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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
A Familiar Point
v4913   52
N a few seconds ago by SimplisticFormulas
Source: EGMO 2023/6
Let $ABC$ be a triangle with circumcircle $\Omega$. Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$). Let $I$ be the incenter of $ABC$. Let $\omega_b$ be the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$, and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$. Show that the line $IN_a$, and the lines through the intersections of $\omega_b$ and $\omega_c$, meet on $\Omega$.
52 replies
v4913
Apr 16, 2023
SimplisticFormulas
a few seconds ago
Tangential quadrilateral and 8 lengths
popcorn1   72
N 4 minutes ago by cj13609517288
Source: IMO 2021 P4
Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]
Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland
72 replies
popcorn1
Jul 20, 2021
cj13609517288
4 minutes ago
An algorithm for discovering prime numbers?
Lukaluce   3
N 22 minutes ago by TopGbulliedU
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
3 replies
Lukaluce
May 18, 2025
TopGbulliedU
22 minutes ago
Random concyclicity in a square config
Maths_VC   5
N 22 minutes ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
5 replies
Maths_VC
Tuesday at 7:38 PM
Royal_mhyasd
22 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   3
N 36 minutes ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
3 replies
Maths_VC
Tuesday at 7:54 PM
Royal_mhyasd
36 minutes ago
Prime number theory
giangtruong13   2
N an hour ago by RagvaloD
Find all prime numbers $p,q$ such that: $p^2-pq-q^3=1$
2 replies
giangtruong13
2 hours ago
RagvaloD
an hour ago
Problem 2
delegat   147
N 2 hours ago by math-olympiad-clown
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
147 replies
delegat
Jul 10, 2012
math-olympiad-clown
2 hours ago
Coloring points of a square, finding a monochromatic hexagon
goodar2006   6
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P1
Prove that for each coloring of the points inside or on the boundary of a square with $1391$ colors, there exists a monochromatic regular hexagon.
6 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Van der Warden Theorem!
goodar2006   7
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P2
Suppose $W(k,2)$ is the smallest number such that if $n\ge W(k,2)$, for each coloring of the set $\{1,2,...,n\}$ with two colors there exists a monochromatic arithmetic progression of length $k$. Prove that


$W(k,2)=\Omega (2^{\frac{k}{2}})$.
7 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Maxi-inequality
giangtruong13   0
2 hours ago
Let $a,b,c >0$ and $a+b+c=2abc$. Find max: $$P= \sum_{cyc} \frac{a+2}{\sqrt{6(a^2+2)}}$$
0 replies
giangtruong13
2 hours ago
0 replies
Isosceles triangles among a group of points
goodar2006   2
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P2
Consider a set of $n$ points in plane. Prove that the number of isosceles triangles having their vertices among these $n$ points is $\mathcal O (n^{\frac{7}{3}})$. Find a configuration of $n$ points in plane such that the number of equilateral triangles with vertices among these $n$ points is $\Omega (n^2)$.
2 replies
goodar2006
Jul 27, 2012
quantam13
2 hours ago
APMO Number Theory
somebodyyouusedtoknow   12
N 2 hours ago by math-olympiad-clown
Source: APMO 2023 Problem 2
Find all integers $n$ satisfying $n \geq 2$ and $\dfrac{\sigma(n)}{p(n)-1} = n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.
12 replies
somebodyyouusedtoknow
Jul 5, 2023
math-olympiad-clown
2 hours ago
My Unsolved Problem
ZeltaQN2008   0
2 hours ago
Source: IDK
Let \( P(x) = x^{2024} + a_{2023}x^{2023} + \cdots + a_1x + a_0 \) be a polynomial with real coefficients.

(a) Suppose that \( 2023a_{2023}^2 - 4048a_{2022} < 0 \). Prove that the polynomial \( P(x) \) cannot have 2024 real roots.

(b) Suppose that \( a_0 = 1 \) and \( 2023(a_1^2 + a_2^2 + \cdots + a_{2023}^2) \leq 4 \). Prove that \( P(x) \geq 0 \) for all real numbers \( x \).
0 replies
ZeltaQN2008
2 hours ago
0 replies
Points of a grid
goodar2006   2
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P4
Prove that from an $n\times n$ grid, one can find $\Omega (n^{\frac{5}{3}})$ points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of $\frac{5}{3}$!
2 replies
goodar2006
Jul 27, 2012
quantam13
2 hours ago
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
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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
684 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
68 posts
#75 • 1 Y
Y by ABYSSGYAT
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
650 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
1748 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
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[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
1556 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
129 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
1925 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_
280 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
3449 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
667 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
3007 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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