \cap BC; B,C)=-1$ (well-known Ceva + Menelaus config) and $-1=(\infty_{\parallel BC}, M;B,C)=(D',E;B,C)=(QE \cap BC, QD'\cap BC; B,C)$. Because $HD \cap BC = QE\cap BC$, we get $XY \cap BC=QD'\cap BC \implies XY, BC, QD'$ is concurrent.
Z K Y
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math_comb01
659 posts
#5 • 2 Y
Y by Noob_at_math_69_level, MathLuis
A non-projective solution (ft. spiral)
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12.5 cm); 
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ 
pen dotstyle = black; /* point style */ 
real xmin = -7.735748180148413, xmax = 7.33858306397737, ymin = -3.3026424134490555, ymax = 5.397974096812354;  /* image dimensions */
pen ccqqqq = rgb(0.8,0,0); pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); 

draw((1.164434181729008,4.916247290725071)--(-0.6502157932064757,-1.8236167460721382)--(5.469784206793524,-1.963616746072138)--cycle, linewidth(0.4) + ccqqqq); 
draw((-1.673635102853051,1.4439718900326037)--(-6.845377604602749,-1.6818973582297398)--(-4.276598773829513,-1.740660272659193)--cycle, linewidth(0.4) + zzttff); 
draw((-1.673635102853051,1.4439718900326037)--(-0.6502157932064757,-1.8236167460721382)--(-0.2717039626462117,-0.41777082011823513)--cycle, linewidth(0.4) + zzttff); 
draw((-1.673635102853051,1.4439718900326037)--(5.469784206793524,-1.963616746072138)--(3.6846021017659614,0.8890688873866683)--cycle, linewidth(0.4) + zzttff); 
draw((-1.673635102853051,1.4439718900326037)--(1.164434181729008,4.916247290725071)--(3.965230973398127,4.852176775882967)--cycle, linewidth(0.4) + zzttff); 
 /* draw figures */
draw((1.164434181729008,4.916247290725071)--(-0.6502157932064757,-1.8236167460721382), linewidth(0.4) + ccqqqq); 
draw((-0.6502157932064757,-1.8236167460721382)--(5.469784206793524,-1.963616746072138), linewidth(0.4) + ccqqqq); 
draw((5.469784206793524,-1.963616746072138)--(1.164434181729008,4.916247290725071), linewidth(0.4) + ccqqqq); 
draw(circle((2.474816347719361,0.9492168429715979), 4.177850164181177), linewidth(0.4) + zzttqq); 
draw((xmin, 0.625789986842355*xmin-1.4167182133967662)--(xmax, 0.625789986842355*xmax-1.4167182133967662), linewidth(0.4)); /* line */
draw((xmin, -0.26924133261860383*xmin-0.4909247570988565)--(xmax, -0.26924133261860383*xmax-0.4909247570988565), linewidth(0.4)); /* line */
draw((xmin, -1.5979801866850798*xmin-2.86264870068581)--(xmax, -1.5979801866850798*xmax-2.86264870068581), linewidth(0.4)); /* line */
draw((xmin, 3.714139988367086*xmin-22.279160996262704)--(xmax, 3.714139988367086*xmax-22.279160996262704), linewidth(0.4)); /* line */
draw((xmin, -0.022875816993464037*xmin + 4.942884673967238)--(xmax, -0.022875816993464037*xmax + 4.942884673967238), linewidth(0.4)); /* line */
draw(circle((2.8260109375406977,0.36816631196387156), 4.626464322853354), linewidth(0.4) + linetype("4 4") + ccqqqq); 
draw(circle((1.0994020408031704,2.073413701681334), 2.8435773234341077), linewidth(0.4) + qqwuqq); 
draw(circle((2.485131564734392,1.4001391867715074), 4.49699959634103), linewidth(0.4) + linetype("4 4") + qqwuqq); 
draw(circle((2.409784206793524,-1.8936167460721383), 3.0608005488760615), linewidth(0.4) + linetype("4 4")); 
draw((xmin, 0.3303181518892607*xmin-0.32802206931594974)--(xmax, 0.3303181518892607*xmax-0.32802206931594974), linewidth(0.4)); /* line */
draw((xmin, -0.022875816993464037*xmin-1.8384909635637896)--(xmax, -0.022875816993464037*xmax-1.8384909635637896), linewidth(0.4)); /* line */
draw((xmin, 1.2234639300582837*xmin + 3.491604070452697)--(xmax, 1.2234639300582837*xmax + 3.491604070452697), linewidth(0.4)); /* line */
draw((xmin, -0.8173514358959167*xmin + 0.07602383554985229)--(xmax, -0.8173514358959167*xmax + 0.07602383554985229), linewidth(0.4)); /* line */
draw((xmin, -5.468232946351302*xmin + 11.28364464711325)--(xmax, -5.468232946351302*xmax + 11.28364464711325), linewidth(0.4)); /* line */
draw(circle((2.3447520658676875,-4.736450335115873), 4.1778501641811765), linewidth(0.4) + linetype("4 4")); 
draw((-1.673635102853051,1.4439718900326037)--(3.965230973398127,4.852176775882967), linewidth(0.4)); 
draw((3.965230973398127,4.852176775882967)--(6.755206752008701,2.8106225310601456), linewidth(0.4)); 
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Let $L = SD \cap BC$ where $D$ is intersection of $(ABC)$ with line through $A$ parallel to $BC$. Notice that it suffices to prove that $Y-Z-L$ To prove this Notice the spiral similarity at $S :$ $\triangle SBL \sim \triangle SAC$ and $\triangle SKL \sim \triangle SEB \sim \triangle SFC \sim \triangle SAD$ then do some ratio chasing to get the desired conclusion.
This post has been edited 1 time. Last edited by math_comb01, Dec 20, 2023, 6:34 PM
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bin_sherlo
661 posts
#6 • 1 Y
Y by ehuseyinyigit
Let $Q'$ be the reflection of $A-$queue point over $BC$ which lies on $(DBCH)$. Since $H$ is the orthocenter of $\triangle DXY$, $Q'$ is $D-$queue point on $\triangle DXY$. $QD',Q'D,BC$ are concurrent and $XY,BC,Q'D$ concur thus, power of point at the concurrency point finishes.$\blacksquare$
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awesomeming327.
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#7
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Note that $\angle HBD =\angle HBC+\angle CBD=\angle HBC+\angle BCA=90^\circ$ and similarly $\angle HCD=90^\circ$. Thus, $H$ is the orthocenter of $\triangle DXY$.

Let $Q'$ be the Queue Point of $\triangle DXY$.

Claim 1: $Q$ and $Q'$ are reflections across $BC$.
Let $H'$ be the reflection of $H$ across $BC$. Then $\triangle H'BC\stackrel{+}{\sim}\triangle HXY$, so there is a spiral similarity from $Q$ taking $\triangle H'BC$ to $HXY$. Thus, since $Q'H$ bisects $XY$, $Q'H'$ must bisect $BC$. Therefore, if $Q''$ is the reflection of $Q'$ across $BC$ then $Q''$ lies on $(ABC)$ and also $HM$ where $M$ is the midpoint of $BC$, so $Q=Q''$, meaning that $Q'$ is the reflection of $Q$ across $BC$.
Note that by Miquel Point of a cyclic quadrilateral, $XY$, $BC$, and $Q'D$ concur. $QD'$ is the reflection of $Q'D$ across $BC$, so it is also concurrent with the other lines. By Radical Center we are done.
This post has been edited 1 time. Last edited by awesomeming327., Yesterday at 9:35 PM
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