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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
IMO 2010 Problem 6
mavropnevma   42
N 10 minutes ago by awesomeming327.
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\]

Proposed by Morteza Saghafiyan, Iran
42 replies
mavropnevma
Jul 8, 2010
awesomeming327.
10 minutes ago
PJ // AC iff BC^2 = AC· QC
parmenides51   1
N an hour ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1998 OMM P5
The tangents at points $B$ and $C$ on a given circle meet at point $A$. Let $Q$ be a point on segment $AC$ and let $BQ$ meet the circle again at $P$. The line through $Q $ parallel to $AB$ intersects $BC$ at $J$. Prove that $PJ$ is parallel to $AC$ if and only if $BC^2 = AC\cdot QC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
an hour ago
Self-evident inequality trick
Lukaluce   10
N an hour ago by ytChen
Source: 2025 Junior Macedonian Mathematical Olympiad P4
Let $x, y$, and $z$ be positive real numbers, such that $x^2 + y^2 + z^2 = 3$. Prove the inequality
\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]When does the equality hold?
10 replies
Lukaluce
Sunday at 3:34 PM
ytChen
an hour ago
Power Of Factorials
Kassuno   181
N 2 hours ago by SomeonecoolLovesMaths
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
181 replies
Kassuno
Jul 17, 2019
SomeonecoolLovesMaths
2 hours ago
No more topics!
Find relation in triangle
Rushil   19
N Aug 6, 2023 by Krishijivi
Source: INMO 1992 Problem 1
In a triangle $ABC$, $\angle A = 2 \cdot \angle B$. Prove that $a^2 = b (b+c)$.
19 replies
Rushil
Oct 3, 2005
Krishijivi
Aug 6, 2023
Find relation in triangle
G H J
Source: INMO 1992 Problem 1
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Rushil
1592 posts
#1 • 2 Y
Y by Adventure10, Mango247
In a triangle $ABC$, $\angle A = 2 \cdot \angle B$. Prove that $a^2 = b (b+c)$.
This post has been edited 1 time. Last edited by Rushil, Oct 4, 2005, 5:39 AM
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shobber
3498 posts
#2 • 2 Y
Y by Adventure10, Mango247
Rushil wrote:
INMO 1992 Problem 1

In a triangle $ABC$ , $\angle A = 2 \times \angle B$. Prove that $a^2 = b (b+c)$
Try to search for Double Angle Theorem.
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Andreas
578 posts
#3 • 3 Y
Y by AmitayasB, Adventure10, Mango247
$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}$ $\Longrightarrow$ $c = b(4\cos^2 x - 1)$ and $a = 2b\cos x$.
$b(b + c) = 4b^2\cos^2 x = a^2$
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Ezbakhe Yassin
146 posts
#4 • 1 Y
Y by Adventure10
Andreas wrote:
$\frac{a}{\sin 2x} = \frac{b}{\sin x} = \frac{c}{\sin 3x}$

How did you get this one?
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mathmanman
1444 posts
#5 • 1 Y
Y by Adventure10
That's just the law of sines ;)
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Elemennop
1421 posts
#6 • 1 Y
Y by Adventure10
$\sin{C}=\sin{(180-A-B)}=\sin{(180-3x)}=\sin{(3x)}$, Because $\sin{x}=\sin{(180-x)}$.
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The QuattoMaster 6000
1184 posts
#7 • 1 Y
Y by Adventure10
Rushil wrote:
In a triangle $ ABC$, $ \angle A = 2 \cdot \angle B$. Prove that $ a^2 = b (b + c)$.
Sorry for awakening an old topic, but here's a neat solution that no-one here mentioned already:
Solution
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earth
99 posts
#8 • 3 Y
Y by Samujjal101, Adventure10, Mango247
Hi,here is a good one ! please try out! :)
Click to reveal hidden text
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Mateescu Constantin
1842 posts
#9 • 2 Y
Y by Adventure10, Mango247
Another solution
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Dranzer
154 posts
#10 • 5 Y
Y by SHREYAS333, Wizard_32, PME2018, Adventure10, and 1 other user
A pure geometric solution
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OmkarDivekar
2 posts
#11 • 3 Y
Y by sayanjoddar, Adventure10, Mango247
Sorry for reviving a very old thread !
We have to prove a^2=b(b+c)
Rearranging the terms we get a/b=(b+c)/a
L.H.S:-
By sine rule, a/b=sinA/sinB=2cosB ...(A=2B)
R.H.S:-
By sine rule, (b+c)/a=(sinB+sinC)/sinA=(2sin2B*cosB)/sin2B=2cosB ...(A=2B)
Hence, L.H.S=R.H.S proved.
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PhysicsMonster_01
1445 posts
#12 • 4 Y
Y by pikazag, MathematicalGiant, Adventure10, Mango247
A bit different solution with the use of trigonometry
This post has been edited 2 times. Last edited by PhysicsMonster_01, Sep 13, 2018, 7:48 AM
Reason: LaTeX
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LoveMaths26102003
84 posts
#13 • 2 Y
Y by Adventure10, Mango247
Similarity
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ftheftics
651 posts
#14 • 2 Y
Y by Adventure10, Mango247
Suppose,$\angle B =b'$ ,$\implies \angle A =2b'$,
and,$\angle C =180-3b'$.

See that,$\frac{C}{\sin 3b'} =\frac{b}{\sin b'}$.

$\implies c=b(3-4\sin ^2 b')$.

$\implies c=b(2\cos2b'+1)$.

$\implies c^2 = bc+2bc \cos 2b'$.

Using cosine rule we have ,

$a^2=b^2+c^2 -2bc \cos 2b'$.

$\implies a^2=b^2 +bc$.

As desired.$\boxed{a^2 = b (b+c)}$
This post has been edited 1 time. Last edited by ftheftics, Feb 15, 2020, 12:38 PM
Reason: BNNn
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sqing
42254 posts
#15 • 1 Y
Y by Adventure10
Rushil wrote:
In a triangle $ABC$, $\angle A = 2 \angle B$. Prove that $a^2 = b (b+c)$.
$$\frac{a}{\sin 2B} = \frac{b}{\sin B}\Leftrightarrow  \frac{a}{ 2b}=cosB=\frac{c^2+a^2-b^2}{ 2ca}
\Leftrightarrow  a^2 = b (b+c)$$
In a triangle $ABC$, $a^2 = b (b+c)$. Prove that $\angle A = 2\angle B.$
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BestChoice123
1119 posts
#16 • 1 Y
Y by Adventure10
mathmanman wrote:
That's just the law of sines ;)

oops that link failed :(
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sqing
42254 posts
#17 • 1 Y
Y by Adventure10
BestChoice123 wrote:
mathmanman wrote:
That's just the law of sines ;)

oops that link failed :(
Law_of_Sines
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S1281
209 posts
#18
Y by
We know that,
$a/sinA=b/sinB=c/sinC$
$a/sin2x=b /sinx=c/sin(180^{\circ{}}-3x)$
$a/2cosx=b$
$c/sin3x=b/sinx$
$c/(3sinx-4sin^3x)=b/sinx$
$c/(3-4sin^2x)=b$
$b(b+c)=b^2+bc=\frac{a^2}{4cos^2x}+\frac{a}{2cosx}(3-4sin^2x)\frac{a}{2cosx}=\frac{a^2}{4 cos^2x}4(1-sin^2x)=\frac{a^2}{4 cos^2x}4cos^2x=a^2$.
This post has been edited 2 times. Last edited by S1281, Apr 19, 2021, 12:13 PM
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SatisfiedMagma
461 posts
#19
Y by
I was just about the bash thing with trigonometry. But here is a geometric solution I guess.

Solution: Denote $D$ to be the foot of angle bisector of $\angle BAC$ on side $BC$. Now it is easy to notice that $\angle CAD = \angle ABD$. So we can deduce that $AC$ is tangent to $\odot(ADB)$. Applying Power of a Point along with Angle Bisector Theorem we get
\begin{align*}
b^2= \frac{ab}{b+c} \cdot a \\
\implies a^2= b(b+c)
\end{align*}which proves the desired. $\blacksquare$
This post has been edited 2 times. Last edited by SatisfiedMagma, May 14, 2022, 6:57 PM
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Krishijivi
100 posts
#20
Y by
Let angle B=x
Then , angle A=2x
By sine rule,
a/ b=sin 2x/ sin x
cos x =a/2b
cos x=c²+a²-b²/2ac
bc²+ba²-b²-a²c=0
(c-b){b(b+c)-a²}=0
If c=b
Then also a²=b(b+c)
If b(b+c)-a²=0
a²=b(b+c)
@ Krishijivi
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