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IMO Shortlist 2010 - Problem G1
Amir Hossein   130
N 7 minutes ago by LeYohan
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
130 replies
Amir Hossein
Jul 17, 2011
LeYohan
7 minutes ago
Nordic 2025 P3
anirbanbz   7
N 8 minutes ago by anirbanbz
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
7 replies
1 viewing
anirbanbz
Mar 25, 2025
anirbanbz
8 minutes ago
nice problem
hanzo.ei   0
39 minutes ago
Source: I forgot
Let triangle $ABC$ be inscribed in the circumcircle $(O)$ and circumscribed about the incircle $(I)$, with $AB < AC$. The incircle $(I)$ touches the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. A line through $I$, perpendicular to $AI$, intersects $BC$, $CA$, and $AB$ at $X$, $Y$, and $Z$, respectively. The line $AI$ meets $(O)$ at $M$ (distinct from $A$). The circumcircle of triangle $AYZ$ intersects $(O)$ at $N$ (distinct from $A$). Let $P$ be the midpoint of the arc $BAC$ of $(O)$. The line $AI$ cuts segments $DF$ and $DE$ at $K$ and $L$, respectively, and the tangents to the circle $(DKL)$ at $K$ and $L$ intersect at $T$. Prove that $AT \perp BC$.
0 replies
hanzo.ei
39 minutes ago
0 replies
Find a given number of divisors of ab
proglote   9
N an hour ago by zuat.e
Source: Brazil MO 2013, problem #2
Arnaldo and Bernaldo play the following game: given a fixed finite set of positive integers $A$ known by both players, Arnaldo picks a number $a \in A$ but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer $b$ (not necessarily in $A$). Then Arnaldo tells the number of divisors of $ab$. Show that Bernaldo can choose $b$ in a way that he can find out the number $a$ chosen by Arnaldo.
9 replies
proglote
Oct 24, 2013
zuat.e
an hour ago
2025 TST 22
EthanWYX2009   1
N an hour ago by hukilau17
Source: 2025 TST 22
Let \( A \) be a set of 2025 positive real numbers. For a subset \( T \subseteq A \), define \( M_T \) as the median of \( T \) when all elements of \( T \) are arranged in increasing order, with the convention that \( M_\emptyset = 0 \). Define
\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]Find the smallest real number \( C \) such that for any set \( A \) of 2025 positive real numbers, the following inequality holds:
\[
P(A) - Q(A) \leq C \cdot \max(A),
\]where \(\max(A)\) denotes the largest element in \( A \).
1 reply
+1 w
EthanWYX2009
4 hours ago
hukilau17
an hour ago
Deriving Van der Waerden Theorem
Didier2   0
an hour ago
Source: Khamovniki 2023-2024 (group 10-1)
Suppose we have already proved that for any coloring of $\Large \mathbb{N}$ in $r$ colors, there exists an arithmetic progression of size $k$. How can we derive Van der Waerden's theorem for $W(r, k)$ from this?
0 replies
Didier2
an hour ago
0 replies
Not so classic orthocenter problem
m4thbl3nd3r   6
N an hour ago by maths_enthusiast_0001
Source: own?
Let $O$ be circumcenter of a non-isosceles triangle $ABC$ and $H$ be a point in the interior of $\triangle ABC$. Let $E,F$ be foots of perpendicular lines from $H$ to $AC,AB$. Suppose that $BCEF$ is cyclic and $M$ is the circumcenter of $BCEF$, $HM\cap AB=K,AO\cap BE=T$. Prove that $KT$ bisects $EF$
6 replies
m4thbl3nd3r
Yesterday at 4:59 PM
maths_enthusiast_0001
an hour ago
Functional equations
hanzo.ei   1
N an hour ago by GreekIdiot
Source: Greekldiot
Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$
1 reply
hanzo.ei
2 hours ago
GreekIdiot
an hour ago
A number theory problem from the British Math Olympiad
Rainbow1971   6
N 2 hours ago by ektorasmiliotis
Source: British Math Olympiad, 2006/2007, round 1, problem 6
I am a little surprised to find that I am (so far) unable to solve this little problem:

[quote]Let $n$ be an integer. Show that, if $2 + 2 \sqrt{1+12n^2}$ is an integer, then it is a perfect square.[/quote]

I set $k := \sqrt{1+12n^2}$. If $2 + 2 \sqrt{1+12n^2}$ is an integer, then $k (=\sqrt{1+12n^2})$ is at least rational, so that $1 + 12n^2$ must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs $(n, k)$ for which this happens are $(0,1), (2,7), (28,97)$ and $(390, 1351)$, and that, for every such pair $(n,k)$, the "next" such pair can be calculated as
$$
\begin{bmatrix}
7 & 2 \\
24 & 7 
\end{bmatrix} \begin{bmatrix}
n \\
k 
\end{bmatrix}
.$$The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?

Thank you.




6 replies
Rainbow1971
Yesterday at 8:39 PM
ektorasmiliotis
2 hours ago
A number theory about divisors which no one fully solved at the contest
nAalniaOMliO   20
N 2 hours ago by Bluecloud123
Source: Belarusian national olympiad 2024
Let's call a pair of positive integers $(k,n)$ interesting if $n$ is composite and for every divisor $d<n$ of $n$ at least one of $d-k$ and $d+k$ is also a divisor of $n$
Find the number of interesting pairs $(k,n)$ with $k \leq 100$
M. Karpuk
20 replies
nAalniaOMliO
Jul 24, 2024
Bluecloud123
2 hours ago
CHKMO 2017 Q3
noobatron3000   7
N 2 hours ago by Entei
Source: CHKMO
Let ABC be an acute-angled triangle. Let D be a point on the segment BC, I the incentre of ABC. The circumcircle of ABD meets BI at P and the circumcircle of ACD meets CI at Q. If the area of PID and the area of QID are equal, prove that PI*QD=QI*PD.
7 replies
noobatron3000
Dec 31, 2016
Entei
2 hours ago
Geometry
Jackson0423   1
N 2 hours ago by ricarlos
Source: Own
In triangle ABC with circumcenter O, if the intersection point of lines BO and AC is N, then BO = 2ON, and BMN = 122 degrees with respect to the midpoint M of AB. Find MNB.
1 reply
Jackson0423
Yesterday at 4:40 PM
ricarlos
2 hours ago
A functional equation from MEMO
square_root_of_3   24
N 2 hours ago by pco
Source: Middle European Mathematical Olympiad 2022, problem I-1
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(x+f(x+y))=x+f(f(x)+y)$$holds for all real numbers $x$ and $y$.
24 replies
square_root_of_3
Sep 1, 2022
pco
2 hours ago
Prove the equality
HHT23   3
N Mar 14, 2015 by TelvCohl
Given convex quadrilateral $ABCD$ inscribed circle $(O)$. Let $I$ be the intersection of $AC$ and $BC$. Line $\Delta$ passes through $I,$ intersects the segments $AB$, $CD$ at $M$, $N$ and intersects $(O)$ at $P$, $Q$, respectively ($M, N$ lie on the segments $IQ, IP$, respectively). Prove that $$\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}$$
3 replies
HHT23
Mar 14, 2015
TelvCohl
Mar 14, 2015
Prove the equality
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HHT23
34 posts
#1 • 2 Y
Y by Adventure10, Mango247
Given convex quadrilateral $ABCD$ inscribed circle $(O)$. Let $I$ be the intersection of $AC$ and $BC$. Line $\Delta$ passes through $I,$ intersects the segments $AB$, $CD$ at $M$, $N$ and intersects $(O)$ at $P$, $Q$, respectively ($M, N$ lie on the segments $IQ, IP$, respectively). Prove that $$\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}$$
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Luis González
4145 posts
#3 • 3 Y
Y by HHT23, Adventure10, Mango247
By Desargues involution theorem, $\Delta$ cuts the opposite sidelines of $ABCD$ and its circumcircle $(O)$ at pairs of points in involution $\Longrightarrow$ $(P,M,I,N)=(Q,N,I,M)$ $\Longrightarrow$

$\frac{PM}{PI} \cdot \frac{NI}{NM}=\frac{QN}{QI} \cdot \frac{MI}{MN} \Longrightarrow \frac{MP}{IM \cdot IP}=\frac{QN}{IN \cdot IQ} \Longrightarrow$

$\frac{IM+IP}{IM \cdot IP}=\frac{IN +IQ}{IN \cdot IQ} \Longrightarrow \frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}.$

P.S. Note that the relation still holds for any $ABCD$ and a conic through $A,B,C,D.$
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HHT23
34 posts
#4 • 2 Y
Y by Adventure10, Mango247
Can you tell me what Desargues involution theorem is and how to prove it? Thank you.
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TelvCohl
2312 posts
#5 • 3 Y
Y by HHT23, Adventure10, Mango247
HHT23 wrote:
Given convex quadrilateral $ABCD$ inscribed circle $(O)$. Let $I$ be the intersection of $AC$ and $B \color{red}{ D }\normalcolor $. Line $\Delta$ passes through $I,$ intersects the segments $AB$, $CD$ at $M$, $N$ and intersects $(O)$ at $P$, $Q$, respectively ($M, N$ lie on the segments $IQ, IP$, respectively). Prove that $$\frac{1}{IM}+\frac{1}{IP}=\frac{1}{IN}+\frac{1}{IQ}$$
Typo corrected :)

My solution:

Let $ X=\Delta \cap \odot (ICD), Y=\Delta \cap \odot (IAB) $ .
Let $ \Psi $ be the inversion with center $ I $ which swaps $ P, Q $ .
Let $ A^*, D^* $ be the reflection of $ A, D $ in the perpendicular bisector $ \ell $ of $ PQ $, respectively .

Since $ \angle AA^*D=180^{\circ}-\angle DCA=180^{\circ}-\angle DXY $ ,
so combine with $ AA^* \parallel XY $ we get $ A^*, D, X $ are collinear .
Similarly we can prove $ Y \in D^*A \Longrightarrow X, Y $ are symmetry WRT $ \ell \Longrightarrow QX=PY $ .

Since $ X, Y, P, Q $ is the image of $ M, N, Q, P $ under $ \Psi $, respectively ,

so from $ IQ+IX=QX=PY=IP+IY \Longrightarrow \frac{1}{IP}+\frac{1}{IM}=\frac{1}{IQ}+\frac{1}{IN} $ .

Q.E.D
____________________________________________________________
P.S. You can find some information about Desargue involution theorem at here :)
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