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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
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0 replies
jwelsh
Jul 1, 2025
0 replies
Are all solutions normal ?
loup blanc   11
N 3 hours ago by GreenKeeper
This post is linked to this one
https://artofproblemsolving.com/community/c7t290f7h3608120_matrix_equation
Let $Z=\{A\in M_n(\mathbb{C}) ; (AA^*)^2=A^4\}$.
If $A\in Z$ is a normal matrix, then $A$ is unitarily similar to $diag(H_p,S_{n-p})$,
where $H$ is hermitian and $S$ is skew-hermitian.
But are there other solutions? In other words, is $A$ necessarily normal?
I don't know the answer.
11 replies
loup blanc
Jul 17, 2025
GreenKeeper
3 hours ago
Group Theory resources
JerryZYang   0
3 hours ago
Can someone give me some resources for group theory. ;)
0 replies
JerryZYang
3 hours ago
0 replies
Geometry / trigonometry
Vice8427   1
N 4 hours ago by vanstraelen
You have a triangle ABC, D and E are points in AC so that AD = DE = 2, and EC = 3, now with that points the following triangles are formed, triangle BDC and triangle BEC.
Now a, b and c are the angles on the opposite vertex of side BC on all three triangles (BAC, BDC and BEC) respectively (a is at vertex A, b at vertex D and c at vertex E).
Given that a+b+c= 90° or π\2
What is the lenght of BC?
The question is supposed to be doable with at most basic trigonometry. Not requiring more advanced or complex contents.
Sorry if I made any sort of grammar mistake or couldnt explain myself correctly, I'm a native spanish speaker.

Also, you can't upload images? I couldnt find a button or tool to upload the image of the problem.
1 reply
Vice8427
Jun 9, 2025
vanstraelen
4 hours ago
Axiomatic real numbers x^0
Safal   2
N 5 hours ago by Safal
Source: Discussion
Here is an interesting question for you all:

Assume $\mathbb{R}$ is an ordered field with all field axioms holding.

$\textbf{Question:}$ Suppose $x>0$ is a real number and $0\in\mathbb{R}$(Set of real numbers). Prove that if $x^0\in\mathbb{R}$. Then it (that is $x^0$) must be $1$.

Also show same thing is true if $x<0$ is a real number.

Proof

$\textbf{Question:}$ If $z\neq 0$ be any complex number and $0\in\mathbb{C}$. Show that if $z^0\in\mathbb{C}$ , then $z^0=1$.

$\textbf{Remark:}$ Order property is not true for all complex numbers.

Hint
2 replies
Safal
Today at 3:20 PM
Safal
5 hours ago
Inequality
Ecrin_eren   1
N 5 hours ago by mathprodigy2011


For positive real numbers a, b, c, prove that

[(a + 2b + 2c)(3a + 2b + 2c)(b + c)(2a + b + c)]
+
[(b + 2c + 2a)(3b + 2c + 2a)(c + a)(2b + c + a)]
+
[(c + 2a + 2b)(3c + 2a + 2b)(a + b)(2c + a + b)]
≥ 105/8




1 reply
Ecrin_eren
Today at 2:13 PM
mathprodigy2011
5 hours ago
Solution set of 2/x>3/3-x
EthanWYX2009   4
N 5 hours ago by mathprodigy2011
The solution set of the inequality \( \frac{2}{x} > \frac{3}{3-x} \) is ______.

Proposed by Baihao Lan, High School Attached to Northwest Normal University
4 replies
EthanWYX2009
Yesterday at 2:17 AM
mathprodigy2011
5 hours ago
D1054 : A measure porblem
Dattier   1
N Today at 5:59 PM by greenturtle3141
Source: les dattes à Dattier
$M=\bigcup\limits_{n\in\mathbb N^*} \{0,1\}^n$, for $m \in M$, $\overline m=\{mx : x\in \{0,1\}^{\mathbb N} \}$

Let $A \subset M$ with $\forall (a,b) \in A,a\neq b$, $\overline a \cap \overline b=\emptyset$.

Is it true that $\sum\limits_{a\in A} 2^{-|a|}\leq 1$ ?

PS : for $m \in M$, $|m|$ is the length of $m$, hence $|0011|=4$
1 reply
Dattier
Today at 4:36 PM
greenturtle3141
Today at 5:59 PM
Basic Inequalities Doubt
JetFire008   8
N Today at 5:14 PM by P0tat0b0y
If $x+y=1$, find the maximum value of $x^2+y^2=1$.
I saw a solution to this question where Titu's lemma was applied and the answer was $\frac{1}{2}$. But my doubt is can't we apply other inequality to get the maximum result? or did they us titu's lemma because the given information can fit only in this lemma?
8 replies
JetFire008
Jul 18, 2025
P0tat0b0y
Today at 5:14 PM
Nothing but a game
AlexCenteno2007   0
Today at 5:11 PM
Let ABCD be a trapeze with AD ∥ BC. M and N are the midpoints of CD and BC
respectively, and P is the common point of the lines AM and DN. If PM/AP = 4, show that
ABCD is a parallelogram.
0 replies
AlexCenteno2007
Today at 5:11 PM
0 replies
Inequality
Ecrin_eren   1
N Today at 4:39 PM by Ecrin_eren
For positive real numbers a, b, c satisfying
ab + ac + bc = 3abc, prove that

bc / (a⁴(b + c)) + ac / (b⁴(a + c)) + ab / (c⁴(a + b)) ≥ 3/2
1 reply
Ecrin_eren
Today at 2:16 PM
Ecrin_eren
Today at 4:39 PM
Inequalities
sqing   12
N Today at 3:28 PM by DAVROS
Let $ a,b,c,ab+bc+ca = 3. $Prove that$$\sqrt[3]{ \frac{1}{a} + 7b} + \sqrt[3]{\frac{1}{b} + 7c} + \sqrt[3]{\frac{1}{c} + 7a } \leq \frac{6}{abc}$$
12 replies
sqing
Jun 18, 2025
DAVROS
Today at 3:28 PM
Inequalities
sqing   17
N Today at 2:36 PM by Bbmr
Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $ and $ abc = -16 $. Prove that$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} -a^2\geq 2$$$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b}-a^2-bc\geq -2$$Equality holds when $a=-4,b=c=2.$
17 replies
1 viewing
sqing
Jul 9, 2025
Bbmr
Today at 2:36 PM
Function
Ecrin_eren   0
Today at 2:26 PM

Is there a function from non-negative real numbers to non-negative real numbers satisfying

f(f(x)) = |x - 1| for all x ≥ 0?











0 replies
Ecrin_eren
Today at 2:26 PM
0 replies
Minimum value
Ecrin_eren   0
Today at 2:08 PM


Let x,y,z be positive real numbers such that
xyz=8
What is the minimum value of the expression

yz / [x²(y + z)] + xz / [y²(x + z)] + xy / [z²(x + y)] ?





0 replies
Ecrin_eren
Today at 2:08 PM
0 replies
Putnam 2013 A5
Kent Merryfield   11
N Jul 7, 2025 by quantam13
For $m\ge 3,$ a list of $\binom m3$ real numbers $a_{ijk}$ $(1\le i<j<k\le m)$ is said to be area definite for $\mathbb{R}^n$ if the inequality \[\sum_{1\le i<j<k\le m}a_{ijk}\cdot\text{Area}(\triangle A_iA_jA_k)\ge0\] holds for every choice of $m$ points $A_1,\dots,A_m$ in $\mathbb{R}^n.$ For example, the list of four numbers $a_{123}=a_{124}=a_{134}=1, a_{234}=-1$ is area definite for $\mathbb{R}^2.$ Prove that if a list of $\binom m3$ numbers is area definite for $\mathbb{R}^2,$ then it is area definite for $\mathbb{R}^3.$
11 replies
Kent Merryfield
Dec 9, 2013
quantam13
Jul 7, 2025
Putnam 2013 A5
G H J
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Kent Merryfield
18574 posts
#1 • 2 Y
Y by Adventure10, Mango247
For $m\ge 3,$ a list of $\binom m3$ real numbers $a_{ijk}$ $(1\le i<j<k\le m)$ is said to be area definite for $\mathbb{R}^n$ if the inequality \[\sum_{1\le i<j<k\le m}a_{ijk}\cdot\text{Area}(\triangle A_iA_jA_k)\ge0\] holds for every choice of $m$ points $A_1,\dots,A_m$ in $\mathbb{R}^n.$ For example, the list of four numbers $a_{123}=a_{124}=a_{134}=1, a_{234}=-1$ is area definite for $\mathbb{R}^2.$ Prove that if a list of $\binom m3$ numbers is area definite for $\mathbb{R}^2,$ then it is area definite for $\mathbb{R}^3.$
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remo
18 posts
#2 • 3 Y
Y by programjames1, Adventure10, Mango247
Projecting to a random 2-dimentional subspace in $R^3$ multiplies the area of a triangle by some constant $c>0$ on average. (We do not need the value, just that by symmetry it is the same for all triangles.) Letting $S$ be the sum and $S'$ the sum after the random projection, we have $E(S')=c S$. If a list is area definite for $R^2$ then $S'\geq0$ and so is its expectation.

[Edit:] To clarify, the first claim is that $E area(P(A)) = c area(A)$ for a triangle $A$ in $R^3$, when $P$ is a random projection independent of $A$. Indeed, this holds for any surface $A$.
This post has been edited 1 time. Last edited by remo, Dec 9, 2013, 1:33 AM
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Kent Merryfield
18574 posts
#3 • 3 Y
Y by KarlMahlburg, Adventure10, Mango247
remo: I have a hard time believing that. What if the points are not coplanar so that the various triangles have various different orientations with respect to your random subspace?
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dnkywin
699 posts
#4 • 2 Y
Y by Adventure10, Mango247
@Kent - I think he means that the expected area when you project to a random plane is a constant time the area of the triangle, so it doesn't really matter what the orientation of any particular triangle is.
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KarlMahlburg
23 posts
#5 • 2 Y
Y by Adventure10, Mango247
But Kent's point is that once a given plane has been chosen, each triangle will have a different constant c. There is not one c that multiplies through as remo wrote: it is not true that $S = cS'$.
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Teki-Teki
553 posts
#6 • 2 Y
Y by Adventure10, Mango247
remo wrote:
... on average. (We do not need the value, just that by symmetry it is the same for all triangles.)
This is the key point of remo's solution. I'll try and explain it in a bit more detail, although the details of this solution are not mine.

Let $A_i$Let $a_{ijk}$ be any area-definite-for-$\mathbb{R}^2$ set of real numbers, and let $P$ be a plane. Let $B_i$ be the projections of the $A_i$ onto $P$. Then clearly the sum $\Sigma a_{ijk}\;\mathrm{area}(\bigtriangleup B_iB_jB_k)$ - which we will denote $\Sigma(P)$ - is non-negative for all planes $P$.

Now integrate over all planes $P$ passing through the origin. (This is just really just integrating over $S^2$, which is why we specify that the planes pass through the origin - we don't want to integrate over a non-compact set.) Then how are $\mathrm{area}(\bigtriangleup A_iA_jA_k)$, $\mathrm{area}(\bigtriangleup B_iB_jB_k)$, and the unit normal vector $\vec{n}$ to $P$ related? It's not hard to show, letting $\vec{v}_{ijk}$ be a unit normal vector to the plane of $A_iA_jA_k$, that \[\mathrm{area}(\bigtriangleup B_iB_jB_k) = |\vec{n}\cdot\vec{v}_{ijk}|\;\mathrm{area}(\bigtriangleup A_iA_jA_k).\] We sum over all $i, j, k$ with weights $a_{ijk}$ and integrate over all planes $P$. Then we get \[\int_{S^2} \Sigma(P) = \sum_{i,j,k} a_{ijk}\;\mathrm{area}(\bigtriangleup A_iA_jA_k) \int_{S^2} |\vec{n}\cdot\vec{v}_{ijk}|.\] What remo is arguing is that $\int_{S^2} |\vec{n}\cdot\vec{v}_{ijk}|$ is some absolute positive constant $C$ independent of $\vec{v}_{ijk}$, which it definitely is - the standard measure on $S^2$ is rotationally symmetric, and the integrand is non-negative, continuous in $\vec{n}$, and not always zero. Hence we have \[\sum_{i,j,k} a_{ijk}\;\mathrm{area}(A_iA_jA_k) = \frac{1}{C}\int_{S^2} \Sigma(P) \ge 0.\] So $a_{ijk}$ is area definite for $\mathbb{R}^3$ as desired.
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Ravi12346
22 posts
#7 • 1 Y
Y by Adventure10
Just to clarify the part of remo's solution that KarlMahlburg was asking about: given a plane, any particular projection scales areas of triangles (or any regions) in that plane by a constant factor, which depends only on the plane and the projection. So averaging over all projections scales areas by the average $c$ of all these constant factors, proving that $c$ depends only on the plane in which the triangle lies. Then, as has been pointed out, symmetry shows that $c$ doesn't even depend on this.

My solution was essentially the same, albeit probably a bit longer than it needed to be. Instead of just averaging over all possible projections, I averaged (equivalently) over all possible rotations of 3-space composed with the standard projection. This meant that I had to integrate over $SO(3)$ instead of $S^2$, which still works.
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darij grinberg
6555 posts
#8 • 2 Y
Y by Adventure10, Mango247
Neat -- a generalization of the averaging method to $2$-forms instead of vectors.

I assume triangles can be generalized to simplices without any complications?
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GeronimoStilton
1521 posts
#9
Y by
By Linearity of Expectation and the observation that the expected value of the area of triangle $A_iA_jA_k$ if projected in a random direction is nonzero, it suffices to show that the inequality holds for any projection of the points $A_i$, but this is trivial, so we are done.

To rigorize this, we can integrate over the area of the triangle projected onto the plane perpendicular to a vector $\vec v$ as it moves over unit vectors.
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yofro
3155 posts
#10 • 1 Y
Y by PRMOisTheHardestExam
Solved with RubiksCube3.1415

Like the above solutions, we project onto planes in all directions and integrate. First, consider what happens to the area of $\triangle A_iA_jA_k$ when projected onto a plane with normal vector $\vec{n}$. There's actually a known formula for this quantity (known as projected area), but we present a proof anyway. Let $\vec{u}$ be $\vec{A_j}-\vec{A_i}$ and $\vec{v}=\vec{A_k}-\vec{A_i}$. Then $[\triangle A_iA_jA_k]=\frac{1}{2}|\vec{u}\times \vec{v}|$. By standard formulas, the projection of $\vec{u}$ onto $\vec{n}$ is $\vec{n}(\vec{u}\cdot \vec{n})$ and hence the projection of $\vec{u}$ onto the plane with normal vector $\vec{n}$ is $\vec{u}-\vec{n}(\vec{u}\cdot \vec{n})$. Thus the projected area is
$$\frac{1}{2}|(\vec{u}-\vec{n}(\vec{u}\cdot \vec{n}))\times (\vec{v}-\vec{n}(\vec{v}\cdot \vec{n}))|$$A quick expansion using the identity $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$ gives that the projected area is
$$\frac{1}{2}|((\vec{u}\times \vec{v})\vec{n})\cdot \vec{n}|=\frac{1}{2}|\vec{u}\times \vec{v}|\cos(\theta_{\text{planes}})$$Where $\theta_{\text{planes}}$ is the angle between the plane containing $A_i, A_j, A_k$ and the plane we're projecting onto. Switching from unsigned area to signed area this becomes $|\cos\theta_{\text{planes}}|$ For a fixed choice of $i,j,k$ we integrate over planes in every direction (specifically, tangency planes to the unit sphere at every point):
$$\iint\limits_{\text{unit sphere}}[\triangle A_iA_jA_k]|\cos\theta|\mathrm{d}A=[\triangle A_iA_jA_k]\iint\limits_{\text{unit sphere}}|\cos\theta|\mathrm{d}A$$Where $\theta$ is the angle between the plane tangent to the piece of area $\mathrm{d}A$ on the unit sphere and the plane of $\triangle A_i A_j A_k$. By symmetry, for all $i,j,k$, we have for some constant $C>0$,
$$\iint\limits_{\text{unit sphere}}|\cos\theta|\mathrm{d}A=C$$Therefore, we have that
$$\sum_{i,j,k}a_{ijk}[\triangle A_i A_j A_k]=\sum_{i,j,k}a_{i,j,k}\frac{\iint_{\text{unit sphere}}A_{\text{projected}}\mathrm{d}A}{C}$$$$=\frac{1}{C}\iint_{\text{unit sphere}}\left(\sum_{i,j,k}a_{i,j,k}A_{\text{projected}}\right)\mathrm{d}A\ge 0$$Because $\{a_{ijk}\}$ is area definite for each individual plane.
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blackbluecar
307 posts
#11
Y by
Suppose $a_{ijk}$ $(1\le i<j<k\le m)$ is area definite and $\mathcal{A} = \{ A_1, A_2, \ldots, A_n \} \subseteq \mathbb{R}^3$. It is well know that if one takes $S \subseteq \mathbb{R}^2$ with finite area $a$ and randomly orient it in space and project onto a fixed plane, then the expected area of the projection is $a/2$. So, if one randomly orients $\mathcal{A}$ in space, then project it onto a fixed plane yielding a set $\mathcal{A}' = \{A_1', A_2', \ldots, A_n' \}$ then
\[ 0 \le \mathbb{E} \left [ \sum_{i < j < k} a_{ijk} \cdot \text{Area}(\triangle A_i'A_j'A_k')  \right ] = 
\sum_{i < j < k} a_{ijk} \cdot \mathbb{E} [ \text{Area}(\triangle A_i'A_j'A_k') ] = \frac 12 \sum_{i<j<k} a_{ijk} \cdot \text{Area}(\triangle A_iA_jA_k) \]\[ \implies \sum_{i<j<k} a_{ijk} \cdot \text{Area}(\triangle A_iA_jA_k) \ge 0 \]implying its area definite.
This post has been edited 1 time. Last edited by blackbluecar, Jun 2, 2025, 10:03 PM
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quantam13
161 posts
#12
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Good problem! The key idea is that if a list of reals is not area definite for $\mathbb{R}^3$, then we take a counterexample in $\mathbb{R}^3$ and project it to a random orientation of $\mathbb{R}^2$ in $\mathbb{R}^3$ and notice that there is an absolute constant $c$ such that the expected area of a projection from some triangle in $\mathbb{R}^3$ to $\mathbb{R}^2$ is $c$ times the area of the original triangle, so its not so hard to see from here we can get some counterexample in $\mathbb{R}^2$, so the list of reals is not area definite for $\mathbb{R}^3$.
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