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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
4 var inequality
sqing   0
4 minutes ago
Source: Own
Let $ a,b,c,d >0 $ and $ abcd=3,a+b+c+d=6 $. Prove that
$$ a^2+b^2+c^2+d^2  \leq 12 $$$$   a^3+b^3+c^3+d^3 \leq 30$$$$   a^4+b^4+c^4+d^4 \leq84$$
0 replies
1 viewing
sqing
4 minutes ago
0 replies
Dou Fang Geometry in Taiwan TST
Li4   1
N 17 minutes ago by Parsia--
Source: 2025 Taiwan TST Round 3 Mock P2
Let $\omega$ and $\Omega$ be the incircle and circumcircle of the acute triangle $ABC$, respectively. Draw a square $WXYZ$ so that all of its sides are tangent to $\omega$, and $X$, $Y$ are both on $BC$. Extend $AW$ and $AZ$, intersecting $\Omega$ at $P$ and $Q$, respectively. Prove that $PX$ and $QY$ intersects on $\Omega$.

Proposed by kyou46, Li4, Revolilol.
1 reply
Li4
5 hours ago
Parsia--
17 minutes ago
4 var inequality
sqing   1
N 37 minutes ago by sqing
Source: https://bbs.emath.ac.cn/thread-39778-1-1.html
Let $ a,b,c,d>0 $ and $ a+b+c+d=4. $ Prove that$$a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}\leq 2(1+\sqrt{abcd})$$Let $ a,b,c,d\geq -1 $ and $ a+b+c+d=0. $ Prove that$$ab+bc+cd\leq \frac{5}{4}$$
1 reply
1 viewing
sqing
44 minutes ago
sqing
37 minutes ago
NT Tourism
B1t   2
N 43 minutes ago by B1t
Source: Mongolian TST 2025 P2
Let $a, n$ be natural numbers such that
\[
\frac{a^n - 1}{(a - 1)^n + 1}
\]is a natural number.


1. Prove that $(a - 1)^n + 1$ is odd.
2. Let $q$ be a prime divisor of $(a - 1)^n + 1$.
Prove that
\[
    a^{(q - 1)/2} \equiv 1 \pmod{q}.
    \]3. Prove that if a is prime and $a \equiv 1 \pmod{4}$, then
\[
    2^{(a - 1)/2} \equiv 1 \pmod{a}.
    \]
2 replies
B1t
3 hours ago
B1t
43 minutes ago
No more topics!
Find an angle
Rushil   15
N Oct 30, 2024 by namanrobin08
Source: Indian RMO 1998 Problem 1
Let $ABCD$ be a convex quadrilateral in which $\angle BAC = 50^{\circ}, \angle CAD = 60^{\circ}$and $\angle BDC = 25^{\circ}$. If $E$ is the point of intersection of $AC$ and $BD$, find $\angle AEB$.
15 replies
Rushil
Oct 26, 2005
namanrobin08
Oct 30, 2024
Find an angle
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Source: Indian RMO 1998 Problem 1
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Rushil
1592 posts
#1 • 3 Y
Y by Biographicsmean, Adventure10, Mango247
Let $ABCD$ be a convex quadrilateral in which $\angle BAC = 50^{\circ}, \angle CAD = 60^{\circ}$and $\angle BDC = 25^{\circ}$. If $E$ is the point of intersection of $AC$ and $BD$, find $\angle AEB$.
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frt
1294 posts
#2 • 6 Y
Y by raltz, mihirb, opptoinfinity, Rushil_india, Adventure10, Mango247
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srulikbd
400 posts
#3 • 1 Y
Y by Adventure10
could you explain me please why does Since $\angle BDC=\frac{1}{2}\angle BAC$, $A$ is the circumcenter of $\triangle BCD$?
thanks :)
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D.P.L
223 posts
#4 • 2 Y
Y by Adventure10, Mango247
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aznluster
570 posts
#5 • 2 Y
Y by Adventure10, Mango247
frt wrote:
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Can you direct me to the proof of your first statement?
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geoistrivial
41 posts
#6 • 2 Y
Y by Excursion, Adventure10
aznluster wrote:
frt wrote:
Click to reveal hidden text

Can you direct me to the proof of your first statement?

It is actually a false statement. It is true that if a central angle and inscribed angle of a circle both cut the same arc, the central angle has twice the value of the inscribed angle. However, his statement is false because there are many other points that satisfy such constraints. Indeed, if you draw a diagram, A is clearly not the circumcenter.
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sunken rock
4384 posts
#7 • 1 Y
Y by Adventure10
Draw the circle $\odot (ABC)$, take $O$ the midpoint of its arc $BC$ containing $A$, draw the circle $(O, OB)$.
From $A$, draw a halfline $(Ax)$, directed to the halfplane not containing $B$, such as $\angle CAx=60^\circ$, it will intersect the second circle at $D$; see that for many $A$ on the first circle will find a $D$ on the second one, and the required angle is not constant.

Best regards,
sunken rock
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geoistrivial
41 posts
#8 • 2 Y
Y by Adventure10, Mango247
I believe this problem does not give enough information. Perhaps this is the real one: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=451779
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10000th User
3049 posts
#9 • 1 Y
Y by Adventure10
Surprisingly, it turns out that $A$ IS indeed the circumcenter of $\triangle BCD$ even with the additional information! :what?:

EDIT: Yes, the additional information is necessary because I was able to construct two different quadrilaterals with Rushil's conditions.
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Sayan
2130 posts
#10 • 1 Y
Y by Adventure10
No extra condition is necessary. A trigonometric approach shows that $\angle AEB = 95^\circ$
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coolcheetah157
139 posts
#11 • 1 Y
Y by Adventure10
@Sayan - what trig approach did you use?
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tst
118 posts
#12 • 2 Y
Y by Adventure10, Mango247
A would be the circumcentre because DC chord and BC chord subtend double angle at A....
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amar_04
1915 posts
#14 • 7 Y
Y by AlastorMoody, Path_to_Almighty, stranger_02, srijonrick, Bumblebee60, Adventure10, Mango247
Not a proper solution has been posted here, so I'm posting, sorry for bumping this thread.

Extend $CA$ to a point $K$ such that $2\angle BKD=\angle BAD=55^\circ$ and as $\angle BCD=125^\circ$. So, $BCDK$ is a cyclic quadrilateral. So, $\angle BKC=25^\circ\implies \angle KBA=25^\circ$. So, $KA=AB$. Similarly $KA=AD$. So, $AB=AD$. So,$\angle ADB=35^\circ\implies \angle AEB=95^\circ$.
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srijonrick
168 posts
#15
Y by
Sorry for bumping this thread again, but in the book An Excursion in Mathematics an extra condition is given: $\angle CBD=30^{\circ}$, which is not there in the question, mods kindly add it.
P.S:- The source (the book) is really authentic. :)
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Hexagon_6-
23 posts
#16
Y by
This is tough for early student , but legend first notice that angle BDC=1/2CAB >> A is centre now you feel easy
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namanrobin08
99 posts
#17
Y by
Please note that if the angle subtended by two points $P, Q$ at $X$ is half of that subtended by them at some other point $Y$, it does NOT imply that $X$ is the circumcentre of $ABY$.

But in the present question, it happens to be the circumcentre. This is NOT because that half angle thing, but becaue that half angle thing happens twice at the same point.
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