ka May Highlights and 2025 AoPS Online Class Information
jlacosta0
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.
Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.
Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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Some users don't want to learn, some other simply ignore advises.
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In an acute triangle the points and are the feet of the altitudes through and respectively. The incenters of the triangles and are and respectively; the circumcenters of the triangles and are and respectively. Prove that and are parallel.
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
Source: Al-Khwarizmi Junior International Olympiad 2025 P7
Let be a cyclic quadrilateral with circumcenter , such that is not a diameter of its circumcircle. The lines and intersect at point , so that lies between and , and lies between and . Suppose triangle is acute and let be its orthocenter. The points and on the lines and , respectively, are such that and . The line through , perpendicular to , intersects at , and the line through , perpendicular to , intersects at . Prove that the points ,, are collinear.
Let be a positive integer and be a positive integer coprime to . Let , and for , define Find, in terms of and , the greatest positive integer for which there exists an index such that is divisible by .
Alice and Bob play, 8x8 table, white red black, minimum n for victory
parmenides5114
N3 hours ago
by Ilikeminecraft
Source: JBMO Shortlist 2018 C3
The cells of a table are initially white. Alice and Bob play a game. First Alice paints of the fields in red. Then Bob chooses rows and columns from the table and paints all fields in them in black. Alice wins if there is at least one red field left. Find the least value of such that Alice can win the game no matter how Bob plays.
Let be an interior point of the acute triangle with so that The point on the segment satisfies the point on the segment satisfies and the point on the line satisfies Let and be the circumcenters of the triangles and respectively. Prove that the lines and are concurrent.
We divide both sides by ...we wil check the case in which a=b=0 after.
So we get
so
we know that is always positive.....
so must be 0.. but there are no integers and for which
but we forgot that a and b can be 0. checking if we find it true so that is 1 solution.Also we can have 1 of a and b be 0 and the other a random integer. So if we have we get whih we already have....the same thing will happen when a=0 and b is a random integer.
Now check all other cases fo which
we get when we substitute s0 therefore the only solution is
We also check wth 1 of or b=0 and the other a radom integer in which case we get (for b=0) whih only occurs when a=c=0...the samewill happen with a=0 and b another random integer...
This post has been edited 1 time. Last edited by basketball9, Apr 4, 2010, 2:14 PM
It is easy to determine that and must all be even by considering the possible parities and working in .
Let . We find . We may continue this process for an infinite amount of time, and find that each of and must be divisible by an infinite amount of s, so by infinite descent there are no nontrivial solutions to this. Thus, the only solution is .
Y byCircleGeometryGang, sabkx, Adventure10, Mango247
basketball9 wrote:
We divide both sides by ....
[1] So we get
[2] so
[1] a) The third term should be .
b) While this equation would have been mathematically correct (see [1] a above), it is misleading,
because the direct next step after dividing by , but befoe reordering terms, would be .
[2] If you type it out horizontally, it has to have grouping symbols around the denominator such as
Otherwise, type out the fractions in a vertical style and avoid the needed use of grouping symbols:
(This post is being neutral towards the correctness of your overall method.)
@basketball: I don't see how that shows that c needs to be 0. As long as c<ab, the right hand side is still positive.
xpmath's solution still works, because the right side will always be 0 mod 4, so all of the terms on the left side need to be 0 mod 2 as well (because a perfect square is only 0 or 1 mod 4, of course), and you can still perform the infinite descent. The coefficient of the right hand side does grow larger, but it is still always 0 mod 4.
I also do not understand Basketball9's logic.
@harekrishna: xpmath got that 2a_1=a, 2b_1=b, 2c_1=c
so we get:
4(a_1)^2+4(b_1)^2+4(c_1)^2=4(a_1)(b_1)
Dividing through by 4, we get:
(a_1)^2+(b_1)^2+(c_1)^2=(a_1)(b_1)
Which is correcy.
Y bymaXplanK, CircleGeometryGang, Adventure10, Mango247, and 3 other users
Rearrange this as . By looking at this mod 4, it can easily be seen that must be even. It is well-known that if , then ,, or . Since is odd and no prime divides 1, we must have that all prime factors of are congruent to 1 modulo 4. Hence, the product of the prime factors of must be conrguent to 1 modulo 4. But , so we have a contradiction. Hence, there are no nontrivial solutions.
Suppose that not all of are zero. Let be the greatest nonnegative integer such that . Let . Then . If , taking the equation mod 4 trivially shows that are all even, contradicting the maximality of . If , taking the equation mod 4 again shows that are all even, again contradicting the maximality of . Hence, our equation has only the trivial solution.
Sorry for bringing this back up but could someone explain what infinite descent means? I understand how the first part of the solution but not the descent part. Also if someone could pm me maybe a link to somewhere that explains the concept well.
It's not really a fancy concept or anything. We start with this equation:
We find that are even and we let . We find .
Using another argument, we can show that are even. So let .
Then we sub this into the equation and find that are even.
And we can repeat this argument an infinite number of times to get that are all even and it would imply that have an infinite number of factors of 2. But this is impossible, so solutions can't exist.
We find that . But since is not a qr mod any number 3 mod 4. and must both be even. In particular (a-1)(a+1) = 2n*(2n+2). So 16 divides . But is not a qr mod , so there are not solutions to the equation. Where did I miss the 0,0,0 trivial solution?
We find that . But since is not a qr mod any number 3 mod 4. and must both be even.
This statement is the first place where your solution breaks, and it feels really fishy to me. In particular, I have no idea how non-QR-ness of two integers imply that they both must be even. (Also keep in mind that any two non-QRs mod a prime must multiply to a QR.)
This post has been edited 1 time. Last edited by djmathman, Mar 13, 2018, 9:05 PM
If were odd, then it is the product of and both odd. In particular, one of these is mod and must be divisible by a prime mod . This prime divides in turn so mod . So is a qr mod some prime congruent to mod , which is not possible so is even.
Is this correct?
This post has been edited 1 time. Last edited by AllenWang314, Mar 14, 2018, 3:22 AM
It is easy to determine that and must all be even by considering the possible parities and working in .
Let . We find . We may continue this process for an infinite amount of time, and find that each of and must be divisible by an infinite amount of s, so by infinite descent there are no nontrivial solutions to this. Thus, the only solution is .
Yup, that's what I did, except that a_1^2+b_1^2+c_1^2=4a_1^2b_1^2, which again is 0 mod 4 on RHS, so LHS must be 0 mod 2 for each of a_1, b_1, and c_1 to have their sum of their squares by 0 mod 4. Also, a bit of an explanation. If both a and b are odd, then the RHS is 1 mod 4, and LHS is 1 (a^2) + 1 (b^2) + c^2 mod 4, must be 1 mod 4, absurd, because this implies c^2 must be 3 mod 4. Now if at least of one of a and b is even, then RHS is 0 mod 4, LHS is 0 mod 4 + b^2 + c^2, or b^2+c^2 must be 0 mod 4, absurd (2+2, 1+3 all impossible) unless both b and c are 0 mod 2. Now a and b must both be 0 mod 2, meaning RHS is even and c must also be even for even + even + even = even.
This post has been edited 1 time. Last edited by huashiliao2020, Apr 14, 2023, 4:16 AM
Write the given equation as By Fermat Christmas, divisors of the RHS must be or 1 mod . On the other hand, the factors on the LHS are either multiples of or mod , unless . Indeed this is the sole solution.
Take mod . This is either or . But the second one requires and to be both odd, contradiction. The first one falls to infinite descent and has solution .
It's clear that and can't be both odd at the same time, because . Now if both and are at least , so will have a prime divisor , and by Fermat's Christmas Theorem will divide both and which is impossible. And now by bashing the small values of and we get the only solution to be:
Note that perfect squares are or . is not possible as this lead both and are . So,
Thus, and ,, are even.
Let ,, and substitute in the original equation. Again ,, are even. Let ,, and we find that ,, are also even.
If we repeat this argument infinitely many times, we see that ,, can be divided by infinitely many times, which is only possible when .
This post has been edited 1 time. Last edited by surpidism., May 20, 2023, 5:56 AM
If or then all three must be 0. If then , which has only solution .
Now, take .
Clearly, is not possible unless . Also, . (i.e., ).
Note that . So, if an odd prime , then . This means . It follows that if is even, then all divisors of are . So, , a contradiction. It follows that must be odd. Similarly, must also be odd.
But then , a contradiction.
Therefore, the only integer solution to the given equation is .
This post has been edited 1 time. Last edited by Pyramix, May 20, 2023, 1:35 PM Reason: Unnecessary bounds removed.
Rearranging, we find
Fermat's Christmas Theorem tells us all factors of the LHS with magnitude greater than 1 must be 1 or 2 modulo 4, contradiction. Thus , giving the only solution .
Simplifying our expression, we get If is odd, then But, if then so is even. Now, if we have that so Hence, Now, is odd, so is even. Thus, one of or is and the other is But, these are divisors of and all divisors of are so this is impossible. Thus, there are no solutions, except for when since then there are no prime factors and when
(a,b,c)=(0,0,0) clearly works,otherwise
"Let O be odd and E be even. The following combinations (a,b,c) – (O,E,E), (E,O,E), (E,E,O), (O,O,E), (O,E,O), (E,O,O), (O,O,O), (E,E,E) easily yield contradictions modulo 4."
We claim that the only triple of solutions is which clearly works. We now show that this is the only solution. First note that if is a triple of solutions since this reduces the equation to which has no real solutions. We also observe that if is a triple of solutions so are all triples of the form . Hence it suffices to solve the equation over positive integers. We rearrange the equation to, If is odd, one of and is divisible by and thus, which is impossible for all positive integers . If is even, one of and is . Thus, there exists a prime which divides the right-hand side. But then, which is impossible by Fermat's Christmas theorem. Hence, cannot take any positive integral value, implying that all solutions are of the claimed forms.
This post has been edited 1 time. Last edited by cursed_tangent1434, Apr 29, 2025, 9:18 AM Reason: fakesolve