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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
AIME score for college apps
Happyllamaalways   56
N 3 hours ago by Countmath1
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.
56 replies
Happyllamaalways
Mar 13, 2025
Countmath1
3 hours ago
MIT Beaverworks Summer Institute
PowerOfPi_09   0
3 hours ago
Hi! I was wondering if anyone here has completed this program, and if so, which track did you choose? Do rising juniors have a chance, or is it mainly rising seniors that they accept? Also, how long did it take you to complete the prerequisites?
Thanks!
0 replies
PowerOfPi_09
3 hours ago
0 replies
Convolution of order f(n)
trumpeter   70
N 4 hours ago by HamstPan38825
Source: 2019 USAMO Problem 1
Let $\mathbb{N}$ be the set of positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the equation \[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.

Proposed by Evan Chen
70 replies
trumpeter
Apr 17, 2019
HamstPan38825
4 hours ago
k HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY
alcumusftwgrind   8
N Today at 10:13 AM by maxamc
rant lol

Imagine a poor senior waiting for their MIT decisions just to have their hopes CRUSHED on 3/14 and they can't even celebrate pi day...

and even worse, this year's pi day is special because this year is a very special number...

8 replies
alcumusftwgrind
Today at 2:11 AM
maxamc
Today at 10:13 AM
No more topics!
Problem #23 2016 AMC 12A:
ddkim08   33
N Oct 12, 2022 by coolmath_2018
Three numbers in the interval [0,1] are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$
33 replies
ddkim08
Feb 3, 2016
coolmath_2018
Oct 12, 2022
Problem #23 2016 AMC 12A:
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ddkim08
8 posts
#1 • 5 Y
Y by dantx5, mathcrazymj, 277546, Adventure10, Mango247
Three numbers in the interval [0,1] are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$
This post has been edited 2 times. Last edited by djmathman, Feb 3, 2016, 3:41 PM
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62861
3564 posts
#2 • 76 Y
Y by RadioActive, r31415, dantx5, spartan168, speck, ythomashu, liopoil, WalkerTesla, Ancy, Galo1s, whatshisbucket, MSTang, chezbgone, Flash12, adihaya, checkmatetang, aadavi, champion999, Shaddoll, yrnsmurf, acegikmoqsuwy2000, Wave-Particle, Mudkipswims42, PiDude314, jam10307, spin8, xwang1, DominicanAOPSer, AKAL3, saagar, ThisIsASentence, YadisBeles, aaa16797, MathSlayer4444, KevinTang2005, Intellectuality123, m1234567, blep, skipiano, wu2481632, Python54, claserken, BobaFett101, AstrapiGnosis, anantmudgal09, GeneralCobra19, 3_14152, coolrg, Order, mathwiz0803, yojan_sushi, GeronimoStilton, Ultroid999OCPN, mathleticguyyy, j-_-, Vietjung, Lol_man000, brainiacmaniac31, Mathcollege, magicarrow, opptoinfinity, etvat, ETS1331, sotpidot, rayfish, math31415926535, celestialphoenix3768, fuzimiao2013, aidan0626, Geometry285, ninjaforce, jeff10, Bradygho, Turtwig113, Adventure10, Jack_w
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.
This post has been edited 2 times. Last edited by 62861, Feb 3, 2016, 5:04 PM
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FTWninja
35 posts
#4 • 3 Y
Y by Vector-13, Adventure10, Mango247
Click to reveal hidden text
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MSTang
6012 posts
#5 • 45 Y
Y by shiningsunnyday, spin8, AstrapiGnosis, acegikmoqsuwy2000, 0x5f3759df, dantx5, r3mark, spartan168, 15Pandabears, ythomashu, jam10307, Deathranger999, Wave-Particle, math101010, aadavi, champion999, yrnsmurf, sjwon3789, elitechicken, trumpeter, bearytasty, High, BarbieRocks, hexagram, 62861, DominicanAOPSer, wtasfias, saagar, mathguy5041, MathSlayer4444, ThisIsASentence, m1234567, muti66, blep, skipiano, sajith, ccx09, Ultroid999OCPN, j-_-, mathleticguyyy, Imayormaynotknowcalculus, sotpidot, fuzimiao2013, Adventure10, Mango247
This is why you do HMMT Monthly
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High
876 posts
#6 • 2 Y
Y by Adventure10, Mango247
Does anyone have a good argument for this?
possible sol
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va2010
1276 posts
#7 • 4 Y
Y by dantx5, pycops, Adventure10, Mango247
beautiful
This post has been edited 3 times. Last edited by va2010, Feb 4, 2016, 3:38 AM
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dongrae
14 posts
#8 • 2 Y
Y by Adventure10, Mango247
I think the answer is (C) 1/2
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DrMath
2130 posts
#9 • 2 Y
Y by MSTang, Adventure10
By 2014 WOOT POTD 9/1, there are 505 ways to choose 0<a,b,c<=10 such that they form a triangle. This makes the probability $\frac{505}{1000}$ which is close to 1/2, so C.

But I didn't realize this during the test because I had little time oops.
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bobthesmartypants
4337 posts
#10 • 6 Y
Y by dongrae, Deathranger999, yrnsmurf, Ultroid999OCPN, Adventure10, Mango247
Although I didn't take the 12, I solved the problem by just drawing a 3-d probability diagram. The areas for which the three numbers cannot form a triangle are just three tetrahedrons with area 1/6th the area of the cube, so the answer is just $1-3\cdot \dfrac{1}{6}=\dfrac{1}{2}$
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62861
3564 posts
#11 • 2 Y
Y by AKAL3, Adventure10
DrMath wrote:
By 2014 WOOT POTD 9/1, there are 505 ways to choose 0<a,b,c<=10 such that they form a triangle. This makes the probability $\frac{505}{1000}$ which is close to 1/2, so C.

But I didn't realize this during the test because I had little time oops.

Wow what an amazing solution! I wonder how you eventually thought of it.
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dongrae
14 posts
#12 • 1 Y
Y by Adventure10
bobthesmartypants wrote:
Although I didn't take the 12, I solved the problem by just drawing a 3-d probability diagram. The areas for which the three numbers cannot form a triangle are just three tetrahedrons with area 1/6th the area of the cube, so the answer is just $1-3\cdot \dfrac{1}{6}=\dfrac{1}{2}$
I solved this exact same way with you. I draw the cube and eliminate 3 volumes of tetrahedrons.
This post has been edited 1 time. Last edited by dongrae, Feb 3, 2016, 4:57 PM
Reason: add
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acegikmoqsuwy2000
767 posts
#13 • 3 Y
Y by Adventure10, Mango247, giratina3
i had this exact same problem on clevermath not too long ago.. :O

except the answer was 3 for that one because you know...
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mishai
1048 posts
#14 • 2 Y
Y by Adventure10, Mango247
WLOG $c\geq b\geq a$, so $c$ has a region of $a$ to be in, so this is $\int_0^1a\ da=1/2$. I think this is wrong, but it gave the right answer.
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speck
1727 posts
#15 • 3 Y
Y by liopoil, Adventure10, Mango247
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

Wow. This is clearly the best solution.
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Deathranger999
123 posts
#16 • 2 Y
Y by Adventure10, Mango247
I just used a geometric argument by envisioning a 1x1x1 cube in a 3-dimensional coordinate grid, then looking at the inequalities $a + b > c$, $b + c > a$, and $c + a > b$. Each of those inequalities, combined with the cube, graphs a pyramid with volume 1/6. Thus the volume of all of them is 1/2, out of the total volume of 1. So the answer is 1/2.
This post has been edited 1 time. Last edited by Deathranger999, Feb 3, 2016, 9:26 PM
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wwwrqnojcm
221 posts
#17 • 2 Y
Y by Adventure10, Mango247
Calculus solution:
When $a>b$, consider two cases:

1) $0<a<\frac{1}{2}$, then
$\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,dbda=\frac{1}{24}$

2)$\frac{1}{2}<a<1$, then
$\int_{\frac{1}{2}}^{1} (\int_{0}^{1-a}2b \,db + \int_{1-a}^{a}1+b-a \,db)da=\frac{5}{24}$

$a<b$ is the same. Thus the answer is $\frac{1}{2}$.
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Galo1s
5 posts
#18 • 2 Y
Y by Adventure10, Mango247
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

Oh my god this works...
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va2010
1276 posts
#19 • 2 Y
Y by Adventure10, Mango247
az_phx_brandon_jiang wrote:
Does anyone have a good argument for this?
possible sol

No, this is an abuse of the notion of signed area.
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62861
3564 posts
#20 • 1 Y
Y by Adventure10
va2010 wrote:
az_phx_brandon_jiang wrote:
Does anyone have a good argument for this?
possible sol

No, this is an abuse of the notion of signed area.

pretty sure az was trolling
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mathguy5041
2659 posts
#21 • 2 Y
Y by Adventure10, Mango247
Galo1s wrote:
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

Oh my god this works...

Since whatever the largest number is, you can scale it to make it 1. Then the solution follows.
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BarbieRocks
1102 posts
#22 • 4 Y
Y by shiningsunnyday, MSTang, Adventure10, Mango247

Speaking of which, the February round has been released at http://www.hmmt.co/tournaments/monthly!
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62861
3564 posts
#23 • 2 Y
Y by Adventure10, Mango247

wait lol

i didn't even do HMMT monthly, I discovered my solution during the test

-_-
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High
876 posts
#24 • 4 Y
Y by Mudkipswims42, m1234567, Adventure10, Mango247
^wow whoa pr0
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Gamma6
39 posts
#25 • 2 Y
Y by Adventure10, Mango247
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

I don't get why we can WLOG and let the largest number be 1...can anyone explain why this is valid? Can't the largest be any value from 0 to 1, why can we just suppose it's 1?

Thanks!
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Gamma6
39 posts
#28 • 2 Y
Y by Adventure10, Mango247
Bump....
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donot
1180 posts
#29 • 2 Y
Y by Gamma6, Adventure10
Gamma6 wrote:
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

I don't get why we can WLOG and let the largest number be 1...can anyone explain why this is valid? Can't the largest be any value from 0 to 1, why can we just suppose it's 1?

Thanks!
Let the numbers be $a,b,d$, with $a\geq b\geq c$, then multiply $a,b,d$ by $\frac{1}{a}$.

If you are wondering why the probability that $b+d > 1$ is the same even though we have multiplied by $\frac{1}{a}$, which is greater than $1$, just note that the range of values for $b,d,b+d$ remains the same!
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VivekA
694 posts
#30 • 1 Y
Y by Adventure10
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

How can you say this is WLOG?
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william122
1576 posts
#31 • 2 Y
Y by Adventure10, Mango247
VivekA wrote:
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

How can you say this is WLOG?

Clearly, if we assign the largest value to be A, the probability is still going to be 1/2. So, just WLOG.
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Gamma6
39 posts
#32 • 2 Y
Y by Adventure10, Mango247
william122 wrote:
VivekA wrote:
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

How can you say this is WLOG?

Clearly, if we assign the largest value to be A, the probability is still going to be 1/2. So, just WLOG.

If it was clear that the probability is still going to be 1/2 then what's the point of WLOG? You would just answer 1/2. clearly this isn't "obvious" rather you just repeating the problem.
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62861
3564 posts
#33 • 3 Y
Y by Gamma6, Adventure10, Mango247
to several above posts asking about the WLOG:

You can scale all the numbers upwards so that the largest number is now $1$. For example, if the numbers were $0.4$, $0.5$, and $0.7$, then you could scale up by $\frac{10}{7}$.
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Gamma6
39 posts
#34 • 2 Y
Y by Adventure10, Mango247
donot wrote:
Gamma6 wrote:
CantonMathGuy wrote:
WLOG the largest number is 1. Then the probability that the other two add up to at least 1 is $1/2$.

Thus the answer is $1/2$.

I don't get why we can WLOG and let the largest number be 1...can anyone explain why this is valid? Can't the largest be any value from 0 to 1, why can we just suppose it's 1?

Thanks!
Let the numbers be $a,b,d$, with $a\geq b\geq c$, then multiply $a,b,d$ by $\frac{1}{a}$.

If you are wondering why the probability that $b+d > 1$ is the same even though we have multiplied by $\frac{1}{a}$, which is greater than $1$, just note that the range of values for $b,d,b+d$ remains the same!

Sorry I am still not finding this clear...let me try explaining what I know and maybe you can find out where I am struggling.

So we have our number $1\ge a\ge b\ge d\ge 0$. Then when we scale each number by $\frac{1}{a}$ they increase and $a\rightarrow 1$, $b\rightarrow \frac{b}{a}$, $d\rightarrow \frac{d}{a}$. Since $a\ge b$ and $a\ge d$ they are both still in the range. So are you saying that it is obvious that all numbers in the domain can be created with $\frac{b}{a}$ and $\frac{d}{a}$? For some reason that doesn't seem obvious. So now we try finding the probability of $\frac{b}{a}+\frac{d}{a}>1$ and thats how you WLOG?

Is this correct?

Thanks,
Gamma6

EDIT: Ok...looking at the above post this logic seems to be correct but confirmation would still be nice.
This post has been edited 2 times. Last edited by Gamma6, Aug 15, 2016, 11:30 PM
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ABCDE
1963 posts
#35 • 2 Y
Y by budu, Adventure10
Consider the following question:

Two numbers in the interval [0,a] are chosen independently and at random where a is a positive real number. What is the probability that the two numbers sum to more than a?

Does the answer depend on the value of a?
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Gamma6
39 posts
#36 • 2 Y
Y by Adventure10, Mango247
ABCDE wrote:
Consider the following question:

Two numbers in the interval [0,a] are chosen independently and at random where a is a positive real number. What is the probability that the two numbers sum to more than a?

Does the answer depend on the value of a?

Ah that makes more sense, thank you so much ABCDE.
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coolmath_2018
2807 posts
#37
Y by
nice problem. I don't think this solution has been posted.

Since it is a triangle, we know $x + y > z$, $y + z >x$ and $x + z > y$ where $x, y, z \in [0, 1]$.

Throw this on to the 3-D plane, and doing so we get a cube from $(0, 0, 0)$ to $(1, 1,1)$.

By symmetry we can solve for $1 - 3 \cdot P(x + y \le z)$.

The volume of $P(x + y \le z)$ is just the tetrahedron formed by $(0, 0, 0), (1, 0, 1), (0, 1, 1), (0, 0, 1)$.

Thus the volume of this tetrahedron is just $\frac{1}{6}$.

So the answer is $1 - 1/2 = 1/2$.
This post has been edited 1 time. Last edited by coolmath_2018, Oct 12, 2022, 3:51 AM
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