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Geometry
Lukariman   10
N 4 minutes ago by Captainscrubz
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.
10 replies
Lukariman
Tuesday at 12:43 PM
Captainscrubz
4 minutes ago
9 ARML Location
deduck   36
N 4 hours ago by idk12345678
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina


Put your USERNAME in the list ONLY IF YOU WANT TO!!!! !!!!!

I'm going to UNR if anyone wants to meetup!!! :D

Current List:
Iowa
UNR
PSU
St Anselm
WCU
36 replies
deduck
Tuesday at 4:19 PM
idk12345678
4 hours ago
how prestigious is hsmc
ConfidentKoala4   3
N 5 hours ago by ConfidentKoala4
been wonderin this for a while

how prestigious is it? ik its not as good as mathily (they rejected me :mad: ) but Idk how good it actually is
3 replies
ConfidentKoala4
Today at 12:46 AM
ConfidentKoala4
5 hours ago
9 Does Mental Health Actually Matter?
heheman   9
N Today at 12:47 AM by maxamc
Looking at the goals I once had, it was all just so silly and stupid

I didn't even reach my "Low" goal for AIME... so pathetic

Missed JMO by a huge margin, after missing by only 12.5 points last year

(BTW i didn't slack off one bit)

I guess the most important thing is just to keep my head up and keep going. I can't let failures stop me. Honestly I don't care about setting goals anymore. They only give me a lot of internal pressure to do well. I think the most important thing is to focus on what I do everyday, consistently, and pay attention to the beautiful things in life (like math).

I'm going to try getting more involved in real life. After coming back from COVID, I had trouble to make as many friends with non-math people. But I was reconnecting with some of my friends that I had prepandemic and I realized how precious those friendships really were.

Now the last thing to do is grind my last bit of nonexistent ego to dust and focus on the present, stop looking back

(Note: This doesn't mean I'm going to quit, I just mean I'm going to do math on my own and try to not feel any pressure to do well. Cause i feel like that pressure really beat me a lot.)

I love this community and am happy for everyone who qualified olympiad but at this point competition math just reminds me only of my failures. (Even if it's my own fault.) So I'm probably going to take a break for a while. Thanks everyone for being nice to me and stuff. Sorry if this sounds cringe (it will in a week)

9 replies
heheman
Mar 8, 2024
maxamc
Today at 12:47 AM
HCSSiM results
SurvivingInEnglish   60
N Today at 12:32 AM by Vivaandax
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
60 replies
SurvivingInEnglish
Apr 5, 2024
Vivaandax
Today at 12:32 AM
Rational sequences
tenniskidperson3   57
N Yesterday at 10:25 PM by OronSH
Source: 2009 USAMO problem 6
Let $s_1, s_2, s_3, \dots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \dots.$ Suppose that $t_1, t_2, t_3, \dots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$. Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.
57 replies
tenniskidperson3
Apr 30, 2009
OronSH
Yesterday at 10:25 PM
1:1 Physics Tutors
DinoDragon186   3
N Yesterday at 7:26 PM by talhee
I am looking for 1:1 physics tutor.
I am a beginner in physics and am in 9th grade.
I want to make it to IPhO in the coming years.
3 replies
DinoDragon186
Dec 10, 2024
talhee
Yesterday at 7:26 PM
Looking for Physics or USAPhO Tutor
physicsplease   4
N Yesterday at 7:20 PM by talhee
Hii I am looking for a USAPhO tutor for next year's season. I think I have tried literally everything possible to improve but I feel like I just hit a massive roadblock right now.

It would be ideal if I can find someone who have a lot of experience with physics olympiads. My goal is medal/gold in usapho next year, and I am very determined & willing to put in a lot of hours, especially more so in the summer. Please recommend anyone or dm in aops, thank you.

Have qualified usapho before (last year), took both physics c and sufficient higher math.
4 replies
physicsplease
Apr 11, 2025
talhee
Yesterday at 7:20 PM
MathILy 2025 Decisions Thread
mysterynotfound   40
N Yesterday at 4:11 PM by bjump
Discuss your decisions here!
also share any relevant details about your decisions if you want
40 replies
mysterynotfound
Apr 21, 2025
bjump
Yesterday at 4:11 PM
Mathcounts state
happymoose666   39
N Yesterday at 1:54 PM by Inaaya
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
39 replies
happymoose666
Mar 24, 2025
Inaaya
Yesterday at 1:54 PM
force overlay inversion vibes
v4913   63
N Yesterday at 1:47 PM by starchan
Source: USAMO 2023/6
Let $ABC$ be a triangle with incenter $I$ and excenters $I_a$, $I_b$, and $I_c$ opposite $A$, $B$, and $C$, respectively. Let $D$ be an arbitrary point on the circumcircle of $\triangle{ABC}$ that does not lie on any of the lines $II_a$, $I_bI_c$, or $BC$. Suppose the circumcircles of $\triangle{DII_a}$ and $\triangle{DI_bI_c}$ intersect at two distinct points $D$ and $F$. If $E$ is the intersection of lines $DF$ and $BC$, prove that $\angle{BAD} = \angle{EAC}$.

Proposed by Zach Chroman
63 replies
v4913
Mar 23, 2023
starchan
Yesterday at 1:47 PM
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
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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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