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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
∑(a-b)(a-c)/(2a^2 + (b+c)^2) >= 0
Zhero   24
N 10 minutes ago by RevolveWithMe101
Source: ELMO Shortlist 2010, A2
Let $a,b,c$ be positive reals. Prove that
\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]

Calvin Deng.
24 replies
Zhero
Jul 5, 2012
RevolveWithMe101
10 minutes ago
i am not abel to prove or disprove
frost23   8
N 13 minutes ago by frost23
Source: made on my own
Let $a_1a_1a_2a_2.............a_na_n$ be a perfect square then, is it true that it must be of the form
$10^{2(n-2)}\cdot7744$
8 replies
frost23
2 hours ago
frost23
13 minutes ago
points on sides of a triangle, intersections, extensions, ratio of areas wanted
parmenides51   1
N 21 minutes ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1997 OMM P5
Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.
1 reply
parmenides51
Jul 28, 2018
FrancoGiosefAG
21 minutes ago
starting with intersecting circles, line passes through midpoint wanted
parmenides51   2
N 40 minutes ago by EmersonSoriano
Source: Peru Ibero TST 2014
Circles $C_1$ and $C_2$ intersect at different points $A$ and $B$. The straight lines tangents to $C_1$ that pass through $A$ and $B$ intersect at $T$. Let $M$ be a point on $C_1$ that is out of $C_2$. The $MT$ line intersects $C_1$ at $C$ again, the $MA$ line intersects again to $C_2$ in $K$ and the line $AC$ intersects again to the circumference $C_2$ in $L$. Prove that the $MC$ line passes through the midpoint of the $KL$ segment.
2 replies
1 viewing
parmenides51
Jul 23, 2019
EmersonSoriano
40 minutes ago
An inequality
Rushil   14
N 40 minutes ago by frost23
Source: Indian RMO 1994 Problem 8
If $a,b,c$ are positive real numbers such that $a+b+c = 1$, prove that \[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]
14 replies
Rushil
Oct 25, 2005
frost23
40 minutes ago
3 var inequality
SunnyEvan   6
N 41 minutes ago by JARP091
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
6 replies
SunnyEvan
May 17, 2025
JARP091
41 minutes ago
collinearity as a result of perpendicularity and equality
parmenides51   2
N an hour ago by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1996 OMM P6
In a triangle $ABC$ with $AB < BC < AC$, points $A' ,B' ,C'$ are such that $AA' \perp BC$ and $AA' = BC, BB' \perp  CA$ and $BB'=CA$, and $CC' \perp AB$ and $CC'= AB$, as shown on the picture. Suppose that $\angle AC'B$ is a right angle. Prove that the points $A',B' ,C' $ are collinear.
2 replies
parmenides51
Jul 28, 2018
FrancoGiosefAG
an hour ago
3 var inequality
JARP091   6
N an hour ago by JARP091
Source: Own
Let \( x, y, z \in \mathbb{R}^+ \). Prove that
\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]without using the Rearrangement Inequality or Chebyshev's Inequality.
6 replies
JARP091
Today at 8:54 AM
JARP091
an hour ago
Helplooo
Bet667   1
N an hour ago by Lil_flip38
Let $ABC$ be an acute angled triangle.And altitudes $AD$ and $BE$ intersects at point $H$.Let $F$ be a point on ray $AD$ such that $DH=DF$.Circumcircle of $AEF$ intersects line $BC$ at $K$ and $L$ so prove that $BK=BL$
1 reply
Bet667
2 hours ago
Lil_flip38
an hour ago
Cyclic sum of 1/(a+1/b+1)
v_Enhance   22
N 2 hours ago by Rayvhs
Source: ELMO Shortlist 2013: Problem A2, by David Stoner
Prove that for all positive reals $a,b,c$,
\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]Proposed by David Stoner
22 replies
v_Enhance
Jul 23, 2013
Rayvhs
2 hours ago
xf(x + xy) = xf(x) + f(x^2)f(y)
orl   14
N 2 hours ago by jasperE3
Source: MEMO 2008, Team, Problem 5
Determine all functions $ f: \mathbb{R} \mapsto \mathbb{R}$ such that
\[ x f(x + xy) = x f(x) + f \left( x^2 \right) f(y) \quad  \forall  x,y \in \mathbb{R}.\]
14 replies
orl
Sep 10, 2008
jasperE3
2 hours ago
Beautiful Number Theory
tastymath75025   34
N 2 hours ago by Adywastaken
Source: 2022 ISL N8
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.
34 replies
tastymath75025
Jul 9, 2023
Adywastaken
2 hours ago
Hard Functional Equation in the Complex Numbers
yaybanana   1
N 2 hours ago by jasperE3
Source: Own
Find all functions $f:\mathbb {C}\rightarrow \mathbb {C}$, s.t :

$f(xf(y)) + f(x^2+y) = f(x+y)x + f(f(y))$

for all $x,y \in \mathbb{C}$
1 reply
yaybanana
Apr 9, 2025
jasperE3
2 hours ago
Find all numbers
Rushil   11
N 2 hours ago by frost23
Source: Indian RMO 1994 Problem 3
Find all 6-digit numbers $a_1a_2a_3a_4a_5a_6$ formed by using the digits $1,2,3,4,5,6$ once each such that the number $a_1a_2a_2\ldots a_k$ is divisible by $k$ for $1 \leq k \leq 6$.
11 replies
Rushil
Oct 25, 2005
frost23
2 hours ago
Arbitrary point on BC and its relation with orthocenter
falantrng   35
N May 11, 2025 by Giant_PT
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
35 replies
falantrng
Apr 27, 2025
Giant_PT
May 11, 2025
Arbitrary point on BC and its relation with orthocenter
G H J
G H BBookmark kLocked kLocked NReply
Source: Balkan MO 2025 P2
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falantrng
252 posts
#1 • 7 Y
Y by farhad.fritl, Frd_19_Hsnzde, ehuseyinyigit, pomodor_ap, Nuran2010, Rounak_iitr, user4747
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
This post has been edited 1 time. Last edited by falantrng, Apr 27, 2025, 4:38 PM
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MuradSafarli
110 posts
#2
Y by
nice problem
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Sadigly
226 posts
#3 • 5 Y
Y by alexanderhamilton124, Nuran2010, Amkan2022, ihatemath123, Funcshun840
MuradSafarli wrote:
nice problem

gurt:yo
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GreekIdiot
251 posts
#4
Y by
Sadigly wrote:
MuradSafarli wrote:
nice problem

gurt:yo

yo:what?
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Assassino9931
1362 posts
#5 • 1 Y
Y by GeorgeRP
Trig setup
This post has been edited 3 times. Last edited by Assassino9931, Apr 27, 2025, 12:26 PM
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ErTeeEs06
65 posts
#6 • 2 Y
Y by khina, Funcshun840
Feels a bit troll, solved it in around 5 minutes.
Simple angle chase gives that $BCPH, AEPF, BDPE, CDPF$ are all cyclic. Let $A'$ be reflection of $A$ in $D$. Then $A'$ is obviously on $(BCPH)$. Also $$\angle BPD=\angle BED=\angle BCA=180^\circ-\angle BHA=\angle BHA'=\angle BPA'$$so $P, D, A'$ are collinear. Now Pascal on $CCPA'HB$ solves.
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wassupevery1
325 posts
#7
Y by
Diagrams

Solution
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alexanderchew
12 posts
#8
Y by
Solution: We first claim the following:
This claim wrote:
The reflection of $P$ over $BC$ is the second intersection of $AD$ and $(ABC)$.
Proof. Let $P'$ be the second intersection of $AD$ and $(ABC)$. Then, since \begin{align*}
\measuredangle PBC &= \measuredangle FBD \\
&= \measuredangle FAD \\
&= \measuredangle CAP' \\
&= \measuredangle CBP' \\
&= -\measuredangle P'BC
\end{align*}then $BP$ and $BP'$ are reflections over $BC$. Note that since $\measuredangle AEP = \measuredangle ADC = \measuredangle ADB = \measuredangle AFP$, then $AEPF$ is cyclic, implying that $\measuredangle BPC = \measuredangle FPE = -\measuredangle BAC = \measuredangle BP'C$, so $P$ and $P'$ are indeed reflections over $BC$.

Now we reflect everything except $A$ over $BC$, without overlaying the new diagram with the old one. We can also do barycentrics on $\triangle ABC$ now.
Let $a$, $b$, $c$, $A$, $B$, $C$, $S_A$, $S_B$, $S_C$ denote $BC$, $CA$, $AB$, $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 1)$, $\frac{-a^2+b^2+c^2}{2}$, $\frac{a^2-b^2+c^2}{2}$, $\frac{a^2+b^2-c^2}{2}$ respectively. (I know that's a lot but they're just common notation anyway)
We first calculate $H$. Let $H=(t:S_C:S_B)$. Then, \begin{align*}
-a^2S_BS_C - b^2S_Ct - c^2tS_B &= 0 \\
\iff t &= -\frac{a^2S_BS_C}{b^2S_B+c^2S_C}
\end{align*}so $H=(-a^2S_BS_C: S_C(b^2S_B+c^2S_C): S_B(b^2S_B+c^2S_C))$. Let $P = (x:y:z)$. Then obviously $-a^2yz-b^2zx-c^2xy=0$. We can also calculate $X=(-a^2S_BS_Cx : -a^2S_BS_Cy : S_Bx(b^2S_B+c^2S_C))$, $D=(0:y:z)$ and $L=(-a^2S_C:b^2S_C:b^2S_B)$. Finally, \begin{align*}
\begin{vmatrix}
0&y&z\\
-a^2S_C&b^2S_C&b^2S_B\\
-a^2S_BS_Cx&-a^2S_BS_Cy&S_Bx(b^2S_B+c^2S_C)\\
\end{vmatrix} &= -a^2S_C
\begin{vmatrix}
0&y&z\\
1&b^2S_C&b^2S_B\\
S_Bx&-a^2S_BS_Cy&S_Bx(b^2S_B+c^2S_C)\\
\end{vmatrix} \\
&=-a^2S_BS_C
\begin{vmatrix}
0&y&z\\
1&b^2S_C&b^2S_B\\
x&-a^2S_Cy&x(b^2S_B+c^2S_C)\\
\end{vmatrix} \\
&= -a^2S_BS_C((-a^2S_Cyz - xy(b^2S_B+c^2S_C))+(b^2S_Bxy-b^2S_Cxz)) \\
&= -a^2S_BS_C(-a^2S_Cyz-b^2S_Czx-c^2S_Cxy) \\
&= 0
\end{align*}so we're done.
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VicKmath7
1390 posts
#9
Y by
Here is non-projective synthetic solution.
Since $\angle BEC=\angle ADB=\angle AFB$, $AEPF$ and $BHPC$ are cyclic and $\angle FPC=\angle BAC=\angle FDC$, so $CFPD$ is cyclic. Now, we claim that $L$ lies on the radical axis of $(BHD)$ and $(CDPF)$, which clearly finishes the problem as this radical axis is $XD$ due to $XH \cdot XB=XP\cdot XC$. Let $AH \cap (BHD)=Q$ and $LC \cap (CPD)=R$. Observe that $\angle LCB=\alpha$ and $\angle DRC=\angle DFC=\beta$, so $\angle RDC=\gamma=\angle BHQ=\angle BDQ$, so $Q, D, R$ are collinear. Then $\angle HQR=\angle HQD=\angle HBC=\angle RCH$, i.e. $HQCR$ is cyclic, i.e. $LH \cdot LQ=LR \cdot LC$ and thus $L$ lies on the radical axis.
This post has been edited 2 times. Last edited by VicKmath7, Apr 27, 2025, 3:38 PM
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mariairam
8 posts
#10 • 2 Y
Y by vi144, Ciobi_
Note that saying points $D,X,L$ lie on the same line is equivalent to saying $BH,CE,DL$ are concurrent lines.

It is then natural to apply Desargues's Theorem on $\triangle LHC$ and $\triangle DBE$.

Let $A'$ and $C'$ be the feet of the heights from $A$ and $C$ respectively.
Since we need to prove that $LH\cap DB, HC\cap BE, LC\cap DE$ are collinear,
and since (by Reim's Theorem) $A'C'\parallel DE$,
then it would be sufficient to prove that $LC$ is parallel to these two lines as well.

As noted before, by rather straightforward angle chasing, $P$ lies on the circle $(BHC)$.

Hence $\angle LCH= \angle HBC=\angle HAC$. And since $\angle HCB= \angle HAB$, we get $\angle LCB= \angle A=\angle EDB$ and the conclusion follows.
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hukilau17
288 posts
#11
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Complex bash with $\triangle ABC$ inscribed in the unit circle, and let $AD$ meet the unit circle again at $U$, so that
$$|a|=|b|=|c|=|u|=1$$$$h = a+b+c$$$$d = \frac{au(b+c) - bc(a+u)}{au-bc}$$Let lines $BF,CE$ intersect the unit circle again at $V,W$ respectively. Now since $A,B,D,F$ are concyclic, we have
$$\frac{(a-d)(b-f)}{(a-f)(b-d)} \in \mathbb{R} \implies \frac{(a-u)(b-v)}{(a-c)(b-c)} = \frac{c^2(a-u)(b-v)}{uv(a-c)(b-c)} \implies c^2 = uv$$So
$$v = \frac{c^2}u$$and similarly
$$w = \frac{b^2}u$$Then
\begin{align*}
p &= \frac{bv(c+w) - cw(b+v)}{bv-cw} \\
&= \frac{\frac{bc^2}u\left(c+\frac{b^2}u\right) - \frac{b^2c}u\left(b+\frac{c^2}u\right)}{\frac{bc^2}u-\frac{b^2c}u} \\
&= \frac{bc^3u+b^3c^2-b^3cu-b^2c^3}{bc^2u-b^2cu} \\
&= \frac{c^2u+b^2c-b^2u-bc^2}{cu-bu} \\
&= \frac{bu+cu-bc}u
\end{align*}(So $P$ is the reflection of $U$ over line $BC$.) Now since $L$ lies on line $HA$, we have
$$\overline{\ell} = \frac{a\ell + bc - a^2}{abc}$$And since $LC$ is tangent to the circumcircle of $\triangle PBC$, we have
$$\frac{(c-\ell)(b-p)}{(b-c)(c-p)} \in \mathbb{R}$$$$\frac{c(c-\ell)(b-u)}{b(b-c)(c-u)} = \frac{\frac1c\left(\frac1c-\frac{a\ell + bc - a^2}{abc}\right)\left(\frac1b-\frac1u\right)}{\frac1b\left(\frac1b-\frac1c\right)\left(\frac1c-\frac1u\right)} = -\frac{(a^2+ab-a\ell-bc)(b-u)}{a(b-c)(c-u)}$$$$ac(c-\ell) = -b(a^2+ab-a\ell-bc) \implies \ell = \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)}$$Now we find the coordinate of $X$. Since $X$ lies on line $BH$, we have
$$\overline{x} = \frac{bx+ac-b^2}{abc}$$Since $X$ lies on line $CP$, we have
$$\frac{c-x}{c-p} \in \mathbb{R}$$$$\frac{u(c-x)}{b(c-u)} = \frac{\frac1u\left(\frac1c - \frac{bx+ac-b^2}{abc}\right)}{\frac1b\left(\frac1c-\frac1u\right)} = -\frac{ab-ac+b^2-bx}{a(c-u)}$$$$au(c-x) = -b(ab-ac+b^2-bx) \implies x = \frac{ab^2-abc+acu+b^3}{au+b^2}$$Now we find the vectors
\begin{align*}
d-\ell &= \frac{au(b+c) - bc(a+u)}{au-bc} - \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)} \\
&= \frac{a(b+c)(abu+acu-abc-bcu) - (au-bc)(a^2b+ab^2+ac^2-b^2c)}{a(b+c)(au-bc)} \\
&= \frac{-a^3bu-a^2bc^2+2a^2bcu+ab^3c+abc^3-abc^2u-b^3c^2}{a(b+c)(au-bc)} \\
&= -\frac{b(a-c)(a^2u+ac^2-acu-b^2c)}{a(b+c)(au-bc)}
\end{align*}and
\begin{align*}
x-\ell &= \frac{ab^2-abc+acu+b^3}{au+b^2} - \frac{a^2b+ab^2+ac^2-b^2c}{a(b+c)} \\
&= \frac{a(b+c)(ab^2-abc+acu+b^3) - (au+b^2)(a^2b+ab^2+ac^2-b^2c)}{a(b+c)(au+b^2)} \\
&= \frac{-a^3bu-a^2b^2u-a^2bc^2+a^2bcu+ab^3c-ab^2c^2+ab^2cu+b^4c}{a(b+c)(au+b^2)} \\
&= -\frac{b(a+b)(a^2u+ac^2-acu-b^2c)}{a(b+c)(au+b^2)}
\end{align*}Then
$$\frac{d-\ell}{x-\ell} = \frac{(a-c)(au+b^2)}{(a+b)(au-bc)}$$which is real. $\blacksquare$
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bin_sherlo
733 posts
#12 • 1 Y
Y by egxa
Let $A'$ be the reflection of $A$ over $BC$. $D$ is the miquel of $AEPF$. Since $A,E,F,P$ are concyclic, $P\in (BHCA')$. Also $\measuredangle (DE,AH)=\measuredangle (AH,DF)$ hence projecting DDIT at $AEPF$ from $D$, there exists an involution $(A,DP\cap AH),(DE\cap AH,DF\cap AH),(AH\cap BC,AH\cap BC)$. This must be reflection over $BC\cap AH$ thus, $D,P,A'$ are collinear. Pascal at $BHA'PCC$ gives the result as desired.$\blacksquare$
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Orestis_Lignos
558 posts
#13
Y by
Proposed by Theoklitos Parayiou, Cyprus :)
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giangtruong13
148 posts
#14
Y by
Orestis_Lignos wrote:
Proposed by Theoklitos Parayiou, Cyprus :)
is that the same guy proposed the 2020 JBMO-P2?
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Marios
24 posts
#15
Y by
giangtruong13 wrote:
Orestis_Lignos wrote:
Proposed by Theoklitos Parayiou, Cyprus :)
is that the same guy proposed the 2020 JBMO-P2?

Yes, It is the same person. He proposed a handful of other geometry problems for Balkan olympiads as well.
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Steve12345
620 posts
#16
Y by
WLOG, solve the analogous problem where $L$ is the intersection of the tangent at $B$ and line $AH$, so $X = BF \cap CH$. Let $H_A = AH \cap BC$. Let $x = \angle CBF = \angle DAC$. By Menelaus applied on the transversal $L,X,D$ on triangle $HH_AC$ it is enough to prove: \[\frac{LH}{LH_A} \cdot\frac{H_AD}{DC} \cdot \frac{CX}{XH} = 1 \]Using the Ratio Lemma on triangle $BH_AH$ we get: \[\frac{LH}{LH_A} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)}\]Using the Ratio Lemma on triangle $BCH$ we get: \[ \frac{CX}{CH} = \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} \]Using the Ratio Lemma on triangle $H_AAC$ we get: \[ \frac{H_AD}{DC} = \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \]Multiplying the three expressions, we get:
\[ \frac{LH}{LH_A} \cdot \frac{H_AD}{DC} \cdot \frac{CX}{XH} = \frac{\cos(\beta)}{\sin(\gamma)\sin(\alpha)} \cdot \frac{\sin(\gamma)\cos(\gamma + x)}{\sin(x)} \cdot \frac{\sin(\alpha)\sin(x)}{\cos(\beta)\cos(\gamma + x)} = 1 \]
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MathLuis
1546 posts
#17
Y by
From miquelpoint we have $BEPD, DPFC$ cyclic and additional trivially from the angles we have $BHPC$ cyclic, let $A'$ reflection of $A$ over $BC$ then from the angles from miquel config at $P$ we trivially have $D,P,A'$ colinear and thus a pascal at $(BHC)$ finishes.
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popmath
71 posts
#19
Y by
This is a quick one I found when I was testing the problem.Same as previous solutions we prove that $H_CH_A\parallel ED\parallel l$, where $l$ is the tangent to $(BHPC)$ at $C$. Now define $L$ as the intersection of $AH$ and $DX$. Apply Desargues's theorem on triangles $\triangle BH_AH_C$ and $\triangle XLC$. Since $LH_A, CH_C$ and $BX$ are concurrent at $H$, we get the intersection of $BH_C$ and $CX$ which is $E$, the intersection of $BH_A$ and $LX$ which is $D$ and the intersection of $LC$ and $H_AH_C$ are collinear. However, since $DE \parallel H_AH_C$, $LC$ is also parallel to these lines, therefore it coincides with the tangent and we are done. Not surprisingly Desargues's theorem on triangle $\triangle BDE$ and $\triangle LXC$ also works.
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Rayvhs
24 posts
#20
Y by
Let $DP\cap AH=Q$.

From ABDF and ACDE being cyclic, we get that BEPDand CDPF are cyclic as well.
Thus, we have
\[\angle BDP = \angle AEC = \angle ADC = \angle CDQ.\]Also, since $AQ\perp BD$, $\bigtriangleup ADQ$ is isosceles.
Therefore, Q is the symmetric point of A wrt BC.
Apply Pascal to BHQPCC and we're done.
This post has been edited 1 time. Last edited by Rayvhs, Apr 28, 2025, 3:55 PM
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jrpartty
44 posts
#21
Y by
Let $A’$ be the image of $A$ under reflection across $BC$.

By cyclic chasing, we obtain that $P$ lies on $(BHC)$ and $B,E,P,D$ are concyclic,

implying $P,D,A’$ are collinear. Note that $A’$ also lies on $(BHC)$.

Applying Pascal on $PCCBHA’$, we are done.
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DeathIsAwe
18 posts
#22
Y by
MMP solution:

Claim: $B$, $H$, $P$, $C$ are cyclic.
Proof: $180 - \angle BPC = \angle PBC + \angle PCB = \angle FAD + \angle EAD = \angle BAC = 180 - \angle BHC \square$

Let $AH$ intersect $(BHPC)$ at $K \neq H$.

Claim: $P$, $D$, $K$ collinear.
Proof: Reflect $P$ over $BC$, name the point $P'$. Notice $P'$ lies on $(ABC)$.
$\angle DAC = \angle DBP = \angle DBP' = \angle P'AC$
Thus $P'$, $D$, $A$ collinear, so $P$, $D$, $K$ collinear $\square$

Now notice that if we let $\deg(D) = 1$, then $P = KD \cap (BHC)$, thus $\deg(P) = 2$, and then by Conic doubling, $\deg(CP) = 1$ and $\deg(X) = 1$. Notice $L$ is fixed since $P$ is on $(BHC)$. Thus we need to check $1 + 1 + 0 + 1 = 3$ cases. Pick $D$ on $AH \cap BC$, $B$ and $C$.
This post has been edited 4 times. Last edited by DeathIsAwe, May 9, 2025, 8:04 PM
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Mapism
19 posts
#23
Y by
We switch the points $E$ and $F$. Notice $D$ is the miquel point of complete quadrilateral $FPEA$. This yields,
$$BEPD,\ CFPD\ \text{cyclic}\ ,\ \ D\in BC \implies FPEA\ \text{cyclic}$$$$180-\angle BHC=\angle BAC=\angle FAC=\angle FDB=\angle FPB=180-\angle BPC \implies BHPC\ \text{cyclic}$$Let $A'$ be the reflection of $A$ across $BC$, it is well known that $A'\in BHPC$. Pascal's theorem on cyclic quadrilateral $CCPA'HB$ gives
$$D,X,L\ \text{collinear} \iff P,D,A' \ \text{collinear}\iff \angle PDC=\angle BDA'$$$$\angle BDA'=\angle BDA=\angle BEA=180-\angle PEC=\angle PDC$$thus we're done $\Box$
This post has been edited 2 times. Last edited by Mapism, Apr 29, 2025, 6:56 AM
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Rotten_
5 posts
#24
Y by
2 days and P1 and P4 are still nowhere to be found
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EVKV
71 posts
#25 • 1 Y
Y by Rotten_
@above they are on AoPS forums but not in contest collections for some reason
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SomeonesPenguin
129 posts
#26
Y by
A quick angle chase yields $\angle BPC=180^\circ-\angle A=\angle BHC$ which means that $BHPC$ is cyclic. Therefore, $L$ is the fixed point on $HA$ such that $\angle BCL=\angle A$. Now, the function $f:BC\mapsto BH$ defined by $D\mapsto E\mapsto X$ is projective since $E=D\infty_{\ell_b}\cap AB$, where $\ell_B$ is the tangent line to $(ABC)$ through $B$ and $X=BH\cap CE$. Notice that $f(B)=B$ so by prism lemma (or Steiner conic) the line $DX$ passes through a fixed point as $D$ moves along $BC$. When $D$ is the foot of the $A$-altitude, $DX$ becomes $HA$ and when $D=C$ we get $\angle BCX=\angle A$, hence the fixed point is indeed $L$.
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EeEeRUT
83 posts
#27
Y by
MMP lets gooo
Consider a variable point $X$ on $BH$.
We let $LX$ intersect $BC$ at $D’$
Notice that deg$E =1$ and deg $D’ =1$.
Perform an inversion at $A$ radius $AC$ that map $E$ to $E_1$ and $D’$ to $D_1$. Hence, the condition $E_1, D_1, C$ are collinear has deg $2$. And since the inversion is projective, the condition $E, D’, A, C$ concyclic has a deg $2$.
Hence, its suffice to check $3$ cases.
Consider $X = B, H, BE \cap LC$
Case $1$: $X=B$
This one is trivial.
Case $2$: $X=H$.
This one is normal orthocenter config.
Case $2$: $X= BE \cap LC = Y$
Let $LC \cap AB = Z$, we need to show that $(AZC)$ is tangent to $BC$
This could be done by angle chasing, hence we are done.
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Yagiz_Gundogan
13 posts
#28
Y by
Solution. We start with a few basic observations.
Claim. $B,H,C,P$ are concyclic.
Proof. This is true since
$$\angle{BPC}=180-\angle{PBC}-\angle{PCB}=180-\angle{FBD}-\angle{ECD}=180-\angle{DAB}-\angle{DAC}=180-A=\angle{BHC}$$holds. $\blacksquare$

Claim. $P,E,B,D$ and $P,F,C,D$ are concyclic.
Proof. The following angle equalities hold.
$$\angle{PBD}=\angle{FBD}=\angle{FAD}=\angle{CAD}=\angle{CED}=\angle{PED}$$One can similiarly prove that $\angle{PCD}=\angle{PFD}$, which implies the claim. $\blacksquare$

Define $A'$ as the reflection of $A$ w.r.t $BC$.
Claim. $\overline{P-D-A'}$ are collinear.
Proof. It is well known that $(BHC)$ is the reflection of $(ABC)$ w.r.t $BC$. From this we obtain $A'\in(BHC)$. The aforementioned claim implies that
$$\angle{BPD}=C=\angle{A'CB} \text{ and } \angle{CPD}=B=\angle{A'BC}$$hold. This is enough to prove the claim. $\blacksquare$

Pascal in $(CCPA'HB)$ implies $\overline{CC\cap A'H-CP\cap HB-PA'\cap BC}\Rightarrow \overline{L-X-D}$ are collinear. $\blacksquare$
This post has been edited 2 times. Last edited by Yagiz_Gundogan, May 1, 2025, 12:54 PM
Reason: grammar
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Baimukh
11 posts
#29
Y by
$\angle BPC=180^\circ-\angle PBC-\angle PCB=180^\circ-\angle FBD-\angle ECD=180-\angle FAD-\angle EAD=180-\angle BAC=\angle BHC\Longrightarrow$ $BCPH$ inscribed $(XH\cdot XB=XP\cdot XC)$. Let $AH\cup (BDH)=G$, $LC\cup (CDP)=I$ and $GD\cup AC=J\Longrightarrow \angle GAJ=\angle HAC=\angle HBC=\angle HBD=\angle HGD=\angle AGJ=\alpha$ $\angle BHG=\angle HAB+\angle HBA=\angle HCB+\angle HCA=\angle ACB=90^\circ-\alpha;$ $\angle DJC=\angle AGJ+\angle GAJ=2\alpha \Longrightarrow 90^\circ-\alpha=\angle CDJ=\angle BDG \Longrightarrow G,D,J$ lie on the same straight line. This means that $\angle ICH=\angle LCH=\angle LCP+\angle PCH=\angle CPB+\angle PBH=\angle CBH=\angle DBH=\angle DGH=\angle IGH\Longrightarrow CIHG$, inscribed in $LH\cdot LG=LI\cdot LC\Longrightarrow D,L,X$, lies on the radical axis of the circumscribed circles $(BDH)$ and $(CDP)$.
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Z4ADies
64 posts
#30
Y by
After some angle chasing, let $CE \cap AH$ at $R$ and connect $CH$ then do ratio lemma with menelaus.
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optimusprime154
23 posts
#31
Y by
Easy problem.
let $deg(D)=1$ move on BC, similar to other solutions we know \(BHPC\) cyclic so $deg(P)=2$ if we let $D=C$ we get $P=C$ so $deg(PC)=1$ since \(BH\) is fixed then $deg(X) = 1$ \(L\) is a fixed point so it suffices to check 3 values of \(D\) we check $D=B$, $D=C$ and $D=V$ where \(V\) is the foot from \(A\) to \(BC\) all are trivial.
This post has been edited 1 time. Last edited by optimusprime154, May 6, 2025, 1:46 PM
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Mathgloggers
90 posts
#32
Y by
AMAZING PROBLEM ,
Let,$AH \cap (BPC) =L'$
$P$ is clearly the Miquel of triangle $ABC$, easily proven by angle chasing into those two cyclic quads.
CLAIM:
$\angle DGB= \angle C$

PROOF:
$\angle PDF =\angle PCF$($X$ being Miquel point). Hence we have $\angle DGB =\angle GDF+\angle GFD=\angle ECD +\angle FCG= \angle C $


CLAIM:
$B,H,P,C$ are concylic points.

PROOF:
$\angle BGD +\angle CGD  =\angle BED +\angle CFD =\angle C +\angle B =180^{0}-\angle A =\angle BHC $
Hence we are done.

Now notice that , $\angle BHL' =\angle BGL'=\angle C$,
$\implies$ $\boxed{G,D,L}$ is collinear.

Now we are easily done by pascals theorem on $B,H,L',P,C,C$
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NZP_IMOCOMP4
30 posts
#33 • 1 Y
Y by Mathgloggers
Fun geometry.

Call the angles of the triangle $\alpha, \beta, \gamma$ and sides $a,b,c$ and $A'$ the foot of the perpendicular from $A$ to $BC$. Angle chasing gives us $x\equiv \measuredangle DAC=\measuredangle DEC=\measuredangle FBC=\measuredangle PCL=\measuredangle ECL$ and therefore $ED\parallel  LC$. It suffices to prove $LC/ED=CX/EX$. We also have $\measuredangle ECB=\alpha-x$ and so $\measuredangle A'CL=\alpha$. We also have $\measuredangle BEC=\gamma+x$, $\measuredangle EBX=90^{o}-\alpha$, $\measuredangle CBX=90^{o}-\gamma$, $\measuredangle BED=\gamma$ and $\measuredangle EDB=\alpha$.

Now, $LC=\frac{A'C}{\cos \alpha}=b\frac{\cos \gamma}{\cos \alpha}$. We have $\triangle EDB\sim \triangle CAB$ and therefore $ED=b\frac{EB}{CB}=b\frac{\sin(\alpha-x)}{\sin(\gamma+x)}$ by applying the Sine theorem to $\triangle ECB$. Applying Sine theorems to $\triangle XEB$ and $\triangle XCB$ we obtain $EX=BX\frac{\cos\alpha}{\sin(\gamma+x)}$ and $CX=BX\frac{\cos\gamma}{\sin(\alpha-x)}$. Combining all the results gives us: $$\frac{LC}{ED}=\frac{\cos(\gamma)\sin(\gamma+x)}{\cos(\alpha)\sin(\alpha-x)}=\frac{CX}{EX}.$$Q.E.D.
This post has been edited 1 time. Last edited by NZP_IMOCOMP4, May 4, 2025, 10:44 AM
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DVDTSB
129 posts
#34 • 1 Y
Y by Mathgloggers
Really interesting sidequest: Prove that the second intersection of the circles $(BHC)$ and $(AEF)$ is the $A$ Humpty Point.

Let $T$ be the second intersection of the circles $(BHC)$ and $(AEF)$, and $M$ the intersection of $AT$ and $BC$. We wish to show that $M$ is the midpoint of $BC$ by proving that $(B,C; M, P^{BC}_\infty)$ is a harmonic.
Now let $S$ be on the circle $(AEF)$ such that $BC \parallel AS$. We can project $(B,C;M,P^{BC}_\infty)$ with respect to $A$ to get $(B,C; M, P^{BC}_\infty)=(E,F; T,S)$.
Let $N$ be the midpoint of $AS$, $A'$ the reflection of $A$ with respect to $BC$, and $K$ the intersection of $A'A$ with $BC$. It's easy to see that $KN\parallel A'S$.
Let $O$ be the center of the circle $AEF$. Notice that $D$ is the Miquel point of $AEPF$, and since $AEPF$ is cyclic, we have that $OD\perp BC$. Now, since $ON\perp AS$ and $AS\parallel BC$, we must have $O,D,N$ colinear and $DN \perp BC$, from where we easily get that $D$ is on $A'S$, so we have $A',D,P,S$ colinear.
Let $V$ be the intersection of $TP$ and $BC$ (possibly on the line at infinity). Now, we can project again with respect to $P$ and get $(E,F;T,S)=(C,B;V,D)$.
Let $U$ be the intersection of $FD$ and $PC$. We want to show that $BU$, $TP$ and $AC$ are concurrent. Let $R$ be the intersection of circles $(ABD)$ and $(BHC)$. Since $FPDC$ is cyclic, by Power of a Point we have $UP\cdot UC = UF \cdot UD$, but notice that $UP \cdot UC$ is the PoP of U with respect to $(BHC)$ and $UF \cdot UD$ is the PoP of U in respect to $(ABD)$, so we have that $U$ is on the radical axis of $(ABD)$ and $(BHC)$, so $B,U,R$ are colinear. Now, its easy to see that $BR$, $TP$ and $AC$ are concurrent by looking at the radical center of circles $(AEF)$, $(BHC)$ and $(ABD)$.
Now, by looking at $(C,B;V,D)$ from $F$, its obvious that $(C,B;V,D)$ is harmonic, so we have that $(B,C;M,P^{BC}_\infty)$ is harmonic, so $M$ is the midpoint of $BC$, and so $T$ is the $A$ Humpty point.
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Thapakazi
67 posts
#35 • 1 Y
Y by ABYSSGYAT
We perform a simple angle chase to show that $(AEPF), (BEPD), (DPFC)$ are all cyclic. Let $A'$ be the reflection of $A$ across $BC$. Then, as

\[\measuredangle PDB = \measuredangle AEP = \measuredangle ADC = -\measuredangle ADB = -\measuredangle BDA'.\]
So, the points $P-D-A'$ are collinear. Finally, Pascal's on hexagon $BCCPA'H$ implies $D-X-L$ collinear, as needed.
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Mamadi
7 posts
#36
Y by
Note that \( P, D, C, F \) and \( P, D, E, B \) are cyclic.
Because \( \angle EBD = \angle DFC \) and \( \angle FCD = \angle DEB \) ( Due to \( AFDB \) and \( AEDC \) being cyclic)
So \( \triangle EBD \sim \triangle CFD \). Now we conclude that \( \triangle BDF \sim \triangle DEC \).

So Now we know \( \angle FBD = \angle PED = \angle ECL \) ( due to \( LC \) being tangent to \( PBC \)) so \( ED \parallel CL \).
We need to prove that \( \frac{EX}{XC} = \frac{ED}{LC} \)
( if this expression is established and \( L, X, D \) are not on the same line we get an obvious contraction )

Now we know \( \frac{EX}{XC} = \frac{BE}{BC} \). \( \frac{\sin \angle ABX}{\sin \angle XBC} = \frac{BE}{BC} \cdot \frac{\sin (90 - A)}{\sin (90 - C)} \)

We also know that \( \frac{BE}{BC} = \frac{DE}{AC} \) and \( \frac{AC}{CL} = \frac{\sin(90 - A)}{\sin(90 - C)} \) because \( \triangle ABC \sim \triangle BDE \) and \( \angle LCB = \angle EDB = \angle BAC \).
So \( \frac{EX}{XC} = \frac{DE}{AC} \cdot \frac{AC}{CL} = \frac{DE}{CL} \)
Done.
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Giant_PT
45 posts
#37
Y by
Let $A'$ be reflection of $A$ across $BC$.
Claim1: $A$, $E$, $P$, and $F$ are concyclic
$$180^\circ = \measuredangle BDA+\measuredangle ADC = \measuredangle BFA + \measuredangle AEC = \measuredangle PFA + \measuredangle PEA$$Which implies the claim. $\square$

This is enough to imply that $B$, $P$, $H$, and $C$ are concyclic since $\measuredangle BPC = \measuredangle BHC = \measuredangle CAB$.

Claim 2: $B$, $E$, $P$, and $D$ are concyclic
$$\measuredangle DBP = \measuredangle DAC = \measuredangle DPE$$Which implies the claim. $\square$

This is enough to imply that $P$, $D$ and $A'$ are collinear since,
$$\measuredangle BDP =\measuredangle AEC= \measuredangle ADC = \measuredangle CDA'.$$Now by applying Pascal's theorem on cyclic hexagon $CCBHA'P$, we see that $L = CC\cap HA'$, $D = CB\cap A'P$, and $X = BH\cap PC$ must be collinear, thus finishing the problem. $\square$
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