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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
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Calculus, sets
wl8418   0
8 minutes ago
Is empty set a proper set of an non empty set? Why or why not? Any clarification or insight is appreciated. Thanks in advance!
0 replies
wl8418
8 minutes ago
0 replies
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Find r_1^2 + r_2^2 + r_3^2
BlackOctopus23   3
N 2 hours ago by Li0nking
Let $r_1$, $r_2$, and $r_3$ be the roots of $3x^3 - 8x^2 + 4x - 13$. Find $r_1^2 + r_2^2 + r_3^2$Solution
3 replies
BlackOctopus23
Yesterday at 1:50 AM
Li0nking
2 hours ago
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Pls help I'm stuck
UtkarshSri   2
N 2 hours ago by Li0nking
So I was solving this problem
Cbrt(x+1) - cbrt(x-1) = pow((x²-1),1/6)


Here, the square root is defined in the domain x≤-1 OR x≥1
I substituted a= cbrt(x+1) and b= cbrt(x-1)
And then the equation turned out to be
a-b= sqrt(ab)
Here, since the RHS is +ve, the LHS must be too, so a>b (Done so as to prevent extraneous roots)
Now, I can write the domain in terms of a and b as ab≥0
Squaring, we get a²+b²-3ab=0
On solving we get a/b = (3±sqrt(5))/2
Now, we know that a>b from the above condition, so the only root is a/b= (3+sqrt(5))/2 and on substituting for x, and applying componendo and dividendo after cubing, I get x= sqrt(5)/2

Now, my question is, since the equation is even i.e f(x)=f(-x), it's symmetric about the y-axis, so x= -sqrt(5)/2 must also satisfy the equation, and indeed it does, so where did I made some irreversible step that caused this loss of the root?
when I plugged the equation to wolframalpha, it showed that the "principle root" has the former solⁿ only where's for a "real valued root", it has both.I also found that on solving the equation
b-a= sqrt(ab), I get the latter root.But I mean where did that extra -ve sign come from?
I'm familiar that f(x)=sqrt(g(x)) is equivalent to f²(x)=g(x) AND f(x)≥0 AND g(x)≥0 but the g(x)≥0 condition is not necessary as f²(x) is already ≥0
and also that abs(f(x))=abs(g(x)) is equivalent to f²(x)=g²(x)

PS: Pardon me I don't know latex well, I'm tried to make this as comprehensive as possible and made sure there's no loss of equivalence in any step.
I hope someone clarifies this for me, it's really bothering me for a while.
2 replies
UtkarshSri
Yesterday at 9:57 AM
Li0nking
2 hours ago
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Geometry
MathsII-enjoy   1
N 3 hours ago by MathsII-enjoy
Given triangle $ABC$ inscribed in $(O)$, $M, N$ are respectively the midpoints of the major and minor arcs $BC$. Let $I$ be the center of the inscribed circle, $R$ be $A-mix$, $D$ is the intersection point of the line through $A$ parallel to $BC$ with $(O)$. $DI$ intersects $AR$ at $K$, take $L$ on $AK$ so that $LI//BC$. $NL$ intersects $(O)$ at $G$, $AG$ intersects $LI$ at $Z$. Prove that: $ZM$ is perpendicular to $KN$.
1 reply
MathsII-enjoy
May 23, 2025
MathsII-enjoy
3 hours ago
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Inequalities
sqing   0
4 hours ago
Let $a,b,c,d>0,a^2 + d^2-ad = (b + c)^2 $ aand $ a^2 + b^2 = c^2 + d^2.$ Prove that$$ \frac{ab+cd}{ad+bc} \geq \frac{ 4}{5}$$
0 replies
sqing
4 hours ago
0 replies
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Largest Prime Factor
P162008   3
N 4 hours ago by maromex
The largest prime factor of the sum $\sum_{k=1}^{11} k^5$ is $\lambda.$ Find the sum of the digits of $\lambda.$
3 replies
P162008
Yesterday at 12:04 AM
maromex
4 hours ago
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Inequalities
sqing   27
N 4 hours ago by sqing
Let $ a,b>0 $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1} \leq \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1} \leq \frac{2}{5} $$
27 replies
sqing
May 13, 2025
sqing
4 hours ago
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Divisors of factorials can't be always products of consecutive integers
Johann Peter Dirichlet   0
5 hours ago
Let $M$ an even number.

Show that $\frac{n!}{M^2}$ is not the product of consecutive integers for infinitely many naturals $n$.
0 replies
Johann Peter Dirichlet
5 hours ago
0 replies
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IOQM P22 2024
SomeonecoolLovesMaths   3
N Yesterday at 10:51 PM by SomeonecoolLovesMaths
In a triangle $ABC$, $\angle BAC = 90^{\circ}$. Let $D$ be the point on $BC$ such that $AB + BD = AC + CD$. Suppose $BD : DC = 2:1$. if $\frac{AC}{AB} = \frac{m + \sqrt{p}}{n}$, Where $m,n$ are relatively prime positive integers and $p$ is a prime number, determine the value of $m+n+p$.
3 replies
SomeonecoolLovesMaths
Sep 8, 2024
SomeonecoolLovesMaths
Yesterday at 10:51 PM
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AP calc?
Thayaden   30
N Yesterday at 9:53 PM by Pengu14
How are we all feeling on AP calc guys?
30 replies
Thayaden
May 20, 2025
Pengu14
Yesterday at 9:53 PM
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Calculate the radius of a circle using sidelengths.
richminer   0
Yesterday at 6:17 PM
Given triangle ABC with incircle (I), with D being the touchpoint of (I) and BC. Let M be the tangent point of the A-Mixtilinear circle (internally tangent). A' is the reflection of A through I. Calculate the radius of the circle (MDA') using the side lengths of the triangle ABC.
0 replies
richminer
Yesterday at 6:17 PM
0 replies
J
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Number of real roots
girishpimoli   0
Yesterday at 5:35 PM
Number of real roots of

$\displaystyle 2\sin(\theta)\cos(3\theta)\sin(5\theta)=-1$
0 replies
girishpimoli
Yesterday at 5:35 PM
0 replies
J
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Factorization Ex.28a Q30
Obvious_Wind_1690   1
N Yesterday at 4:43 PM by Lankou
Please help with factorization. Given is the question


\begin{align*} a(a+1)x^2+(a+b)xy-b(b-1)y^2\\ \end{align*}
And the given answer is


\begin{align*} [(a+1)x-(b-1)y][ax+by]\\ \end{align*}
But I am unable to reach the answer.
1 reply
Obvious_Wind_1690
Yesterday at 4:17 AM
Lankou
Yesterday at 4:43 PM
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Polynomials
P162008   4
N Yesterday at 4:19 PM by HAL9000sk
If $f(x)$ is a polynomial function such that $f(x) = x\sqrt{1 + (x + 1)\sqrt{1 + (x + 2)\sqrt{1 + (x + 3)\sqrt{1 + \cdots}}}}$ then

A) Degree of $f(x)$ must be greater than $2$

B) $f(-2) = 0$

C) $\sum_{r=1}^{5} \frac{1}{f(r)} = \frac{25}{42}$

D) $\sum_{r=1}^{n} \frac{1}{f(r)} = \frac{n(3n + 5)}{4(n+1)(n+2)}$
4 replies
P162008
Monday at 11:18 PM
HAL9000sk
Yesterday at 4:19 PM
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concurrency in hexagon, <A=<B,<C=<D,<E =<F (2018 AESC MSU Internet Olympiad 7.3)
parmenides51   0
Feb 20, 2021
In a convex hexagon $ABCDEF$, all angles of which are obtuse, $\angle A = \angle B$, $\angle C = \angle D$, $\angle E =\angle F$. Prove that the perpendicular bisectors of its sides $AB, CD, EF$ intersect at one point.
0 replies
parmenides51
Feb 20, 2021
0 replies
J
concurrency in hexagon, <A=<B,<C=<D,<E =<F (2018 AESC MSU Internet Olympiad 7.3)
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parmenides51
#1 h  Feb 20, 2021, 7:27 AM
Y by
In a convex hexagon $ABCDEF$, all angles of which are obtuse, $\angle A = \angle B$, $\angle C = \angle D$, $\angle E =\angle F$. Prove that the perpendicular bisectors of its sides $AB, CD, EF$ intersect at one point.
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