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Heptagon in Taiwan TST!!!
Hakurei_Reimu   1
N 6 minutes ago by CrazyInMath
Source: 2025 Taiwan TST Round 3 Independent Study 2-G
Let $ABCDEFG$ be a regular heptagon with its center $O$. $H$ is the orthocenter of triangle $CDF$, $I$ is the incenter of triangle $ABD$. Let $M$ be the midpoint of $IG$ and $X$ be the intersection point of $OH$ and $FG$. Assume $P$ is the circumcenter of triangle $BCI$. Prove that $CF, MP, XB$ concur at a single point.

Proposed by HakureiReimu.
1 reply
Hakurei_Reimu
3 hours ago
CrazyInMath
6 minutes ago
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Perpendicular if and only if Centre
shobber   3
N 8 minutes ago by Tonne
Source: Pan African 2004
Let $ABCD$ be a cyclic quadrilateral such that $AB$ is a diameter of it's circumcircle. Suppose that $AB$ and $CD$ intersect at $I$, $AD$ and $BC$ at $J$, $AC$ and $BD$ at $K$, and let $N$ be a point on $AB$. Show that $IK$ is perpendicular to $JN$ if and only if $N$ is the midpoint of $AB$.
3 replies
shobber
Oct 4, 2005
Tonne
8 minutes ago
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$5^t + 3^x4^y = z^2$
Namisgood   0
14 minutes ago
Source: JBMO shortlist 2017
Solve in nonnegative integers the equation $5^t + 3^x4^y = z^2$
0 replies
Namisgood
14 minutes ago
0 replies
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Concurrent lines
MathChallenger101   2
N 19 minutes ago by pigeon123
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
2 replies
MathChallenger101
Feb 8, 2025
pigeon123
19 minutes ago
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Find all natural numbers $n$
ItsBesi   7
N 24 minutes ago by justaguy_69
Source: Kosovo Math Olympiad 2025, Grade 9, Problem 2
Find all natural numbers $n$ such that $\frac{\sqrt{n}}{2}+\frac{10}{\sqrt{n}}$ is a natural number.
7 replies
ItsBesi
Nov 17, 2024
justaguy_69
24 minutes ago
J
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diopantine 5^t + 3^x4^y = z^2 , solve in nonnegative
parmenides51   10
N 41 minutes ago by Namisgood
Source: JBMO Shortlist 2017 NT4
Solve in nonnegative integers the equation $5^t + 3^x4^y = z^2$.
10 replies
parmenides51
Jul 25, 2018
Namisgood
41 minutes ago
J
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\pi(n) is a good divisor
sefatod628   1
N 42 minutes ago by kiyoras_2001
Source: unknown/given at a math club
Prove that there is an infinity of $n\in \mathbb{N^*}$ such that $\pi(n) \mid n$.

Hint :Click to reveal hidden text
1 reply
sefatod628
Yesterday at 2:37 PM
kiyoras_2001
42 minutes ago
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Easy Functional Inequality Problem in Taiwan TST
chengbilly   3
N an hour ago by shanelin-sigma
Source: 2025 Taiwan TST Round 3 Mock P4
Let $a$ be a positive real number. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $af(x) - f(y) + y > 0$ and
\[ f(af(x) - f(y) + y) \leq x + f(y) - y, \quad \forall x, y \in \mathbb{R}^+. \]
proposed by chengbilly
3 replies
chengbilly
4 hours ago
shanelin-sigma
an hour ago
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4 var inequality
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c,d >0 $ and $ abcd=3,a+b+c+d=6 $. Prove that
$$ a^2+b^2+c^2+d^2 \leq 12 $$$$ a^3+b^3+c^3+d^3 \leq 30$$$$ a^4+b^4+c^4+d^4 \leq84$$
1 reply
sqing
an hour ago
sqing
an hour ago
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Dou Fang Geometry in Taiwan TST
Li4   1
N 2 hours ago by Parsia--
Source: 2025 Taiwan TST Round 3 Mock P2
Let $\omega$ and $\Omega$ be the incircle and circumcircle of the acute triangle $ABC$, respectively. Draw a square $WXYZ$ so that all of its sides are tangent to $\omega$, and $X$, $Y$ are both on $BC$. Extend $AW$ and $AZ$, intersecting $\Omega$ at $P$ and $Q$, respectively. Prove that $PX$ and $QY$ intersects on $\Omega$.

Proposed by kyou46, Li4, Revolilol.
1 reply
Li4
Today at 5:03 AM
Parsia--
2 hours ago
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4 var inequality
sqing   1
N 2 hours ago by sqing
Source: https://bbs.emath.ac.cn/thread-39778-1-1.html
Let $ a,b,c,d>0 $ and $ a+b+c+d=4. $ Prove that$$a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}\leq 2(1+\sqrt{abcd})$$Let $ a,b,c,d\geq -1 $ and $ a+b+c+d=0. $ Prove that$$ab+bc+cd\leq \frac{5}{4}$$
1 reply
sqing
2 hours ago
sqing
2 hours ago
J
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easy functional
B1t   5
N 2 hours ago by Haris1
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[ f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)). \]
5 replies
B1t
4 hours ago
Haris1
2 hours ago
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orz otl fr
Hip1zzzil   8
N Mar 31, 2025 by jerrome2685
Source: FKMO 2025 P3
An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$.
$I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$. $AD$ and $BE$ meet at $P$. Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$, and define $l_{2}$ and $l_{3}$ on $E$ and $F$, respectively.
Prove $l_{1},l_{2},l_{3}$ meet at one point.
8 replies
Hip1zzzil
Mar 29, 2025
jerrome2685
Mar 31, 2025
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orz otl fr
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Reveal topic tags
geometry
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G H BBookmark kLocked kLocked NReply
Source: FKMO 2025 P3
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Hip1zzzil
14 posts
Hip1zzzil
#1 h  Mar 29, 2025, 10:23 AM
Y by
An acute triangle $\bigtriangleup ABC$ is given which $BC>CA>AB$.
$I$ is the interior and the incircle of $\bigtriangleup ABC$ meets $BC, CA, AB$ at $D,E,F$. $AD$ and $BE$ meet at $P$. Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$, and define $l_{2}$ and $l_{3}$ on $E$ and $F$, respectively.
Prove $l_{1},l_{2},l_{3}$ meet at one point.
This post has been edited 4 times. Last edited by Hip1zzzil, Mar 29, 2025, 10:12 PM
Reason: Mistake
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whwlqkd
97 posts
whwlqkd
#2 h  Mar 29, 2025, 10:32 AM
Y by
Why is geometry in p3/6
It is very unexpected and soooooo hard
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EthanWYX2009
855 posts
EthanWYX2009
#3 h  Mar 29, 2025, 11:35 AM
Y by
Complex bash should be working, ig 30 minute calculation
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kmh1
105 posts
kmh1
#4 h  Mar 29, 2025, 12:47 PM • 1 Y
Y by Retemoeg
I assume you mean $I$ is the incenter, not just any point in the interior.

solution
This post has been edited 2 times. Last edited by kmh1, Mar 29, 2025, 12:57 PM
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Acorn-SJ
59 posts
Acorn-SJ
#5 h  Mar 29, 2025, 12:50 PM
Y by
Yes, $I$ is supposed to be the incenter. Also, the original statement contained the fact that $BC > CA > AB$.
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LoloChen
478 posts
LoloChen
#6 h  Mar 29, 2025, 1:25 PM • 3 Y
Y by qwerty123456asdfgzxcvb, whwlqkd, Resolut1on07
But this can be solved instantly using cubic curve. :maybe:
Generalization (using basically the same method): $I$ is the Miquel point of $\bigtriangleup ABC$ and $D,E,F$ which lies on $BC, CA, AB$ respectively. Furthermore, $AD$, $BE$ and $CF$ meet at $P$. Let $l_{1}$ be a tangent from D to the circumcircle of $\bigtriangleup DIP$, and define $l_{2}$ and $l_{3}$ on $E$ and $F$, respectively. Then $l_{1},l_{2},l_{3}$ meet at one point.
Sketch of proof: Consider the Qa-Cu1 cubic curve of $ABCDEFP$. $l_{1},l_{2},l_{3}$ are the tangents of Qa-Cu1. Consider its Abelian group, we only need $D+D=E+E=F+F$. But that's pretty obvious because $A+P+D=B+P+E=C+P+F$ and $A+A=B+B=C+C$. See here for details.
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space10
8 posts
space10
#7 h  Mar 30, 2025, 11:29 AM • 8 Y
Y by qwedsazxc, ehuseyinyigit, segment, GuvercinciHoca, Mop2018, Kingsbane2139, Lufin, Gaussss
Very good problem. This is my solution.
Attachments:
This post has been edited 1 time. Last edited by space10, Mar 30, 2025, 1:54 PM
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sansae
119 posts
sansae
#8 h  Mar 30, 2025, 3:24 PM
Y by
Complex bash should definitely work in contest hour.

Let the incircle be the unit circle, with $D=a, E=b, F=c$. Then $P=K$ is the symmedian point, whose complex coordinate is given in Post #3 by
$$k=\frac{2(\sum a^{2}b^{2} - \sum a^{2}bc)}{\sum_{sym}a^{2}b-6abc}$$(All the summations without index will denote a cyclic sum.)

Let $Q=q$ be the intersection of $l_{1}$ and $l_{2}$. It is enough to show that $q$ is symmetric w.r.t. $a, b, c$.

Since $\angle QDO = \angle OKD$,
$$ \frac{k(q-a)}{a(k-a)} = \frac{a\overline{k}(\overline{q}-\overline{a})}{\overline{k}-\overline{a}}, \quad \overline{q}-\overline{a} = \frac{k(q-a)(\overline{k}-\overline{a})}{a^{2}\overline{k}(k-a)}$$
Note that by definition of $K=k$, $D=a$, and $A=\frac{2bc}{b+c}$ are collinear, hence
$$ \frac{\overline{k}-\overline{a}}{k-a} = \frac{b+c-2a}{a(ab+ac-2bc)}$$
Using the formula for $k$, one can easily see that
$$ \frac{k}{\overline{k}} = \frac{\sum a^{2}b^{2} - \sum a^{2} bc}{\sum a^{2} - \sum ab}$$.

Plugging these into the equation for $q$, and subtracting with the similar equation for the $E=b$-side gives a linear equation on $q$:
$$\overline{b}-\overline{a} = \frac{\sum a^{2}b^{2} - \sum a^{2} bc}{\sum a^{2} - \sum ab}\left ( \frac{(b+c-2a)(q-a)}{a^{3}(ab+ac-2bc)} - \frac{(c+a-2b)(q-b)}{b^{3}(bc+ba-2ca)}\right )$$
Solving this equation gives
$$q=\frac{abc(\sum ab) (\sum_{sym}a^{2}b - 6abc)}{(\sum_{sym}a^{2}b) (\sum a^{2}b^{2} - \sum a^{2}bc)},$$which is symmetric. Hence the claim.
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jerrome2685
46 posts
jerrome2685
#9 h  Mar 31, 2025, 7:05 AM
Y by
Same problem with https://artofproblemsolving.com/community/q2h2407674p19737244 (Note that $P$ is a symmedian point of $DEF$)
This post has been edited 1 time. Last edited by jerrome2685, Mar 31, 2025, 7:05 AM
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