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geometry
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Right angles
USJL   10
N an hour ago by bin_sherlo
Source: 2018 Taiwan TST Round 3
Let $I$ be the incenter of triangle $ABC$, and $\ell$ be the perpendicular bisector of $AI$. Suppose that $P$ is on the circumcircle of triangle $ABC$, and line $AP$ and $\ell$ intersect at point $Q$. Point $R$ is on $\ell$ such that $\angle IPR = 90^{\circ}$.Suppose that line $IQ$ and the midsegment of $ABC$ that is parallel to $BC$ intersect at $M$. Show that $\angle AMR = 90^{\circ}$

(Note: In a triangle, a line connecting two midpoints is called a midsegment.)
10 replies
USJL
Apr 2, 2020
bin_sherlo
an hour ago
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Inequality with a,b,c
GeoMorocco   0
an hour ago
Source: Morocco Training
Let $a,b,c$ be positive real numbers. Prove that:
$$\sqrt[3]{a^3+b^3}+\sqrt[3]{b^3+c^3}+\sqrt[3]{c^3+a^3}\geq \sqrt[3]{2}(a+b+c)$$
0 replies
GeoMorocco
an hour ago
0 replies
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Abusing surjectivity
Sadigly   7
N 2 hours ago by Sadigly
Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ and $g:\mathbb{Q}\rightarrow\mathbb{Q}$ such that

$$f(f(x)+yg(x))=(x+1)g(y)+f(y)$$
for any $x;y\in\mathbb{Q}$
7 replies
Sadigly
4 hours ago
Sadigly
2 hours ago
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Ratio of lengths
Sadigly   1
N 2 hours ago by Frd_19_Hsnzde
In a triangle $ABC$, $I$ is the incenter. Line $CI$ intersects circumcircle of $ABC$ at $L$, and it is given that $CI=2IL$. $M;N$ are points chosen on $AB$ such that $\angle AIM=\angle BIN=90$. Prove that $AB=2MN$
1 reply
Sadigly
3 hours ago
Frd_19_Hsnzde
2 hours ago
J
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Mock 22nd Thailand TMO P4
korncrazy   1
N 2 hours ago by YaoAOPS
Source: own
Let $n$ be a positive integer. In an $n\times n$ table, an upright path is a sequence of adjacent cells starting from the southwest corner to the northeast corner such that the next cell is either on the top or on the right of the previous cell. Find the smallest number of grids one needs to color in an $n\times n$ table such that there exists only one possible upright path not containing any colored cells.
1 reply
korncrazy
4 hours ago
YaoAOPS
2 hours ago
J
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Sets With a Given Property
oVlad   4
N 2 hours ago by oVlad
Source: Romania TST 2025 Day 1 P4
Determine the sets $S{}$ of positive integers satisfying the following two conditions:
[list=a]
[*]For any positive integers $a, b, c{}$, if $ab + bc + ca{}$ is in $S$, then so are $a + b + c{}$ and $abc$; and
[*]The set $S{}$ contains an integer $N \geqslant 160$ such that $N-2$ is not divisible by $4$.
[/list]
Bogdan Blaga, United Kingdom
4 replies
oVlad
Apr 9, 2025
oVlad
2 hours ago
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Mock 22nd Thailand TMO P10
korncrazy   2
N 2 hours ago by aidan0626
Source: own
Prove that there exists infinitely many triples of positive integers $(a,b,c)$ such that $a>b>c,\,\gcd(a,b,c)=1$ and $$a^2-b^2,a^2-c^2,b^2-c^2$$are all perfect square.
2 replies
korncrazy
4 hours ago
aidan0626
2 hours ago
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pairwise coprime sum gcd
InterLoop   22
N 2 hours ago by Nuran2010
Source: EGMO 2025/1
For a positive integer $N$, let $c_1 < c_2 < \dots < c_m$ be all the positive integers smaller than $N$ that are coprime to $N$. Find all $N \ge 3$ such that
$$\gcd(N, c_i + c_{i+1}) \neq 1$$for all $1 \le i \le m - 1$.
22 replies
InterLoop
Today at 12:34 PM
Nuran2010
2 hours ago
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one cyclic formed by two cyclic
CrazyInMath   15
N 3 hours ago by ThatApollo777
Source: EGMO 2025/3
Let $ABC$ be an acute triangle. Points $B, D, E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Suppose triangle $ADE$ is acute, and let $H$ be its orthocentre. Points $P$ and $Q$ lie on lines $BM$ and $CN$, respectively, such that $D, H, M,$ and $P$ are concyclic and pairwise different, and $E, H, N,$ and $Q$ are concyclic and pairwise different. Prove that $P, Q, N,$ and $M$ are concyclic.
15 replies
CrazyInMath
Today at 12:38 PM
ThatApollo777
3 hours ago
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Mock 22nd Thailand TMO P3
korncrazy   2
N 3 hours ago by korncrazy
Source: own
Find all triples of positive integers $(a,b,c)$ such that $$a|b+c,\,b|c+a, c|a+b$$and $\gcd(a,b,c)=1$.
2 replies
korncrazy
4 hours ago
korncrazy
3 hours ago
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IMO ShortList 1998, geometry problem 1
orl   25
N 3 hours ago by cj13609517288
Source: IMO ShortList 1998, geometry problem 1
A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.
25 replies
orl
Oct 22, 2004
cj13609517288
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
J
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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