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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Thursday at 11:16 PM
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0 replies
jlacosta
Thursday at 11:16 PM
0 replies
A very beautiful geo problem
TheMathBob   4
N 12 minutes ago by ravengsd
Source: Polish MO Finals P2 2023
Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$Prove that $YI$ is the angle bisector of $XYA$.
4 replies
TheMathBob
Mar 29, 2023
ravengsd
12 minutes ago
Inspired by old results
sqing   6
N 14 minutes ago by Jamalll
Source: Own
Let $ a,b>0 , a^2+b^2+ab+a+b=5 . $ Prove that
$$ \frac{ 1 }{a+b+ab+1}+\frac{6}{a^2+b^2+ab+1}\geq \frac{7}{4}$$$$ \frac{ 1 }{a+b+ab+1}+\frac{1}{a^2+b^2+ab+1}\geq \frac{1}{2}$$$$  \frac{41}{a+b+2}+\frac{ab}{a^3+b^3+2} \geq \frac{21}{2}$$
6 replies
sqing
Apr 29, 2025
Jamalll
14 minutes ago
A Duality Operation on Decreasing Integer Sequences
Ritangshu   0
23 minutes ago
Let \( S \) be the set of all sequences \( (a_1, a_2, \ldots) \) of non-negative integers such that
(i) \( a_1 \geq a_2 \geq \cdots \); and
(ii) there exists a positive integer \( N \) such that \( a_n = 0 \) for all \( n \geq N \).

Define the dual of the sequence \( (a_1, a_2, \ldots) \in S \) to be the sequence \( (b_1, b_2, \ldots) \), where, for \( m \geq 1 \),
\( b_m \) is the number of \( a_n \)'s which are greater than or equal to \( m \).

(i) Show that the dual of a sequence in \( S \) belongs to \( S \).

(ii) Show that the dual of the dual of a sequence in \( S \) is the original sequence itself.

(iii) Show that the duals of distinct sequences in \( S \) are distinct.
0 replies
Ritangshu
23 minutes ago
0 replies
Property of a function
Ritangshu   0
31 minutes ago
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

0 replies
Ritangshu
31 minutes ago
0 replies
Sum of digits is 18
Ecrin_eren   4
N 3 hours ago by maxamc
How many 5 digit numbers are there such that sum of its digits is 18
4 replies
Ecrin_eren
6 hours ago
maxamc
3 hours ago
Inequality
tom-nowy   0
4 hours ago
Let $0<a,b,c,<1$. Show that
$$ \frac{3(a+b+c)}{a+b+c+3abc} > \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} .$$
0 replies
tom-nowy
4 hours ago
0 replies
Logarithm of a product
axsolers_24   2
N 4 hours ago by axsolers_24
Let $x_1=97 ,$ $x_2=\frac{2}{x_1} ,$ $x_3=\frac{3}{x_2} ,$$... , $ $x_8=\frac{8}{x_7}$
then
$ \log_{3\sqrt{2}} \left(\prod_{i=1}^8 x_i-60\right)$
2 replies
axsolers_24
Today at 10:42 AM
axsolers_24
4 hours ago
Inequalities
sqing   1
N 4 hours ago by sqing
Let $ a,b>0 , a^2 + 2b^2 =  a + 2b $. Prove that $$\sqrt{\frac{a}{b( a+2)}} + \sqrt{\frac{b}{a(2b+1)}}  \geq \frac {2}{\sqrt{3}} $$Let $ a,b>0 , a^3 + 2b^3 =  a + 2b $. Prove that $$\sqrt[3]{\frac{a}{b( a+2)}} + \sqrt[3]{\frac{b}{a(2b+1)}}  \geq \frac {2}{\sqrt[3]{3}} $$
1 reply
sqing
5 hours ago
sqing
4 hours ago
Coprime sequence
Ecrin_eren   4
N 4 hours ago by Pal702004
"Let N be a natural number. Show that any two numbers from the following sequence are coprime:

2^1 + 1, 2^2 + 1, 2^4+ 1,2^8+1 ..., 2^(2^N )+ 1."
4 replies
Ecrin_eren
Thursday at 8:53 PM
Pal702004
4 hours ago
Hard Inequality
William_Mai   0
5 hours ago
Given $a, b, c \in \mathbb{R}$ such that $a^2 + b^2 + c^2 = 1$.
Find the minimum value of $P = ab + 2bc + 3ca$.

Source: Pham Le Van
0 replies
William_Mai
5 hours ago
0 replies
Find the minimum
Ecrin_eren   5
N 5 hours ago by Jackson0423
The polynomial is given by P(x) = x^4 + ax^3 + bx^2 + cx + d, and its roots are x1, x2, x3, x4. Additionally, it is stated that d ≥ 5.Find the minimum value of the product:

(x1^2 + 1)(x2^2 + 1)(x3^2 + 1)(x4^2 + 1).

5 replies
Ecrin_eren
Thursday at 9:03 PM
Jackson0423
5 hours ago
Inequalities
sqing   8
N 5 hours ago by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
8 replies
sqing
Jul 12, 2024
sqing
5 hours ago
Angle AEB
Ecrin_eren   2
N 6 hours ago by sunken rock
In triangle ABC, the lengths |AB|, |BC|, and |CA| are proportional to 4, 5, and 6, respectively. Points D and E lie on segment [BC] such that the angles ∠BAD, ∠DAE, and ∠EAC are all equal. What is the measure of angle ∠AEB in degrees?

2 replies
Ecrin_eren
Yesterday at 9:26 AM
sunken rock
6 hours ago
Where to check solutions??
math_gold_medalist28   0
Today at 12:42 PM
I'm studying MONT by aditya khurmi and pathfinder by vikash tiwari...but the problem is there isn't given the solutions means the ans. So how can I be sure that my ans is correct or not ?Please help!!!
0 replies
math_gold_medalist28
Today at 12:42 PM
0 replies
Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB
Valentin Vornicu   16
N Dec 22, 2013 by dizzy
Source: Kazakhstan international contest 2006, Problem 6
Let $ ABCDEF$ be a convex hexagon such that $ AD = BC + EF$, $ BE = AF + CD$, $ CF = DE + AB$. Prove that:
\[ \frac {AB}{DE} = \frac {CD}{AF} = \frac {EF}{BC}.
\]
16 replies
Valentin Vornicu
Jan 22, 2006
dizzy
Dec 22, 2013
Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB
G H J
Source: Kazakhstan international contest 2006, Problem 6
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Valentin Vornicu
7301 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ ABCDEF$ be a convex hexagon such that $ AD = BC + EF$, $ BE = AF + CD$, $ CF = DE + AB$. Prove that:
\[ \frac {AB}{DE} = \frac {CD}{AF} = \frac {EF}{BC}.
\]
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Sung-yoon Kim
324 posts
#2 • 2 Y
Y by Adventure10 and 1 other user
It can be proved by using vectors. Actually the same idea was used in IMO 2003 #3.
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perfect_radio
2607 posts
#3 • 5 Y
Y by BSJL, Adventure10, Crazy4Hitman, pokpokben, and 1 other user
I had a similar idea, but I was very DUMB not to see what happened if I went on this path.

I got the INEQUALITIES, but I thought that adding them would never give something useful:

$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$. (I don't know why, but this inequality seemed too natural for me to yield anything; really don't know why)

Squaring and adding: $\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF} \right)^2 \leq 0$. Thus, $BC \| EF$ etc. Also, $BC \| DA$. The conclusion is now obvious.

Thanks, Sung-yoon Kim.
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Sung-yoon Kim
324 posts
#4 • 2 Y
Y by Adventure10 and 1 other user
Yeah I think you're on the right way..
Quote:
$\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$
Consider a parellelogram BCFX. Then $BC+EF=XF+FE \geq XE=|\overrightarrow{XB}+\overrightarrow{BE}|=|\overrightarrow{CF}+\overrightarrow{BE}|$. The rest is same as yours.
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Tiks
1144 posts
#5 • 1 Y
Y by Adventure10
Guys I think we shoulld find sinthetical solution for this one,I havn't do it still :( ,but I will try again ;).
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Davron
484 posts
#6 • 2 Y
Y by Adventure10, Mango247
Yes it is a good idea to find a synthetic proof for this problem.

Davron
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jin
383 posts
#7 • 1 Y
Y by Adventure10
That's really hard to find a such solution.
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Tiks
1144 posts
#8 • 2 Y
Y by Adventure10, Mango247
Yes,it is realy hard job,if that can help you I know that ther are sintetic solution,I know a man who has found it,but I want find it from misalf,so I havn't ask him about that solution. :)
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barasawala
124 posts
#9 • 2 Y
Y by Adventure10, Mango247
perfect_radio wrote:
The conclusion is now obvious.

Well, after proving all these parallels, I still can't see why it's obvious...
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perfect_radio
2607 posts
#10 • 3 Y
Y by Adventure10, Crazy4Hitman, Mango247
Let $AD \cap CF = \left\{ T \right\}$. The following statements hold:
- $ABCT$ is a parallelogram;
- $DEFT$ is a parallelogram;
- $\triangle ATF \sim \triangle DTC$;
- the conclusion.
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maggot
72 posts
#11 • 1 Y
Y by Adventure10
perfect_radio wrote:
Also, $BC \| DA$.

And why is this?
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perfect_radio
2607 posts
#12 • 2 Y
Y by Adventure10 and 1 other user
$\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}\right)^{2}\leq 0$ implies $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow 0$. Also note that $\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow{CB}+\overrightarrow{EF}$.
Hence,
\[\overrightarrow{AD}= \overrightarrow{BC}+\overrightarrow{FE}.\]
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vittasko
1327 posts
#13 • 1 Y
Y by Adventure10
A simple construction of the configuration, as the problem states, is as follows:

A triangle $\bigtriangleup KLM$ is given and let $A$ be, a fixed point on the extension of the sideline $KL$ $($ $K,$ between $A,$ $L$ $).$

The line through the $A$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $B.$

The line through the $B$ and parallel to $KL,$ intersects the sideline $LM$ at a point, so be it $C.$

The line through the $C$ and parallel to $KM,$ intersects the sideline $KL$ at a point, so be it $D.$

The line through the $D$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $E.$

Through $A,$ $E$ now, we draw to lines parallel to $KM\equiv BE$ and $KL\equiv AD$ respectively, which intersect the sideline $LM,$ at points $F,$ $F'$ and we will prove that $F'\equiv F.$

It is easy to show that $CF = CM+MF = DE+AB$ $,(1)$ and $CF' = CL+LF' = AB+DE$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $CF' = CF$ $\Longrightarrow$ $F'\equiv F.$

So, it has already been constructed the configuration as the problem states.

$\bullet$ From $AB\parallel DE$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE}$ $,(3)$

But, $\frac{KA}{KD}= \frac{EF}{BC}$ $,(4)$ and $\frac{KB}{KE}= \frac{CD}{AF}$ $,(5)$

From $(3),)$ $(4),$ $(5)$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC}$ and the proof is completed.

Kostas Vittas.
Attachments:
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prowler
312 posts
#14 • 2 Y
Y by Adventure10, Mango247
Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$.



Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is
proved.


Now we will solve the problem.

Denote $AB=a,BC=b,CD=c, DE=d, EF=e, AF=f, AC=x_{1}, BD=x_{2}, CE=x_{3}, DF=x_{4}, EA=x_{5}, FB=x_{6}, AD=l_{1}, BE=l_{2}, CF=l_{3}$.

We have
$l_{1}=b+e, l_{2}= c+f, l_{3}=a+f$. Apply our theorem for $ABCD$ and
$DEFA$ we have $a^{2}+c^{2}+2bl_{1}\geq x_{1}^{2}+x_{2}^{2}$ and
$d^{2}+f^{2}+2el_{1}\geq x_{4}^{2}+x_{5}^{2}$. Adding all of them:
\[a^{2}+c^{2}+d^{2}+f^{2}+2l_{1}^{2}\geq x_{1}^{2}+x_{2}^{2}+x_{4}^{2}+x_{5}^{2}\]
Summing all such quadrilaterals we get
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})\geq 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}) \]
Next step is to apply our lemma for quadrilateral $BCEF$:
$x_{3}^{2}+x_{6}^{2}+2be\geq l_{2}^{2}+l_{3}^{2}$. Summing with inequalities for
$CDAF$, $DEAB$:
\[(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2})+2ad+2be+2cf\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \]
Multiplying second inequality by $2$ and adding with first one
we get
\[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+4(ad+be+cf)\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \]
or just $2(a+d)^{2}+2(b+e)^{2}+2(c+f)^{2}\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$

As we see the equality is hold and thus from lemma's case
of equality we get $BC \parallel AD \parallel EF$ and similarly other lines are
parallel. Let a line through $C$ parallel to $AB$ intersects $AD$ at
$P$, then $AP=b$ and thus $PD=e$ and $PD\parallel EF$. Hence $PDEF$ is
parallelogram and $C,P,F$ are collinear as are $A,P,D$. So triangles
$APF$ and $DPC$ are similar and from here
\[\frac{AP}{DP}=\frac{PF}{PC}=\frac{AF}{CD}\]
Rewriting in hexagon's sides:
\[\frac{BC}{EF}=\frac{DE}{AB}=\frac{AF}{CD}\]
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prowler
312 posts
#15 • 2 Y
Y by Adventure10, Mango247
This was problem was proposed for Mathematical Reflections 2 in 2006,
by Nairi Sedrakyan (Armenia).

Mathematical Reflections: http://www.awesomemath.org
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ISLAMIDIN
5 posts
#16 • 1 Y
Y by Adventure10
in a convex hexagon with AD=BC+EF, CF=DE+AB,BE=AF+CD.THERE WE CAN FIND EASILY.A simple construction of the configuration, as the problem states, is as follows:

A triangle \bigtriangleup KLM is given and let A be, a fixed point on the extension of the sideline KL ( K, between A, L ).

The line through the A and parallel to LM, intersects the sideline KM at a point, so be it B.

The line through the B and parallel to KL, intersects the sideline LM at a point, so be it C.

The line through the C and parallel to KM, intersects the sideline KL at a point, so be it D.

The line through the D and parallel to LM, intersects the sideline KM at a point, so be it E.

Through A, E now, we draw to lines parallel to KM\equiv BE and KL\equiv AD respectively, which intersect the sideline LM, at points F, F' and we will prove that F'\equiv F.

It is easy to show that CF = CM+MF = DE+AB ,(1) and CF' = CL+LF' = AB+DE ,(2)

From (1), (2) \Longrightarrow CF' = CF \Longrightarrow F'\equiv F.

So, it has already been constructed the configuration as the problem states.

\bullet From AB\parallel DE \Longrightarrow \frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE} ,(3)

But, \frac{KA}{KD}= \frac{EF}{BC} ,(4) and \frac{KB}{KE}= \frac{CD}{AF} ,(5)

From (3),) (4), (5) \Longrightarrow \frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC} and the proof is completed.
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dizzy
181 posts
#17 • 2 Y
Y by Adventure10, Mango247
prowler wrote:
Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote
$a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$.



Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from
triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will
remember Euler theorem for quadrilateral and midpoint of opposite
sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is
proved.

Sorry for reviving the old topic, but your lemma is not true, and therefore the whole solution as well. From Euler theorem we have not as you wrote $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}$ but $ a^{2}+b^{2}+c^{2}+d^{2}=e^{2}+f^{2}+4MN^{2} $. So your lemma rewrites as $ e^2+f^2+2bd\geq a^2+c^2 $. :wink:
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