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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

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0 replies
jlacosta
Feb 2, 2025
0 replies
P6 Geo Finale
math_comb01   7
N 2 minutes ago by GuvercinciHoca
Source: XOOK 2025/6
Let $ABC$ be a triangle with incenter $I$ and excenters $I_A$, $I_B$, $I_C$ opposite to $A,B,C$ respectively. Suppose $BC$ meets the circumcircle of $I_AI_BI_C$ at points $D$ and $E$. $X$ and $Y$ lie on the incircle of $\triangle ABC$ so that $DX$ and $EY$ are tangents to the incircle (different from $BC$). Prove that the circumcircles of $\triangle AXY$ and $\triangle ABC$ are tangent.

Proposed by Anmol Tiwari
7 replies
math_comb01
Feb 10, 2025
GuvercinciHoca
2 minutes ago
A functional equation
super1978   1
N 13 minutes ago by pco
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(y-x)-xf(y))+f(x)=y(1-f(x)) $$for all $x,y \in \mathbb R$
1 reply
super1978
an hour ago
pco
13 minutes ago
Sequences Handout
M11100111001Y1R   4
N 17 minutes ago by MR.1
Source: Own
Hi everyone, I wrote this handout about sequences in NT.
Hope you enjoy!
4 replies
+2 w
M11100111001Y1R
Oct 19, 2022
MR.1
17 minutes ago
[Handout] 50 non-traditional functional equations
gghx   2
N 44 minutes ago by GreekIdiot
Sup guys,

I'm retired. I love FEs. So here's 50 of them. Yea...

Functional equations have been one of the least enjoyed topics of math olympiads in recent times, mostly because so many techniques have been developed to just bulldoze through them. These chosen problems do not fall in that category - they require some combi-flavoured creativity to solve (to varying degrees).

For this reason, this handout is aimed at more advanced problem solvers who are bored of traditional FEs and are up for a little challenge!

In some sense, this is dedicated to the "covid FE community" on AoPS who got me addicted to FEs, people like EmilXM, hyay, IndoMathXdZ, Functional_equation, GorgonMathDota, BlazingMuddy, dangerousliri, Mr.C, TLP.39, among many others: thanks guys :). Lastly, thank you to rama1728 for suggestions and proofreading.

Anyways...
2 replies
gghx
Sep 23, 2023
GreekIdiot
44 minutes ago
No more topics!
IMO 2010 Problem 4
mavropnevma   126
N Feb 17, 2025 by cj13609517288
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Proposed by Marcin E. Kuczma, Poland
126 replies
mavropnevma
Jul 8, 2010
cj13609517288
Feb 17, 2025
IMO 2010 Problem 4
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mavropnevma
15142 posts
#1 • 24 Y
Y by Batman007, Davi-8191, tenplusten, FlakeLCR, test20, AlastorMoody, Purple_Planet, vsamc, ilovepizza2020, megarnie, pog, EthanTAoPS, jhu08, THEfmigm, Jupiter_is_BIG, sleepypuppy, EpicBird08, Adventure10, Mango247, Aopamy, ItsBesi, cubres, Rounak_iitr, and 1 other user
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Proposed by Marcin E. Kuczma, Poland
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m.candales
186 posts
#2 • 6 Y
Y by jhu08, THEfmigm, mathmax12, Adventure10, sximoz, cubres
$SA \cdot SB=SC^2=SP^2$. Then $SP$ is tangent to the circuncircle of $\triangle{ABP}$. Then $\angle{SPA}=\angle{SBP}$.
Let $R, T$ be the intersections of $SP$ with $\Gamma$ ($R$ is in the $\widehat{LA}$, and $T$ is in the $\widehat{KB}$)
$\frac{1}{2}(\widehat{RA}+\widehat{KT})=\angle{RPA}=\angle{LBA}=\frac{1}{2}\widehat{LA}=\frac{1}{2}(\widehat{LR}+\widehat{RA})$
Then $\widehat{KT}=\widehat{LR}$
Then $\angle{LPR}=\frac{1}{2}(\widehat{LR}+\widehat{TB})=\frac{1}{2}(\widehat{KT}+\widehat{TB})=\frac{1}{2}KB=\angle{PAB}$

$\angle{SPC}=\angle{LPR}+\angle{LPC}=\angle{PAB}+\angle{BPM}=\angle{PAB}+\angle{MCB}+\angle{PBC}$

$\angle{SCP}=\angle{SCA}+\angle{ACM}=\angle{SBC}+\angle{ACM}=\angle{SBL}+\angle{PBC}+\angle{ACM}$

But $\angle{SPC}=\angle{SCP}$. Then $\angle{SBL}+\angle{ACM}=\angle{MCB}+\angle{PAB}$

Then $\widehat{LM}=\widehat{KM}$
Then $LM=KM$
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swaqar
208 posts
#3 • 3 Y
Y by jhu08, Adventure10, cubres
Solution (if it's correct please tell me)
Call the circles $(S,SC) $ $\Omega$. Now the circles $\Omega$ and $ \Gamma$ are orthogonal so the point of intersection of $AB$ and $KL$ lies on the circle $ \Omega$ and call this point $S'$. Assume that $A $ belong to $[S,B]$. Now the line $SP$ is tangent to the circumcircle of $ \triangle APB$ since $SA\cdot SB=SC^{2}=SP^{2} $. Now $SC$ is the external symmedian of the vertex $C$. Calling the intersection of the angle bisector of $C$ with the line $ AB$, $P'$, we have the points $ (S',P',A,B) $ forming a harmonic range and this implies that $ SC = SP'$. So the circle $\Omega$ is is the appollonian circle of the vertex $ C $ of the $\triangle ABC$ . Define the point $N$ on the line segment $KL$ so that the line $CN$ is the bisector of the $ \angle LCM$. Let the external symmedian meet the the $S'C$ in $J$. We have $JN=JC$ and $ SC = SP$ so the traingle $ SCP $ and $ JCN $ are homothetic so the points $ A , P $ and $ N $ are collinear and we have $M$ being the midpoint of the arc $ \overarc{KMN}$.
This post has been edited 11 times. Last edited by swaqar, Jul 8, 2010, 2:12 PM
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leejho20
1 post
#4 • 3 Y
Y by jhu08, Adventure10, cubres
here's my proof

Since, angle BPC = angle BAK + angle CAK + angle ABL + angle ACL and angle SPC = angle SCP

To see angle MLK = angle ACM + angle ABL = angle BCM + angle BAK = angle MKL

it is sufficient to prove angle BPS = angle BPK.

But we have SA * SB = SC^2 = SP^2, from this it is clear that triangle BPS is similar to triangle PAS

this proves angle BPS = angle APS.
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Justanotherone
10 posts
#5 • 3 Y
Y by jhu08, Adventure10, cubres
we know,as SC is tangent to the circle SC^2 =SA*SB
so by the problem SP^2=SA*SB
so SP/SA=SB/SP and as triangle SPA and triangle SPB have one angle in common,they are similar.so angleSPA and anglePBA(both equal to a) are equal.
extending SP to meet the circle at x,we also have angleXPM=anglePAB(both equal to b)
also,angleSCA=angleCBA(both equal to c)
again ,being on the same arc, angleACM=angleABM(both equal to d)
similarly,angleCAK=angleCMK(both equal to e)
now in triangle APM, angleAPM=angleCPK=angleCAK+angleACM=e+d
also, angleSPC=angleSCP=c+d,and angleSPA=a
so adding,180=a+c+2d+e -(1)
now, angleKLM=angleKLB+angleBLM=b+angleBLM
also, angleBLM=angleBAM=180-b-c-d-e,after calculating from triangle CAM
From (1)this turns out to be equal to a+d-b
so angleKLM=a+d
also angleLKM=angleLKA+angleAKM=a+d
so in triangle KLM we have angleKLM=angleLKM
so ML=MK................
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Pirshtuk
17 posts
#6 • 8 Y
Y by DHu, Sush0, jhu08, Adventure10, Mango247, cubres, and 2 other users
My proof:

Because triangles $APC$ and $MPK$ are similar, we have that $\frac{MK}{MP}=\frac{AC}{AP}.$ Analogously we have that $\frac{ML}{MP}=\frac{BC}{BP}.$ So it's enough to prove that $\frac{AP}{BP}=\frac{AC}{BC}$.

We have that $SP^2=SC^2=SA \cdot SB,$ so $\frac{SP}{SA}=\frac{SB}{SP}$. Using this result and because angle $CSA$ is common for triangles $SPA$ and $SBP$, we conclude that these triangles are similar. So $\frac{AP}{BP}=\frac{SA}{SP}=\frac{SA}{SC}.$ But triangles $SCA$ and $SBC$ are similar too. So $\frac{SA}{SC}=\frac{AC}{BC}.$ Therefore $\frac{AP}{BP} = \frac{AC}{BC},$ and we have done!
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silouan
3952 posts
#7 • 10 Y
Y by UK2019Project, AllanTian, jhu08, ericxyzhu, Adventure10, Mango247, cubres, and 3 other users
Let me add my solution too. I think it is a little different.

Well by the powe point theorem we have that $SP^2=SA\cdot SB$ which means that the triangles
SPB, PAS are similar. Using this fact a bit angle chasing shows that $\angle{SPL}=\angle{BAP}=\angle{BLK}$ so $SP//LK$ (1). But $OC\bot CS$ and $\angle{SPC}=\angle{SCP}$, so $OM\bot SP$ (2)
From (1) and (2) we have that $OM\bot LK$ and we are done.
Attachments:
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KoolNerd
3 posts
#8 • 4 Y
Y by jhu08, Adventure10, Mango247, cubres
As SA.SB = SC^2=SP^2, SP is tangent to the circumcircle of triangle ABP.

=> angle BPS = angle BAK = angle BLK.
Let KL intersect BC at R and SP intersect BC at Q.

In triangles BRL and BQP, angle RBL is common. angle BPQ = angle BPR. So, angle PQB = angle LRB.
=> SP is parallel to KL.

As SP = SC, angle SPC = angle SCP = angle MLC = angle NPM ( N is the intersection point of SP and the circle, N is on smaller are LA)
So,
In triangles MLC and MPD where D is the intersection point of ML and SR,
angle LMC is common, angle MPD = angle MLC. So, angle MDP = angle LCM and we know angle LCM = angle MKL.

As, SR is parallel to KL, angle MDP = angle MLK = angle MKL

We Are Done!
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YamoSky
209 posts
#9 • 3 Y
Y by jhu08, Adventure10, cubres
KoolNerd wrote:
As SA.SB = SC^2=SP^2, SP is tangent to the circumcircle of triangle ABP.

=> angle BPS = angle BAK = angle BLK.
Let KL intersect BC at R and SP intersect BC at Q.

In triangles BRL and BQP, angle RBL is common. angle BPQ = angle BPR. So, angle PQB = angle LRB.
=> SP is parallel to KL.

As SP = SC, angle SPC = angle SCP = angle MLC = angle NPM ( N is the intersection point of SP and the circle, N is on smaller are LA)
So,
In triangles MLC and MPD where D is the intersection point of ML and SR,
angle LMC is common, angle MPD = angle MLC. So, angle MDP = angle LCM and we know angle LCM = angle MKL.

As, SR is parallel to KL, angle MDP = angle MLK = angle MKL

We Are Done!

My solution is quite similar to yours...It's really an easy problem...
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April
1270 posts
#10 • 4 Y
Y by jhu08, Adventure10, Mango247, cubres
mavropnevma wrote:
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.
We have $MK=ML\iff \angle PCA + \angle PBA = \angle PCB + \angle PAB\iff \angle PAB - \angle PBA = \angle PCA - \angle PCB$

From $SC=SP$ we conclude that $SP^2= SC^2=SA\cdot SB$ $\Longrightarrow SP$ is tangent to the circumcircle of triangle $ABP$ at $P$. Therefore $\angle PAB - \angle PBA = \angle PAB - \angle APS = \angle PSB\quad (1)$.

On the other hand, $\angle PCA - \angle PCB=\angle SCP - \angle SCA-\angle PCB=\angle SPC-\angle SBC - \angle PCB=\angle PSB\quad (2)$.

From $(1)$ and $(2)$ we have the conclusion.
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Mithril
28 posts
#11 • 4 Y
Y by jhu08, Adventure10, Mango247, cubres
Let the second tangent from $S$ meet $\Gamma$ at $C'$. Notice that $(A,B;C,C')$ is harmonic, because projecting from $S$ fixes $C$ and $C'$, and swaps $A$ and $B$.

Now, let $C'P$ meet $\Gamma$ again at $M'$. Projecting from $P$ gives $(A,B;C,C') = (K,L;M,M') = -1$, thus $LM/MK = LM'/M'K$. Now, if we prove that $M$ and $M'$ are antipodal, we're done, because the only way for those fractions to be equal is that $M$ is the midpoint of arc $LK$ (if we move $M$, one fraction grows while the other is smaller).

That is the same as saying $\angle C'CP + \angle CC'M' = 90^{o} - \angle C'BC$. But we know that $SC' = SP = SC$, thus $S$ is the circumcenter of $C'PC$, and $\angle C'CP + \angle CC'M = \angle C'SC / 2$.

Now, if $O$ is the circumcenter of $ABC$, we have $\angle C'SC / 2 = (180^{o} - \angle C'OC)/2 = 90^{o} - \angle C'BC$, as we wanted.
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m.candales
186 posts
#12 • 5 Y
Y by jhu08, Adventure10, Mango247, cubres, and 1 other user
An even simpler solution:

$SA \cdot SB=SC^2=SP^2$. Then $SP$ is tangent to the circuncircle of $\triangle{ABP}$. Then $\angle{SPA}=\angle{SBP}$.
Let $R, T$ be the intersections of $SP$ with $\Gamma$ ($R$ is in $\widehat{LA}$, and $T$ is in $\widehat{KB}$)

$\frac{1}{2}(\widehat{LR}+\widehat{RA})=\frac{1}{2}\widehat{LA}=\angle{LBA}=\angle{RPA}=\frac{1}{2}(\widehat{RA}+\widehat{KT})$. Then $\widehat{LR}=\widehat{KT}$

$\frac{1}{2}(\widehat{RC}+\widehat{RM})=\frac{1}{2}\widehat{CM}=\angle{SCP}=\angle{SPC}=\frac{1}{2}(\widehat{RC}+\widehat{TM})$. Then $\widehat{RM}=\widehat{TM}$

Then $\widehat{LM}=\widehat{KM}$. Then $LM=KM$
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SnowEverywhere
801 posts
#13 • 6 Y
Y by ayan_mathematics_king, jhu08, Adventure10, Mango247, cubres, and 1 other user
Yay. I liked this problem. I think that my solution is also a little bit different. Looks like there were quite a number of different solutions to this problem...

Solution

Let the tangent at $M$ to $\Gamma$ intersect $SC$ at $X$. We now have that since $\triangle{XMC}$ and $\triangle{SPC}$ are both isosceles, $\angle{SPC}=\angle{SCP}=\angle{XMC}$. This yields that $MX \| PS$.

Now consider the power of point $S$ with respect to $\Gamma$.

\[SC^2 = SP^2 =SA \cdot SB \quad \Rightarrow \quad \frac{SP}{SA}=\frac{SB}{SP}\]
Hence $\triangle{SPA} \sim \triangle{SBP}$. Combining this with the arc angle theorem yields that $\angle{SPA}=\angle{SBP}=\angle{PKL}$. Hence $PS \| LK$.

This implies that the tangent at $M$ is parallel to $LK$ and therefore that $M$ is the midpoint of arc $LK$. Hence $MK=ML$.
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JG
19 posts
#14 • 4 Y
Y by jhu08, Adventure10, Mango247, cubres
Let $SP$ cut the circumcircle at $X$ and $Y$, then $SC^2=SP^2=SX\cdot SY$ so if $Q$ is the reflection of $P$ over $S$ then $X$, $Y$ are harmonic conjugates of $P$, $Q$. Notice also that $\angle QCP=90^{\circ}$ then $\angle XCP=\angle PCY$, this means $MX=MY$, but since $SP$ is tangent to the circumcircle of $PAB$ ($SP^2=SC^2=SA\cdot SB$) we have $\angle SPA=\angle PBA=\angle PKL$ so $XY$ is parallel to $KL$ and the result follows.
This post has been edited 1 time. Last edited by JG, Jan 16, 2013, 5:00 PM
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Jorge Miranda
157 posts
#15 • 4 Y
Y by jhu08, Adventure10, Mango247, cubres
A straightforward angle chase
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