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PQ = r and 6 more conditions
avisioner   41
N 8 minutes ago by ezpotd
Source: 2023 ISL G2
Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.
41 replies
avisioner
Jul 17, 2024
ezpotd
8 minutes ago
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Functional equations in IMO TST
sheripqr   50
N 17 minutes ago by megahertz13
Source: Iran TST 1996
Find all functions $f: \mathbb R \to \mathbb R$ such that $$ f(f(x)+y)=f(x^2-y)+4f(x)y $$ for all $x,y \in \mathbb R$
50 replies
sheripqr
Sep 14, 2015
megahertz13
17 minutes ago
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JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   1
N 33 minutes ago by Maths_VC
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$≥3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
1 reply
FishkoBiH
Today at 1:38 PM
Maths_VC
33 minutes ago
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IMO Shortlist 2009 - Problem C3
nsato   25
N an hour ago by popop614
Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll} a_{i+1} = \begin{cases} 2a_{i-1} + 3a_i, \\ 3a_{i-1} + a_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_i = 0, \\ \text{if } \varepsilon_i = 1, \end{array} & \text{for each } i = 1, \dots, n - 1, \\[15pt] b_{i+1}= \begin{cases} 2b_{i-1} + 3b_i, \\ 3b_{i-1} + b_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_{n-i} = 0, \\ \text{if } \varepsilon_{n-i} = 1, \end{array} & \text{for each } i = 1, \dots, n - 1. \end{array}\] Prove that $a_n = b_n$.

Proposed by Ilya Bogdanov, Russia
25 replies
nsato
Jul 6, 2010
popop614
an hour ago
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JBMO TST Bosnia and Herzegovina 2024 P4
FishkoBiH   1
N an hour ago by TopGbulliedU
Source: JBMO TST Bosnia and Herzegovina 2024 P4
Let $m$ and $n$ be natural numbers. Every one of the $m*n$ squares of the $m*n$ board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this:
- a white square becomes black;
-a black square becomes blue;
-a blue square becomes white.
For which $m$ and $n$ can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
1 reply
FishkoBiH
5 hours ago
TopGbulliedU
an hour ago
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Van der Corput Inequality
EthanWYX2009   1
N an hour ago by grupyorum
Source: en.wikipedia.org/wiki/Van_der_Corput_inequality
Let $V$ be a real or complex inner product space. Suppose that ${\displaystyle v,u_{1},\dots ,u_{n}\in V} $ and that ${\displaystyle \|v\|=1}.$ Then$${\displaystyle \displaystyle \left(\sum _{i=1}^{n}|\langle v,u_{i}\rangle |\right)^{2}\leq \sum _{i,j=1}^{n}|\langle u_{i},u_{j}\rangle |.}$$
1 reply
EthanWYX2009
Today at 3:36 AM
grupyorum
an hour ago
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two sequences of positive integers and inequalities
rmtf1111   52
N an hour ago by Kappa_Beta_725
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
52 replies
rmtf1111
Apr 10, 2019
Kappa_Beta_725
an hour ago
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side ratio in triangle, angle bisectors form cyclic quadrilateral
jasperE3   3
N 2 hours ago by lpieleanu
Source: Mongolia MO 2000 Grade 10 P6
In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then
$$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$
3 replies
jasperE3
Apr 22, 2021
lpieleanu
2 hours ago
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Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   5
N 2 hours ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
5 replies
OgnjenTesic
May 22, 2025
JARP091
2 hours ago
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JBMO TST Bosnia and Herzegovina 2024 P3
FishkoBiH   2
N 2 hours ago by Ianis
Source: JBMO TST Bosnia and Herzegovina 2024 P3
Let $ABC$ be a right-angled triangle where ∠$ACB$=90°.Let $CD$ be an altitude of that triangle and points $M$ and $N$ be the midpoints of $CD$ and $BC$, respectively.If $S$ is the circumcenter of the triangle $AMN$, prove that $AS$ and $BC$ are paralel.
2 replies
FishkoBiH
6 hours ago
Ianis
2 hours ago
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A line parallel to the asymptote of a cubic
kosmonauten3114   0
2 hours ago
Source: My own, but uninspiring...
Given a scalene triangle $\triangle{ABC}$, let $P$ be a point ($\neq \text{X(4)}$). Let $P'$ be the anticomplement of $P$, and let $Q$ be the $\text{X(1)}$-anticomplementary conjugate of $P$. Prove that the line $P'Q$ is parallel to the real asymptote of the circular pivotal isocubic with pivot $P$.
0 replies
kosmonauten3114
2 hours ago
0 replies
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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
9045 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
921 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
604 posts
pi_quadrat_sechstel
#3 h  Apr 23, 2025, 6:47 AM
Y by
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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