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collinearity wanted, 2 intersecting circles tangent to 3rd circle and it's chord
parmenides51   1
N 12 minutes ago by SuperBarsh
Source: 2008 Italy TST 1.3
Let $ABC$ be an acute triangle, let $AM$ be a median, and let $BK$ and $CL$ be the altitudes. Let $s$ be the line perpendicular to $AM$ passing through $A$. Let $E$ be the intersection point of $s$ with $CL$, and let $F$ be the intersection point of $s$ with $BK$.
(a) Prove that $A$ is the midpoint of $EF$.
(b) Let $\Gamma$ be the circumscribed circle of the triangle $MEF$ , and let $\Gamma_1$ and $\Gamma_2$ be any two circles that have two points $P$ and $Q$ in common, and are tangent to the segment $EF$ and the arc $EF$ of $\Gamma$ not containing the point $M$. Prove that points $M, P, Q$ are collinear.
1 reply
parmenides51
Sep 25, 2020
SuperBarsh
12 minutes ago
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Diophantine equation with primes
BR1F1SZ   6
N an hour ago by ririgggg
Source: Argentina IberoAmerican TST 2024 P1
Find all positive prime numbers $p$, $q$ that satisfy the equation
$$p(p^4+p^2+10q)=q(q^2+3).$$
6 replies
BR1F1SZ
Aug 9, 2024
ririgggg
an hour ago
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one variable function
youochange   1
N an hour ago by Fishheadtailbody
$f:\mathbb R-\{0,1\} \to \mathbb R$


$f(x)+f(\frac{1}{1-x})=2x$
1 reply
youochange
an hour ago
Fishheadtailbody
an hour ago
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Tangent circles
Vlados021   3
N an hour ago by fearsum_fyz
Source: 2017 Belarus Team Selection Test 6.3
Given an isosceles triangle $ABC$ with $AB=AC$. let $\omega(XYZ)$ be the circumcircle of a triangle $XYZ$. Tangents to $\omega(ABC)$ at $B$ and $C$ meet at $D$. Point $F$ is marked on the arc $AB$ (opposite to $C$). Let $K$, $L$ be the intersection points of $AF$ and $BD$, $AB$ and $CF$, respectively.
Prove that if circles $\omega(BTS)$ and $\omega(CFK)$ are tangent to each other, the their tangency point belongs to $AB$. (Here $T$ and $S$ are the centers of the circles $\omega(BLC)$ and $\omega(BLK)$, respectively.)
3 replies
Vlados021
Mar 31, 2019
fearsum_fyz
an hour ago
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Inspired by SunnyEvan
sqing   0
an hour ago
Source: Own
Let $ a,b,c $ be reals such that $ a^2+b^2+c^2=2(ab+bc+ca). $ Prove that$$ \frac{1}{12} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{5}{36} $$Let $ a,b,c $ be reals such that $ a^2+b^2+c^2=5(ab+bc+ca). $ Prove that$$ -\frac{1}{25} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{197}{675} $$
0 replies
sqing
an hour ago
0 replies
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Prove that the triangle is isosceles.
TUAN2k8   5
N an hour ago by JARP091
Source: My book
Given acute triangle $ABC$ with two altitudes $CF$ and $BE$.Let $D$ be the point on the line $CF$ such that $DB \perp BC$.The lines $AD$ and $EF$ intersect at point $X$, and $Y$ is the point on segment $BX$ such that $CY \perp BY$.Suppose that $CF$ bisects $BE$.Prove that triangle $ACY$ is isosceles.
5 replies
TUAN2k8
Yesterday at 9:55 AM
JARP091
an hour ago
J
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Israel Number Theory
mathisreaI   64
N an hour ago by Jlzh25
Source: IMO 2022 Problem 5
Find all triples $(a,b,p)$ of positive integers with $p$ prime and \[ a^p=b!+p. \]
64 replies
mathisreaI
Jul 13, 2022
Jlzh25
an hour ago
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Find the minimum
sqing   7
N 2 hours ago by sqing
Source: China Shandong High School Mathematics Competition 2025 Q4
Let $ a,b,c>0,abc>1$. Find the minimum value of $ \frac {abc(a+b+c+8)}{abc-1}. $
7 replies
sqing
Today at 9:12 AM
sqing
2 hours ago
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Insspired by Shandong 2025
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a,b,c>0,abc>1$. Prove that$$ \frac {abc(a+b+c+ab+bc+ca+3)}{ abc-1}\geq \frac {81}{4}$$$$ \frac {abc(a+b+c+ab+bc+ca+abc+2)}{ abc-1}\geq 12+8\sqrt{2}$$
5 replies
sqing
Today at 9:23 AM
sqing
2 hours ago
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Bound of number of connected components
a_507_bc   3
N 3 hours ago by MmdMathLover
Source: St. Petersburg 2023 11.7
Let $G$ be a connected graph and let $X, Y$ be two disjoint subsets of its vertices, such that there are no edges between them. Given that $G/X$ has $m$ connected components and $G/Y$ has $n$ connected components, what is the minimal number of connected components of the graph $G/(X \cup Y)$?
3 replies
a_507_bc
Aug 12, 2023
MmdMathLover
3 hours ago
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A circle tangent to the circumcircle, excircles related
kosmonauten3114   0
3 hours ago
Source: My own, maybe well-known
Let $ABC$ be a scalene triangle with excircles $\odot(I_A)$, $\odot(I_B)$, $\odot(I_C)$. Let $\odot(A')$ be the circle which touches $\odot(I_B)$ and $\odot(I_C)$ and passes through $A$, and whose center $A'$ lies outside of the excentral triangle of $\triangle{ABC}$. Define $\odot(B')$ and $\odot(C')$ cyclically. Let $\odot(O')$ be the circle externally tangent to $\odot(A')$, $\odot(B')$, $\odot(C')$.

Prove that $\odot(O')$ is tangent to the circumcircle of $\triangle{ABC}$ at the anticomplement of the Feuerbach point of $\triangle{ABC}$.
0 replies
kosmonauten3114
3 hours ago
0 replies
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Reflected point lies on radical axis
Mahdi_Mashayekhi   3
N Apr 19, 2025 by Mahdi_Mashayekhi
Source: Iran 2025 second round P4
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
3 replies
Mahdi_Mashayekhi
Apr 19, 2025
Mahdi_Mashayekhi
Apr 19, 2025
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Reflected point lies on radical axis
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Reveal topic tags
geometry
circumcircle
reflection
power of a point
radical axis
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G H BBookmark kLocked kLocked NReply
Source: Iran 2025 second round P4
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Mahdi_Mashayekhi
695 posts
Mahdi_Mashayekhi
#1 h  Apr 19, 2025, 11:01 AM
Y by
Given is an acute and scalene triangle $ABC$ with circumcenter $O$. $BO$ and $CO$ intersect the altitude from $A$ to $BC$ at points $P$ and $Q$ respectively. $X$ is the circumcenter of triangle $OPQ$ and $O'$ is the reflection of $O$ over $BC$. $Y$ is the second intersection of circumcircles of triangles $BXP$ and $CXQ$. Show that $X,Y,O'$ are collinear.
This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Apr 19, 2025, 11:25 AM
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ItzsleepyXD
147 posts
ItzsleepyXD
#2 h  Apr 19, 2025, 11:26 AM
Y by
Angle chasing :
$\angle BPX = \angle BPA + \angle APX = 180^{\circ} - \angle BAP - \angle ABO + 90^{\circ} + \angle ACB = \angle BAC + 90^{\circ} = 180^{\circ} - \angle O'BC $
and from $OX//BC$
Let $O'B \cap OX=E , O'B \cap OX = F$
so $\angle BEX + \angle BPX = 180^{\circ}$
thus $E,B,P,X$ concyclic .
in the same way $F,C,P,X$ concyclic.
so $O'$ is on the radical axis of $(BXP),(CXQ)$
implies that $X,Y,O'$ collinear . done $\square$
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gghx
1072 posts
gghx
#3 h  Apr 19, 2025, 11:35 AM • 1 Y
Y by blackidea
Define $Y'$ as the second intersection of $O'X$ and $(BO'C)$. We claim that $Y=Y'$. Note that $$\angle BY'X=\angle BCO'=\angle BCO=90^\circ-\angle PQO=\angle XPO,$$hence $BPYX'$ is concyclic. Similarly, $CXQY$ is concyclic, and we are done.
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Mahdi_Mashayekhi
695 posts
Mahdi_Mashayekhi
#4 h  Apr 19, 2025, 4:34 PM • 2 Y
Y by Parsia--, sami1618
Note that $\angle XYC = \angle XQC =\angle XQO = 90 - \frac{\angle QXO}{2} = 90 - \angle QPO = 90 - (90 - \angle C + 90 - \angle B) = 90-\angle A$ and $\angle XYB = \angle XPO = 90 - \frac{\angle PXO}{2} = 90 - \angle PQO = 90 - (90 - \angle C + 90 - \angle B) = 90 - \angle A$ so $X$ lies on angle bisector of $BYC$.
Now note that $\angle BYC = 2(90 - \angle A) = 180 - 2\angle A = 180 - \angle BOC = 180 - \angle BO'C$ so $BYCO'$ is cyclic and since $BO'=CO'$ then $O'$ also lies on angle bisector of $BYC$ so $Y,X,O'$ are collinear as wanted.
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