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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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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
Inequality, inequality, inequality...
Assassino9931   9
N 4 minutes ago by ZeroHero
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
9 replies
2 viewing
Assassino9931
Today at 9:38 AM
ZeroHero
4 minutes ago
Vectors in a tilted square
mathwizard888   20
N 7 minutes ago by HamstPan38825
Source: 2016 IMO Shortlist A3
Find all positive integers $n$ such that the following statement holds: Suppose real numbers $a_1$, $a_2$, $\dots$, $a_n$, $b_1$, $b_2$, $\dots$, $b_n$ satisfy $|a_k|+|b_k|=1$ for all $k=1,\dots,n$. Then there exists $\varepsilon_1$, $\varepsilon_2$, $\dots$, $\varepsilon_n$, each of which is either $-1$ or $1$, such that
\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]
20 replies
mathwizard888
Jul 19, 2017
HamstPan38825
7 minutes ago
Combi Geo
Adywastaken   0
17 minutes ago
Source: NMTC 2024/8
A regular polygon with $100$ vertices is given. To each vertex, a natural number from the set $\{1,2,\dots,49\}$ is assigned. Prove that there are $4$ vertices $A, B, C, D$ such that if the numbers $a, b, c, d$ are assigned to them respectively, then $a+b=c+d$ and $ABCD$ is a parallelogram.
0 replies
Adywastaken
17 minutes ago
0 replies
Japan MO Finals 2023
parkjungmin   1
N 21 minutes ago by EvansGressfield
It's hard. Help me
1 reply
parkjungmin
Yesterday at 2:35 PM
EvansGressfield
21 minutes ago
Non-homogenous Inequality
Adywastaken   0
33 minutes ago
Source: NMTC 2024/7
$a, b, c\in \mathbb{R}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
0 replies
Adywastaken
33 minutes ago
0 replies
Calculus
youochange   1
N 33 minutes ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
1 reply
youochange
2 hours ago
youochange
33 minutes ago
Classic Diophantine
Adywastaken   0
36 minutes ago
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
0 replies
Adywastaken
36 minutes ago
0 replies
Indian Geo
Adywastaken   0
39 minutes ago
Source: NMTC 2024/5
$\triangle ABC$ has $\angle A$ obtuse. Let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$ respectively. Let $A_1$, $B_1$, $C_1$ be arbitrary points on $BC$, $CA$, $AB$ respectively. The circles with diameter $AA_1$, $BB_1$, $CC_1$ are drawn. Show that the lengths of the tangents from the orthocentre of $ABC$ to these circles are equal.
0 replies
Adywastaken
39 minutes ago
0 replies
Interesting inequalities
sqing   2
N 41 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
2 replies
sqing
3 hours ago
sqing
41 minutes ago
Cyclic ine
m4thbl3nd3r   0
41 minutes ago
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
0 replies
m4thbl3nd3r
41 minutes ago
0 replies
Angle ratio
Adywastaken   0
an hour ago
Source: NMTC Junior 2024/4
An acute angle triangle $\triangle PQR$ is inscribed in a circle. Given $\angle P=\frac{\pi}{3}$ and $\angle R>\angle Q$. Let $H$ and $I$ be the orthocentre and incentre of $\triangle PQR$ respectively. Find the ratio of $\angle PHI$ to $\angle PRQ$.
0 replies
Adywastaken
an hour ago
0 replies
concyclic , touchpoints of incircle related
parmenides51   2
N an hour ago by Blackbeam999
Source: All-Russian MO 1994 Regional (R4) 11.3
A circle with center $O$ is tangent to the sides $AB$, $BC$, $AC$ of a triangle $ABC$ at points $E,F,D$ respectively. The lines $AO$ and $CO$ meet $EF$ at points $N$ and $M$. Prove that the circumcircle of triangle $OMN$ and points $O$ and $D$ lie on a line.
2 replies
parmenides51
Aug 26, 2024
Blackbeam999
an hour ago
8 degree polynomial
Adywastaken   0
an hour ago
Source: NMTC Junior 2024/3
Let $a, b, c, d, e, f\in \mathbb{R}$ such that the polynomial $p(x)=x^8-4x^7+7x^6+ax^5+bx^4+cx^3+dx^2+ex+f$ has 8 linear factors of the form $x-x_i$ with $x_i>0$ for $i=1, 2, 3, 4, 5, 6, 7, 8$. Find all possible values of the constant $f$.
0 replies
Adywastaken
an hour ago
0 replies
System of equations
Adywastaken   0
an hour ago
Source: NMTC 2024/2
$(1+4^{2x-y})5^{1-2x+y}=1+2^{2x-y+1}$
$y^3+4x+1+\log(y^2+2x)=0$
0 replies
Adywastaken
an hour ago
0 replies
RMM 2019 Problem 2
math90   79
N Apr 29, 2025 by lpieleanu
Source: RMM 2019
Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$. Prove that $PE$ is tangent to $\Omega$.

Jakob Jurij Snoj, Slovenia
79 replies
math90
Feb 23, 2019
lpieleanu
Apr 29, 2025
RMM 2019 Problem 2
G H J
G H BBookmark kLocked kLocked NReply
Source: RMM 2019
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StefanSebez
53 posts
#71
Y by
Seems like this is a new solution

Redefine $P$ such that $PE, PD$ are tangent to $(CDE)$
Draw point $F\in (ABCD)$ such that $AD=AF$
Both $\Delta AFD$ and $\Delta PED$ are isosceles and $\angle DFA=\angle DCA=\angle DEP$
$\implies \Delta AFD \sim \Delta PED$
So a spiral similarity at $D$ takes $\Delta AFD$ to $\Delta PED$, so we get $\Delta DPA \sim DEF$
$AF=AD=BC$ so $ACBF$ is isosceles trapezoid $\implies \angle AEF=\angle BEC$ (because $EA=EC$)
Now we can angle chase the rest
Let $\angle BAC=\alpha, \angle CBA=\beta, \angle EDC=\theta, \angle EBA=\phi$
$\angle DPA=\angle DEF=\angle DEA+\angle AEF=\theta+\alpha+\angle BEC=\theta+\phi+2\alpha$
$\angle EAP=\angle EAD-\angle PAD=(\beta-\alpha)-(180-\angle ADP-\angle DPA)
\newline =\beta-\alpha-180+(180-\beta-\alpha-\theta)+(\theta+\phi+2\alpha)=\phi$
So $PA$ is tangent to $(AEB)$ as desired
Attachments:
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GrantStar
821 posts
#72 • 1 Y
Y by OronSH
Let $F$ be the midpoint of $BC$, on both circles. Let $G$ be the point on $\Omega$ such that $(GF;ED)=-1$. Let $GF$ hit $ED$ at $J$.
Redefine $P$ as the intersections of the tangents at $E,D$ to $\Omega$. Thus $P,F,J,G$ are collinear.

Claim: $FP$ is tangent to $\omega$.
Proof. Let $G'=CG\cap EF$. As $-1=(GF;ED)\overset{C}{=}(G'F;E\infty)$ so $E$ is the midpoint of $G'F$. Thus $AFCG'$ is a parallelogram, so $CG \parallel AF$. Reim finishes. $\blacksquare$

Let $PE$ intersect $AF$ at $I$ and $\omega$ at $H$.

Claim: $EFIJ$ is cyclic
Proof. The proof is just angles. \[\measuredangle JFI=\measuredangle PFI=\measuredangle FBA=\measuredangle FDC =\measuredangle FDE+\measuredangle EDC=\measuredangle FEI+\measuredangle JEF=\measuredangle JEI\]$\blacksquare$
Therefore, by reims on $\omega$ and $\Omega$, $FH\parallel EG$. Similarly, reims on $(EFJI)$ and $\omega$ gives $IJ\parallel FH$. Thus, \[-1=(PJ;FG)\overset{\infty_{FH}}{=}(PI;HE)\overset{F}{=}(FA;HE)\]Hence $EH$ is a symmedian and thus $P$ is the intersection of the tangents at $A$ and $F$ to $\omega$. We thus conclude the result.
Remark: Very good.
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Leo.Euler
577 posts
#73
Y by
Let $K = \overline{AC} \cap \overline{BD}$ and $F$ denote the midpoint of $BD$.

Claim: Let $Q$ be the second intersection of $(KAF)$ and $(KED)$. Then $\overline{QK} \parallel \overline{CD}$.
Proof. Let $f(\bullet) = \text{Pow}(\bullet, (KAF)) - \text{Pow}(\bullet, (KED))$. Then it suffices to show that $f(C)=f(D)$. Compute \[ f(C)-f(D) = (CK \cdot CA - CE \cdot CK) - (DF \cdot DK - 0) = CK \cdot EA - DF \cdot DK= CK \cdot EC - DF \cdot DK = 0, \]as desired.
:yoda:

Claim: In fact, $P=Q$.
Proof. Note that as $ABEF$ is an isosceles trapezoid, the center of $(ABEF)$ lies on $(KAF)$. Thus, $P$ lies on $(KAF)$. I contend that $\overline{PK} \parallel \overline{FE}$, which would imply $P=Q$ by the prior claim. By simple angle chasing we have \[ \angle PKA = \angle PFA = \angle FEA, \]which implies the desired parallelism.
:yoda:

Claim: Finally, $PD=PE$.
Proof. Note that $P$ is the center of the spiral similarity mapping $\overline{AF} \mapsto \overline{ED}$. Thus, $\triangle FPA \sim \triangle DPE$, so as $FP=AP$, we have $DP=EP$, as desired.
:yoda:

We conclude from the final claim.
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Shreyasharma
682 posts
#74
Y by
Set $(DEC)$ as the unit circle and redefine $P$ as the point such that $\overline{PD}$ and $\overline{PE}$ are tangent to $\Omega$. Orient so that $\overline{CD}$ is perpendicular to the real axis. Define $F$ as the midpoint of $\overline{BD}$. Compute,
\begin{align*}
d &= 1/c\\
f &= 1/e\\
a &= 2e - c\\
p &= \frac{2de}{d+e} = \frac{2e/c}{1/c + e} = \frac{2e}{ce + 1}
\end{align*}However clearly we have,
\begin{align*}
\frac{p - a}{f - a} &= \frac{\frac{2e - (2e - c)(ce + 1)}{ce + 1}}{1/e - 2e + c}\\
&= \frac{\frac{2e - 2e - 2e^2c + c^2e + c}{ce + 1}}{1/e(1 - 2e^2 + ce)}\\
&= \frac{\frac{c(2e^2 + ce + 1)}{ce + 1}}{1/e(1 - 2e^2 + ce)}\\
&= \frac{ce}{ce + 1}
\end{align*}Also we have,
\begin{align*}
\frac{a - c}{d - c} &= \frac{2e - 2c}{1/c - c}\\
&= \frac{2c(e - c)}{1 - c^2}
\end{align*}Then dividing we have,
\begin{align*}
\frac{2(e-c)(ec + 1)}{e(1 - c^2)}
\end{align*}Taking the conjugate this is simply,
\begin{align*}
\frac{2/c^2e^2(c - e)(ec + 1)}{1/c^2e(c^2 - 1)} = \frac{2(e - c)(ec+1)}{e(1-c^2)}
\end{align*}and so we find $\angle PAF = \angle DCA = \angle FEA$ proving the tangency.
This post has been edited 1 time. Last edited by Shreyasharma, Feb 20, 2024, 4:55 PM
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Thapakazi
61 posts
#75
Y by
Let $F$ be the other intersection of $(ABC)$ and $(CDE)$, redefine $P$ to be the intersection of tangents through $D$ and $E$ of $(DEC).$ Then, $B,F,E$ are collinear as:

\[\measuredangle BFE = \measuredangle BAE = \measuredangle ACD = -\measuredangle EFD.\]
Furthermore, this also implies $EF \parallel BC$ as

\[\measuredangle BFE = \measuredangle BAC = \measuredangle BDC = \measuredangle FDC.\]
Now, we use complex numbers with $(DFEC)$ as the unit circle.

We see that $p = \frac{2de}{d+e}$, $f = cd/e$, and $a = 2e - c.$ We compute

\begin{align*}
        \frac{a-p}{a-f} \div \frac{e-a}{e-f} &= \frac{2e-c-\frac{2de}{d+e}}{2e-c-\frac{cd}{e}} \times \frac{e-\frac{cd}{e}}{c-e} \\
        &= \frac{e}{d+e} \times \frac{(d+e)(2e-c)-2de}{2e^2-ce-cd} \times \frac{e^2-cd}{e(c-e)} \\
        &= \frac{1}{d+e} \times \frac{2e^2 - cd - ce}{2e^2 - ce - cd} \times \frac{e^2-cd}{c-e}\\
        &= \frac{e^2-cd}{(d+e)(c-e).}
    \end{align*}
This has conjugate

\[\frac{\frac{1}{e^2} - \frac{1}{cd}}{(\frac{1}{d} + \frac{1}{e})(\frac{1}{c}-\frac{1}{e})} = \frac{cd-e^2}{(e+d)(e-d)} = \frac{a-p}{a-f} \div \frac{e-a}{e-f}.\]
Hence, $\frac{a-p}{a-f} \div \frac{e-a}{e-f} \in \mathbb{R}$ giving us $\measuredangle FAP = \measuredangle FEA$ as needed.
This post has been edited 1 time. Last edited by Thapakazi, Mar 28, 2024, 1:56 PM
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cursed_tangent1434
625 posts
#76
Y by
Set $\Omega$ be the unit circle. WLOG, rotate so that $CD$ is parallel to the real axis. Then, $d=-\frac{1}{c}$. We can then obtain,
\[a=2e-c\]and,
\[b=-\overline{a}=-\overline{2e-c}=\frac{e-2c}{ce}\]Let $P'$ (denoted by the complex number $p$) be the intersection of the tangents to $\Omega$ at $D$ and $E$.Then, we can also easily obtain that,
\begin{align*}
    p &= \frac{2de}{d+e}\\
    &= \frac{2\left(-\frac{1}{c}\right)e}{e-\frac{1}{c}}\\
    &= \frac{-\frac{2e}{c}}{\frac{ce-1}{c}}\\
    &= \frac{2e}{1-ec}
\end{align*}Now, we wish to show $\measuredangle PAE = \measuredangle ABE$. Thus, we require,
\[\arg \left(\frac{e-a}{p-a}\right)=\arg\left(\frac{e-b}{a-b}\right)\]Thus, we must have $A = \frac{(e-a)(a-b)}{(p-a)(e-b)} \in \mathbb{R}$. Note that,
\begin{align*}
    A &= \frac{(e-a)(a-b)}{(p-a)(e-b)}\\
    &= \frac{(e-2e+c)(2e-c+\frac{2c-e}{ce})}{\left( \frac{2e}{1-ce}-2e+c \right)\left( e+\frac{2c-e}{ce} \right)}\\
    &= \frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)}
\end{align*}Now, we also have that,
\begin{align*}
    \overline{A} &= \overline{\left(\frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)}\right)}\\
    &= \frac{\left(\frac{1}{c}-\frac{1}{e}\right)\left(\frac{2}{ce^2}-\frac{1}{c^2e}+\frac{2}{c}-\frac{1}{e}\right)\left(1-\frac{1}{ce}\right)}{\left( \frac{2}{ce^2}-\frac{1}{c^2e}+\frac{1}{c} \right)\left(\frac{2}{c}+\frac{1}{ce^2}-\frac{1}{e}\right)}\\
    &= \frac{\frac{(e-c)}{ec}\left( \frac{2c-e+2ce^2-c^2e}{c^2e^2} \right)\left(\frac{ce-1}{ce}\right)}{\left(\frac{2c-e+ce^2}{c^2e^2}\right)\left( \frac{2e^2-ce+1}{ce^2} \right)}\\
    &= \frac{c^3e^4(c-e)(1-ec)(2ce^2-c^2e+2c-e)}{c^4e^4(ce^2+2c-e)(2e^2+1-ec)}\\
    &=\frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)}
\end{align*}Thus, $\overline{A}=A$ and indeed, $\frac{(e-a)(a-b)}{(p-a)(e-b)} \in \mathbb{R}$ which implies that $\measuredangle PAE = \measuredangle ABE$. Thus, $\overline{PA}$ is tangent to $\Gamma$ at $A$ implying that $P'=P$. Thus, indeed $\overline{PE}$ is tangent to $\Omega$ as claimed.
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OronSH
1745 posts
#77
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Let $F$ be the midpoint of $BD.$ Redefine $P$ as the center of spiral similarity sending $DF$ to $EA.$ Since $DF=EA$ we have $\triangle DPF\cong\triangle EPA$ so $DP=EP,FP=AP.$ Let $X$ be the intersection of $DF$ and $EA.$ Then we have $PXED$ cyclic so $\measuredangle DPE=\measuredangle DXE=2\measuredangle DCA,$ but this implies $PD,PE$ are tangent to $(CDFE).$ Similarly we get $PA,PF$ are tangent to $(ABEF),$ so we are done.
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Ywgh1
139 posts
#78
Y by
[asy]
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[/asy]

Sketch for the solution:

First of we redefine $P$ as the intersection of the tangents $PD$, $PE$.

1- Show that $PDEH$ cyclic.

2- Show that $PED$ congruent to $PAF$

Then a spiral similarity centered at $P$ sending $DE$ to $AF$ does the job.
This post has been edited 5 times. Last edited by Ywgh1, Jun 9, 2024, 3:08 PM
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MathLuis
1524 posts
#79 • 3 Y
Y by OronSH, teomihai, kingu
Evil MathLuis be like:
Let $F$ midpoint of $BD$, now take $(ECDF)$ as the unit circle and let $E=e, C=c, D=d, F=f$
$$CD \parallel EF \implies \frac{d-c}{f-e} \in \mathbb R \implies \frac{d-c}{f-e}=\frac{\overline{d}-\overline{c}}{\overline{f}-\overline{e}}=\frac{\frac{c-d}{cd}}{\frac{e-f}{ef}} \implies f=\frac{cd}{e}$$Now redefine $P=p=\frac{2ed}{e+d}$ and due to reflection $A=a=2e-c$, now we compute $\text{arg}(\angle FAP)$
$$\frac{f-a}{p-a}=\frac{\frac{cd}{e}-(2e-c)}{\frac{2ed}{e+d}-(2e-c)}=\frac{e+d}{e} \cdot \frac{cd-2e^2+ec}{2cd-2e^2+ec-2cd+cd}=\frac{e+d}{e}$$$$\angle FAP=\angle DCE \iff \frac{d-c}{e-c} : \frac{f-a}{p-a} \in \mathbb R \iff \frac{d-c}{e-c} \cdot \frac{e}{e+d}=\frac{d-c}{e-c} \cdot \frac{ec}{dc} \cdot \frac{d}{e+d}=\frac{d-c}{e-c} \cdot \frac{e}{e+d}$$Therefore $\angle FAP=\angle DCE=\angle FEA$ thus $AP$ is tangent to $(ABE)$ as desired thus we are done :diablo: :diablo: :diablo:
This post has been edited 1 time. Last edited by MathLuis, Jul 6, 2024, 12:24 AM
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SomeonesPenguin
128 posts
#80 • 1 Y
Y by teomihai
Let $AC\cap BD=\{O\}$ and let $a:=\angle ACD$.

Claim: $\angle APD=\angle DEC+\angle AEB$

Proof: $\angle DEC=180^\circ-\angle EDC-a$ but notice that $\angle EDC=\angle PDO$ so $\angle DEC=180^\circ-\angle PDO-a$ we similarly get $\angle AEB=180^\circ-\angle PAO-a$, hence their sum is equal to $360^\circ-\angle PDO-\angle PAO-2a$ and the conclusion follows from the quadrilateral $PDOA$ and noting that $\angle AOD=2a$.

Claim: $\frac{\sin(\angle DPO)}{\sin(\angle APO)}=\frac{\sin(\angle DEC)}{\sin(\angle AEB)}$

Proof: Trig Ceva in $\triangle APD$ with point $O$ gives: $$\frac{\sin(\angle DPO)}{\sin(\angle APO)}=\frac{\sin(\angle DAO)}{\sin(\angle PAO)}\cdot\frac{\sin(\angle PDO)}{\sin(\angle ADO)}$$
We also have that $\angle PDO=\angle EDC$ and $\angle PAO=\angle ABE$ and $\frac{\sin(\angle DAO)}{\sin(\angle ADO)}=\frac{DO}{AO}=\frac{DC}{AB}$ so LOS in $\triangle DEC$ and $\triangle ABE$ gives the desired conclusion.

Finally, from our claims we can conclude that $\angle DPO=\angle DEC$ and $\angle APO=\angle AEB$ so $PDOE$ is cyclic. This gives $a=\angle DCA=\angle PDE=\angle POA$ so $\angle POD=\angle PED=a$, therefore $PE$ is tangent to $\Omega$. $\blacksquare$
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cj13609517288
1916 posts
#81
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I bashed this on paper, so here's a summary:

Redefine $P$ as the pole of $DE$ wrt $(CDE)$. Then we want to show that $\angle PAF=\angle ABF$. Now we can just use complex with $(CDE)$ as the unit circle. Eventually we get that we want to prove
\[\frac{de(d-f)}{(d+e)(2de-ef-2df+d^2)}\in\mathbb{R}.\]This is not hard to do.
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peppapig_
281 posts
#82 • 1 Y
Y by teomihai
Let $(CDE)$ be the unit circle so that $c=\frac{1}{d}$.

Note

* $a=2e-c$,
* $b=\overline a=\frac{2c-e}{ec}$.

Let $P'=DD\cap EE$. Note
\[p'=\frac{2de}{d+e}=\frac{2e}{1+ec}.\]Then,
\[P=P'\iff P'\in AA \iff \angle P'AE=\angle ABE,\]so
\[\frac{\left(\frac{p'-a}{e-a}\right)}{\left(\frac{a-b}{e-b}\right)}=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.\]Conjugating,
\[\left(\overline{\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}}\right)=\frac{\left(\frac{1}{c}\right)\left(\frac{e^2c+e-2c}{e^2c}\right)\left(\frac{-2e^2+ec+1}{e^2c}\right)}{\left(\frac{1+ec}{ec}\right)\left(\frac{e-c}{ec}\right)\left(\frac{2c-e-2e^2c+ec^2}{e^2c^2}\right)},\]\[=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.\]They are equal, so $P=P'$.
This post has been edited 2 times. Last edited by peppapig_, Jan 1, 2025, 5:34 PM
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ihatemath123
3446 posts
#83 • 1 Y
Y by OronSH
This problem is a corollary of the main claim in the solution to 2019 TST #1. See the diagram of Evan Chen's solution for reference. In that diagram, we claim $X$ is in fact the spiral center from $A \cup (AMN)$ to $M \cup (MBY)$. Indeed, this is easy to check by adding in the centers $O_1$ and $O_2$ of $(AMN)$ and $(MBY)$ and showing with lengths that $\triangle XAO_1 \sim \triangle XMO_2$. In the referenced solution, it is shown with symmedians that the second tangent from $X$ to $(AMN)$ lies on $(BMY)$ as well.

Analogously, in the RMM problem, it follows that $P$ is the center of spiral similarity sending $A \cup (ABE)$ to $E \cup (CED)$. Moreover, the second tangent from $P$ to $(ABE)$ is $F$, the second intersection of $(ABE)$ and $(CED)$. Clearly, line $EF$ is the midline of the trapezoid. But since $\angle ECD = \angle FBA$, arcs $ECD$ and $ABF$ are similiar, implying $\triangle PAF \sim \triangle PED$, which finishes.
This post has been edited 1 time. Last edited by ihatemath123, Mar 18, 2025, 4:07 PM
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Ilikeminecraft
626 posts
#84
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Let $O, O_{\Gamma}, O_\Omega$ be the center of $(ABCD), (AEB)$ and $(EDC).$ First, I claim that the midpoint of $\overline{O_\Gamma O_\Omega}$ is $O.$ Since the perpendicular bisector of $\overline{AB}$ is the same as the perpendicular bisector of $\overline{EC},$ we know that $O  \in \overline{O_\Omega O_{\Gamma}}.$ We also have that the distance from $O$ to the perpendicular bisectors of $\overline{AE}$ and $\overline{EC}$ are the same, and hence $O$ is the midpoint of $\overline{O_\Gamma O_\Omega}.$

Let $C, D, E$ be $c, d, e$ such that $(CDE)$ forms the unit circle. Thus, we have that $A$ is $2e - c.$ We can compute $O$ by using the fact that the foot to $\overline{EC}$ is $E$ and that it lies on the perpendicular bisector of $\overline{DC}.$ Let $O$ be $o = k\cdot \frac{d + c}{2}$ for some $k\in \mathbb R.$ Then, we have that $e = \frac12(e + c + o - ec\overline o) \implies 2e - 2c = k \cdot (d + c) - k \cdot \frac{(d + c)e}{d} \implies k = \frac{2d(e - c)}{(d + c)(d - e)}.$ Thus, $o = \frac{d(e - c)}{d - e},$ and hence $O_\Gamma$ is $\frac{2d(e - c)}{d - e}.$ Now, define $P'$ such that $\overline{P'E}, \overline{P'D}$ are both tangent to $O_\Omega.$ We know that $P'$ is $\frac{2de}{d + e}$.

We will prove that $P = P'.$ We will do this by proving that $\overline{AP'} \perp \overline{AO_\Gamma}.$ We have that:
\begin{align*}
	\frac{AP'}{AO_\Gamma}\frac{\frac{2de}{d + e} - 2e + c}{\frac{2de - 2dc}{d - e} - 2e + c} & = \frac{\frac{cd + ec -2e^2}{d + e}}{\frac{2e^2 - dc - ec}{d - e}} \\
	& = \frac{e - d}{d + e} \\
	\overline{\left(\frac{e - d}{d + e}\right)} & = \frac{d - e}{e + d}
\end{align*}Thus, we have that they are perpendicular, and hence we are done.
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lpieleanu
3001 posts
#85
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Solution
This post has been edited 1 time. Last edited by lpieleanu, Apr 30, 2025, 8:34 PM
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