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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
BMO 2015 #1: Inequality on a,b,c.
MathKnight16   25
N 9 minutes ago by Rayvhs
Source: BMO 2015 problem 1
If ${a, b}$ and $c$ are positive real numbers, prove that

\begin{align*}
 a ^ 3b ^ 6 + b ^ 3c ^ 6 + c ^ 3a ^ 6 + 3a ^ 3b ^ 3c ^ 3 &\ge{ abc \left (a ^ 3b ^ 3 + b ^ 3c ^ 3 + c ^ 3a ^ 3 \right) + a ^ 2b ^ 2c ^ 2 \left (a ^ 3 + b ^ 3 + c ^ 3 \right)}.
\end{align*}

(Montenegro).
25 replies
MathKnight16
May 5, 2015
Rayvhs
9 minutes ago
4 lines concurrent
Zavyk09   5
N 11 minutes ago by tomsuhapbia
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 $OH$.
5 replies
Zavyk09
Apr 9, 2025
tomsuhapbia
11 minutes ago
SL 2015 G1: Prove that IJ=AH
Problem_Penetrator   136
N 41 minutes ago by Mathgloggers
Source: IMO 2015 Shortlist, G1
Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.
136 replies
Problem_Penetrator
Jul 7, 2016
Mathgloggers
41 minutes ago
4 variables with quadrilateral sides 2
mihaig   2
N 44 minutes ago by mihaig
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
2 replies
mihaig
Tuesday at 8:47 PM
mihaig
44 minutes ago
(14n+25)/(2n+1) 'is a perfect square - Portugal OPM 2017 p1
parmenides51   4
N Today at 10:03 AM by Namisgood
Determine all integer values of n for which the number $\frac{14n+25}{2n+1}$ 'is a perfect square.
4 replies
parmenides51
May 15, 2024
Namisgood
Today at 10:03 AM
Inequalities
sqing   4
N Today at 9:46 AM by sqing
Let $ a,b,c>0 . $ Prove that
$$ \left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq 4\left(\frac{a+b}{b+c}+ \frac{b+c}{a+b}\right)$$$$ \left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq \frac{32}{9}\left(\frac{a+b}{b+c}+ \frac{c+a}{a+b}\right)$$$$ \left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq  \frac{8}{3}\left(  \frac{a+b}{b+c}+ \frac{b+c}{c+a}+ \frac{c+a}{a+b}\right)$$$$ \left(1 +\frac{a^2}{b^2}\right)\left(1+\frac{b^2}{c^2}\right)\left(1+\frac{c^2}{a^2}\right )\geq \frac{8}{3}\left(  \frac{a^2+bc}{b^2+ca}+\frac{b^2+ca}{c^2+ab}+\frac{c^2+ab}{a^2+bc}\right)$$
4 replies
sqing
Today at 12:20 AM
sqing
Today at 9:46 AM
Inequalities
sqing   15
N Today at 9:28 AM by sqing
Let $ a,b>0  $ and $ a+ b^2=\frac{3}{4} $.Prove that
$$  \frac{1}{a^3(a+b)} + \frac{2}{b^3(2b+1)} + \frac{16}{2a+1}    \geq 24$$Let $ a,b>0  $ and $a^2+b^2=\frac{1}{2} $.Prove that
$$   \frac{1}{a^3(a+b)} + \frac{2}{b^3(2b+1)} + \frac{16}{2a+1}    \geq 24$$
15 replies
sqing
Nov 29, 2024
sqing
Today at 9:28 AM
Inequalities
sqing   6
N Today at 8:00 AM by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
6 replies
sqing
Jul 12, 2024
sqing
Today at 8:00 AM
(3x-1)^2/x+(3y-1)^2/y >=1, for x,y>0, x+y=1 Austria Beginners' 2010 p3
parmenides51   22
N Today at 6:38 AM by justaguy_69
Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that
$$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds?

(K. Czakler, GRG 21, Vienna)
22 replies
parmenides51
Oct 3, 2021
justaguy_69
Today at 6:38 AM
New but easy
ZETA_in_olympiad   2
N Today at 6:26 AM by jasperE3
Find all functions $f:\mathbb R\to \mathbb R$ such that $$f(f(x)+f(y))=xf(y)+yf(x)$$for all $x,y\in \mathbb R.$
2 replies
ZETA_in_olympiad
Oct 1, 2022
jasperE3
Today at 6:26 AM
2025 CMIMC team p7, rephrased
scannose   4
N Today at 6:21 AM by scannose
In the expansion of $(x^2 + x + 1)^{2024}$, find the number of terms with coefficient divisible by $3$.
4 replies
scannose
Apr 18, 2025
scannose
Today at 6:21 AM
DA + AE = KC +CM = AB=BC=CA - All-Russian MO 1997 Regional (R4) 8.3
parmenides51   2
N Today at 5:47 AM by sunken rock
On sides $AB$ and $BC$ of an equilateral triangle $ABC$ are taken points $D$ and $K$, and on the side $AC$ , points $E$ and $M$ so that $DA + AE = KC +CM = AB$. Prove that the angle between lines $DM$ and $KE$ is equal to $60^o$.
2 replies
parmenides51
Sep 23, 2024
sunken rock
Today at 5:47 AM
Frankenstein FE
NamelyOrange   3
N Today at 3:53 AM by jasperE3
[quote = My own problem]Solve the FE $f(x)+f(-x)=2f(x^2)$ over $\mathbb{R}$. Ignore "pathological" solutions.[/quote]

How do I solve this? I made this while messing around, and I have no clue as to what to do...
3 replies
NamelyOrange
Jul 19, 2024
jasperE3
Today at 3:53 AM
Find the domain and range of $f(x)=\frac{1}{1-2\cos x}.$
Vulch   1
N Today at 2:22 AM by aidan0626
Find the domain and range of $f(x)=\frac{1}{1-2\cos x}.$
1 reply
Vulch
Today at 2:06 AM
aidan0626
Today at 2:22 AM
Polynomial approximation and intersections
egxa   2
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.6
What is the smallest value of \( k \) such that for any polynomial \( f(x) \) of degree $100$ with real coefficients, there exists a polynomial \( g(x) \) of degree at most \( k \) with real coefficients such that the graphs of \( y = f(x) \) and \( y = g(x) \) intersect at exactly $100$ points?
2 replies
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
Polynomial approximation and intersections
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Source: All Russian 2025 10.6
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egxa
209 posts
#1
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What is the smallest value of \( k \) such that for any polynomial \( f(x) \) of degree $100$ with real coefficients, there exists a polynomial \( g(x) \) of degree at most \( k \) with real coefficients such that the graphs of \( y = f(x) \) and \( y = g(x) \) intersect at exactly $100$ points?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
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MathLuis
1519 posts
#2
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Recall the well known analysis fact that if a polynomial has exactly $n$ real roots counting multiplicity then its derivative has exactly $n-1$ real roots counting multiplicity. So if $k \le 97$ the considering the $98$-th derivative of $f-g$ gives that for any three given reals $a,b,c$ we must have that $ax^2+bx+c$ always has two real root if $f-g$ had exactly $100$ different roots, thus a contradiction by picking suitable $a,b,c$ (my example would just be $a,c>0$ and $b=0$ lol).
Now we will show $k=98$ works, first clean all the polynomial of degree $98$ inside $f$ using $g$ and from there many a suitable pick, in fact $100$ does not matter so we can prove a slightly stronger claim using induction.
We show that it is possible to construct a polynomial of degree $n$ where only coefficients $x^{n}, x^{n-1}$ are not in our control, so that the polynomial has all real roots with no multiplicity and none of them are zero.
For $n=2$ we can obviously shift the parabola by a constant acording to what is needed due to discriminant criteria so we should be fine here, due to continuity in fact you can just avoid a root being zero or the resulting parabola being tangent to the $x$ axis, for $n=3$ you have something unfree of the form $x^2(ax+b)$ and so you pick some $c(ax+b)+d$ on the free part where $c,d$ have freedom acording to needs like if $b=0$ then $d \ne 0$ and could just be suficiently small given we put $c<0$ to have two roots and that in order to not have repeated roots or the cubic being tangent to the $x$ axis at any moment (continuity yay).
For $n=4$ you basically have control over a depressed cubic upon doing a similar setup as seen above just shift everything by a small constant at the end to satisfy all the given conditions needed and now we will show this is the kind of work you have to do.
Suppose it was true for $n=\ell$ then we prove it for $n=\ell+1$, notice the not free term is $x^{\ell}(ax+b)$ so we just use the free term to focusing on a polynomial of the kind $x^{\ell}+q(x)$ for $\deg q \le \ell-2$, notice here it might happen that $-b \cdot a^{-1}$ is a root twice then we just shift by a very small constant as needed whether is above or below, one of them should work by continuity and IVT, since have $0$ as a root is only a one thing thing on a small interval we can also use continuity to avoid this and still satisfy the no double roots or more condition, hence by using the inductive hypothesis this is also completed and thus our claim is proven.
Just throw the claim and use a suitable $g$ to win, and thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 19, 2025, 3:49 AM
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iliya8788
8 posts
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We claim that the answer is $\boxed{k=98}$.
Claim 1: $k>97$
Assume the contrary. there exists a polynomial $g(x)$ with degree less than $98$ such that $f(x)-g(x)$ has 100 roots for $f(x)=x^100$. Define $k$ as the degree of $g(x)$
Define a_{1},...,a_{k} such that $f(x)-g(x)=x^{100}+a_{k}x^k+...+a_{0}$. Define $r_{1},...,r_{100}$ as the real roots of $f(x)-g(x)$.
By vieta's formula since the coefficient of $x^{99}$ and $x^{98}$ is equal to $0$: $\sum_{i=1}^{100}r_{i} = 0$ and $\sum_{i<j}^{}r_{i}r_{j} = 0 \implies \sum_{i=1}^{100}r_{i}^2 = 0 \implies r_{i}=0 \implies f(x)-g(x)=0$ which is a contradiction. It follows that $k>97$.
Claim 2: $k=98$
We can basically alter all the coefficients of $f(x)-g(x)$ except the coefficients of $x^{100}$ and $x^{99}$ so by vieta's formula our only restriction is the sum of the roots of the polynomial. So we just need to pick 100 arbitrary pairwise different real numbers such that their sum is equal to $-\frac{a_{99}}{a_{100}}$ with $a_{99}$ being the coefficient of $x^{99}$ and $a_{100}$ being the leading coefficient. Obviously this is possible. From here we just alter the other coefficients so that each one of the coefficients becomes equal to the coefficients of the polynomial that has all these $100$ numbers as roots multiplied by $a_{n}$ and so we are done. $\blacksquare$.
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