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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

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Be sure to mark your calendars for the following events:
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0 replies
jlacosta
Mar 2, 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
Where is the equality?
AndreiVila   2
N 19 minutes ago by MihaiT
Source: Romanian District Olympiad 2025 9.3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2 replies
AndreiVila
Mar 8, 2025
MihaiT
19 minutes ago
Truth or lie at a table
SinaQane   7
N 26 minutes ago by Oksutok
Source: 239 2019 S4
There are $n>1000$ people at a round table. Some of them are knights who always tell the truth, and the rest are liars who always tell lies. Each of those sitting said the phrase: “among the $20$ people sitting clockwise from where I sit there are as many knights as among the $20$ people seated counterclockwise from where I sit”. For what $n$ could this happen?
7 replies
SinaQane
Jul 31, 2020
Oksutok
26 minutes ago
2 var inquality
sqing   4
N 27 minutes ago by sqing
Source: Own
Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{3}{2}. $ Show that$$ a+b+ab\geq1$$Let $ a ,  b\geq 0 $ and $ \frac{1}{a^2+1}+\frac{1}{b^2+1}\le   \frac{5}{6}. $ Show that$$ a+b+ab\geq2$$
4 replies
sqing
Today at 4:06 AM
sqing
27 minutes ago
Functional inequality on N
BartSimpsons   21
N 28 minutes ago by Tony_stark0094
Source: European Mathematical Cup 2017 Problem 1
Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that the inequality $$f(x)+yf(f(x))\le x(1+f(y))$$holds for all positive integers $x, y$.

Proposed by Adrian Beker.
21 replies
BartSimpsons
Dec 27, 2017
Tony_stark0094
28 minutes ago
A colouring game on a rectangular frame
Tintarn   0
32 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 4
For integers $m,n \ge 3$ we consider a $m \times n$ rectangular frame, consisting of the $2m+2n-4$ boundary squares of a $m \times n$ rectangle.

Renate and Erhard play the following game on this frame, with Renate to start the game. In a move, a player colours a rectangular area consisting of a single or several white squares. If there are any more white squares, they have to form a connected region. The player who moves last wins the game.

Determine all pairs $(m,n)$ for which Renate has a winning strategy.
0 replies
Tintarn
32 minutes ago
0 replies
Find the value
sqing   5
N 33 minutes ago by sqing
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
5 replies
sqing
Today at 5:02 AM
sqing
33 minutes ago
A touching question on perpendicular lines
Tintarn   0
34 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
0 replies
Tintarn
34 minutes ago
0 replies
The last nonzero digit of factorials
Tintarn   0
36 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 2
For each integer $n \ge 2$ we consider the last digit different from zero in the decimal expansion of $n!$. The infinite sequence of these digits starts with $2,6,4,2,2$. Determine all digits which occur at least once in this sequence, and show that each of those digits occurs in fact infinitely often.
0 replies
Tintarn
36 minutes ago
0 replies
Fridolin just can't get enough from jumping on the number line
Tintarn   0
38 minutes ago
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 1
Fridolin the frog jumps on the number line: He starts at $0$, then jumps in some order on each of the numbers $1,2,\dots,9$ exactly once and finally returns with his last jump to $0$. Can the total distance he travelled with these $10$ jumps be a) $20$, b) $25$?
0 replies
Tintarn
38 minutes ago
0 replies
D1016 : A strange result about the palindrom polynomials
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $Q\in \{-1,1,0\}[x]$ with $Q(1)=0$.

Is it true that $\exists Q \in \mathbb Z[x]$ palindrom with $Q | P$ ?
0 replies
Dattier
an hour ago
0 replies
EMC last problem
GreekIdiot   1
N an hour ago by Tintarn
Source: EMC 2024 P4, Ioannis Galamatis
Find all functions $f: \mathbb{R_+} \rightarrow \mathbb{R_+}$ such that $f(x+yf(x))=xf(y+1) \: \forall\: x, \:y \in \mathbb{R_+}$.
Note: With the symbol $\mathbb{R_+}$ we denote the set of positive real numbers.
1 reply
GreekIdiot
Yesterday at 7:02 PM
Tintarn
an hour ago
Functional equation with a parameter
Tintarn   8
N an hour ago by NicoN9
Source: Baltic Way 2024, Problem 1
Let $\alpha$ be a non-zero real number. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that
\[
xf(x+y)=(x+\alpha y)f(x)+xf(y)
\]for all $x,y\in\mathbb{R}$.
8 replies
Tintarn
Nov 16, 2024
NicoN9
an hour ago
D1015 : A strange EF for polynomials
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
1 reply
Dattier
Yesterday at 8:37 PM
Dattier
an hour ago
Functional Equations Marathon March 2025
Levieee   19
N an hour ago by Jupiterballs
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form $\text{"find all functions:"}$
19 replies
Levieee
Today at 1:03 AM
Jupiterballs
an hour ago
IMO 2012 P5
mathmdmb   122
N Today at 2:51 AM by KevinYang2.71
Source: IMO 2012 P5
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.

Show that $MK=ML$.

Proposed by Josef Tkadlec, Czech Republic
122 replies
mathmdmb
Jul 11, 2012
KevinYang2.71
Today at 2:51 AM
IMO 2012 P5
G H J
Source: IMO 2012 P5
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peace09
5409 posts
#114
Y by
'nother fact that holds true:

Claim. $\tfrac{C_DL}{CL}=\tfrac{C_DK}{CK}$, where $C_D$ is the reflection of $C$ across $AB$.

Proof. Let $\omega_A,\omega_B$ be the circles centered at $A$ passing through $C,L$ and centered at $B$ passing through $C,K$, respectively. Additionally, define $K_2,L_2$ to be the second intersections of $BK,AL$ with $\omega_A,\omega_B$ respectively, and for completeness' sake add $C_A,C_B$, the antipodes of $C$ with respect to $\omega_A,\omega_B$.

The trick is to invert at $C$ to produce the following (awfully low resolution, sorry) diagram:

https://cdn.artofproblemsolving.com/attachments/e/0/b8aaaae6a8cae5288aee2c656e87225910cd25.png

Here, $\omega_A'\perp\omega_B'$ implies that $C_D'$ is the midpoint of both $K'K_2'$ and $L'L_2'$. But $C_D'$ lies on the radical axis of $(A'CX')$ and $(B'CX')$, meaning
\[C_D'K'{}^2=C_D'K'\cdot C_D'K_2'=C_D'L'\cdot C_D'L_2'=C_D'L'{}^2.\]So $C_D'K'=C_D'L'$, which becomes $\tfrac{C_DK\cdot r^2}{CC_D\cdot CK}=\tfrac{C_DL\cdot r^2}{CC_D\cdot CL}$ where $r$ is the radius of inversion. Rearranging gives the desired. $\Box$
This post has been edited 1 time. Last edited by peace09, Jun 25, 2023, 8:01 PM
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minusonetwelth
225 posts
#115
Y by
The main idea is to see $MK$ and $ML$ as tangents to some circle. Call $\gamma_A$ the circle through $A$ with radius $AC$ and define $\gamma_B$ similarly. Let $\gamma=(ABC)$. Define $L'=BX\cap \gamma_A$ and $K'=AX\cap \gamma_B$. Let $C'$ be the reflection of $C$ over $AB$, which clearly lies on $\gamma$, $\gamma_A$ and $\gamma_B$. Then
\[XK'\cdot XK=XC\cdot XC'=XL\cdot XL',\]so $L'KLK'$ is cyclic. Then, $BK^2=BC^2=BL\cdot BL'$, so $BK$ and thus $MK$ is tangent to $(L'KLK')$. Similarly, $ML$ is tangent to this circle, implying the result by equal tangents.
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cursed_tangent1434
548 posts
#116
Y by
Let $S$ be the reflection of $C$ over $AB$. Since $\measuredangle ASB = 90^\circ = \measuredangle ACB$, it is easy to see that $S$ lies on $(ABC)$. Now, let $\omega_K$ be the circle centered at $B$ with radius $BK$ and $\omega_L$ the circle centered at $A$ with radius $AL$. Since,
\[AC=AL=AS \text{ and } BC=BK=BS\]we also have that $C$ and $S$ lie on both $\omega_B$ and $\omega_C$. Let $K'= \overline{AK} \cap \omega_K \neq K$ and $L' = \overline{BL} \cap \omega_L \neq L$. Since,
\[XL\cdot XL' = XC \cdot XS = XK \cdot XK'\]we also have that $KLK'L'$ is cyclic (let its circumcircle be $\Gamma$). Now, note that
\[BL \cdot BL' = \text{Pow}_{\omega_L}(B)= BA^2-AC^2 = BC^2\]Thus, $BC$ is a tangent to $\omega_L$. Similarly, $AC$ is a tangent to $\omega_K$. Now, note that,
\[BK^2=BC^2=BL \cdot BL'\]Thus, $BK$ is a tangent to $\Gamma$. Similarly, $AL$ is also a tangent to $\Gamma$. Now, this means, that $M$ is the intersection of the tangents to $\Gamma$ at $K$ and $L$ and thus, we have
\[MK=ML\]which was the required conclusion.
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stevenmeow
554 posts
#117
Y by
Here is a cheesy computational solution which uses is nearly limited to addition and multiplication. This necromantic post is partly for my own benefit. The following is probably hard to find compared to other computational solutions here. We use the notation aa for the square of a, following the style of Riemann (see wiki: On_the_Number_of_Primes_Less_Than_a_Given_Magnitude )

In fact, we are using X as the main angle. Define $a = AX, b = BX, k = KX, l = LX$. Let $c = -\cos(\angle AXB)$. Note that $[X, K, A]$ are collinear in that order as are $[X, L, B]$.

[1] $AL^2 = aa+ll+2alc.$
[2] $BK^2 = bb+kk+2bkc.$
[3] Given that $AL^2 + BK^2 = AC^2 + BC^2 = AB^2 = aa+bb+2abc,$
[4] $ll+kk=2c(ab-al-bk).$
From a standard lemma about perpendicular lines and lengths applied to $AB$ and $CX$,
[5] $aa + BK^2 = AX^2 + BC^2 = BX^2 + AC^2 = bb + AL^2.$
Substituting,
[6] $kk-ll=2c(al-bk).$

Now working backwards, it suffices to show the following, using the law of sines on triangles MAK and MBL (excuse the fractions)
[7 STS] $ AK / \sin( \angle AMK ) * \sin ( \angle MAK ) = BL / \sin ( \angle BML ) * \sin ( \angle MBL ).$
We cancel the vertical angles, substitute (a, b, k, l), and use that a > k; b > l
[8 STS] $ (a-k) \sin ( \angle LAX ) = (b-l) \sin( \angle BKX ).$
Using the the law of sines on triangles LAX and BKX,
[9 STS] $ (a-k) l \sin ( \angle X) / AL = (b-l) k \sin( \angle X) / BK.$
Since both sides are positive, we can square both sides. Also we cancel out sin(X), clear denominators (I was getting worried about the fractions there) and substitute [1] and [2].
[10 STS] $ (a-k)^2 ll (bb + kk + 2bkc) = (b-l)^2 kk (aa+ll-2alc). $

[4 + 6] $kk = abc - 2bkc.$
[4 - 6] $ll = abc - 2alc.$

We use [4 + 6] and [4 - 6] to replace kk and ll in [10] (but not in $(a-k)^2, (b-l)^2$).
[11 STS] $ (a-k)^2 (abc - 2alc) (bb+abc) = (b-l)^2 (abc - 2bkc) (aa+abc).$
Cancelling factors of a, b, and c,
[12 STS] $ (a-k)^2 (b-2l) (b+ac) = (b-l)^2 (a-2k) (a+bc).$

Here is the computational simplification that comes out of nowhere for a true geometry problem.

We observe $(b-2al)(b+ac) = bb-2bl+abc-2alc = bb-2bl=ll = (b-l)^2,$ using [4-6]. Similarly, $(a-2k)(a+bc) = (a-k)^2,$ so [12] is true and we are done.
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IAmTheHazard
5000 posts
#118
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Let $\omega_A$ and $\omega_B$ be the circles centered at $A$ and $B$ (resp.) passing through $C$, so $K$ and $L$ lie on $\omega_B$ and $\omega_A$ respectively. Let $K'$ and $L'$ be the other intersections of $\overline{AX}$ with $\omega_B$ and $\overline{BX}$ with $\omega_A$.

By the converse of the radical axis theorem, $KLK'L'$ is cyclic. Furthermore, since $\angle BCA=90^\circ$, $\omega_A$ and $\omega_B$ are orthogonal, so inverting at $A$ swaps $K$ and $K'$ (and obvious fixes $L$ and $L'$). Thus $\omega_A$ and $(KLK'L')$ are orthogonal, so by definition $\overline{AL}$ is tangent to $(KLK'L')$. Likewise, $\overline{BK}$ is tangent to $(KLK'L')$, hence $M$ is the intersection of the tangents from $K$ and $L$ to $(KLK'L')$ and $MK=ML$. $\blacksquare$
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IAmTheHazard
5000 posts
#119
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CyclicISLscelesTrapezoid wrote:
My solution is very similar to everyone else's, but I want to note something that I think is really cool. Let $\overline{AK}$ intersect the circle centered at $B$ through $C$ again at $P$, and let $\overline{BL}$ intersect the circle centered at $A$ through $C$ again at $Q$. Notice that lines $KQ$ and $LP$ intersect on $\overline{AB}$, as do lines $KL$ and $PQ$. There's a projective proof to this, but I challenge you to find the synthetic proof. Hint

Given the tangency facts above, this is just Brocard on $KLK'L'$ since $A$ is the pole of $\overline{KK'}$ and $B$ is the pole of $\overline{LL'}$, so any proof of Brocard will work here and any proof of this should just prove Brocard's theorem
This post has been edited 1 time. Last edited by IAmTheHazard, Dec 27, 2023, 10:40 PM
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trk08
614 posts
#120
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Note that $K$ lies on the circle with center $B$ that goes through $C$. Also, $L$ lies on the circle with center $A$ that goes through $C$. Call these circles $\omega, \gamma$, respectively. Also, let $T=(BDK)\cap (ADL)$

Claim:
$X$ is the radical center of $\omega, \gamma, (BDK), (ADL)$.
Proof:
Note that $AC$ is tangent to $\omega$. Therefore, using the given right angle condition:
\[Pow_{\omega}(A)=AC^2=AD\cdot AB=Pow_{(BDK)}(A),\]so $AKX$ is the radical axis of $\omega$ and $(BDK)$.

Similarly, $BLX$ is the radical axis of $(LDA)$ and $\gamma$. Therefore, the radical center is:
\[AKX\cap BLX=X\]$\square$

Claim:
$T$ lies on the radical axis of $\omega,\gamma$
Proof:
It suffices to show that $X$ lies on $TD$. Note that $TD$ is the radical axis of $(ADL),(BDK)$. Using our first claim, we are done $\square$

Claim:
The circle with center $T$ and goes through $L$, goes through $K$, and is tangent to $ML$ and $MK$
Proof:
Note:
\[\angle TLA=\angle TDA=\angle TKB=90.\]Therefore:
\[Pow_{\omega}(T)=TK^2=Pow_{\gamma}(T)=TL^2,\]so $TL=TK$, implying that the circle exists. As $\angle TLM=\angle TKM=90$, the tangency condition is implied $\square$

Therefore, $ML=MK$, as desired $\blacksquare$
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shendrew7
788 posts
#121
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Let $K^*$ and $L^*$ be the inverses of $K$ and $L$ wrt $\omega_A$ and $\omega_B$, the circles through $C$ centered at $A$ and $B$, respectively.
  • $\omega_A$ and $\omega_B$ are orthogonal, so $K^*$ lies on $\omega_B$ and $L^*$ lies on $\omega_A$.
  • $X$ lies on the radax of $\omega_A$, and $\omega_B$, so $KLK^*L^*$ is cyclic. Let its circumcircle be $\gamma$.
  • $KLK^*L^*$ is also harmonic, with tangents from $K$ and $K^*$ meeting at $B$ and tangents from $L$ and $L^*$ meeting at $A$, as
    \[AK \cdot AK^* = AC^2 = AL^2.\]

Thus $MK$ and $ML$ are tangents to $\gamma$, which finishes. $\blacksquare$
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ihatemath123
3427 posts
#122
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Let $\omega_A$ be the circle centered at $A$ passing through $C$ and $L$; define $\omega_B$ likewise. Clearly, $\omega_A$ and $\omega_B$ are orthagonal. Let $A_1$ be the second intersection of line $KX$ with $\omega_A$; define $B_1$ likewise. Let $L_1$ and $K_1$ be the second intersections of line $LK$ with $\omega_A$ and $\omega_B$, respectively.

Inversion about $W_A$ sends $L_1$ and $L$ to itself, as well as sending $K$ to $A_1$. Hence, $ALA_1L_1$ is cyclic, so
\[\angle ALL_1 = \angle AL_1L = \angle AA_1L\]Similarly, $\angle BKK_1 = \angle BB_1K$.

But by the radical axis theorem, $ABA_1B_1$ is cyclic, so $\angle AA_1L = \angle BB_1K$, hence $\angle ALL_1 = \angle BKK_1$. So, $\triangle MKL$ is isosceles as desired.
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CyclicISLscelesTrapezoid
371 posts
#124 • 1 Y
Y by GeoKing
Let $\omega_A$ and $\omega_B$ be the circles through $C$ centered at $A$ and $B$, and let $C'$ be the reflection of $C$ over $\overline{AB}$. Choose $Y$ on $\overline{CC'}$ such that $(C,C';X,Y)=-1$. By projecting through $K$, we have $(C,C';\overline{KX} \cap \omega_B,\overline{KY} \cap \omega_B)=-1$, so $\overline{KY}$ is tangent to $\omega_B$. Analogously, $\overline{LY}$ is tangent to $\omega_C$.

By radical axis, there exists a circle tangent to $\omega_A$ at $L$ and $\omega_B$ at $K$. The center of this circle lies on $\overline{AL}$ and $\overline{BK}$, so it must be $M$. Therefore, we have $MK=ML$, as desired. $\square$
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SenorSloth
37 posts
#125 • 1 Y
Y by OronSH
Diagram

Let $P$ be the foot of the perpendicular from $B$ to $AX$, and let $Q$ be the foot of the perpendicular from $A$ to $BX$. Let the circle centered at $B$ through $C$ be $\omega_1$, and let the circle centered at $A$ through $C$ be $\omega_2$. Define $Y$ to be the second intersection of $AX$ with $\omega_1$, and let $Z$ be the second intersection of $BX$ with $\omega_2$.

Note that $P$ and $Q$ lie on $(ABC)$ because of the right angles. Let $F=BP\cap AQ$. We claim that $F$ lies on line $XD$. By Power of a Point, $FP\cdot FB = FQ\cdot FA$, which means $F$ has equal power to the circles $(BDXP)$ and $(ADXQ)$. Thus, $F$ must lie on their radical axis, which is the line $XD$.

Since $X$ lies on the radical axis of $\omega_1$ and $\omega_2$, $LX\cdot XZ = KX \cdot XY$, which implies that $KLYZ$ is cyclic. Note that since $F$ is the intersection of two perpendicular bisectors of sides of $KLYZ$, it must be the center.

Note that $BP\cdot BF= BD\cdot BA=BC^2=BK^2$, so $BK\perp KF$. Similarly, $AQ\cdot AL=AD\cdot AB=AC^2=AL^2$, so $AL\perp LF$. Thus $\angle FLM = \angle FKM = 90^{\circ}$. We also know $FL=FK$ since $F$ is the center of $(KLYZ)$, and the side $FM$ is shared, so then $FLM\cong FKM$ and $MK=ML$, as desired.
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Ywgh1
136 posts
#126
Y by
IMO 2012 p5
Let $\omega_1$ and $\omega_2 $ be circles Centered at $A$ and $B$ passing through $C$.
Let $K’$ and $L’$ be the intersections of $AK$ and $BL$ with circles $\omega_1$ and $\omega_2$.
By radical axis we have that $KLK’L’$ is cyclic.

Now since $\omega_1$ and $\omega_2$ are orthogonal, we get that $KLK’L’$ is harmonic, hence $MK$ and $ML$ are tangent to $(KLK’L’)$. Hence we are done.
This post has been edited 1 time. Last edited by Ywgh1, Aug 13, 2024, 8:52 AM
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Eka01
204 posts
#127 • 2 Y
Y by Sammy27, AaruPhyMath
Let $BX$ and $AX$ meet $(A,AC)$ and $(B,BC)$ again at $V$ and $U$ respectively.
It is obvious that the above circles are orthogonal and that $X$ lies on the radical axis of these two circles, hence by converse of power of point
$(KLUV)$ is cyclic. Also $AC^2=AK.AU=AL^2$ so $AL$ and similarly $BK$ is tangent to $(KLUV)$ so $M$ is the pole of $KL$ which implies what we needed to show.
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dolphinday
1310 posts
#128
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Let $\omega_A$ be the circle with center $A$ passing through $C$ and define $\omega_B$ similarly.
Let the second intersections of $AK$ and $BL$ with $\omega_B$ and $\omega_A$ be $M$ and $N$ respectively.
Since $\angle ACB = 90^\circ$ we have that $AC$ and $BC$ are tangents to $\omega_B$ and $\omega_A$ respectively. Then by Radical Axis since $KM$ and $LN$ intersect on the radical axis of $\omega_A$ and $\omega_B$, $KLMN$ is cyclic.
Since $\omega_A$ and $\omega_B$ are orthogonal, $K$ and $M$ are inverses wrt $\omega_A \implies AK \cdot AM = AL^2$ so $AL$ is a tangent to $(KLMN)$ and $BK$ is as well. So this implies that $MK$ and $ML$ are tangents so $MK = ML$, done.
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KevinYang2.71
393 posts
#129
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We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$. Note that $a^2+b^2=c^2$ and $D=(a^2:b^2:0)$. Let $X=:(a^2:b^2:t)$ and suppose $K=(k:b^2:t)=\left(\frac{k}{k+b^2+t},\,\frac{b^2}{k+b^2+t},\,\frac{t}{k+b^2+t}\right)$. Then $\overrightarrow{BK}=\left(\frac{k}{k+b^2+t},\,-\frac{k+t}{k+b^2+t},\,\frac{t}{k+b^2+t}\right)$, so from $BK=BC$ we get
\begin{align*}
a^2&=\frac{a^2t(k+t)-b^2kt+c^2k(k+t)}{(k+b^2+t)^2}\\
&=\frac{a^2t(k+t)-b^2kt+(a^2+b^2)k(k+t)}{(k+b^2+t)^2}\\
&=\frac{a^2(k+t)^2+b^2k}{(k+b^2+t)^2}\\
\implies b^2k^2&=a^2((k+b^2+t)^2-(k+t)^2)\\
&=a^2b^2(b^2+2k+2t).
\end{align*}Hence $k^2-2a^2k-a^2(b^2+2t)=0$ so
\[
k=\frac{2a^2+\sqrt{4a^4+4a^2(b^2+2t)}}{2}=a^2+a\sqrt{c^2+2t}.
\]Similarly, $L=\left(a^2:b^2+b\sqrt{c^2+2t}:t\right)$. Now
\[
M=\left(a^2+a\sqrt{c^2+2t}:b^2+b\sqrt{c^2+2t}:t\right)=\left(\frac{a^2+a\sqrt{c^2+2t}}{\alpha},\,\frac{b^2+b\sqrt{c^2+2t}}{\alpha},\,\frac{t}{\alpha}\right),
\]where $\alpha:=c^2+t+(a+b)\sqrt{c^2+2t}$. Let $\beta:=k+b^2+t=c^2+t+a\sqrt{c^2+2t}$ so
\[
\overrightarrow{KM}=\left(\left(a^2+a\sqrt{c+2t}\right)\left(\frac{1}{\alpha}-\frac{1}{\beta}\right),\,b^2\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)+\frac{b\sqrt{c+2t}}{\alpha},\,t\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\right).
\]Noting that $\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{b\sqrt{c^2+2t}}{\alpha\beta}$, we get
\[
\overrightarrow{KM}=\frac{b\sqrt{c^2+2t}}{\alpha\beta}\left(-\left(a^2+a\sqrt{c+2t}\right),\,\beta-b^2,\,-t\right)
\]Note that $\frac{\sqrt{c^2+2t}}{\alpha}$ is independent of $a$ and $b$. Now
\begin{align*}
\left(\frac{\alpha}{\sqrt{c^2+2t}}KM\right)^2&=\frac{b^2}{\beta^2}\left(a^2t(\beta-b^2)-b^2t\left(a^2+a\sqrt{c^2+2t}\right)+c^2(\beta-b^2)\left(a^2+a\sqrt{c^2+2t}\right)\right)\\
&=\frac{ab^2}{\beta^2}\left((\beta-b^2)\left(at+ac^2+c^2\sqrt{c^2+2t}\right)-b^2t\left(a+\sqrt{c^2+2t}\right)\right)\\
&=\frac{ab^2}{\beta^2}\left((\beta-b^2)\left(a\beta+b^2\sqrt{c^2+2t}\right)-b^2t\left(a+\sqrt{c^2+2t}\right)\right)\\
&=\frac{ab^2}{\beta^2}\left(a\beta(\beta-b^2)+b^2\sqrt{c^2+2t}\left(a^2+t+a\sqrt{c^2+2t}\right)-b^2t\left(a+\sqrt{c^2+2t}\right)\right)\\
&=\frac{a^2b^2}{\beta^2}\left(\beta(\beta-b^2)+b^2\sqrt{c^2+2t}\left(a+\sqrt{c^2+2t}\right)-b^2t\right)\\
&=\frac{a^2b^2}{\beta^2}\left(\beta^2+b^2\left(\sqrt{c^2+2t}\left(a+\sqrt{c^2+2t}\right)-t-\beta\right)\right)\\
&=\frac{a^2b^2}{\beta^2}\left(\beta^2+b^2\left(a\sqrt{c^2+2t}+c^2+2t-t-\beta\right)\right)\\
&=a^2b^2,
\end{align*}which is independent of $a$ and $b$. Thus $KM=LM$. $\square$
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