easy olympiad problem

by kjhgyuio, Apr 17, 2025, 2:00 PM

Find all positive integer values of \( x \) such that
\[
\sqrt{x - 2011} + \sqrt{2011 - x} + 10
\]is an integer.
This post has been edited 2 times. Last edited by kjhgyuio, Thursday at 2:01 PM
Reason: nil

Website to learn math

by hawa, Apr 9, 2025, 2:52 AM

Hi, I'm kinda curious what website do yall use to learn math, like i dont find any website thats fun to learn math
L

An algebra math problem

by AVY2024, Apr 8, 2025, 10:55 AM

Solve for a,b
ax-2b=5bx-3a

State target p8 sol

by EaZ_Shadow, Apr 6, 2025, 3:53 PM

Math and AI 4 Girls

by mkwhe, Apr 5, 2025, 11:24 PM

Hey everyone!

The 2025 MA4G competition is now open!

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Visit https://www.mathandai4girls.org/ to get started!

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Mathcounts Nationals Roommate Search

by iwillregretthisnamelater, Mar 31, 2025, 8:36 PM

Does anybody want to be my roommate at nats? Every other qualifier in my state is female. :sob:
Respond quick pls i gotta submit it in like a couple of hours.

The daily problem!

by Leeoz, Mar 21, 2025, 10:01 PM

Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! :)

Please hide solutions and answers, hints are fine though! :)

Problems usually get harder throughout the week, so Sunday is the easiest and Saturday is the hardest!

Past Problems!
This post has been edited 7 times. Last edited by Leeoz, Apr 14, 2025, 4:51 AM

100 post!

by JohannIsBach, Mar 10, 2025, 4:07 PM

this is my 100th post! i cant believe it. :wow:
P.S. can everyone who reads this give it an upvote? thx!
This post has been edited 1 time. Last edited by JohannIsBach, Mar 10, 2025, 4:08 PM

300 MAP Goal??

by Antoinette14, Jan 30, 2025, 10:30 PM

Hey, so as a 6th grader, my big goal for MAP this spring is to get a 300 (ambitious, i know). I'm currently at a 285 (288 last year though). I'm already taking a intro to counting and probability course (One of my weak points), but is there anything else you recommend I focus on to get a 300?

0!??????

by wizwilzo, Jul 6, 2016, 5:35 PM

why is 0! "1" ??!

IMO Shortlist 1995 Combo #3

by KingSmasher3, Aug 20, 2013, 11:12 PM

Determine all integers $ n > 3$ for which there exist $ n$ points $ A_{1},\cdots ,A_{n}$ in the plane, no three collinear, and real numbers $ r_{1},\cdots ,r_{n}$ such that for $ 1\leq i < j < k\leq n$, the area of $ \triangle A_{i}A_{j}A_{k}$ is $ r_{i} + r_{j} + r_{k}$.
_______

I claim that for $n \ge 5,$ the given statement is impossible. Consider 5 points in the plane with corresponding numbers $r_1, r_2, r_3, r_4,$ and $r_5.$ We have 3 possible arrangements of these 5 points.

Arrangement 1: They form a convex pentagon. By triangulating the pentagon, we see that the area is $2r_i+2r_{i+2}+2r_{i+3}+r_{i+1}+r_{i+4},$ for $1 \le r \le 5.$ Thus $r_i+r_{i+3}$ is constant. It follows that $r_i=r_{i+1},$ meaning that all 5 numbers are equal. But this is impossible since it would imply that $A_3, A_4, A_5$ are collinear.

Now note that if we have three points with a fourth inside the triangle, the fourth point has a value equal to the additive inverse of the average of the three points. Additionally, given convex quadrilateral $A_1A_2A_3A_4,$ we have $r_1+r_3=r_2+r_4.$

Arrangement 2: Four points form a convex quadrilateral with the fifth point inside. The four outer points form 4 triangles. The fifth point must lie within two of these triangles. Then by the above note, two adjacent vertices on the quadrilateral must have equal values. This makes the quadrilateral a trapezoid, so the two other points must be equal as well. However, the fifth point forms a convex quadrilateral with three of the four points of the original quadrilateral. From the above note, this implies that the fifth point must lie on one of the parallel sides of the trapezoid.

Arrangement 3: Two points lie inside a triangle formed by the other three points. By the above note, the two points inside must have equal values. Hence, the segments they form must be parallel to all three sides of the triangle, which is obviously impossible.

When $n=4,$ we can arrange the four points in a square and assign equal values to all four.
This post has been edited 1 time. Last edited by KingSmasher3, Oct 10, 2013, 3:09 AM

USAMO 2006 Problem #2

by KingSmasher3, Apr 13, 2013, 3:54 AM

For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\frac{N}{2}.$
_______

Let the $2k+1$ distinct positive integers be $a_1, a_2, a_3, ..., a_{2k+1}$ in increasing order. Since the integers are distinct,

\[\frac{N}{2} \ge \sum_{i=k+2}^{2k+1} a_i \ge \sum_{i=1}^{k-1} a_i + k(k+1).\]
We are also given,

\[ a_{k+1} + \sum_{i=k+2}^{2k+1} a_i + \sum_{i=1}^{k-1} a_i > N.\]
These two inequalities tell us that $a_{k+1}>k(k+1),$ or $a_{k+1}\ge k(k+1)+1.$ Hence,

\[N \ge 2\sum_{i=k+2}^{2k+1} a_i \ge \sum_{i=1}^{k} (a_{k+1}+i) \ge 2k^3+3k^2+3k.\]
We can check that this minimal value does indeed hold for the set $\{k^2+1, k^2+2, k^2+3, ..., k^2+2k+1\}.$ We are done.

Main Point(s): The fact that the $2k+1$ integers are distinct is very useful in determining the bounds for $N.$ Compare max subset of $k$ and min subset of $k.$
This post has been edited 1 time. Last edited by KingSmasher3, Apr 13, 2013, 3:57 AM

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  • orz blog!!!

    by KevinChen_Yay, Dec 29, 2024, 1:39 AM

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  • Bumpity bump

    by mathboy282, Dec 13, 2020, 9:38 PM

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    by Pluto1708, Mar 20, 2019, 2:25 PM

  • dude,your blog is awesome.Please don't stop and continue your posts!! :)

    by Jiminhio 10, Jan 16, 2014, 2:11 PM

  • thanks haha

    by KingSmasher3, Sep 6, 2013, 3:15 AM

  • happy birthday

    by cire_il, Sep 3, 2013, 9:10 PM

  • btw im totally not trolling

    dude problem 2s are so hard
    what is this madness
    what is going on
    hehe
    we are not spamming up your blog like it might seem at first
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    by applepi2000, Jun 25, 2013, 12:31 AM

  • Hmm USAMO so hard

    by antimonyarsenide, Jun 25, 2013, 12:28 AM

  • dude you do problem 2s?
    dude so legit man
    i am not trolling

    by applepi2000, Jun 25, 2013, 12:28 AM

  • Hmm USAMO so hard

    by antimonyarsenide, Jun 25, 2013, 12:28 AM

  • dude you do problem 2s?
    dude so legit man
    i am not trolling

    by applepi2000, Jun 25, 2013, 12:27 AM

  • Hey look an excellent problem blog.

    It contains a bunch of USAMO problems that are familiar to me because I did them a couple months ago.

    by yugrey, Apr 5, 2013, 1:55 AM

  • Are you my mommy?

    by meisepic, Mar 26, 2013, 6:47 AM

  • too much math

    by cire_il, Mar 26, 2013, 3:15 AM

  • Yay you made a blog!

    by dinoboy, Mar 19, 2013, 4:29 AM

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