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Cyclic quads jigsaw
Miquel-point   0
2 hours ago
Source: Kürschák József Competition 2024/1
The quadrilateral $ABCD$ is divided into cyclic quadrilaterals with pairwise disjoint interiors. None of the vertices of the cyclic quadrilaterals in the decomposition is an interior point of a side of any cyclic quadrilateral in the decomposition or of a side of the quadrilateral $ABCD$. Prove that $ABCD$ is also a cyclic quadrilateral.
0 replies
Miquel-point
2 hours ago
0 replies
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A cyclic inequality
KhuongTrang   3
N 2 hours ago by paixiao
Source: own-CRUX
IMAGE
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
3 replies
KhuongTrang
Apr 21, 2025
paixiao
2 hours ago
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Perfect polynomials
Phorphyrion   5
N 3 hours ago by Davdav1232
Source: 2023 Israel TST Test 5 P3
Given a polynomial $P$ and a positive integer $k$, we denote the $k$-fold composition of $P$ by $P^{\circ k}$. A polynomial $P$ with real coefficients is called perfect if for each integer $n$ there is a positive integer $k$ so that $P^{\circ k}(n)$ is an integer. Is it true that for each perfect polynomial $P$, there exists a positive $m$ so that for each integer $n$ there is $0<k\leq m$ for which $P^{\circ k}(n)$ is an integer?
5 replies
Phorphyrion
Mar 23, 2023
Davdav1232
3 hours ago
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Finding all integers with a divisibility condition
Tintarn   14
N 4 hours ago by Assassino9931
Source: Germany 2020, Problem 4
Determine all positive integers $n$ for which there exists a positive integer $d$ with the property that $n$ is divisible by $d$ and $n^2+d^2$ is divisible by $d^2n+1$.
14 replies
Tintarn
Jun 22, 2020
Assassino9931
4 hours ago
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Geometry Handout is finally done!
SimplisticFormulas   2
N 4 hours ago by parmenides51
If there’s any typo or problem you think will be a nice addition, do send here!
handout, geometry
2 replies
SimplisticFormulas
Yesterday at 4:58 PM
parmenides51
4 hours ago
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IMO ShortList 2002, number theory problem 2
orl   57
N 5 hours ago by Maximilian113
Source: IMO ShortList 2002, number theory problem 2
Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.
57 replies
orl
Sep 28, 2004
Maximilian113
5 hours ago
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"Mistakes were made" -Luke Rbotaille
a1267ab   10
N 5 hours ago by Martin.s
Source: USA TST 2025
Let $a_1, a_2, \dots$ and $b_1, b_2, \dots$ be sequences of real numbers for which $a_1 > b_1$ and
\begin{align*} a_{n+1} &= a_n^2 - 2b_n\\ b_{n+1} &= b_n^2 - 2a_n \end{align*}for all positive integers $n$. Prove that $a_1, a_2, \dots$ is eventually increasing (that is, there exists a positive integer $N$ for which $a_k < a_{k+1}$ for all $k > N$).

Holden Mui
10 replies
a1267ab
Dec 14, 2024
Martin.s
5 hours ago
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Problem 4 (second day)
darij grinberg   92
N 5 hours ago by cubres
Source: IMO 2004 Athens
Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.
92 replies
darij grinberg
Jul 13, 2004
cubres
5 hours ago
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Perpendicularity with Incircle Chord
tastymath75025   31
N 6 hours ago by cj13609517288
Source: 2019 ELMO Shortlist G3
Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$. The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$. The circumcircles of $ABC$ and $A'EF$ meet at $G$, and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$, where $M$ is the midpoint of $EF$. Prove that if $GH$ and $EF$ meet at $T$, then $DT\perp EF$.

Proposed by Ankit Bisain
31 replies
tastymath75025
Jun 27, 2019
cj13609517288
6 hours ago
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\frac{1}{5-2a}
Havu   2
N 6 hours ago by arqady
Let $a\ge b\ge c \ge \frac{1}{2}$ and $a^2+b^2+c^2=3$. Find minimum:
\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]
2 replies
Havu
Wednesday at 9:56 AM
arqady
6 hours ago
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Geometry Problem in Taiwan TST
chengbilly   3
N Mar 29, 2025 by Hakurei_Reimu
Source: 2025 Taiwan TST Round 2 Independent Study 2-G
Given a triangle $ABC$ with circumcircle $\Gamma$, and two arbitrary points $X, Y$ on $\Gamma$. Let $D$, $E$, $F$ be points on lines $BC$, $CA$, $AB$, respectively, such that $AD$, $BE$, and $CF$ concur at a point $P$. Let $U$ be a point on line $BC$ such that $X$, $Y$, $D$, $U$ are concyclic. Similarly, let $V$ be a point on line $CA$ such that $X$, $Y$, $E$, $V$ are concyclic, and let $W$ be a point on line $AB$ such that $X$, $Y$, $F$, $W$ are concyclic. Prove that $AU$, $BV$, $CW$ concur at a single point.

Proposed by chengbilly
3 replies
chengbilly
Mar 27, 2025
Hakurei_Reimu
Mar 29, 2025
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Geometry Problem in Taiwan TST
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Source: 2025 Taiwan TST Round 2 Independent Study 2-G
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chengbilly
8 posts
chengbilly
#1 h  Mar 27, 2025, 5:09 AM
Y by
Given a triangle $ABC$ with circumcircle $\Gamma$, and two arbitrary points $X, Y$ on $\Gamma$. Let $D$, $E$, $F$ be points on lines $BC$, $CA$, $AB$, respectively, such that $AD$, $BE$, and $CF$ concur at a point $P$. Let $U$ be a point on line $BC$ such that $X$, $Y$, $D$, $U$ are concyclic. Similarly, let $V$ be a point on line $CA$ such that $X$, $Y$, $E$, $V$ are concyclic, and let $W$ be a point on line $AB$ such that $X$, $Y$, $F$, $W$ are concyclic. Prove that $AU$, $BV$, $CW$ concur at a single point.

Proposed by chengbilly
This post has been edited 1 time. Last edited by chengbilly, Mar 27, 2025, 5:18 AM
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Tamam
12 posts
Tamam
#2 h  Mar 27, 2025, 6:26 AM • 1 Y
Y by ehuseyinyigit
From Ratio Lemma $\frac{BU}{UC}=\frac{\frac{XB}{XC}\cdot \frac{YB}{YC}}{\frac{DB}{DC}}$.It is similar for $V,W$. So the numerator is obviously 1 from the cyclic product and the denominator is 1 from Ceva $\frac{BU}{BC}\cdot \frac{VC}{VA}\cdot \frac{WA}{WB}=1$.From Ceva the question is done.
This post has been edited 2 times. Last edited by Tamam, Mar 31, 2025, 9:54 AM
Reason: ceva
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Li4
43 posts
Li4
#3 h  Mar 27, 2025, 6:43 AM • 1 Y
Y by MS_asdfgzxcvb
This problem has the following generalization:

Let $\triangle P_aP_bP_c$, $\triangle Q_aQ_bQ_c$ be the cevian triangle of $P$, $Q$ with respect to $\triangle ABC$. Let $\triangle U_AU_BU_C$, $\triangle V_AV_BV_C$ be the circumcevian triangle of $U$, $V$ with respect to $\triangle ABC$. Suppose that $P_b$, $Q_b$, $U_B$, $V_B$ and $P_c$, $Q_c$, $U_C$, $V_C$ are concyclic. Then $P_a$, $Q_a$, $U_A$, $V_A$ is concyclic.

Proof. Click to reveal hidden text
This post has been edited 1 time. Last edited by Li4, Mar 27, 2025, 7:15 AM
Reason: typo
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Hakurei_Reimu
40 posts
Hakurei_Reimu
#4 h  Mar 29, 2025, 2:06 PM
Y by
Lemma: $\dfrac{BU}{UC}=\dfrac{BX}{XC}\cdot\dfrac{BY}{YC}\cdot\dfrac{CD}{DB}$
Proof.
By Ceva's Theorem, it's equivalent to show $\dfrac{BU}{UC}\cdot\dfrac{CV}{VA}\cdot\dfrac{AW}{WB}=1$.
Using lemma with symmetry, LHS will be$$\left(\dfrac{BX}{XC}\cdot\dfrac{BY}{YC}\cdot\dfrac{CD}{DB}\right)\cdot\left(\dfrac{CX}{XA}\cdot\dfrac{CY}{YA}\cdot\dfrac{AE}{EC}\right)\cdot\left(\dfrac{AX}{XB}\cdot\dfrac{AY}{YB}\cdot\dfrac{BF}{FA}\right)=\dfrac{CD}{DB}\cdot\dfrac{AE}{EC}\cdot\dfrac{BF}{FA}$$By Ceva's Theorem again, it's clear that $\dfrac{CD}{DB}\cdot\dfrac{AE}{EC}\cdot\dfrac{BF}{FA}=1$, so we're done. $\blacksquare$
This post has been edited 1 time. Last edited by Hakurei_Reimu, Mar 29, 2025, 2:10 PM
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