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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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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
Something nice
KhuongTrang   33
N 2 minutes ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
33 replies
+1 w
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
2 minutes ago
Geometry hard
Lukariman   2
N 2 minutes ago by Primeniyazidayi
Given triangle ABC inscribed in circle (O). The bisector of angle A intersects (O) at D. Let M, N be the midpoints of AB, AC respectively. OD intersects BC at P and AD intersects MN at S. The circle circumscribed around triangle MPS intersects BC at Q different from P. Prove that QA is tangent to (O).
2 replies
Lukariman
43 minutes ago
Primeniyazidayi
2 minutes ago
help!!!!!!!!!!!!
Cobedangiu   3
N 3 minutes ago by sqing
help
3 replies
Cobedangiu
Mar 23, 2025
sqing
3 minutes ago
Inequality
nguyentlauv   1
N 4 minutes ago by NguyenVanHoa29
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
1 reply
nguyentlauv
Yesterday at 12:19 PM
NguyenVanHoa29
4 minutes ago
Great similarity
steven_zhang123   3
N 17 minutes ago by Lil_flip38
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
3 replies
steven_zhang123
43 minutes ago
Lil_flip38
17 minutes ago
AD=BE implies ABC right
v_Enhance   117
N 2 hours ago by cj13609517288
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
117 replies
v_Enhance
Apr 10, 2013
cj13609517288
2 hours ago
Geometry
gggzul   6
N 3 hours ago by Captainscrubz
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
6 replies
gggzul
Yesterday at 8:22 AM
Captainscrubz
3 hours ago
Geometry
Lukariman   5
N 3 hours ago by Lukariman
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
5 replies
Lukariman
Yesterday at 12:43 PM
Lukariman
3 hours ago
Aime type Geo
ehuseyinyigit   4
N 5 hours ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
5 hours ago
n variables with n-gon sides
mihaig   1
N 5 hours ago by mihaig
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
1 reply
mihaig
Apr 25, 2025
mihaig
5 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   3
N Today at 5:48 AM by EeEeRUT
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
3 replies
BR1F1SZ
Monday at 9:45 PM
EeEeRUT
Today at 5:48 AM
area of O_1O_2O_3O_4 <=1, incenters of right triangles outside a square
parmenides51   2
N Today at 4:35 AM by Solilin
Source: Thailand Mathematical Olympiad 2012 p4
Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.
2 replies
parmenides51
Aug 17, 2020
Solilin
Today at 4:35 AM
Geo metry
TUAN2k8   3
N Today at 4:34 AM by TUAN2k8
Help me plss!
Given an acute triangle $ABC$. Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively. Lines $BD$ and $CE$ intersect at point $F$. The circumcircles of triangles $BDF$ and $CEF$ intersect at a second point $P$. The circumcircles of triangles $ABC$ and $ADE$ intersect at a second point $Q$. Point $K$ lies on segment $AP$ such that $KQ \perp AQ$. Prove that triangles $\triangle BKD$ and $\triangle CKE$ are similar.
3 replies
TUAN2k8
Yesterday at 10:33 AM
TUAN2k8
Today at 4:34 AM
China South East Mathematical Olympiad 2013 problem 2
s372102   3
N Today at 2:11 AM by AGCN
$\triangle ABC$, $AB>AC$. the incircle $I$ of $\triangle ABC$ meet $BC$ at point $D$, $AD$ meet $I$ again at $E$. $EP$ is a tangent of $I$, and $EP$ meet the extension line of $BC$ at $P$. $CF\parallel PE$, $CF\cap AD=F$. the line $BF$ meet $I$ at $M,N$, point $M$ is on the line segment $BF$, the line segment $PM$ meet $I$ again at $Q$. Show that $\angle ENP=\angle ENQ$
3 replies
s372102
Aug 10, 2013
AGCN
Today at 2:11 AM
Prove excircle is tangent to circumcircle
sarjinius   8
N Apr 24, 2025 by Lyzstudent
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
8 replies
sarjinius
Mar 9, 2025
Lyzstudent
Apr 24, 2025
Prove excircle is tangent to circumcircle
G H J
Source: Philippine Mathematical Olympiad 2025 P4
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sarjinius
240 posts
#1 • 4 Y
Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
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ilovemath0402
187 posts
#2
Y by
bump bump this problem is so nice
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sarjinius
240 posts
#3
Y by
ilovemath0402 wrote:
bump bump this problem is so nice

Thanks, I proposed this problem :)
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SimplisticFormulas
113 posts
#4
Y by
what’s the solution? I am completely stuck
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MathLuis
1524 posts
#5 • 4 Y
Y by drago.7437, sarjinius, Mysteriouxxx, radian_51
Well this geo is really amazing I have to say...solved in around 30 mins but I think this could even be around 30-35 MOHS because the way to find things on this problem requires deep intuition.
Let $BI \cap (ABC)=M_B$ and $CI \cap (ABC)=M_C$, also let $N_A$ be midpoint of arc $BAC$ on $(ABC)$, now let reflections of $D$ over $EX, FY, BI, CI, Y \infty_{\perp CI}, X \infty_{\perp BI}$ be $D_B, D_C, L', K', K, L$ respectively now let reflection of $D_B$ over $AX$ be $L_1$ and reflection of $D_C$ over $AY$ be $K_1$.
Using the paralelogram we can easly see from the direction of the reflections that $KK'$ and $LL'$ are diameters on $(Y, YD), (X, XD)$ respectively, now let $I_B, I_C$ be the $B,C$ excenters of $\triangle ABC$ then notice we have $\measuredangle II_CA=\measuredangle CBI=\measuredangle CDY=\measuredangle YK'C$ which implies $I_CAK'Y$ cyclic and similarily $I_BAL'X$ is cyclic however since $\measuredangle CDY=\measuredangle YD_CF$ we also get that $I_CAK'YD_C$ is cyclic and similarily $I_BAL'XD_B$ is cyclic, however it doesn't end here...
Now notice that $YK'=YD_C$ so $Y$ is midpoint of arc $K'D_C$ on $(I_CAK')$ however $D, K'$ are symetric in $CI$ which means both $I_CD, I_CD'$ are reflections of $I_CK'$ over $CI$ and thus $I_C, D, D_C$ are colinear, and similarily $I_B, D, D_B$ are colinear.
Now $\measuredangle CDY=\measuredangle YD_CA=\measuredangle AK_1Y$ which means $CK_1YD$ is cyclic and similarily we have $L_1BXD$ cyclic, but also note that $\measuredangle L_1DL=\measuredangle L_1L'L=\measuredangle AI_BX=\measuredangle ACI=\measuredangle K_1DY$ which means that $L_1, D, K_1$ are colinear.
Now from here notice that $\measuredangle DL_1A=\measuredangle DXI=\measuredangle IYD=\measuredangle AK_1D$ which does in fact show that $\triangle L_1AK_1$ is isosceles and therefore $AK_1=AL_1$, and from reflections this gives $AD_B=AD_C$, but notice from other reflections we have $D_BG=DG=D_CG$ where $EX \cap FY=G$ (clearly then $G$ is A-excenter of $\triangle EAF$), but now also note that we have $\measuredangle AD_BG=\measuredangle GDE=\measuredangle AD_CG$ which means that $AD_BGD_C$ is cyclic but by summing arcs we end up realising $AG$ is diameter and in fact now this means $(D_BDD_C)$ is $\omega$ from the tangencies.
To finish let $J$ be the miquelpoint of $L_1BCK_1$ then $J$ lies on $(ABC)$ but also from Reim's we get $N_A, D, J$ colinear and then Reim's twice gives $M_CX \cap M_BY=J$ and from double Reim's once again we have that $(AL'X) \cap (AK'Y)=J$ and this is excellent news because now we can note that $\measuredangle D_CJD_B=\measuredangle D_CJA+\measuredangle AJD_B=\measuredangle D_CI_CA+\measuredangle AI_BD_B=\measuredangle D_CDD_B$ which shows that $J$ lies on $\omega$ as well, but since $N_A, D, J$ are colinear from the converse of Archiemedes Lemma (or just shooting Lemma/homothety) we have that $\omega, (ABC)$ are tangent at $J$ as desired thus we are done :cool:
This post has been edited 1 time. Last edited by MathLuis, Mar 13, 2025, 8:16 PM
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AndreiVila
210 posts
#6 • 1 Y
Y by Lyzstudent
Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$. Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$. Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$.

Claim 1. The $A$-excircle of $\triangle AEF$ is tangent to $EF$ in $D$.
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$. This is equivalent to $$BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$$Simplifying yields $AE+ED=AF+FD$, which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$.
Proof: By Casey's Theorem, we need to prove that $$b\cdot BD + c\cdot CD = a\cdot AK.$$But $$AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$$With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.
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SimplisticFormulas
113 posts
#7
Y by
I found that $X,Y$ are in centres, $XE$ meets $YF$ in $Z=$$A$- excentre of $AEF$ and that$A$ appears to be Miquel point of $IXYZ$.
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markam
4 posts
#10
Y by
sarjinius, what solution did you have in mind at first, when you proposed this problem?
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Lyzstudent
1 post
#11
Y by
AndreiVila wrote:
Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$. Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$. Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$.

Claim 1. The $A$-excircle of $\triangle AEF$ is tangent to $EF$ in $D$.
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$. This is equivalent to $$BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$$Simplifying yields $AE+ED=AF+FD$, which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$.
Proof: By Casey's Theorem, we need to prove that $$b\cdot BD + c\cdot CD = a\cdot AK.$$But $$AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$$With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.
Excellent!!!Much better than the solution above.
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