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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 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
A nice triangle center on line OI
proglote   23
N Apr 28, 2025 by Ilikeminecraft
Source: Brazil MO 2013, problem #6
The incircle of triangle $ABC$ touches sides $BC, CA$ and $AB$ at points $D, E$ and $F$, respectively. Let $P$ be the intersection of lines $AD$ and $BE$. The reflections of $P$ with respect to $EF, FD$ and $DE$ are $X,Y$ and $Z$, respectively. Prove that lines $AX, BY$ and $CZ$ are concurrent at a point on line $IO$, where $I$ and $O$ are the incenter and circumcenter of triangle $ABC$.
23 replies
proglote
Oct 24, 2013
Ilikeminecraft
Apr 28, 2025
A nice triangle center on line OI
G H J
Source: Brazil MO 2013, problem #6
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proglote
958 posts
#1 • 5 Y
Y by Davi-8191, anantmudgal09, mijail, Adventure10, Mango247
The incircle of triangle $ABC$ touches sides $BC, CA$ and $AB$ at points $D, E$ and $F$, respectively. Let $P$ be the intersection of lines $AD$ and $BE$. The reflections of $P$ with respect to $EF, FD$ and $DE$ are $X,Y$ and $Z$, respectively. Prove that lines $AX, BY$ and $CZ$ are concurrent at a point on line $IO$, where $I$ and $O$ are the incenter and circumcenter of triangle $ABC$.
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vslmat
154 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $DG, EK, FL$ are the altitudes of $\Delta EFD$. First we are going to show that $X$ lies on $AG, Y$ on $BK, Z$ on $CL$. Let $AO$ meet the circumcircle again at $M$, then by a well known lemma that $M, I, G$ are collinear. Let $N\equiv EF\cap AD$, as $(A, P; N, D) = -1$ and $GD\perp GN, GN$ is the angle bisector of $\angle PGA$, this meens that $X$, the reflection of $P$ over $EF$ lies on $AG$. Similarly with $Y$ and $Z$.
So to show $AX, BY, CZ$ concur is equivalent to show that $AG, BK, CL$ concur. This can be done by trig Ceva. Easy to see that:
$\frac{sinFAG}{sinGAE} = \frac{FG}{GE}$ and so on …

As $\frac{FG}{GE}.\frac{EL}{LD}.\frac{KD}{KF} = 1$, $AG, CL, BY$ indeed concur at a point, say $Q$. Now we are going to show that $Q, H, I$ are collinear. Notice that $\angle FGK = \angle FDE = \angle AFE$, hence $GK\parallel AB$, but $HK\parallel IB$ and $HG\parallel AI$, so $\Delta HGK\sim \Delta IAB$ and so $QK/QB = GK/AB = KH/BI$, this means that $Q, H, I$ are collinear.
But by another well known lemma, the Euler line of $\Delta DEF$, that is $HI$ goes through $O$, so $Q$ lies on $OI$.
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mathocean97
606 posts
#3 • 3 Y
Y by Adventure10, Mango247, ehuseyinyigit
I would just like to mention that this concurrence point is the same point as this RMM problem.

So after we get that (using the notation of the above post) $A, X, G$ are collinear, once again by simple Trig-Ceva on triangle $AEF$, we can get $I_CF, I_BE, AG$ are collinear, where $I_C$, etc. are the excenters. (it reduces to $\frac{AI_C}{AI_B} = \frac{FG}{GE}$, which is simple to verify). Now it reduces to the well-known problem that $I_AD, I_BE, I_CF$ concur on $OI$.
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IDMasterz
1412 posts
#4 • 2 Y
Y by Adventure10, Mango247
Hang on, but if the altitude from $D$ to $EF$ is $X$, then $XF$ bisects $\angle AXP$ hence $AXP_D$ is a line ($P_D$ is the reflection of $P$ over $EF$). Now if we define $Y, Z$ similarly, then $\triangle XYZ$ and $\triangle ABC$ are homothetic. By an inversion, if $N$ is the ninepoint centre of $\triangle DEF$ then $O, I, N$ are collinear, and since $O, N$ are corresponding we are done.
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jayme
9801 posts
#5 • 1 Y
Y by Adventure10
Dear Mathlinkers,
P is the Gergonne point of ABC.
Sincerely
Jean-Louis
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jayme
9801 posts
#6 • 2 Y
Y by Adventure10, ehuseyinyigit
Dear Mathlinkers,
I come back..
note that ABC is the P-anticevian of DEF and XYZ the P-symmetric triangle wrt DEF.
After my source thuis result come from Emelyanov L. and T. (2002).
Sincerely
Jean-Louis
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IDMasterz
1412 posts
#7 • 2 Y
Y by Adventure10, Mango247
jayme wrote:
Dear Mathlinkers,
I come back..
note that ABC is the P-anticevian of DEF and XYZ the P-symmetric triangle wrt DEF.
After my source thuis result come from Emelyanov L. and T. (2002).
Sincerely
Jean-Louis

The concurrence point is the isogonal conjugate of the mitenpunkt point of $\triangle ABC$. Do you think this problem con be generalised?

Note that $P$ is on the Feuerbach hyperbola, and $OI$ is the tangent... $P$ is the perspector of $\triangle ABC$ and the pedal triangle of $I$.
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WJ.JamshiD
29 posts
#8 • 2 Y
Y by Adventure10, Mango247
jayme wrote:
Dear Mathlinkers,
P is the Gergonne point of ABC.
Sincerely
Jean-Louis

What is Gergonne point of triangle?
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tkhalid
965 posts
#9 • 1 Y
Y by Adventure10
If $D, E,$ and $F$ are the contact points of the incircle with the sides of $\triangle{ABC}$ then the lines $AD, BE,$ and $CF$ are concurrent at the Gergonne Point of $\triangle{ABC}$.
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WJ.JamshiD
29 posts
#10 • 2 Y
Y by Adventure10, Mango247
Thanks :)
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mathcool2009
352 posts
#11 • 2 Y
Y by Adventure10, Mango247
Note. In barycentric coordinates, the line \[ux+vy+wz = 0\]will be denoted as \[(u \, ; \, v \, ; \, w ).\]
Note that $\overline{AD}, \overline{BE}, \overline{CF}$ concur at the Gergonne point $G$ of $\triangle ABC$.

We will use barycentric coordinates to characterize the line $\overline{IO}$. In particular, it's easy to check that this line is \[(\frac{b-c}{a}(-a+b+c) \, ; \, \frac{c-a}{b}(a-b+c) \, ; \, \frac{a-b}{c} (a+b-c))\]
Let's characterize the cevian $\overline{AX}$. By Law of Sines we see that \[\frac{\sin FAX}{\sin XAE} =  \frac{FX \cdot \frac{\sin AFX}{AX}}{EX \cdot \frac{\sin AEX}{AX}} = \frac{FG}{EG} \cdot \frac{\sin AFX}{\sin AEX}.\]A bit of angle chasing yields \[\measuredangle AFX = \measuredangle AFE + \measuredangle EFX = \measuredangle FEA + \measuredangle CFE = \measuredangle FEC + \measuredangle CFE = \measuredangle FCE \](in directed angles) and thus \[\frac{\sin FAX}{\sin XAE} = \frac{FG}{EG} \cdot \frac{\sin GCE}{\sin FBG} = \frac{(\frac{\sin GCE}{EG})}{(\frac{\sin FBG}{FG})} = \frac{(\frac{\sin EGC}{CE})}{(\frac{\sin BGF}{BF})} = \frac{BF}{CE} = \frac{a-b+c}{a+b-c}.\]It follows that $X$ has coordinates \[(x : \frac{b}{a-b+c} : \frac{c}{a+b-c}).\]Thus $\overline{AX}, \overline{BY}, \overline{CZ}$ concur at the point \[Q = (\frac{a}{-a+b+c}:\frac{b}{a-b+c} : \frac{c}{a+b-c}) \]which is on line $\overline{IO}$, as desired. $\blacksquare$
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anantmudgal09
1980 posts
#12 • 2 Y
Y by Adventure10, Mango247
proglote wrote:
The incircle of triangle $ABC$ touches sides $BC, CA$ and $AB$ at points $D, E$ and $F$, respectively. Let $P$ be the intersection of lines $AD$ and $BE$. The reflections of $P$ with respect to $EF, FD$ and $DE$ are $X,Y$ and $Z$, respectively. Prove that lines $AX, BY$ and $CZ$ are concurrent at a point on line $IO$, where $I$ and $O$ are the incenter and circumcenter of triangle $ABC$.

Construct parallelogram $PEX'F$. Define $Y', Z'$ analogously. First we prove the

Claim. $\overline{AX'}, \overline{BY'}, \overline{CZ'}$ concur at a point $Q$.

(Proof) Observe that $$P=\left(\frac{1}{s-a}: \frac{1}{s-b}: \frac{1}{s-c}\right)=\left(\frac{(s-b)(s-c)}{\sum ab -s^2}:--:--\right).$$Now $\overrightarrow{X'}=\overrightarrow{E}+\overrightarrow{F}-\overrightarrow{P}$ so we compute $$(X')_y=\left(\frac{s-a}{c}-\frac{(s-a)(s-c)}{\sum ab -s^2}\right)=\left(\frac{(s-a)}{\sum ab -s^2}\right) \cdot \left(b\left((c+a)(c+b)-s(s+c)\right)\right)$$and similarly we have $(X')_z$. Consequently,
$$X'=\left(--:b\left((c+a)(c+b)-s(s+c)\right):c\left((b+a)(b+c)-s(s+b)\right)\right).$$It is now clear that the claim must hold. As an immediate corollary, we see $Q=\left(\frac{a}{(a+b)(a+c)-s(s+a)}:--:--\right)$. $\blacksquare$

Now let $R$ be the isogonal conjugate of $Q$. It is sufficient to show $RIO$ is a line. Note that $R=\left(a\left((a+b)(a+c)-s(s+a)\right):--:--\right)$.

Recall that $\overline{IO}$ has equation $$\sum_{\text{cyc}} x \cdot \frac{(b-c)(b+c-a)}{a} =0.$$Now we undertake the arduous task of plugging in $R$. It suffices to show $$\sum_{\text{cyc}} a(b-c)\left((a+b)(a+c)-s(s+a)\right)=\sum_{\text{cyc}} (b^2-c^2)\left((a+b)(a+c)-s(s+a)\right).$$Compute the RHS as \begin{align*} \sum_{\text{cyc}} (b^2-c^2)\left((a+b)(a+c)-s(s+a)\right) &= \prod_{\text{cyc}} (b+c) \cdot \sum_{\text{cyc}} (b-c)-s\sum_{\text{cyc}} (b^2-c^2)(s+a) \\ &=s\sum_{\text{cyc}} a^2(b-c).\end{align*}On the other hand, compute LHS as \begin{align*} \sum_{\text{cyc}} a(b-c)\left((a+b)(a+c)-s(s+a)\right) &=\sum_{\text{cyc}} a^3(b-c)+\sum_{\text{cyc}} a(b-c)(ab+bc+ca)-s^2\sum_{\text{cyc}} a(b-c)-s\sum_{\text{cyc}} a^2(b-c) \\ &= \sum_{\text{cyc}} a^3(b-c)-s\sum_{\text{cyc}} a^2(b-c).\end{align*}Consequently, it suffices to check $$\sum_{\text{cyc}} a^3(b-c)=(a+b+c)\cdot \sum_{\text{cyc}} a^2(b-c)$$which is clearly true. $\blacksquare$

Remark. Again the parallelogram trick (alongside isogonal conjugates) saved the day! It is motivated because computing $X$ directly is otherwise too bad since $P$ is semi-decent at best.

Edit: Apparently the post just above shows $X$ isn't that bad. Oops.
This post has been edited 2 times. Last edited by anantmudgal09, Sep 29, 2017, 5:13 AM
Reason: oops
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WizardMath
2487 posts
#13 • 4 Y
Y by amar_04, Arabian_Math, Adventure10, Mango247
Turns out the fact that the perspector of the reflection triangle of the symmedian point amd the tangential triangle lies on the Euler line trivialises the problem.
This post has been edited 2 times. Last edited by WizardMath, Sep 29, 2017, 9:02 AM
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omarius
91 posts
#14 • 1 Y
Y by Adventure10
IDMasterz wrote:
Hang on, but if the altitude from $D$ to $EF$ is $X$, then $XF$ bisects $\angle AXP$ hence $AXP_D$ is a line ($P_D$ is the reflection of $P$ over $EF$). Now if we define $Y, Z$ similarly, then $\triangle XYZ$ and $\triangle ABC$ are homothetic. By an inversion, if $N$ is the ninepoint centre of $\triangle DEF$ then $O, I, N$ are collinear, and since $O, N$ are corresponding we are done.

Why does $(XF)$ bisects $<AXP$ ?
This post has been edited 1 time. Last edited by omarius, Jan 14, 2018, 12:42 PM
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WizardMath
2487 posts
#15 • 1 Y
Y by Adventure10
@above, project $D, EF\cap BC, B,C$ from $P$ onto $AD$, and use harmonic conjugates and the $90^\circ$ you can see.
This post has been edited 1 time. Last edited by WizardMath, Jan 14, 2018, 1:08 PM
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Aryan-23
558 posts
#16 • 3 Y
Y by spartacle, Mr_Guy, Nathanisme
Solved with pinetree1 and Psyduck909 :love: :love:

Let $D'$ denote the foot of altitude from $D$ onto $EF$. Define $E',F,$ similarly. We will prove that $D'\in AX$, etc.

Suppose $S = AP\cap EF$, then we have $(AP; SD) = -1$, and $\angle DD'S = 90^{\circ}$, so this implies $D'S$ bisects $\angle AD'P$. This, however implies that $D' \in AX$. So the claim holds good.$\square$

Note that by angle-chasing $\triangle D'E'F' \sim \triangle ABC$, so $AD',BE',CF'$ concur at the homothetic centre of the two triangles, say $T$. Note that $T$ lies on the line joining the circumcenter of the two triangles, which is $OI$.(since it's the Euler Line of $\triangle DEF$) This completes the proof. $\square$
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amar_04
1916 posts
#17 • 5 Y
Y by Kar-98k, A-Thought-Of-God, Mr_Guy, Mathematicsislovely, ehuseyinyigit
This is a very well known Property. From here we get that $\overline{AX},\overline{BY},\overline{CZ}$ are concurrent on the Isogonal Mittenpunkt Point $(X_{57})$ of $\Delta ABC$ which lies on $\overline{OI}$. $\blacksquare$
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mathlogician
1051 posts
#18 • 3 Y
Y by Bhakti, Muaaz.SY, ehuseyinyigit
Slowly but surely I am starting to dislike geometry more and more :(

Relabel $P$ as $G$, the Gerogonne point of $\triangle ABC$. Suppose that $L$ is the foot of the altitude from $D$ to $\overline{EF}$. Suppose that $\overline{AG}$ meets $\overline{EF}$ at a point $K$, and lines $\overline{EF}$ and $\overline{BC}$ intersect at a point $T$. Note that $-1 = (TD;BC) \stackrel{E}{=} (AG;KD)$ and moreover $\angle DLK = 90^\circ$, so $\overline{AL}$ and $\overline{LG}$ are reflections about $\overline{EF}$. This implies that $\overline{AXL}$ is collinear. Now we know that $\overline{AX}, \overline{BY}, \overline{CZ}$ concur at a point $P$ by Cevian Nest Theorem.

I claim that $P$ and $O$ lies on the Euler Line of $\triangle DEF$. The latter is well-known by incircle inversion. Moreover, note that $P$ is the center of homothety sending $(ABC)$ to the nine-point circle of $\triangle DEF$, so it must lie on the Euler Line of $\triangle DEF$. This concludes the proof.
This post has been edited 1 time. Last edited by mathlogician, Oct 11, 2020, 2:44 AM
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amuthup
779 posts
#19 • 2 Y
Y by Mango247, Mango247
Let $\overline{AX},\overline{BY},\overline{CZ}$ intersect $\overline{EF},\overline{FD},\overline{DE}$ at $P,Q,R$ respectively.

$\textbf{Claim: }$ $P$ is the foot of the altitude from $D$ to $\overline{EF}.$

$\emph{Proof: }$ Let $A'$ be the reflection of $A$ across $\overline{EF},$ and let $P'$ be the foot of the altitude from $D$ to $\overline{EF}.$ It suffices to show that $A',G,P'$ are collinear.

Let $M$ be the midpoint of $\overline{DP'}$ and let $N$ be the midpoint of $\overline{EF}.$ By definition, $G$ is the symmedian point of $\triangle DEF,$ so $G,M,N$ collinear. Furthermore, $A,G,D$ collinear and $\overline{AN}\parallel\overline{DM}.$

Therefore, there exists a homothety centered at $G$ sending $\overline{AN}$ to $\overline{DM}.$ It is easy to see that this homothety sends $A'$ to $P',$ so we are done. $\blacksquare$

Similarly, we can show $Q,R$ are the feet of the altitudes from $E,F$ to $\overline{FD},\overline{DE}$ respectively. Now the concurrency follows from the Cevian Nest Theorem, so let $T$ be the concurrency point.

Note that $\angle QPF=\angle FDE=\angle AFP,$ so $\overline{PQ}\parallel\overline{AB}.$ Similarily, $\overline{QR}\parallel\overline{BC}$ and $\overline{RP}\parallel\overline{CA},$ so triangles $PQR$ and $ABC$ are homothetic.

Therefore, since the circumcenter of $\triangle ABC$ is $O,$ the circumcenter of $\triangle PQR$ is the nine-point center of $\triangle DEF,$ and $O$ lies on the Euler line of $\triangle DEF,$ we know $T$ lies on $\overline{OI},$ as desired.
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rcorreaa
238 posts
#20
Y by
Firstly, we will prove the following claim:

Claim: Let $\Delta ABC$ a triangle and $\Gamma$ its circumcircle and $H_A$ the altitude from $A$ to $BC$. Suppose that the tangents through $B,C$ to $\Gamma$ intersect at $T$. Also, let $K$ the symmedian point of $\Delta ABC$ and $K_A$ its reflection WRT $BC$. Then, $T,K_A,H_A$ are collinear.
Proof: Let $M$ the midpoint of $BC$ and $D$ the midpoint of $AH_A$. It's well known that $D,K,M$ are collinear, AKA the $A$-Schwatt line of $\Delta ABC$. Then, observe that $$-1=(A,H_A;D,\infty_{AH_A})\stackrel{M}=(A,\{AT \cap BC\}=L;K,T)$$$\implies$ since $\angle AH_AL= 90º \implies H_AL$ bissects $\angle KH_AT$, hence $K_A$ lies on $H_AT$, as desired. $\square$

Now, from the claim, we have that $AX$ cuts $EF$ on the point $H_D$ such that $DH_D \perp EF$. Define $H_E,H_F$ similarly. An easy angle chasing gives us that $ABC, H_DH_EH_F$ are homothetic $\implies AH_D, BH_E, CH_F$ are concurrent (as well as $AX,BY,CZ$) on the line joining its incenter, which is line $HI$, where $H$ is the ortocenter of $DEF$. But it is well known that $O,H,I$ are collinear. This implies that $AX,BY,CZ$ are concurrent on $OI$, as desired.

$\blacksquare$
This post has been edited 1 time. Last edited by rcorreaa, Oct 29, 2020, 4:10 AM
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v_Enhance
6878 posts
#21 • 2 Y
Y by HamstPan38825, bhan2025
Let $\triangle JKL$ be the orthic triangle of $\triangle DEF$.
Claim: Points $A$, $J$, $X$ are collinear.
Proof. Let $N = \overline{EF} \cap \overline{AD}$. Then $-1 = (A,B; F, ED \cap AB) = (A,G;T,D)$, so from $\angle DJN = 90^{\circ}$, we get that $EF$ is the bisector of $\angle GJA$. $\blacksquare$
Thus $A$, $J$, $X$ are collinear, and we are interested now in $AJ$, $BK$, $CL$ concurrent on $\overline{IO}$ (i.e.\ we can now forget about the points $X$, $Y$, $Z$).

[asy]
size(12cm);
pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair G = extension(B, E, C, F); pair X = -G+2*foot(G, E, F); pair Y = -G+2*foot(G, F, D); pair Z = -G+2*foot(G, D, E); pair J = foot(D, E, F); pair K = foot(E, F, D); pair L = foot(F, D, E);
filldraw(A--B--C--cycle, invisible, blue); filldraw(incircle(A, B, C), invisible, blue); filldraw(D--E--F--cycle, invisible, red); draw(D--J, red); draw(E--K, red); draw(F--L, red); draw(A--D, blue); draw(B--E, blue); draw(C--F, blue); draw(A--J, deepgreen); draw(B--K, deepgreen); draw(C--L, deepgreen); draw(G--X, dashed+orange); draw(G--Y, dashed+orange); draw(G--Z, dashed+orange);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$G$", G, dir(280)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$J$", J, dir(120)); dot("$K$", K, dir(K)); dot("$L$", L, dir(315));
/* -----------------------------------------------------------------+ |                 TSQX: by CJ Quines and Evan Chen                  | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 130 B = dir 210 C = dir 330 I = incenter A B C D = foot I B C E = foot I C A F = foot I A B G 280 = extension B E C F X = -G+2*foot G E F Y = -G+2*foot G F D Z = -G+2*foot G D E J 120 = foot D E F K = foot E F D L 315 = foot F D E A--B--C--cycle / 0.1 lightcyan / blue incircle A B C / 0.1 lightcyan / blue D--E--F--cycle / 0.1 yellow / red D--J / red E--K / red F--L / red A--D / blue B--E / blue C--F / blue A--J / deepgreen B--K / deepgreen C--L / deepgreen G--X / dashed orange G--Y / dashed orange G--Z / dashed orange */
[/asy]
But note that $\triangle JKL$ and $\triangle ABC$ are homothetic, since \[ \measuredangle JKF = \measuredangle JKD = \measuredangle JED = \measuredangle FED = \measuredangle BFD = \measuredangle BFK \implies \overline{BA} \parallel \overline{JK}. \]Now the incenters of $\triangle JKL$ and $\triangle ABC$ are the orthocenter of $\triangle DEF$ and the point $I$, respectively. Using the well-known fact that the Euler line of the intouch triangle is exactly line $IO$, it follows the homothety center lies on this line as well.
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aleksijt
44 posts
#22 • 1 Y
Y by GeoKing
Here is a generalization:
Generalized problem wrote:
Let $P$ be on the Euler line of $\triangle ABC$ and $Q$ it's isogonal conjugate. Let $A'$, $B'$ and $C'$ are the reflections of $Q$ in $BC$, $CA$ and $AB$, and let $\triangle T_AT_BT_C$ be the tangential triangle. Then $T_AA'$, $T_BB'$ and $T_CC'$ concur on the Euler line.
One proof is by considering the rectangular circumhyperbola of $O$.
This post has been edited 1 time. Last edited by aleksijt, Oct 6, 2023, 6:17 PM
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cursed_tangent1434
650 posts
#24
Y by
We start off with the proof of the following claim.

Claim : In an acute triangle $ABC$, let $D$ be the foot of the $A-$altitude, $T$ the intersections of the tangents to the circumcircle of triangle $ABC$ at $B$ and $C$ and $K'$ the reflection of the symmedian point of triangle $ABC$ across $BC$. Points $D$ , $K'$ and $T$ are collinear.

Proof : Let $T'$ be the reflection of $T$ across line $BC$. It suffices to show that points $D$ , $K$ and $T'$ are collinear. Let $M$ denote the midpoint of segment $BC$ and $K_b$ the intersection of line $KB$ with the circumcircle of triangle $ABC$. Let $L$ denote the intersection of line $AK$ with line $BC$. Now,
\[-1=(AC;BK_b) \overset{B}{=}(AL;TK)\overset{D}{=}(\infty,M;T,DK \cap MT)\]But this implies that $M$ is the midpoint of the line segment joining points $DK\cap MT$ and $T$. Thus, $DK\cap MT=T'$ which proves the claim.

Now, looking at the above claim with the intouch triangle as our reference, it is well known that the Gergonne point $G$ is the symmedian point of the intouch triangle. Thus, if $P$ , $Q$ and $R$ are the feet of the altitudes from $D$ , $E$ and $F$ respectively to the opposite sides of the contact triangles, lines $AX$ , $BY$ and $CZ$ are simply lines $AD$ , $BE$ and $CF$ which must concur at the homothety center of the tangential and orthic triangles of $\triangle DEF$ which lies on the Euler Line of $\triangle DEF$ which is simply line $IO$.
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Ilikeminecraft
667 posts
#25
Y by
Note that $G$ is the gergonne point, so $A,G,D$ are collinear.
Define $D’$ the foot from $D$ to $EF.$
Define $S$ to be the intersection of $AD$ with $EF.$
Ceva-menelaus implies that $(AG;SD) = -1.$
Since $\angle DD’S = 90,$ it follows that by right angles and bisectors, $DE$ bisects angle $AD’G,$ which implies that $A, G_A, D$ are collinear.
Now, note that $\angle D’E’F’ = \angle D’E’E + \angle EE’F’ = \angle D’DE + \angle F’FE = 90 - \angle FED + 90 - \angle FED = 180-2\angle FED = \angle ABC,$ implying that $D’E’F’$ and $ABC$ are homothetic.
Define $N_9$ the nine point center of $DEF.$
Inversion about incircle implies $N_9, I, O$ are collinear.
This finishes, as our homothety implies $N_9, O$ are collinear.
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