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diophantine with factorials and exponents
skellyrah   1
N 2 hours ago by pingpongmerrily
find all positive integers $a,b,c$ such that $$ a! + 5^b = c^3 $$
1 reply
skellyrah
2 hours ago
pingpongmerrily
2 hours ago
A circle tangent to AB,AC with center J!
Noob_at_math_69_level   6
N 2 hours ago by awesomeming327.
Source: DGO 2023 Team P2
Let $\triangle{ABC}$ be a triangle with a circle $\Omega$ with center $J$ tangent to sides $AC,AB$ at $E,F$ respectively. Suppose the circle with diameter $AJ$ intersects the circumcircle of $\triangle{ABC}$ again at $T.$ $T'$ is the reflection of $T$ over $AJ$. Suppose points $X,Y$ lie on $\Omega$ such that $EX,FY$ are parallel to $BC$. Prove that: The intersection of $BX,CY$ lie on the circumcircle of $\triangle{BT'C}.$

Proposed by Dtong08math & many authors
6 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
2 hours ago
Easy functional equation
fattypiggy123   15
N 3 hours ago by ariopro1387
Source: Singapore Mathematical Olympiad 2014 Problem 2
Find all functions from the reals to the reals satisfying
\[f(xf(y) + x) = xy + f(x)\]
15 replies
fattypiggy123
Jul 5, 2014
ariopro1387
3 hours ago
Iran TST Starter
M11100111001Y1R   5
N 3 hours ago by DeathIsAwe
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
5 replies
M11100111001Y1R
May 27, 2025
DeathIsAwe
3 hours ago
Very odd geo
Royal_mhyasd   1
N 4 hours ago by Royal_mhyasd
Source: own (i think)
Let $\triangle ABC$ be an acute triangle with $AC>AB>BC$ and let $H$ be its orthocenter. Let $P$ be a point on the perpendicular bisector of $AH$ such that $\angle APH=2(\angle ABC - \angle ACB)$ and $P$ and $C$ are on different sides of $AB$, $Q$ a point on the perpendicular bisector of $BH$ such that $\angle BQH = 2(\angle ACB-\angle BAC)$ and $R$ a point on the perpendicular bisector of $CH$ such that $\angle CRH=2(\angle ABC - \angle BAC)$ and $Q,R$ lie on the opposite side of $BC$ w.r.t $A$. Prove that $P,Q$ and $R$ are collinear.
1 reply
Royal_mhyasd
4 hours ago
Royal_mhyasd
4 hours ago
Calculating sum of the numbers
Sadigly   5
N 4 hours ago by aokmh3n2i2rt
Source: Azerbaijan Junior MO 2025 P4
A $3\times3$ square is filled with numbers $1;2;3...;9$.The numbers inside four $2\times2$ squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $24,24,25,25$

$\text{b)}$ $20,23,26,29$
5 replies
Sadigly
May 9, 2025
aokmh3n2i2rt
4 hours ago
Swap to the symmedian
Noob_at_math_69_level   7
N 4 hours ago by awesomeming327.
Source: DGO 2023 Team P1
Let $\triangle{ABC}$ be a triangle with points $U,V$ lie on the perpendicular bisector of $BC$ such that $B,U,V,C$ lie on a circle. Suppose $UD,UE,UF$ are perpendicular to sides $BC,AC,AB$ at points $D,E,F.$ The tangent lines from points $E,F$ to the circumcircle of $\triangle{DEF}$ intersects at point $S.$ Prove that: $AV,DS$ are parallel.

Proposed by Paramizo Dicrominique
7 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
4 hours ago
Find (AB * CD) / (AC * BD) & prove orthogonality of circles
Maverick   15
N 4 hours ago by Ilikeminecraft
Source: IMO 1993, Day 1, Problem 2
Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)
15 replies
Maverick
Jul 13, 2004
Ilikeminecraft
4 hours ago
f(x+f(x)+f(y))=x+f(x+y)
dangerousliri   10
N 5 hours ago by jasperE3
Source: FEOO, Shortlist A5
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for any positive real numbers $x$ and $y$,
$$f(x+f(x)+f(y))=x+f(x+y)$$Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
10 replies
dangerousliri
May 31, 2020
jasperE3
5 hours ago
n-variable inequality
ABCDE   66
N 5 hours ago by ND_
Source: 2015 IMO Shortlist A1, Original 2015 IMO #5
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
66 replies
ABCDE
Jul 7, 2016
ND_
5 hours ago
PAMO Problem 4: Perpendicular lines
DylanN   11
N Apr 23, 2025 by ATM_
Source: 2019 Pan-African Mathematics Olympiad, Problem 4
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
11 replies
DylanN
Apr 9, 2019
ATM_
Apr 23, 2025
PAMO Problem 4: Perpendicular lines
G H J
G H BBookmark kLocked kLocked NReply
Source: 2019 Pan-African Mathematics Olympiad, Problem 4
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DylanN
194 posts
#1 • 2 Y
Y by Adventure10, Mango247
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
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pad
1671 posts
#3 • 2 Y
Y by Adventure10, Mango247
We know $AD$ is the $A$-symmedian of $\triangle ABC$, so it is well known that $B,C,D,O$ are concyclic. So $(BCD)=(BCO)$. Then $\angle EAO = \angle CAO=90-B$, and $\angle AEF=180-\angle CEF=180-\angle CBF=B$ since $E,C,F,B$ are concyclic. Therefore, $AO\perp EF$.
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DylanN
194 posts
#4 • 2 Y
Y by Adventure10, Mango247
It's not important that $D$ is defined as the intersection of the tangents. We can replace the circumcircle of $\triangle BCD$ with any circle that passes through $B$ and $C$. Indeed, we know that $\angle CAO = 90^\circ - \angle B$, and since $BCEF$ is cyclic, we have that $\angle B = \angle CEF$, from which the result follows.
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khanhnx
1618 posts
#5 • 1 Y
Y by Adventure10
This problem has been posted in here
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AshAuktober
1013 posts
#6
Y by
Observe that we can obtain $\Delta AEF$ by reflecting $\Delta ABC$ about its $A-$angle bisector and taking a homothety at $A$. Now since the circumcentre and orthocentre are isogonal conjugates, $O$ has to lie on the $A-$altitude of $\Delta AEF$, and therefore we are done. $\square$
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Aiden-1089
302 posts
#7
Y by
Let $H$ be the orthocenter of $ABC$.
$\measuredangle BEC = \measuredangle BDC = \measuredangle BFC \implies BCEF$ is concyclic $\implies \Delta ABC \sim \Delta AEF$.
Consider taking a suitable homothety about A then reflecting across the angle bisector of $\angle BAC$, such that $BC$ goes to $EF$. Under this transformation, line $AH$ goes to line $AO$ because they are isogonal in $\angle BAC$. Since $AH \perp BC$, we have $AO \perp EF$.
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Jishnu4414l
155 posts
#8
Y by
We know that $\measuredangle OAC=90^{\circ}-\measuredangle CBA=\measuredangle KAE$
Also, $\measuredangle CBA=\measuredangle CEF=\measuredangle AEK$
Now, $\measuredangle KAE+\measuredangle AEK+\measuredangle EKA=0$
$\implies 90^{\circ}-\measuredangle CBA+\measuredangle CBA +\measuredangle EKA=0$
$\implies \measuredangle EKA=90^{\circ}$
And we are done!
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fasttrust_12-mn
118 posts
#9
Y by
Claim: $O$ lies on the circumcircle of $\triangle BDC$
proof:
$$\measuredangle BAC =\measuredangle BCD \implies \measuredangle BDC= 180^{\circ}- 2\times \measuredangle BAC$$, we know that
$$\measuredangle BOC = 2\times \measuredangle BAC$$so
$$OBDC  \text{ is cyclic quad }$$
$$\measuredangle FBC=\measuredangle AEF $$$$\measuredangle OAC=\measuredangle OCA$$$$\measuredangle OCB=90^{\circ}$$we know that
$$\measuredangle AEG=90^{\circ}-\measuredangle BAO-\measuredangle OAE + \measuredangle BAO\implies 90^{\circ}-\measuredangle OAE$$$$\measuredangle AGE =180^{\circ}-(90^{\circ}-\measuredangle OAE+\measuredangle  OAE)=90^{\circ}$$hence we are done $\blacksquare$
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Tony_stark0094
69 posts
#10
Y by
$$\angle EBF=\angle BAF+\angle AFB=\angle A + \angle BDC=\angle A + 180 - 2\angle A = 180 - \angle A$$$$\implies FA=FB$$$$by\ symmetry\ we \ can\ get \ that\ EA=EC$$hence $F$ lies on perpendicular bisector of $AB$ and $E$ lies on perpendicular bisector of $AC$ moreover $G$ also lies on these lines
$\implies G$ is the orthocentre of $\Delta AEF$ hence $AG$ is perpendicular to $EF$
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Rayanelba
20 posts
#11 • 1 Y
Y by ATM_
To prove that $AO$ perpendicular to $EF$ , it suffies to prove that: $OAC=\frac{\pi}{2}-\measuredangle CFE$
We know that : $\measuredangle OAC=\frac{\pi}{2}-\measuredangle CBA$
And : $\measuredangle CBA=\pi-\measuredangle CBE=\measuredangle CFE$ (because CFEB is cyclic)
$\implies \measuredangle OAC=\frac{\pi}{2}-\measuredangle CFE$
Q.E.D
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ATM_
23 posts
#12
Y by
nice solution*
This post has been edited 1 time. Last edited by ATM_, Apr 23, 2025, 11:08 PM
Reason: .
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ATM_
23 posts
#13
Y by
here is a cute sketch
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