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A circle tangent to the circumcircle, excircles related
kosmonauten3114   0
an hour 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
an hour ago
0 replies
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Bounds on degree of polynomials
Phorphyrion   4
N an hour ago by Kingsbane2139
Source: 2020 Israel Olympic Revenge P3
For each positive integer $n$, define $f(n)$ to be the least positive integer for which the following holds:

For any partition of $\{1,2,\dots, n\}$ into $k>1$ disjoint subsets $A_1, \dots, A_k$, all of the same size, let $P_i(x)=\prod_{a\in A_i}(x-a)$. Then there exist $i\neq j$ for which
\[\deg(P_i(x)-P_j(x))\geq \frac{n}{k}-f(n)\]
a) Prove that there is a constant $c$ so that $f(n)\le c\cdot \sqrt{n}$ for all $n$.

b) Prove that for infinitely many $n$, one has $f(n)\ge \ln(n)$.
4 replies
+1 w
Phorphyrion
Jun 11, 2022
Kingsbane2139
an hour ago
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A point on BC
jayme   7
N 2 hours ago by jayme
Source: Own ?
Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. D the pole of BC wrt 0
4. B', C' the symmetrics of B, C wrt AC, AB
5. 1b, 1c the circumcircles of the triangles BB'D, CC'D
6. T the second point of intersection of the tangent to 1c at D with 1b.

Prove : B, C and T are collinear.

Sincerely
Jean-Louis
7 replies
jayme
Today at 6:08 AM
jayme
2 hours ago
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Zack likes Moving Points
pinetree1   73
N 2 hours ago by NumberzAndStuff
Source: USA TSTST 2019 Problem 5
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle AEF$, and suppose line $AK$ intersects $\Gamma$ again at a point $D$. Prove that line $HK$ and the line through $D$ perpendicular to $\overline{BC}$ meet on $\Gamma$.

Gunmay Handa
73 replies
pinetree1
Jun 25, 2019
NumberzAndStuff
2 hours ago
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Domain and Inequality
Kunihiko_Chikaya   1
N 2 hours ago by Mathzeus1024
Source: 2018 The University of Tokyo entrance exam / Humanities, Problem 1
Define on a coordinate plane, the parabola $C:y=x^2-3x+4$ and the domain $D:y\geq x^2-3x+4.$
Suppose that two lines $l,\ m$ passing through the origin touch $C$.

(1) Let $A$ be a mobile point on the parabola $C$. Let denote $L,\ M$ the distances between the point $A$ and the lines $l,\ m$ respectively. Find the coordinate of the point $A$ giving the minimum value of $\sqrt{L}+\sqrt{M}.$

(2) Draw the domain of the set of the points $P(p,\ q)$ on a coordinate plane such that for all points $(x,\ y)$ over the domain $D$, the inequality $px+qy\leq 0$ holds.
1 reply
Kunihiko_Chikaya
Feb 25, 2018
Mathzeus1024
2 hours ago
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JBMO TST Bosnia and Herzegovina 2020 P1
Steve12345   3
N 2 hours ago by AylyGayypow009
Determine all four-digit numbers $\overline{abcd}$ which are perfect squares and for which the equality holds:
$\overline{ab}=3 \cdot \overline{cd} + 1$.
3 replies
Steve12345
Aug 10, 2020
AylyGayypow009
2 hours ago
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Problem3
samithayohan   116
N 2 hours ago by fearsum_fyz
Source: IMO 2015 problem 3
Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Proposed by Ukraine
116 replies
samithayohan
Jul 10, 2015
fearsum_fyz
2 hours ago
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geometry problem
invt   0
2 hours ago
In a triangle $ABC$ with $\angle B<\angle C$, denote its incenter and midpoint of $BC$ by $I$, $M$, respectively. Let $C'$ be the reflected point of $C$ wrt $AI$. Let the lines $MC'$ and $CI$ meet at $X$. Suppose that $\angle XAI=\angle XBI=90^{\circ}$. Prove that $\angle C=2\angle B$.
0 replies
invt
2 hours ago
0 replies
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the locus of $P$
littletush   10
N 3 hours ago by SuperBarsh
Source: Italy TST 2009 p2
$ABC$ is a triangle in the plane. Find the locus of point $P$ for which $PA,PB,PC$ form a triangle whose area is equal to one third of the area of triangle $ABC$.
10 replies
littletush
Mar 10, 2012
SuperBarsh
3 hours ago
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Abelkonkurransen 2025 3b
Lil_flip38   3
N 3 hours ago by Adywastaken
Source: abelkonkurransen
An acute angled triangle \(ABC\) has circumcenter \(O\). The lines \(AO\) and \(BC\) intersect at \(D\), while \(BO\) and \(AC\) intersect at \(E\) and \(CO\) and \(AB\) intersect at \(F\). Show that if the triangles \(ABC\) and \(DEF\) are similar(with vertices in that order), than \(ABC\) is equilateral.
3 replies
Lil_flip38
Mar 20, 2025
Adywastaken
3 hours ago
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I got stuck in this combinatorics
artjustinhere237   3
N 3 hours ago by GreekIdiot
Let $S = \{1, 2, 3, \ldots, k\}$, where $k \geq 4$ is a positive integer.
Prove that there exists a subset of $S$ with exactly $k - 2$ elements such that the sum of its elements is a prime number.
3 replies
artjustinhere237
May 13, 2025
GreekIdiot
3 hours ago
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Prove a geometry ratio
nhhlqd   1
N Mar 29, 2021 by PROF65
Given triangle $ABC$ with $AB<AC$ inscribed in a circle centered $O$, denoted as $\Gamma$. Let $I$ be an arbitrary point on the circumcircle of triangle $BOC$. Given $E$, $F$ be the moving points on $\Gamma$ such that $BE\parallel CF$. Suppose $BI\cap AF = P$; $CI\cap AE = Q$ and $EF\cap PQ = K$. Shown that the ratio $\dfrac{KE}{KF}$ is not change as $E$ and $F$ move.

This problem created when I tried to generalize a problem. However, I couldn't solve this problem with the way I've done with the previous one.
Any help is appreciated.
Thanks a lot.
1 reply
nhhlqd
Mar 29, 2021
PROF65
Mar 29, 2021
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Prove a geometry ratio
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nhhlqd
59 posts
nhhlqd
#1 h  Mar 29, 2021, 8:45 AM • 1 Y
Y by Mango247
Given triangle $ABC$ with $AB<AC$ inscribed in a circle centered $O$, denoted as $\Gamma$. Let $I$ be an arbitrary point on the circumcircle of triangle $BOC$. Given $E$, $F$ be the moving points on $\Gamma$ such that $BE\parallel CF$. Suppose $BI\cap AF = P$; $CI\cap AE = Q$ and $EF\cap PQ = K$. Shown that the ratio $\dfrac{KE}{KF}$ is not change as $E$ and $F$ move.

This problem created when I tried to generalize a problem. However, I couldn't solve this problem with the way I've done with the previous one.
Any help is appreciated.
Thanks a lot.
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PROF65
2016 posts
PROF65
#2 h  Mar 29, 2021, 1:17 PM
Y by
Let $BF \cap CE= K$ since $BE\parallel CF$ then $ \angle BKE=2 \angle BAC$ i.e.
$K\in (BOC) \implies \angle ICE= \angle IBF$ ,we have also $BF=CE $ because $BCFE$ is trapezoid $ (^*)$.
By Menelaus's it suffices to show that $\frac{PF}{PA}.\frac{QA}{AE} $ is constant .

$\frac{PF}{PA}.\frac{QA}{AE} =\frac{BF}{BA}.\frac{\sin \angle IBF}{\sin\angle IBA}.\frac{CA}{CE}.\frac{\sin \angle ICA}{\sin \angle ICE} \overset{(^*)}{=} $

$\frac{CA}{BA}.\frac{\sin \angle ICA}{\sin \angle IBA} $ which obviously depends only on the choice of $I$
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