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AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
AMC- IMO preparation
asyaela.   9
N 21 minutes ago by Schintalpati
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
9 replies
+3 w
asyaela.
3 hours ago
Schintalpati
21 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   29
N 28 minutes ago by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
29 replies
TennesseeMathTournament
Mar 9, 2025
NashvilleSC
28 minutes ago
AIME score for college apps
Happyllamaalways   75
N 38 minutes ago by hashbrown2009
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
75 replies
Happyllamaalways
Mar 13, 2025
hashbrown2009
38 minutes ago
AMC 8 discussion
Jaxman8   42
N an hour ago by mpcnotnpc
Discuss the AMC 8 below!
42 replies
Jaxman8
Jan 29, 2025
mpcnotnpc
an hour ago
No more topics!
nice geometry
zhoujef000   26
N Today at 3:34 AM by smbellanki
Source: 2025 AIME I #14
Let $ABCDE$ be a convex pentagon with $AB=14,$ $BC=7,$ $CD=24,$ $DE=13,$ $EA=26,$ and $\angle B=\angle E=60^{\circ}.$ For each point $X$ in the plane, define $f(X)=AX+BX+CX+DX+EX.$ The least possible value of $f(X)$ can be expressed as $m+n\sqrt{p},$ where $m$ and $n$ are positive integers and $p$ is not divisible by the square of any prime. Find $m+n+p.$
26 replies
1 viewing
zhoujef000
Feb 7, 2025
smbellanki
Today at 3:34 AM
nice geometry
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Source: 2025 AIME I #14
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zhoujef000
284 posts
#1 • 1 Y
Y by OronSH
Let $ABCDE$ be a convex pentagon with $AB=14,$ $BC=7,$ $CD=24,$ $DE=13,$ $EA=26,$ and $\angle B=\angle E=60^{\circ}.$ For each point $X$ in the plane, define $f(X)=AX+BX+CX+DX+EX.$ The least possible value of $f(X)$ can be expressed as $m+n\sqrt{p},$ where $m$ and $n$ are positive integers and $p$ is not divisible by the square of any prime. Find $m+n+p.$
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zhoujef000
284 posts
#2
Y by
best problem on the test by far
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popop614
264 posts
#3 • 1 Y
Y by aidensharp
exactly what i expected ngl
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Aaron_Q
22 posts
#4
Y by
this one was so fun yet I got it wrong rip
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a_smart_alecks
51 posts
#5
Y by
ends my streak of getting cooked by final 5 geo :o

sketch
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Jack_w
102 posts
#6
Y by
By Ptolemy's Inequality on $ABCX$ and $AXDE$,
$$\frac{1}{\sqrt{3}}AX + \frac{2}{\sqrt{3}}CX \geq BX,$$and $$\frac{1}{\sqrt{3}}AX + \frac{2}{\sqrt{3}}DX \geq EX.$$Adding, $BX + EX \leq \frac{2}{\sqrt{3}}(AX + CX + DX)$, so $f(X) \leq \frac{\sqrt{3}+2}{\sqrt{3}}(AX+CX+DX)$, with equality iff the aforementioned quadrilaterals are cyclic -- but, by $120^{\circ}$ angles, this means $X$ is the Fermat point of $\triangle{ACD}$. Since that point minimizes $AX + CX + DX$, it must be the minimum.

Now compute the sum of the lengths of the distances to the Fermat point using Law of Cosines and sine area. This gives $38 + 19\sqrt{3} \implies \boxed{060}$.
unfortunate story
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john0512
4170 posts
#7
Y by
Rephrase the problem as follows.

Let $\triangle ACD$ be a triangle with $AC=7\sqrt{3}$, $AD=13\sqrt{3}$, and $CD=24$. Let $B$ be such that $\angle CBA=60$ and $\angle BCA=90$, and define $E$ similarly.

Let $X_{13}$ denote the Fermat point of $\triangle ACD$. This point satisfies $\angle CX_{13}D=\angle AX_{13}C=\angle AX_{13}D=120$, and it is well known that it minimizes $AX+CX+DX$ (proof: smoothing). The main claim is the following.

Claim: $X_{13}$ lies on $BE$, and thus also minimizes $BX+XE$. Note that $AX_{13}BC$ and $AX_{13}DE$ are cyclic due to the $60$ and $120$ angles. Thus,
$$\angle AX_{13}B=\angle ACB=90$$and similarly for $\angle AX_{13}E$ so $B,X_{13},E$ are collinear.

Let $P$ be the point on the other side of $CD$ as $A$ such that $PCD$ is equilateral. Let $$AX_{13}=a,CX_{13}=c,DX_{13}=d.$$However, by Ptolemy on $CPDX_{13}$,
$$c+d=X_{13}P \rightarrow a+c+d=AP.$$By Ptolemy on $AX_{13}CB$ and $AX_{13}DE$,
$$BX_{13}+EX_{13}=\frac{7a+14c}{7\sqrt{3}}+\frac{13a+26d}{13\sqrt{3}}=\frac{2}{\sqrt{3}}(a+c+d),$$which means the answer is
$$(a+c+d)(1+\frac{2}{\sqrt{3}}).$$We know that $a+c+d=AP$, so coordinate bash to find that $AP=19\sqrt{3}$, so the answer is ${38+19\sqrt{3}}$.
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DottedCaculator
7303 posts
#8 • 3 Y
Y by SY_Man_Utd, centslordm, OronSH
Let $X$ be the Fermat point of $ACD$. Then, $\angle AXE+\angle AXB=\angle ADE+\angle ACB=180^\circ$, so $X$ lies on $BE$. Therefore, we need to find $AX+CX+DX+BE$. Since $\angle ADC=30^\circ$, the point $A'$ such that $A'CD$ is an equilateral triangle satisfies $\angle ADA'=90^\circ$, so $AX+CX+DX=AA'=\sqrt{(13\sqrt3)^2+24^2}=19\sqrt3$. By Ptolemy, $XE\sqrt3=AX+2DX$ and $XB\sqrt3=AX+2CX$, so $BE\sqrt3=2(AX+CX+DX)=38\sqrt3$, implying $BE=38$. Therefore, the minimum is $\boxed{38+19\sqrt3}$.
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john0512
4170 posts
#9
Y by
Some intuition here - in general, finding the minimum total distance to a set of (5 or more) points in general position is very difficult. so the only way really the problem can be doable is if five lengths can be split into groups where some point minimizes each group at the same time.

This is what motivated the solution for me in test, {A,C,D} {B,E} is probably the most natural partition which then leads you to guess that the Fermat point lies on BE. Once you guess this, it is in fact easy to prove by cyclic quads and two perpendiculars.
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OronSH
1718 posts
#10 • 4 Y
Y by centslordm, GrantStar, ihatemath123, scannose
best solution to this problem by far
This post has been edited 2 times. Last edited by OronSH, Feb 7, 2025, 5:47 PM
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BS2012
889 posts
#11
Y by
buh what is this problem
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sixoneeight
1128 posts
#12
Y by
Best AIME Problem ever ngl ngl

From the given condition, we can guess that if $X'$ be the Fermat Point of $ACD$, $X'$ lies on $BE$. By definition, we have that $\angle AX'C = \angle AX'D = 120^{\circ}$. Therefore, $AX'CB$ and $AX'DE$ are cyclic so $\angle AX'B+\angle AX'E = 90^\circ+90^\circ=180^\circ$.

Thus, the minimum occurs at $X = X'$. Now, we find $BE$. We can use law of cosines to find $\cos \angle CAD = \frac17$, so $\cos BAE = -\frac{11}{14}$. Thus, by law of cosines again $BE = 38$.

Then, we can find $AX+CX+DX$ by constructing an equilateral triangle $CDP$ with $P$ and $A$ on opposite sides of $CD$. Note that $AX$ passes through $P$ and furthermore $XCPD$ is cyclic. By Ptolemy, $AX+CX+DX = AX+XP = AP$. To find this length, we can drop the feet from $A$ and $P$ to $CD$ to get $H_A$ and $H_P$. $PH_P$ has length $12\sqrt3$ and $H_P$ is the midpoint of $CD$. We already found the cosine, so it is easy to see that $\sin \angle CAD = \frac{4\sqrt3}7$, so we can find the area of $ACD$ and divide by half of $CD$ to find that the altitude from $A$ is $\frac{13\sqrt3}2$. Then, $CH_A = \frac92$ so $H_AH_P = \frac{15}2$ (I spent like 15 minutes trying to figure out why my answer was wrong; apparently 24-9 is not 13). Finally, we have $AP = 19\sqrt3$ by Pythag.

Thus, the answer is $38+19\sqrt3 \rightarrow \boxed{60}$.
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Tetra_scheme
92 posts
#13
Y by
incredible problem. I claim the desired point is the intersections of the circumcircles of $ABC$ and $ADE$. First note that these two triangles are right, and that $X$ is the Fermat Point of $ACD$. Therefore this point indeed minimizes $AX+CX+DX$. It remains to prove it also minimizes $BX+EX$. Thus the next claim is that $B,X,E$ are collinear. Note that both $AXE$ and $AXB$ are right angles, and our claim is proven. The rest of the problem now proceeds using $2$ methods. First we use cosine addition on $A+60$ to use LOC to find $BE$, and use law of Cosines on the $3$ triangles formed by the Fermat point, also equating sine area formula, we can solve for the sum of these $3$ lengths as well to be $38+19\sqrt{3}$
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Giant_PT
15 posts
#14
Y by
I literally added wrong at the end
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GrantStar
812 posts
#15 • 1 Y
Y by OronSH
I agree with Oron. Great problem
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gonewiththewind
3 posts
#16
Y by
The Fermat point of triangle ACD lies on BE.
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Bluesoul
870 posts
#17 • 1 Y
Y by OronSH
If you know Fermat Point it might be guessable(I guessed and solved from there)? But Ptolemy is apparently a better way

Assume $AX=a, BX=b, CX=c$, by Ptolemy inequality we have $a+2b\geq \sqrt{3}XE; a+2c\geq \sqrt{3}BX$, while the inequality is reached when both $CXAB$ and $AXDE$ are concyclic. Since $\angle{BCA}=\angle{BXA}=\angle{EXA}=\angle{ADE}=90^{\circ}$, so $B,X,E$ lie on the same line. Thus, the desired value is then $(1+\frac{\sqrt{3}}{2})BE$.

Note $\cos(\angle{DAC})=\frac{1}{7}, \cos (\angle{EAB})=-\frac{11}{14}, BE=38$ by LOC< the answer is then $38+19\sqrt{3}\implies \boxed{060}$
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Scilyse
385 posts
#18
Y by
most original MAA problem
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shendrew7
787 posts
#19 • 1 Y
Y by a_smart_alecks
"Minimizing distances, 60 degree angles, 'sus'"

Denote $X$ as the foot from $A$ to $BE$, which we claim is the optimal point. Note that we have minimized $BX+EX$, and the cyclicities of $ABCX$ and $AEDX$ gives $\angle AXC = \angle AXD = 120$, making $X$ the Fermat point of $\triangle ACD$ and thus minimizing $AX+CX+DX$ too.

Let $F = CX \cap (ADE)$. Notice we have the spiral similarity $\triangle ABE \sim \triangle ACF$, and Fermat point properties gives
\[f(X) = (BX+EX)+(AX+CX+DX) = BE+CF = BE\left(1+\frac{\sqrt 3}{2}\right) = \boxed{38+19\sqrt3}.\]
[asy]
size(300);
pair A, B, C, D, E, F, X;
A = (0,10);
C = (-4,0);
D = (11,0);
B = C + sqrt(3)/3*(A-C)*dir(90);
E = D + sqrt(3)/3*(A-D)*dir(-90);
F = A + (D-A)*dir(60);
X = foot(A, B, E);

draw(A--B--C--D--E--cycle^^C--A--D^^A--F);
draw(B--E^^C--F^^A--X^^D--X, dashed);
draw(circumcircle(A, B, C)^^circumcircle(A, D, E), blue);

dot("$A$", A, 2*dir(105));
dot("$B$", B, dir(225));
dot("$C$", C, dir(250));
dot("$D$", D, dir(290));
dot("$E$", E, dir(0));
dot("$F$", F, dir(45));
dot("$X$", X, 2*dir(270));

label("$14$", A--B, dir(135));
label("$7 \sqrt 3$", A--C, .1*dir(150));
label("$26$", A--E, dir(75));
label("$13 \sqrt 3$", A--D, .01*dir(45));
label("$24$", C--D, dir(270));
[/asy]
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eibc
595 posts
#20
Y by
Note that $\triangle ACB$ and $\triangle ADE$ are both 30-60-90 triangles, and $AC = 7\sqrt{3}$ and $AD = 13\sqrt{3}$.

Let $F$ be the fermat point of $\triangle ACD$. Then $AX + CX + DX$ is minimized for $X = F$. Moreover, since $\angle AXD = \angle AXC = 120^{\circ}$, we find that $ABFC$ and $AEDF$ are cyclic. Thus, $90^{\circ} = \angle ACB = \angle AFB = \angle ADE = \angle AFE$, so $F$ lies on $\overline{BE}$ and thus $f(X)$ is minimized for $X = F$.

Now, let $a = AF, c = CF$, and $d = DF$. By the law of cosines on $\triangle AFC$, $\triangle AFD$, and $\triangle CFD$, we find that
\begin{align*}
a^2 + c^2 + ac &= 147 \\
c^2 + d^2 + cd &= 576 \\
d^2 + a^2 + da &= 507.
\end{align*}Summing all three equations yields $2(a^2 + c^2 + d^2) + ac + cd + da = 1230$. Additionally, from computing the area of $\triangle ACD$ in two ways, we find that $\tfrac{ac\sqrt{3} + cd\sqrt{3} + da\sqrt{3}}{4} = 78\sqrt{3}$, so $ac + cd + da = 312$. We then find that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ac + 2cd + 2da = 1083$, so $a + b + c = 19\sqrt{3}$.

Now, we find $BF + DF$. From Ptolemy's theorem on $ABCF$, we have $BF \cdot 7\sqrt{3} = 7a + 14c$, so $BF\sqrt{3} = a + 2c$. Similarly, $DF\sqrt{3} = a + 2b$, so $BF + DF = \tfrac{2a + 2b + 2c}{\sqrt{3}} = 38$. Thus, $f(F) = 38 + 19\sqrt{3} \implies \boxed{060}$.
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Mathkiddie
322 posts
#21
Y by
Definitely the best problem in this year's AIME I (I almost solved this problem in-comp but ran out of time :(), part of it is actually very similar to a problem I made a year ago here.
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Mr.Sharkman
486 posts
#22
Y by
Oops I got that $X = X_{13}(ACD)$ and showed that it was on $BE$ by inversion but due to a massive skissue was not able to find $XA, XC, XD$ after that.
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apple143
62 posts
#23
Y by
i wish i could see the beauty in geo...
This post has been edited 1 time. Last edited by apple143, Feb 8, 2025, 1:39 AM
Reason: sef
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AngeloChu
471 posts
#24
Y by
what bash?
let AX=a, BX=b, CX=c
be a geo main and do too much oly geo and fermat point stuff we solve systems of equations from law of cosines, get $ab+bc+ca=624$, $a^2+b^2+c^2=459$
then $a+b+c=\sqrt{927}=19\sqrt3$
now do ptolemy for BE
$13\sqrt3 * \sqrt{26^2-a^2}=13a+26b$
$7\sqrt3 * \sqrt{14^2-a^2}=7a+14c$
then we solve and get $(\sqrt{26^2-a^2}+\sqrt{14^2-a^2})*\sqrt3=2a+2b+2c=38\sqrt3$
then BE=38 and literally $38+19\sqrt3$ rfwbuh
hopefully a ton of geo like this on aime ii
This post has been edited 1 time. Last edited by AngeloChu, Feb 8, 2025, 2:32 AM
Reason: latex not texing
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Tetra_scheme
92 posts
#25 • 1 Y
Y by GrantStar
I think the optimal point is not super obvious from the perspective of the full pentagon. Its a lot easier to thing of it as a triangle with $2$ $30-60-90$ extensions which makes it much more obvious that this problem is about the triangle $ACD$, not some random pentagon.
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L13832
248 posts
#26 • 1 Y
Y by Pengu14
This problem is so good
solution (same as @above)
This post has been edited 2 times. Last edited by L13832, Feb 14, 2025, 4:32 AM
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smbellanki
165 posts
#27
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Idk if someone alr said this but the problem looks similar to https://artofproblemsolving.com/community/c5h1409444p7910363 (except that this one's 3d)
This post has been edited 1 time. Last edited by smbellanki, Today at 3:35 AM
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