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IMO Shortlist 2012, Geometry 3
lyukhson   75
N 21 minutes ago by numbertheory97
Source: IMO Shortlist 2012, Geometry 3
In an acute triangle $ABC$ the points $D,E$ and $F$ are the feet of the altitudes through $A,B$ and $C$ respectively. The incenters of the triangles $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.
75 replies
lyukhson
Jul 29, 2013
numbertheory97
21 minutes ago
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FE with devisibility
fadhool   2
N an hour ago by ATM_
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
2 replies
fadhool
3 hours ago
ATM_
an hour ago
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Japan MO Finals 2023
parkjungmin   2
N an hour ago by parkjungmin
It's hard. Help me
2 replies
parkjungmin
Yesterday at 2:35 PM
parkjungmin
an hour ago
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Iranian geometry configuration
Assassino9931   2
N an hour ago by Captainscrubz
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear.

Amir Parsa Hosseini Nayeri, Iran
2 replies
Assassino9931
Today at 9:39 AM
Captainscrubz
an hour ago
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f(m + n) >= f(m) + f(f(n)) - 1
orl   30
N 2 hours ago by ezpotd
Source: IMO Shortlist 2007, A2, AIMO 2008, TST 2, P1, Ukrainian TST 2008 Problem 8
Consider those functions $ f: \mathbb{N} \mapsto \mathbb{N}$ which satisfy the condition
\[ f(m + n) \geq f(m) + f(f(n)) - 1 \]
for all $ m,n \in \mathbb{N}.$ Find all possible values of $ f(2007).$

Author: Nikolai Nikolov, Bulgaria
30 replies
orl
Jul 13, 2008
ezpotd
2 hours ago
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Classic Diophantine
Adywastaken   3
N 2 hours ago by Adywastaken
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
3 replies
Adywastaken
4 hours ago
Adywastaken
2 hours ago
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Add d or Divide by a
MarkBcc168   25
N 3 hours ago by Entei
Source: ISL 2022 N3
Let $a > 1$ be a positive integer and $d > 1$ be a positive integer coprime to $a$. Let $x_1=1$, and for $k\geq 1$, define
$$x_{k+1} = \begin{cases} x_k + d &\text{if } a \text{ does not divide } x_k \\ x_k/a & \text{if } a \text{ divides } x_k \end{cases}$$Find, in terms of $a$ and $d$, the greatest positive integer $n$ for which there exists an index $k$ such that $x_k$ is divisible by $a^n$.
25 replies
MarkBcc168
Jul 9, 2023
Entei
3 hours ago
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Alice and Bob play, 8x8 table, white red black, minimum n for victory
parmenides51   14
N 3 hours ago by Ilikeminecraft
Source: JBMO Shortlist 2018 C3
The cells of a $8 \times 8$ table are initially white. Alice and Bob play a game. First Alice paints $n$ of the fields in red. Then Bob chooses $4$ rows and $4$ columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of $n$ such that Alice can win the game no matter how Bob plays.
14 replies
parmenides51
Jul 22, 2019
Ilikeminecraft
3 hours ago
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GEOMETRY GEOMETRY GEOMETRY
Kagebaka   71
N 3 hours ago by bin_sherlo
Source: IMO 2021/3
Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\angle DAB = \angle CAD.$ The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD,$ the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC,$ and the point $X$ on the line $AC$ satisfies $CX = BX.$ Let $O_1$ and $O_2$ be the circumcenters of the triangles $ADC$ and $EXD,$ respectively. Prove that the lines $BC, EF,$ and $O_1O_2$ are concurrent.
71 replies
Kagebaka
Jul 20, 2021
bin_sherlo
3 hours ago
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Equation of integers
jgnr   3
N 3 hours ago by KTYC
Source: Indonesia Mathematics Olympiad 2005 Day 1 Problem 2
For an arbitrary positive integer $ n$, define $ p(n)$ as the product of the digits of $ n$ (in decimal). Find all positive integers $ n$ such that $ 11p(n)=n^2-2005$.
3 replies
jgnr
Jun 2, 2008
KTYC
3 hours ago
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Divisibility..
Sadigly   4
N 3 hours ago by Solar Plexsus
Source: another version of azerbaijan nmo 2025
Just ignore this
4 replies
Sadigly
Yesterday at 7:37 AM
Solar Plexsus
3 hours ago
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A touching question on perpendicular lines
Tintarn   2
N Apr 23, 2025 by pi_quadrat_sechstel
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
2 replies
Tintarn
Mar 17, 2025
pi_quadrat_sechstel
Apr 23, 2025
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A touching question on perpendicular lines
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Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
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Tintarn
9042 posts
Tintarn
#1 h  Mar 17, 2025, 12:23 PM
Y by
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
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Mathzeus1024
866 posts
Mathzeus1024
#2 h  Mar 22, 2025, 9:19 AM
Y by
Let $k$ be the semicircle $y=\sqrt{R^2-x^2}$ such that we have the points: $A(-R,0); M(0,0); B(R,0); P(R\cos(t), R\sin(t))$ for $t \in (0,\pi)$. The centers of inscribed circles $k_{A}, k_{B}$ shall respectively lie on the angular bisectors $\overline{MC}, \overline{ME}$ through $M$:

$\overline{MC}: y = -\tan\left(\frac{\pi-t}{2}\right)x = -\cot\left(\frac{t}{2}\right)x$ (i);

$\overline{ME}: y = \tan\left(\frac{t}{2}\right)x$ (ii)

which in turn yield the points:

$C\left(-R\sin\left(\frac{t}{2}\right), R\cos\left(\frac{t}{2}\right)\right)$ (iii);

$E\left(R\cos\left(\frac{t}{2}\right), R\sin\left(\frac{t}{2}\right)\right)$ (iv).

If $k_{A}$ has radius $r<R$, contains (iii) above, and is modeled according to $\left[x + \tan\left(\frac{t}{2}\right)r\right]^2 + (y-r)^2=r^2$, then we obtain:

$x^2 +2rx\tan\left(\frac{t}{2}\right) + r^2\tan^{2}\left(\frac{t}{2}\right) + y^2-2ry+r^2=r^2$;

or $R^2 - 2Rr\sin\left(\frac{t}{2}\right)\tan\left(\frac{t}{2}\right)-2Rr\cos\left(\frac{t}{2}\right) + r^2\tan^{2}\left(\frac{t}{2}\right) =0$;

or $R^{2}\cos^{2}\left(\frac{t}{2}\right) - 2Rr\cos\left(\frac{t}{2}\right)\left[\sin^{2}\left(\frac{t}{2}\right) + \cos^{2}\left(\frac{t}{2}\right)\right] +r^{2}\sin^{2}\left(\frac{t}{2}\right)=0$;

or $R^{2}\cos^{2}\left(\frac{t}{2}\right) - 2Rr\cos\left(\frac{t}{2}\right) +r^{2}\sin^{2}\left(\frac{t}{2}\right)=0$;

or $r = \frac{2R\cos(t/2) \pm \sqrt{4R^{2}\cos^{2}(t/2)-4R^2\cos^{2}(t/2)\sin^{2}(t/2)}}{2\sin^{2}(t/2)} = R\cos(t/2)\left[\frac{1 \pm \cos(t/2)}{1-\cos^{2}(t/2)}\right]$;

or $r = R\cos(t/2)\left[\frac{1 \pm \cos(t/2)}{[1-\cos(t/2)][1+\cos(t/2)]}\right]$ (v).

Since $0 < r < R$, we only admit the negative root in the numerator of (v) to ultimate yield $r = \left[\frac{\cos(t/2)}{1+\cos(t/2)}\right]R$ (vi), which also yields the tangency point $D\left(-\left[\frac{\sin(t/2)}{1+\cos(t/2)}\right]R,0\right)$ (vii).

Next, we compute the slopes $m_{AE}, m_{CD}$:

$m_{AE} = \frac{R\sin(t/2)-0}{R\cos(t/2)+R} = \frac{\sin(t/2)}{\cos(t/2)+1}$ (viii);

$m_{CD} = \frac{R\cos(t/2)-0}{-R\sin(t/2)+ \left[\frac{\sin(t/2)}{1+\cos(t/2)}\right]R} = \frac{\cos(t/2)}{-\sin(t/2)+ \left[\frac{\sin(t/2)}{1+\cos(t/2)}\right]} = -\frac{\cos(t/2)+1}{\sin(t/2)}$ (ix).

and upon inspection $m_{AE} \cdot m_{CD} = -1 \Rightarrow \textcolor{red}{\overline{AE} \perp \overline{CD}}$.
This post has been edited 18 times. Last edited by Mathzeus1024, Mar 22, 2025, 10:23 AM
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pi_quadrat_sechstel
601 posts
pi_quadrat_sechstel
#3 h  Apr 23, 2025, 6:47 AM
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Tintarn wrote:
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.

Complete $k$ to the full circle and let $S$ be the midpoint of the arc $AB$ on the new half.

The homothety at $C$ sending $k_A$ to $k$ must send the "lowest" point $D$ to the "lowest" point $S$. So $CD=CS$.

We have congruences $AMC\sim PMC$ and $PME\sim BME$. Thus $\angle EMC=\frac{1}{2}\angle BMA=90^{\circ}$.

So a $90^{\circ}$-rotation at $M$ sends $A$ to $S$ and $E$ to $C$. We get $AE\perp CS=CD$.
This post has been edited 1 time. Last edited by pi_quadrat_sechstel, Apr 23, 2025, 6:48 AM
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