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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Sharygin 2025 CR P2
Gengar_in_Galar   4
N a few seconds ago by FKcosX
Source: Sharygin 2025
Four points on the plane are not concyclic, and any three of them are not collinear. Prove that there exists a point $Z$ such that the reflection of each of these four points about $Z$ lies on the circle passing through three remaining points.
Proposed by:A Kuznetsov
4 replies
Gengar_in_Galar
Mar 10, 2025
FKcosX
a few seconds ago
2 var inquality
sqing   5
N 2 minutes ago by ionbursuc
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$$ a+b+ab\geq2$$
5 replies
+1 w
sqing
Today at 4:06 AM
ionbursuc
2 minutes ago
100 Selected Problems Handout
Asjmaj   32
N 5 minutes ago by John_Mgr
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
32 replies
Asjmaj
Dec 31, 2024
John_Mgr
5 minutes ago
Where is the equality?
AndreiVila   2
N an hour 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
an hour ago
Inequality with roots of cubic
Michael Niland   0
Today at 7:23 AM
The equation $x^3+px+q =0$ has real roots $ a_1 ,a_2 , a_3 $ where $a_1 \leq a_2 \leq a_3$.

Similarly the equation $x^3 +rx +s=0$ has real roots $ b_1, b_2, b_3 $ where $b_1 \leq b_2,\leq b_3$.

Prove that if $ \frac{a_1}{b_1} \leq \frac{a_2}{b_2} \leq \frac{a_3}{b_3}$ (s non zero), then $(\frac{p}{r})^3 =(\frac{q}{s})^2$
0 replies
Michael Niland
Today at 7:23 AM
0 replies
Inequalities
nhathhuyyp5c   3
N Today at 6:36 AM by giangtruong13
Let $a,b,c$ be positive reals such that $a+b+c+2=abc$. Find the maximum value of $$\frac{a+1}{a^2+2}+\frac{b+1}{b^2+2}+\frac{c+1}{c^2+2}$$Given $n\ge 2$ non-zero reals $x_1,x_2,\cdots x_n$ such that their sum is $100$. Prove that there exists two numbers $x_i,x_j$ such that $\frac{1}{2}\le \left|\frac{x_i}{x_j}\right|\le 2$
3 replies
nhathhuyyp5c
Jan 10, 2025
giangtruong13
Today at 6:36 AM
Algebra-1
JetFire008   3
N Today at 5:11 AM by BackToSchool
Find real numbers $p$,$q$ if $1+i$ is a root of $x^3+px^2+qx+6=0$ and solve the equation.
3 replies
JetFire008
Yesterday at 3:19 PM
BackToSchool
Today at 5:11 AM
Cool problem
jkim0656   3
N Today at 4:09 AM by RedFireTruck
Hey AoPS!
I came across a problem recently and it goes like this:
Which is greater:
$$2 ^ {100!} $$or $$2^{100}!$$Soo... can u guys help? thx!
:yoda: and may the force be with u!
Notes: I don't exactly know where this problem came from, but if u find that u are the orig maker of this problem feel free to drop me a PM and ill add u to my post :)
3 replies
jkim0656
Today at 3:04 AM
RedFireTruck
Today at 4:09 AM
help me at this pls i'm so confused
myname17042005   12
N Today at 3:07 AM by sqing
if a, b, c are real numbers and a, b, c >0, prove that
pls help meee
12 replies
myname17042005
Jun 12, 2020
sqing
Today at 3:07 AM
Trig Multiplication
szhang7   1
N Today at 2:24 AM by joeym2011
Find the exact value of $(2-\sin^2(\frac{\pi}{7}))(2-\sin^2(\frac{3\pi}{7}))(2-\sin^2(\frac{3\pi}{7}))$.
1 reply
szhang7
Yesterday at 3:50 PM
joeym2011
Today at 2:24 AM
Algebra-2
JetFire008   1
N Today at 2:11 AM by joeym2011
Prove that if $f(x)$ is a polynomial such that $f(x^n)$ is divisible by $x-1$, then $f(x^n)$ is divisible by $x^n-1$.
1 reply
JetFire008
Yesterday at 3:23 PM
joeym2011
Today at 2:11 AM
a/b + b/a never integer ?
MTA_2024   3
N Yesterday at 10:35 PM by ohiorizzler1434
Let $a$ and $b$ be 2 distinct positive integers.
Can $\frac a b +\frac b a $ be in an integer. Prove why ?
3 replies
MTA_2024
Yesterday at 3:08 PM
ohiorizzler1434
Yesterday at 10:35 PM
2014 Community AIME / Marathon ... Algebra Medium #1 quartic
parmenides51   5
N Yesterday at 7:17 PM by CubeAlgo15
Let there be a quartic function $f(x)$ with maximums $(4,5)$ and $(5,5)$. If $f(0) = -195$, and $f(10)$ can be expressed as $-n$ where $n$ is a positive integer, find $n$.

proposed by joshualee2000
5 replies
parmenides51
Jan 21, 2024
CubeAlgo15
Yesterday at 7:17 PM
interesting problem
sausagebun   1
N Yesterday at 4:05 PM by mathprodigy2011
Six points, labeled A, B, C, D, E, and F, are positioned consecutively on a straight line. Let G be a point not located on this line. The following distances are given: AC = 26, BD = 22, CE = 31, DF = 33, AF = 73, CG = 40, and DG = 30. Determine the area of triangle BGE.
I brute forced this with trig, was wondering if theres a more elegant way of doing this
1 reply
sausagebun
Yesterday at 3:21 PM
mathprodigy2011
Yesterday at 4:05 PM
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
Y by
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
Y by
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
Y by
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
Y by
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
Y by
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
Y by
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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