It's February and we'd love to help you find the right course plan!

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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!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Feb 2, 2025
0 replies
I need this. Pls help me
Giahuytls2326   12
N 6 minutes ago by Giahuytls2326
Source: Own
Given $\triangle ABC$ , orthocenter $H$, $E$ is the center of Euler's circle of $\triangle ABC$. $X,Y,Z$ is the midpoint of $AH,BH,CH$. From $X,Y,Z$ draw the tangent of $(E)$, cut a line through $E$ perpendicular to sides $BC, CA,AB$ at $M,N,P$ .Prove that $AM,BN,CP$ concurrent
12 replies
Giahuytls2326
Thursday at 4:59 PM
Giahuytls2326
6 minutes ago
Thanks u!
Ruji2018252   4
N 7 minutes ago by sqing
Let $a,b,c>0$ and $a+b+c=abc$
Find minimum (and prove)
\[P=(ab-1)(bc+1)^6(ca-1)\]
4 replies
Ruji2018252
Yesterday at 2:38 PM
sqing
7 minutes ago
Functional Equation
AnhQuang_67   4
N 7 minutes ago by gordian.knot
Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$x^2f(x)+f(1-x)=2x-x^4, \forall x \in \mathbb{R} $$
4 replies
AnhQuang_67
Yesterday at 4:02 PM
gordian.knot
7 minutes ago
thanks u!
Ruji2018252   1
N 27 minutes ago by sqing
Let $a,b,c\in \mathbb{R},a,b,c\ne 0$ and $a+b+c=0.$ Find minimum (and prove)
\[C=\dfrac{b^2+c^2-a^2}{b^2+c^2}+\dfrac{c^2+a^2-b^2}{c^2+a^2}+\dfrac{a^2+b^2-c^2}{a^2+b^2}\]
1 reply
1 viewing
Ruji2018252
3 hours ago
sqing
27 minutes ago
No more topics!
Incenter and intersection of lines making rhombus
falantrng   7
N Dec 20, 2024 by Ianis
Source: Azerbaijan NMO 2023. Junior P2
Let $I$ be the incenter in the acute triangle $ABC.$ Rays $BI$ and $CI$ intersect the circumcircle of triangle $ABC$ at points $S$ and $T,$ respectively. The segment $ST$ intersects the sides $AB$ and $AC$ at points $K$ and $L,$ respectively. Prove that $AKIL$ is a rhombus.
7 replies
falantrng
Aug 24, 2023
Ianis
Dec 20, 2024
Incenter and intersection of lines making rhombus
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G H BBookmark kLocked kLocked NReply
Source: Azerbaijan NMO 2023. Junior P2
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falantrng
249 posts
#1
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Let $I$ be the incenter in the acute triangle $ABC.$ Rays $BI$ and $CI$ intersect the circumcircle of triangle $ABC$ at points $S$ and $T,$ respectively. The segment $ST$ intersects the sides $AB$ and $AC$ at points $K$ and $L,$ respectively. Prove that $AKIL$ is a rhombus.
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InterLoop
226 posts
#2
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Draw $U$, the midpoint of arc $BC$. Note that $I$ is the orthocenter of $STU$, which implies $AU \perp ST = KL$. However, this implies, since $AU$ is both angle bisector and altitude of $AKL$, that $AK = AL$.

At this point, we shall prove that $I$ is the reflection of $A$ with respect to $ST$, which is true since $\angle AST = \angle TSB$, $\angle CTS = \angle STA$, and further by Fact 5 / Incenter-Excenter Lemma, that $AT = TI$, $AS = SI$, which together implies congruence between $\triangle ATS$, $\triangle ITS$, and reflection.

Thus $IK = AK = AL = IL$ and we are done.
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ismayilzadei1387
219 posts
#3
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My sol from exam:
1.$BSKI$ cyclic $\implies{AL // IK}$
2.$CTLI$ cyclic $\implies {AK // LI}$
3.$KL$ perpendicular to $AI$
rest is easy.
This post has been edited 1 time. Last edited by ismayilzadei1387, Aug 29, 2023, 3:02 PM
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parmenides51
30618 posts
#4
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Any solution is better than no solution

The key point is that TKIB is cyclic

Notation: (XY) stands for small arc XY

Let <BAI=<IAC=x, <ACI=<ICB=y, <CBI=<IBA= z
By angles triangle sum 2x+ 2y+2z=180^o => x+y+z =90^o

<STC=(SC)/2= <SBC=<ABI=x => TKIB is cyclic
So < KIT= (KB)/2= <KBT=(AT)/2= <ACT=z
<TIB = <ICB + <IBC = y+ z

From angles triangle sum in triangle KIB we get that < BKI= 2x = <KAL = > KI // AL
Similarly IL // AK and combining KI // AL we have that ALIK is parallelogram

<TBI = <TBK + <KAB = y + z = <TIB => TB = TI
And because TB =TA (because arcs AT, TB are equal due to C- angle bisector)
We get that AT = TI so T lies on perpendicular bisector of AI
Similarly we get that S lies on the same perpendicular bisector
So combining those , we conclude that TS is perpendicular bisector of AI
So TS _I_ AI , which means KL _I_AI
So parallelogram ALIK becomes a rhombus
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Sadigly
103 posts
#5
Y by
Let $D$ be the intersection point of $AI$ and $KL$
$B;K;A$ colinear
$C;L;A$ colinear
$B;I;S$ colinear
$C;I;T$ colinear
$T;K;L;S;D$ colinear
$\measuredangle CBS=\measuredangle SBA=\alpha$
$\measuredangle ACT=\measuredangle TCB=\beta$
$\measuredangle KAD=\measuredangle DAL=90-(\alpha+\beta)$
Claim:$BTKI$ is cyclic
Proof:
$BTSC$ cyclic:$\measuredangle CBS=\measuredangle CTS=\measuredangle ITK$
$\measuredangle SBA=\measuredangle IBK$
$\measuredangle SBA=\measuredangle CBS\Rightarrow \measuredangle ITK=\measuredangle IBK$

Do this for $CSLI$ too.

$\measuredangle SBA=\measuredangle IBK=\measuredangle ITK=\alpha$
$\measuredangle ACT=\measuredangle LCI=\measuredangle LSI=\beta$

$\measuredangle CBI+\measuredangle  ICB+\measuredangle BIC=180\Rightarrow\measuredangle BIC=180-(\alpha+\beta)$
$\measuredangle BIC=\measuredangle BIT=\measuredangle BKT=\measuredangle AKD=180-(\alpha+\beta)\Rightarrow \measuredangle DKA=\alpha+\beta$

Do this to get $\measuredangle DLA=180-(\alpha+\beta)\Rightarrow \measuredangle ALD=\alpha+\beta$ too

$\measuredangle AKD+\measuredangle KDA+\measuredangle DAK=180\Rightarrow\measuredangle ADK=\measuredangle ADL=\measuredangle IDK=\measuredangle IDL=90$


$ATBC$ cyclic:$\measuredangle ACT=\measuredangle ABT=\measuredangle KBT=\measuredangle KIT=\beta$
$ABCS$ cyclic:$\measuredangle SBA=\measuredangle SCA=\measuredangle SCL=\measuredangle SIL=\alpha$

$\measuredangle ITK+\measuredangle TKI+\measuredangle KIT=180\Rightarrow\measuredangle TKI=\measuredangle DKI=180-(\alpha+\beta)\Rightarrow \measuredangle IKD=\alpha+\beta$
$\measuredangle LSI+\measuredangle SIL+\measuredangle ILS=180\Rightarrow\measuredangle ILS=\measuredangle ILD=180-(\alpha+\beta)\Rightarrow\measuredangle DLI=\alpha+\beta$

$\measuredangle IKD+\measuredangle KDI+\measuredangle DIK=180\Rightarrow\measuredangle DIK=90-(\alpha+\beta)$
$\measuredangle DLI+\measuredangle LID+\measuredangle IDL=180\Rightarrow\measuredangle LID=90-(\alpha+\beta)$



Look at $AKIL$. We got $$\measuredangle IKL=\measuredangle IKD=\measuredangle DKA=\measuredangle LKA=\alpha+\beta$$$$\measuredangle ALK=\measuredangle ALD=\measuredangle DLI=\measuredangle KLI=\alpha+\beta$$$$\measuredangle KAI=\measuredangle KAD=\measuredangle DAL=\measuredangle IAL=90-(\alpha+\beta)$$$$\measuredangle LIA=\measuredangle LID=\measuredangle DIK=\measuredangle AIK=90-(\alpha+\beta)$$
$AKIL$ is a quadrilateral which its diagonals are angle bisectors,so $AKIL$ is a rhombus.
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Atilla
59 posts
#6
Y by
Sadigly wrote:
Let $D$ be the intersection point of $AI$ and $KL$
$B;K;A$ colinear
$C;L;A$ colinear
$B;I;S$ colinear
$C;I;T$ colinear
$T;K;L;S;D$ colinear
$\measuredangle CBS=\measuredangle SBA=\alpha$
$\measuredangle ACT=\measuredangle TCB=\beta$
$\measuredangle KAD=\measuredangle DAL=90-(\alpha+\beta)$
Claim:$BTKI$ is cyclic
Proof:
$BTSC$ cyclic:$\measuredangle CBS=\measuredangle CTS=\measuredangle ITK$
$\measuredangle SBA=\measuredangle IBK$
$\measuredangle SBA=\measuredangle CBS\Rightarrow \measuredangle ITK=\measuredangle IBK$

Do this for $CSLI$ too.

$\measuredangle SBA=\measuredangle IBK=\measuredangle ITK=\alpha$
$\measuredangle ACT=\measuredangle LCI=\measuredangle LSI=\beta$

$\measuredangle CBI+\measuredangle  ICB+\measuredangle BIC=180\Rightarrow\measuredangle BIC=180-(\alpha+\beta)$
$\measuredangle BIC=\measuredangle BIT=\measuredangle BKT=\measuredangle AKD=180-(\alpha+\beta)\Rightarrow \measuredangle DKA=\alpha+\beta$

Do this to get $\measuredangle DLA=180-(\alpha+\beta)\Rightarrow \measuredangle ALD=\alpha+\beta$ too

$\measuredangle AKD+\measuredangle KDA+\measuredangle DAK=180\Rightarrow\measuredangle ADK=\measuredangle ADL=\measuredangle IDK=\measuredangle IDL=90$


$ATBC$ cyclic:$\measuredangle ACT=\measuredangle ABT=\measuredangle KBT=\measuredangle KIT=\beta$
$ABCS$ cyclic:$\measuredangle SBA=\measuredangle SCA=\measuredangle SCL=\measuredangle SIL=\alpha$

$\measuredangle ITK+\measuredangle TKI+\measuredangle KIT=180\Rightarrow\measuredangle TKI=\measuredangle DKI=180-(\alpha+\beta)\Rightarrow \measuredangle IKD=\alpha+\beta$
$\measuredangle LSI+\measuredangle SIL+\measuredangle ILS=180\Rightarrow\measuredangle ILS=\measuredangle ILD=180-(\alpha+\beta)\Rightarrow\measuredangle DLI=\alpha+\beta$

$\measuredangle IKD+\measuredangle KDI+\measuredangle DIK=180\Rightarrow\measuredangle DIK=90-(\alpha+\beta)$
$\measuredangle DLI+\measuredangle LID+\measuredangle IDL=180\Rightarrow\measuredangle LID=90-(\alpha+\beta)$



Look at $AKIL$. We got $$\measuredangle IKL=\measuredangle IKD=\measuredangle DKA=\measuredangle LKA=\alpha+\beta$$$$\measuredangle ALK=\measuredangle ALD=\measuredangle DLI=\measuredangle KLI=\alpha+\beta$$$$\measuredangle KAI=\measuredangle KAD=\measuredangle DAL=\measuredangle IAL=90-(\alpha+\beta)$$$$\measuredangle LIA=\measuredangle LID=\measuredangle DIK=\measuredangle AIK=90-(\alpha+\beta)$$
$AKIL$ is a quadrilateral which its diagonals are angle bisectors,so $AKIL$ is a rhombus.

Yeah i did similar also here all angles that you say are equals then we see that ATI is equaliteral and it is easy to finish the prove
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wizixez
167 posts
#7
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Easy:::
$(ABC)\in \mathbb{S}^1$,$a=x^2,b=y^2,c=z^2$ so
$s=-xz$ and $t=-xy$.
$k=ST\cap AB$ And $l=ST \cap AC$ first prove that $a+j=k+l$ then $AI\bot KL$ then it is obvious that the figure that satisfies these conditions is indeed a rhombus.$\boxed{\lambda}$
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Ianis
388 posts
#8
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From $SA=SI$ and $TA=TI$ we have that $ST$ is the perpendicular bisector of $AI$, so $AK=KI$, $AL=LI$ and $KL\perp AI$. Since $AI$ is also the angle bisector of $\angle KAL$, we conclude that $AI$ is the perpendicular bisector of $KL$, so $AK=AL$. Hence $KI=AK=AL=LI$, so $AKIL$ is a rhombus.
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