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Tiling rectangle with smaller rectangles.
MarkBcc168   61
N 30 minutes ago by YaoAOPS
Source: IMO Shortlist 2017 C1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore
61 replies
MarkBcc168
Jul 10, 2018
YaoAOPS
30 minutes ago
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A magician has one hundred cards numbered 1 to 100
Valentin Vornicu   49
N 35 minutes ago by YaoAOPS
Source: IMO 2000, Problem 4, IMO Shortlist 2000, C1
A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience draws two cards from two different boxes and announces the sum of numbers on those cards. Given this information, the magician locates the box from which no card has been drawn.

How many ways are there to put the cards in the three boxes so that the trick works?
49 replies
+1 w
Valentin Vornicu
Oct 24, 2005
YaoAOPS
35 minutes ago
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Nice inequality
sqing   2
N 39 minutes ago by Seungjun_Lee
Source: WYX
Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be real numbers . Prove that : There exist positive integer $k\in \{1,2,\cdots,n\}$ such that $$\sum_{i=1}^{n}\{kx_i\}(1-\{kx_i\})<\frac{n-1}{6}.$$Where $\{x\}=x-\left \lfloor x \right \rfloor.$
2 replies
sqing
Apr 24, 2019
Seungjun_Lee
39 minutes ago
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Concurrency
Dadgarnia   27
N 41 minutes ago by zuat.e
Source: Iranian TST 2020, second exam day 2, problem 4
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
27 replies
Dadgarnia
Mar 12, 2020
zuat.e
41 minutes ago
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nice geo
Melid   1
N an hour ago by Melid
Source: 2025 Japan Junior MO preliminary P9
Let ABCD be a cyclic quadrilateral, which is AB=7 and BC=6. Let E be a point on segment CD so that BE=9. Line BE and AD intersect at F. Suppose that A, D, and F lie in order. If AF=11 and DF:DE=7:6, find the length of segment CD.
1 reply
Melid
an hour ago
Melid
an hour ago
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product of all integers of form i^3+1 is a perfect square
AlastorMoody   3
N an hour ago by Assassino9931
Source: Balkan MO ShortList 2009 N3
Determine all integers $1 \le m, 1 \le n \le 2009$, for which
\begin{align*} \prod_{i=1}^n \left( i^3 +1 \right) = m^2 \end{align*}
3 replies
AlastorMoody
Apr 6, 2020
Assassino9931
an hour ago
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IMO ShortList 1998, combinatorics theory problem 1
orl   44
N an hour ago by YaoAOPS
Source: IMO ShortList 1998, combinatorics theory problem 1
A rectangular array of numbers is given. In each row and each column, the sum of all numbers is an integer. Prove that each nonintegral number $x$ in the array can be changed into either $\lceil x\rceil $ or $\lfloor x\rfloor $ so that the row-sums and column-sums remain unchanged. (Note that $\lceil x\rceil $ is the least integer greater than or equal to $x$, while $\lfloor x\rfloor $ is the greatest integer less than or equal to $x$.)
44 replies
orl
Oct 22, 2004
YaoAOPS
an hour ago
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A game optimization on a graph
Assassino9931   2
N an hour ago by dgrozev
Source: Bulgaria National Olympiad 2025, Day 2, Problem 6
Let \( X_0, X_1, \dots, X_{n-1} \) be \( n \geq 2 \) given points in the plane, and let \( r > 0 \) be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points \( X_0, X_1, \dots, X_{n-1} \), i.e., she connects some of the points with edges so that from any point you can reach any other point by moving along the edges.Then, Alice assigns to each vertex \( X_i \) a non-negative real number \( r_i \), for \( i = 0, 1, \dots, n-1 \), such that $\sum_{i=0}^{n-1} r_i = 1$. Bob then selects a sequence of distinct vertices \( X_{i_0} = X_0, X_{i_1}, \dots, X_{i_k} \) such that \( X_{i_j} \) and \( X_{i_{j+1}} \) are connected by an edge for every \( j = 0, 1, \dots, k-1 \). (Note that the length $k \geq 0$ is not fixed and the first selected vertex always has to be $X_0$.) Bob wins if
\[ \frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r; \]otherwise, Alice wins. Depending on \( n \), determine the largest possible value of \( r \) for which Bobby has a winning strategy.
2 replies
Assassino9931
Apr 8, 2025
dgrozev
an hour ago
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Composite sum
rohitsingh0812   39
N an hour ago by YaoAOPS
Source: INDIA IMOTC-2006 TST1 PROBLEM-2; IMO Shortlist 2005 problem N3
Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ be positive integers and let $ S = a+b+c+d+e+f$.
Suppose that the number $ S$ divides $ abc+def$ and $ ab+bc+ca-de-ef-df$. Prove that $ S$ is composite.
39 replies
rohitsingh0812
Jun 3, 2006
YaoAOPS
an hour ago
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Problem 1
SpectralS   145
N an hour ago by IndexLibrorumProhibitorum
Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC.$ Prove that $M$ is the midpoint of $ST.$

(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Proposed by Evangelos Psychas, Greece
145 replies
SpectralS
Jul 10, 2012
IndexLibrorumProhibitorum
an hour ago
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The intersection of two lines lies on circumcircle
LeVietAn   1
N Jun 20, 2023 by Paramizo_Dicrominique
Source: OWN
Let $ABC$ be a triangle with $I$ be the incenter of its. Let $\omega$ be a circle passing through $I$ and intersects $AI$, $BI$, $CI$ and circumcircle of triangle $IBC$ at $D$, $E$, $F$ and $G$, respectively ($I$ different from $D$, $E$, $F$, $G$). The line $DG$ intersects $CA$, $AB$ at $M$, $N$, respectively. The perpendicular bisector of $AD$ intersects $DE$, $DF$ at $K$, $L$, respectively. Prove that the intersection of $KM$ and $LN$ is on circumcircle of triangle $ABC$.
1 reply
LeVietAn
Dec 16, 2016
Paramizo_Dicrominique
Jun 20, 2023
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The intersection of two lines lies on circumcircle
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LeVietAn
375 posts
LeVietAn
#1 h  Dec 16, 2016, 4:52 PM • 3 Y
Y by baopbc, anhtaitran, Adventure10
Let $ABC$ be a triangle with $I$ be the incenter of its. Let $\omega$ be a circle passing through $I$ and intersects $AI$, $BI$, $CI$ and circumcircle of triangle $IBC$ at $D$, $E$, $F$ and $G$, respectively ($I$ different from $D$, $E$, $F$, $G$). The line $DG$ intersects $CA$, $AB$ at $M$, $N$, respectively. The perpendicular bisector of $AD$ intersects $DE$, $DF$ at $K$, $L$, respectively. Prove that the intersection of $KM$ and $LN$ is on circumcircle of triangle $ABC$.
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Paramizo_Dicrominique
275 posts
Paramizo_Dicrominique
#2 h  Jun 20, 2023, 3:27 PM
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$\textbf{Solution:}$
We will solve for the case when $\triangle{ABC}$ is acute, the obtuse case is the same.
Let $(AKL)$ cut $(ABC)$ again at $W$. We will prove $NL$ passes through $W$ and similarly $MK$ passes through $W$.
Let $DK$ cut $AB$ at $U$, $DL$ cut $AC$ at $V$.
We have $\angle{UDL}= \angle{EDF} = 180^{\circ} - \angle{EIF} = 180^{\circ} - \angle{BIC} = 90^{\circ} - \frac{1}{2}\angle{BAC} = 90^{\circ} - \angle{UAD}$
Since $L$ lie on the perpendicular bisector of $AD$ so $L$ is the center of $(AUD)$. Similarly $K$ is the center of $(AVD)$.
We have $\angle{DKV} = 90^{\circ} - \angle{DAV} = 90^{\circ} - \angle{DAU} = \angle{ULD}$ $\implies A,K,U,L,V,W$ lie on a circle.
Let $AI$ cut $(AUV)$ again at $H$ so $H$ is the midpoint of arc $UV$ of $(AUV)$.
Let $WH$ cut $UV$ at $P$. We will prove $\overline{D,G,P}.$
Let $DG$ cut $UV$ at $P'$ so $\frac{P'U}{P'V} = \frac{sin(P'DU)}{sin(P'DV)} \cdot \frac{DU}{DV} = \frac{sin(GDE)}{sin(GDF)} \cdot \frac{DU}{DV}=\frac{sin(GFE)}{sin(GEF)} \cdot \frac{DU}{DV} = \frac{GE}{GF} \cdot \frac{DU}{DV} = \frac{BE}{CF} \cdot \frac{DU}{DV} $ (Since $\triangle{GEB} \overset{+}{\sim} \triangle{GFC}$)
Back working on the ratio of $P$,we have $\frac{PU}{PV} = \frac{WU}{WV} = \frac{BU}{CV}$ (Since $\triangle{WUB} \overset{+}{\sim} \triangle{WVC}$)
To get the claim $\overline{D,G,P}$ we will show $P \equiv P'$ or $\frac{PU}{PV}=\frac{P'U}{P'V}$
It is equivalent to $\frac{BU}{CV} = \frac{BE}{CF} \cdot \frac{DU}{DV} \Longleftrightarrow \frac{BU}{BE} \cdot \frac{CF}{CV} = \frac{DU}{DV} \Longleftrightarrow \frac{sin(BEU)}{sin(BUE)} \cdot \frac{sin(CVF)}{sin(CFV)} = \frac{sin(DVU)}{sin(DUV)} \Longleftrightarrow \frac{sin(DEI)}{sin(BUE)} \cdot \frac{sin(CVF)}{sin(DFI)} = \frac{sin(DVU)}{sin(DUV)} \Longleftrightarrow \frac{sin(CVF)}{sin(BUE)} = \frac{sin(DVU)}{sin(DUV)}$
But since $\angle{UDV} = \angle{EDF} = 180^{\circ} - \angle{EIF} = 180^{\circ} - \angle{BIC} = 90^{\circ} - \frac{1}{2}\angle{BAC}$ and $D$ lie on the bisector of $\angle{UAV}$ so $D$ is the $A-$excenter of $\triangle{AUV}$ so $\angle{BUE}=\angle{DUV},\angle{CVF} = \angle{DVU}$ which implies the equation above so that $\overline{D,G,P}.$
Let $S$ be the second intersection of $(WDV)$ with $AD$.
Let $DW$ cut $(AUV)$ again at $W'$ ,observe $H$ is the midpoint of arc $UV$ which is the center of $(UDV)$ so $\angle{ADG} = \angle{ADP} =\angle{HDP}=\angle{W'WH}=\angle{W'AH}$ but since $KL$ is the perpendicular bisector of $AD$ so $AW'$ cut $DP$ at a point $J$ lie on $KL$.
We will prove $\triangle{ASV} \overset{+}{\sim} \triangle{AND}$.
Since $\angle{AND} = 180^{\circ} - \angle{BND} = 180^{\circ} - \angle{BAD} - \angle{ADN} = 180^{\circ} - \angle{CAD} - \angle{JAD} = 180^{\circ} - \angle{JAV} = 180^{\circ} - \angle{W'AV} = 180^{\circ} - \angle{W'WV} = 180^{\circ} - \angle{DWV} = 180^{\circ} - \angle{DSV} = \angle{ASV}$
$\implies \angle{AND} = \angle{ASV}$ also since $\angle{NAD} = \angle{SAV}$ so we get the claim.
By spiral homothety so if $Q$ is the intersection of $SV,ND$ so $Q$ lie on $(ASN)$ and $(AVD)$.
We will use a well known lemma
$\textbf{Lemma:}$ Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$. Suppose diagonals $AC,BD$ cut at $P$. Prove that $(APB),(DPC),(OAD)$ has a common point.
Using the lemma for cyclic quadrilateral $QAVD$ with circumcenter $K$ so $W$ lie on $(QASN)$.
We have $\angle{LAD} = \angle{LDA} = \angle{VDA} = \angle{VQA} \implies$ $(QASNW)$ is tangent to $AL$ so that $\angle{WNA} = 180^{\circ} - \angle{WQA} = 180^{\circ} - \angle{WAL} = 180^{\circ} - \angle{WAL} = \angle{WUL}$ and since $L$ is the midpoint of arc $AU$ of $(AUW)$ so it is well known that $\overline{W,N,L}$.
Similarly we can prove $\overline{M,W,K}$ $\implies MK$ cut $NL$ at $W$ lie on $(ABC)$.
$Q.E.D.$
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This post has been edited 1 time. Last edited by Paramizo_Dicrominique, Jun 20, 2023, 3:28 PM
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