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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.
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[*]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!
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0 replies
jlacosta
Mar 2, 2025
0 replies
Problem 5
blug   1
N 11 minutes ago by WallyWalrus
Source: Polish Junior Math Olympiad Finals 2025
Each square on a 5×5 board contains an arrow pointing up, down, left, or right. Show that it is possible to remove exactly 20 arrows from this board so that no two of the remaining five arrows point to the same square.
1 reply
blug
Mar 15, 2025
WallyWalrus
11 minutes ago
Cool Number Theory
Fermat_Fanatic108   6
N 31 minutes ago by epl1
For an integer with 5 digits $n=abcde$ (where $a, b, c, d, e$ are the digits and $a\neq 0$) we define the \textit{permutation sum} as the value $$bcdea+cdeab+deabc+eabcd$$For example the permutation sum of 20253 is $$02532+25320+53202+32025=113079$$Let $m$ and $n$ be two fivedigit integers with the same permutation sum.
Prove that $m=n$.
6 replies
Fermat_Fanatic108
4 hours ago
epl1
31 minutes ago
Incenter geometry with parallel lines
nAalniaOMliO   1
N 39 minutes ago by LenaEnjoyer
Source: Belarusian MO 2023
Let $\omega$ be the incircle of triangle $ABC$. Line $l_b$ is parallel to side $AC$ and tangent to $\omega$. Line $l_c$ is parallel to side $BC$ and tangent to $\omega$. It turned out that the intersection point of $l_b$ and $l_c$ lies on circumcircle of $ABC$
Find all possible values of $\frac{AB+AC}{BC}$
1 reply
nAalniaOMliO
Apr 16, 2024
LenaEnjoyer
39 minutes ago
Function equation
Dynic   0
44 minutes ago
Find all function $f:\mathbb{Z}\to\mathbb{Z}$ satisfy all conditions below:
i) $f(n+1)>f(n)$ for all $n\in \mathbb{Z}$
ii) $f(-n)=-f(n)$ for all $n\in \mathbb{Z}$
iii) $f(a^3+b^3+c^3+d^3)=f^3(a)+f^3(b)+f^3(c)+f^3(d)$ for all $n\in \mathbb{Z}$
0 replies
Dynic
44 minutes ago
0 replies
No more topics!
Inequality => square
Rushil   12
N Mar 16, 2025 by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
Mar 16, 2025
Inequality => square
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Source: INMO 1998 Problem 4
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Rushil
1592 posts
#1 • 2 Y
Y by Adventure10, Mango247
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
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shobber
3498 posts
#2 • 2 Y
Y by Adventure10, Mango247
Is $f(x)=\sin{x}$ concave on $[0, \pi]$? If so, then this problem can be proved by $l=2R\sin{\theta}$ then AM-GM then jensen.
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Arne
3660 posts
#3 • 5 Y
Y by AKS_9_54_61, srijonrick, Adventure10, Mango247, and 1 other user
Hm... I'd say that by Ptolemy we have \[ 4 \geq AC \cdot BD = AB \cdot CD + AD \cdot BC \geq 2 \sqrt{AB \cdot BC \cdot CD \cdot DA} \geq 4 \] (since $AC$ and $AB$ are not longer than a diameter of the circle).

So equality must hold everywhere. Hence $AC$ and $BD$ are diameters, $ABCD$ is a rectangle, and also $AB \cdot CD = AD \cdot BC$ which implies that $AB^2 = AD^2$, so $ABCD$ is a square, and we're done.
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shobber
3498 posts
#4 • 2 Y
Y by Adventure10, Mango247
Oh...... I didn't think about using Ptolemy. Nice proof Arne.

Anyway, is my method correct?
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Arne
3660 posts
#5 • 2 Y
Y by Adventure10, Mango247
Yeah, I think so...

Could you write a full solution? Then it will be easier to judge :)
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shobber
3498 posts
#6 • 2 Y
Y by Adventure10, Mango247
OK.

Let $\angle{AOB}=2a$, $\angle{BOC}=2b$, $\angle{COD}=2c$, $\angle{DOA}=2d$. Then $a+b+c+d=180^o$.
Since we also have: $AB=2R \sin{a}$ and etc, Hence:

\[ AB \cdot BC \cdot CD \cdot DA=16R^2 \cdot \sin{a} \sin{b} \sin{c} \sin{d} \]
By AM-GM: $\sin{a} \sin{b} \sin{c} \sin{d} \leq (\dfrac{\sin{a}+\sin{b}+\sin{c}+\sin{d}}{4})^4$.

Then jensen: $\dfrac{\sin{a}+\sin{b}+\sin{c}+\sin{d}}{4} \leq \sin{(\dfrac{a+b+c+d}{4})}=\dfrac{\sqrt{2}}{2}$.

Thus $16R^2\cdot \sin{a} \sin{b} \sin{c} \sin{d} \leq 16 \cdot \frac14=4.$
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Arne
3660 posts
#7 • 2 Y
Y by Adventure10, Mango247
That looks fine to me! :)
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mathbuzz
803 posts
#8 • 1 Y
Y by Adventure10
using Ptolemy's theorem,we have $AB.DC+AD.BC=AC.BD\le4$ with equality iff $AC$ and $BD$ are diameters.
so , we have , by AM-GM ,$AB.BC.CD.AD\le (\frac{AB.DC+AD.BC}{2})^2 \le(4/2)^2=4$ with equality iff $AB.DC=AD.BC$
so , from the given condition in the problem , we must have , $AB.BC.CD.DA=4.$
so , from the equality conditions and simple geometry , it is obvious that ABCD is a square.
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SRIDEV
729 posts
#9 • 2 Y
Y by Adventure10, Mango247
Dear @shobber ,

You wrote AB = 2RSina
Thus AB.BC.CD.DA should = 16R^4. Sina.Sinb.Sinc.Sind

But you wrote
AB.BC.CD.DA = 16R^2. Sina.Sinb.Sinc.Sind

Please do clarify, how R^4 becomes R^2 ?
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Wizard_32
1566 posts
#10 • 2 Y
Y by Adventure10, Mango247
SRIDEV wrote:
Dear @shobber ,

You wrote AB = 2RSina
Thus AB.BC.CD.DA should = 16R^4. Sina.Sinb.Sinc.Sind

But you wrote
AB.BC.CD.DA = 16R^2. Sina.Sinb.Sinc.Sind

Please do clarify, how R^4 becomes R^2 ?
Don't forget that $R=1$ ;)
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mqoi_KOLA
54 posts
#12
Y by
some 20 years later :gleam:
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lksb
163 posts
#13
Y by
shobber wrote:
Is $f(x)=\sin{x}$ concave on $[0, \pi]$? If so, then this problem can be proved by $l=2R\sin{\theta}$ then AM-GM then jensen.

take $(\sin(x))''$
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ohiorizzler1434
709 posts
#14
Y by
Bro! I can help prove that sin(x) is concave! The derivative of sin(x) is cos(x)! The derivative of cos(x) is -sin(x)! But because -sin(x) is below 0 from 0 to pi, we know that sin(x) is concave!

We can also prove it geometrically! Consider sin(a) and sin(b), which are the heights formed from a point to the x-axis on the unit circle, for 0<=a,b <= 180. Now, a linear combination of sin(a) and sin(b) represents the line between (a,sin(a)) and (b,sin(b)) on the graph of sin(x). However, sin(c) for c between a,b has higher value than any point on the line as can be seen on the circle! Thus sin(x) is concave from 0 to pi! Now that's rizz!
This post has been edited 1 time. Last edited by ohiorizzler1434, Mar 16, 2025, 11:18 PM
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