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Non-linear Recursive Sequence
amogususususus   3
N an hour ago by SunnyEvan
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
3 replies
amogususususus
Jan 24, 2025
SunnyEvan
an hour ago
Inspired by 2025 Beijing
sqing   6
N an hour ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
6 replies
sqing
Yesterday at 4:56 PM
sqing
an hour ago
Serbian selection contest for the IMO 2025 - P4
OgnjenTesic   2
N 2 hours ago by sqing-inequality-BUST
Source: Serbian selection contest for the IMO 2025
For a permutation $\pi$ of the set $A = \{1, 2, \ldots, 2025\}$, define its colorfulness as the greatest natural number $k$ such that:
- For all $1 \le i, j \le 2025$, $i \ne j$, if $|i - j| < k$, then $|\pi(i) - \pi(j)| \ge k$.
What is the maximum possible colorfulness of a permutation of the set $A$? Determine how many such permutations have maximal colorfulness.

Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
sqing-inequality-BUST
2 hours ago
Nice "if and only if" function problem
ICE_CNME_4   14
N 2 hours ago by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
14 replies
ICE_CNME_4
Friday at 7:23 PM
wh0nix
2 hours ago
2-var inequality
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b> 0 , ab(a+b+1) =3.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{24}{(a+b)^2} \geq 8$$$$ \frac{a}{b^2}+\frac{b}{a^2}+\frac{49}{(a+  b)^2} \geq \frac{57}{4}$$Let $ a,b> 0 ,  (a+b)(ab+1) =4.$ Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{40}{(a+b)^2} \geq 12$$$$\frac{a}{b^2}+\frac{b}{a^2}+\frac{76}{(a+ b)^2}  \geq 21$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
Balkan Mathematical Olympiad
ABCD1728   1
N 2 hours ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
ABCD1728
Yesterday at 11:27 PM
ABCD1728
2 hours ago
area of quadrilateral
AlanLG   1
N 2 hours ago by Altronrren
Source: 3rd National Women´s Contest of Mexican Mathematics Olympiad 2024 , level 1+2 p5
Consider the acute-angled triangle \(ABC\). The segment \(BC\) measures 40 units. Let \(H\) be the orthocenter of triangle \(ABC\) and \(O\) its circumcenter. Let \(D\) be the foot of the altitude from \(A\) and \(E\) the foot of the altitude from \(B\). Additionally, point \(D\) divides the segment \(BC\) such that \(\frac{BD}{DC} = \frac{3}{5}\). If the perpendicular bisector of segment \(AC\) passes through point \(D\), calculate the area of quadrilateral \(DHEO\).
1 reply
AlanLG
Jun 14, 2024
Altronrren
2 hours ago
Inspired by 2025 SXTB
sqing   1
N 3 hours ago by sqing
Source: Own
Let $ a,b  $ be real number such that $ a^2+b^2=\frac12. $ Prove that
$$-\frac{\sqrt{9+6\sqrt 3}}{2}\leq(a+1)^2- (b-1)^2\leq\frac{\sqrt{9+6\sqrt 3}}{2}$$Let $ x $ be real number . Prove that
$$-\frac{2\sqrt 2}{3}\leq \frac{x}{x^2+1}+ \frac{ 2x}{x^2+4} \leq\frac{2\sqrt 2}{3}$$
1 reply
sqing
3 hours ago
sqing
3 hours ago
IMO Shortlist 2014 G2
hajimbrak   14
N 3 hours ago by ezpotd
Let $ABC$ be a triangle. The points $K, L,$ and $M$ lie on the segments $BC, CA,$ and $AB,$ respectively, such that the lines $AK, BL,$ and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM, BMK,$ and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$.

Proposed by Estonia
14 replies
hajimbrak
Jul 11, 2015
ezpotd
3 hours ago
Divisiblity...
TUAN2k8   0
3 hours ago
Source: Own
Let $m$ and $n$ be two positive integer numbers such that $m \le n$.Prove that $\binom{n}{m}$ divides $lcm(1,2,...,n)$
0 replies
TUAN2k8
3 hours ago
0 replies
Lines AD, BE, and CF are concurrent
orl   48
N Apr 7, 2025 by Avron
Source: IMO Shortlist 2000, G3
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.
48 replies
orl
Aug 10, 2008
Avron
Apr 7, 2025
Lines AD, BE, and CF are concurrent
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Source: IMO Shortlist 2000, G3
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orl
3647 posts
#1 • 8 Y
Y by Davi-8191, mathematicsy, Adventure10, chessgocube, HWenslawski, ImSh95, Mango247, Amir Hossein
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.
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mr.danh
635 posts
#2 • 6 Y
Y by Adventure10, chessgocube, HWenslawski, ImSh95, Mango247, ehuseyinyigit
Solution
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Valentin Vornicu
7301 posts
#3 • 8 Y
Y by SherlockBond, raknum007, megarnie, chessgocube, ImSh95, Adventure10, Mango247, and 1 other user
We know that the orthocenter reflects over the sides of the triangle on the circumcircle. Therefore the minimal distance $ OD+HD$ equals $ R$. Obviously we can achieve this on all sides, so we assume that $ D,E,F$ are the intersection points between $ A',B',C'$ the reflections of $ H$ across $ BC,CA,AB$ respectively. All we have to prove is that $ AD$, $ BE$ and $ CF$ are concurrent.

In order to do that we need the ratios $ \dfrac {BD}{DC}$, $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$, and then we can apply Ceva's theorem.

We know that the triangle $ ABC$ is acute, so $ \angle BAH = 90^\circ- \angle B = \angle OAC$, therefore $ \angle HAO = |\angle A - 2(90^\circ -\angle B)| = |\angle B- \angle C|$. In particular this means that $ \angle OA'H = |\angle B-\angle C|$. Since $ \angle BA'A = \angle C$ and $ \angle AA'C = \angle B$, we have that $ \angle BA'D = \angle B$ and $ \angle DA'C = \angle C$.

By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get
\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]

Using the similar relationships for $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$ we get that those three fractions multiply up to 1, and thus by Ceva's, the lines $ AD, BE$ and $ CF$ are concurrent.
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Zhero
2043 posts
#4 • 7 Y
Y by p1a, chessgocube, ImSh95, Adventure10, Mango247, ehuseyinyigit, and 1 other user
Lemma 1: If an ellipse is inscribed triangle $ABC$, tangent to $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively, then $AD$, $BE$, and $CF$ are concurrent.
Proof: Scale the ellipse about its major axis so that it becomes a circle; here, $A'D'$, $B'E'$, and $C'F'$ concur at the Gergonne point of $A' B' C'$. Scaling preserves incidence, so $AD$, $BE$, and $CF$ concur.

Lemma 2: Let $\ell$ be any line, and let $X$ and $Y$ be any points on the same side of $\ell$. Let $Z$ be the reflection of $X$ across $\ell$, and let $W$ be the intersection of $YZ$ and $\ell$. Then the ellipse with foci $X$ and $Y$ that passes through $W$ is tangent to $\ell$.
Proof: Suppose that the ellipse intersected line $\ell$ at another point, $W'$. Then $XW' + YW' = YW' + ZW' > YZ$ by the triangle inequality, since $W'$, $Y$, and $Z$ are not collinear. On the other hand, by definition of ellipse, $XW' + YW' = XW + YW = ZW + YW = YZ$, so $YZ > YZ$, which is a contradiction.

By lemma 1, it is sufficient to show that there exists an ellipse with foci $O$ and $H$ that are tangent to the sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.

Reflect $H$ across $BC$, $CA$, and $AC$ to get $A'$, $B'$, and $C'$, respectively, and let $OA'$ hit $BC$ at $D$, $OB'$ hit $CA$ at $E$, and $OC'$ hit $AB$ at $F$. . $A',B',C'$ lie on the circumcircle of $ABC$, so $R = OA' = OB' = OC' = HD + DO = HE + EO = HF + FO$, where $R$ is the circumradius of $\triangle ABC$ (since $A', B', C'$ are the reflections of $H$ across the sides of the triangle.)

Consider the ellipse with foci $O$ and $H$ with a major axis of length $R$. By definition, the ellipse passes through $D$, $E$, and $F$. However, by lemma 2, it is tangent to sides $BC$, $CA$, and $AB$, so by lemma 1, we are done.

Some motivation
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Rofler
802 posts
#5 • 2 Y
Y by ImSh95, Adventure10
@Zhero: Instead of scaling, it is more elegant to just use Brainchon's theorem.

Cheers,

Rofler
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Zhero
2043 posts
#6 • 2 Y
Y by ImSh95, Adventure10
Probably. I believe we could also just centrally project it to a circle as well (my first approach). :P I chose to scale because I felt it was the most elementary solution; anyone who knows anything about ellipses should be able to understand the solution I gave.
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goodar2006
1347 posts
#7 • 5 Y
Y by mrdriller, ImSh95, Adventure10, and 2 other users
since $H$ and $O$ are two isogonal conjugate points, there exists an ellipse which these two points are its foci and its tangent to sides of the triangle.
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StefanS
149 posts
#8 • 3 Y
Y by ImSh95, Adventure10, Mango247
Valentin Vornicu wrote:
By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get
\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]

By the law of sines for triangles $~$ $ \triangle{BA'D} $ $~$ and $~$ $ \triangle{DA'C} $ $~$ we get:
\[ \frac { BD } { \sin B } = \frac { A'D } { \sin \angle{DBA'} } \quad \wedge \quad \frac { DC } { \sin C } = \frac { A'D } { \sin \angle{DCA'} } \]
So for:
\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]
to be true, the following:
\[ \sin \angle{DBA'} = \sin \angle{DCA'} \]
should be true as well. I don't think that's correct. :maybe:
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Virgil Nicula
7054 posts
#9 • 3 Y
Y by ImSh95, Adventure10, Mango247
You are right. See here PP8.
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StefanS
149 posts
#10 • 3 Y
Y by ImSh95, Adventure10, Mango247
Virgil Nicula wrote:
You are right. See here PP8.

Wow you had a math blog!!! :w00t: I unfortunately can't read the entire problems. Only half of the picture is displayed. Do you know why? :(
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Virgil Nicula
7054 posts
#11 • 3 Y
Y by ImSh95, Adventure10, Mango247
See click on the title of message.
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exmath89
2572 posts
#12 • 2 Y
Y by ImSh95, Adventure10
Solution
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IDMasterz
1412 posts
#13 • 5 Y
Y by DPS, ImSh95, Adventure10, Mango247, and 1 other user
Sol
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sayantanchakraborty
505 posts
#14 • 2 Y
Y by ImSh95, Adventure10
orl wrote:
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent.

This problem also appeared in one of the Indian TSTs!!!
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bonciocatciprian
41 posts
#15 • 3 Y
Y by ImSh95, Adventure10, silouan
Take $H_a, H_b, H_c$ be the mirror images of $H$ about $BC, AC, $ and $AB$ respectively. They lie on the circumcircle of $\Delta ABC$. Now, take $\{D\} = OH_a \cap BC$, $\{E\} = OH_b \cap AC$ and $\{F\} = OH_c \cap AB$. Now, using Pool's Theorem, we get that $OD + DH = OE + EH = OF + FH$. Moreover, in this case the sum $OD+DH$ is minimal, so the points $D, E, F$ are situated on an ellipse of foci $O$ and $H$, having the constant sum equal to $R$ (the ray of the circumcircle of $\Delta ABC$), and which is tangent to the edges of the triangle. Now, this is just Gergonne's theorem, under a projective transformation.
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