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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
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.

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0 replies
jlacosta
May 1, 2025
0 replies
Docked 4 points Help
sadas123   8
N 35 minutes ago by bjump
In school we had this beginners like middle school contest, but we had to right down our solution kind of like usajmo except no proofs. It was also graded out of 7 but I got 4 Points docked for this question. what was my problem??? But I kind of had to rush the solution on this question because there was another problem before this that was like 1000x times harder.

Question:The solutions to the equation x^3-13x^2+ax−48=0 are all positive whole numbers. What is $a$?


Solution: We can see that we can use Vieta's formulas to find that the product of the roots is $48$, and the sum of the roots is $13$. So we need to find a combination of integers that multiply to $48$ and add up to $13$. Let's call the roots of the equation p, q, and r. From Vieta's, we get that $p+q+r=-13$ and $pqr = -48$. Looking at the factors of $48$, which is $2^4*3$, we try to split the numbers in a way that gives us the correct sum and product. Trying 3, -2, and -8, we see that they add up to $-13$ and multiply to $-48$, so they work. That means the roots of the polynomial are -3, -2, and -8, and the factorization is $(x-3)(x-2)(x-8)$. Multiplying it out, we get $x^3-13x^2+46x-48$, so we find that a = 46.
8 replies
sadas123
Yesterday at 4:06 PM
bjump
35 minutes ago
You are invited to BROOM 2025!
puffypundo   10
N an hour ago by anyuhang
You are invited to BROOM 2025!

BROOM (Building Resolve and Opportunity for Oncoming MOPpers) is a collaborative, highly intensive online math program modeled after MOP, open to students entering 9th grade and above. The program is designed by many past and current MOPpers to bring the MOP experience to everyone. It will take place from June 11th to July 2nd for 6 to 10 hours a day, with activities running in perfect parallel with MOP.

The program will include a structured schedule of student-led classes, mock tests, and community events to get to know your fellow sweepers. Just like MOP this year, there will be 3 practice tests, 2 ELMO-style tests, and 3 TSTST-style tests. Classes will range in difficulty, and more details regarding color groups and tests will be sent to students who register.

To achieve a more immersive experience, BROOM will be hosted on a Minecraft server where players can interact just like in real life, featuring classrooms for classes, lecture halls for tests, and dorms/dining halls for fun! Proximity chat will also be installed to imitate in-person conversation.

For over 150 hours of activities, the program is only $90, and financial aid is available. A copy of Minecraft will be included with your registration. Note that we do not run for profit - all funds are used for running the program itself.

Register for BROOM by June 1st! Extra details are available here. :D

Note
10 replies
puffypundo
Yesterday at 7:07 PM
anyuhang
an hour ago
EGMO help
mathprodigy2011   18
N 2 hours ago by pingpongmerrily
If we have a quadrilateral with 1 pair of parallel sides but the parallel sides are also equal, is that sufficient to stating the quadrilateral is a parallelogram. if it's not, please give a counter-example.
18 replies
mathprodigy2011
4 hours ago
pingpongmerrily
2 hours ago
Cyclic Quad
worthawholebean   131
N 2 hours ago by mathwiz_1207
Source: USAMO 2008 Problem 2
Let $ ABC$ be an acute, scalene triangle, and let $ M$, $ N$, and $ P$ be the midpoints of $ \overline{BC}$, $ \overline{CA}$, and $ \overline{AB}$, respectively. Let the perpendicular bisectors of $ \overline{AB}$ and $ \overline{AC}$ intersect ray $ AM$ in points $ D$ and $ E$ respectively, and let lines $ BD$ and $ CE$ intersect in point $ F$, inside of triangle $ ABC$. Prove that points $ A$, $ N$, $ F$, and $ P$ all lie on one circle.
131 replies
+2 w
worthawholebean
May 1, 2008
mathwiz_1207
2 hours ago
No more topics!
Sort of additive function
tenniskidperson3   112
N Apr 27, 2025 by anudeep
Source: 2015 USAJMO problem 4
Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ such that\[f(x)+f(t)=f(y)+f(z)\]for all rational numbers $x<y<z<t$ that form an arithmetic progression. ($\mathbb{Q}$ is the set of all rational numbers.)
112 replies
tenniskidperson3
Apr 29, 2015
anudeep
Apr 27, 2025
Sort of additive function
G H J
Source: 2015 USAJMO problem 4
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Arrowhead575
2281 posts
#102
Y by
Is that equation citable? I simplified it down to Cauchy's
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jasperE3
11352 posts
#103
Y by
Evan Chen A.3 says
Quote:
In general it is usually okay to cite a result that is (i) named, and (ii) does not trivialize
the given problem.
so it should be fine.
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Eagle_Student5
5 posts
#104 • 2 Y
Y by centslordm, megarnie
Solved with HamstPan38825.

We claim the only solutions are $f(x)=kx+c$, and it is easy to see that this works.

Now, define $x=a$, $y=a+d$, $z=a+2d$, and $t=a+3d$. Putting these into the equation gives us
\[f(a)+f(a+3d)=f(a+d)+f(a+2d)\]If we subtract $d$ from everything, this simplifies to
\[f(a-d)+f(a+2d)=f(a)+f(a+d)\]Then, if we add the two equations together and cancel, we get
\[f(a-d)+f(a+3d)=2f(a+d)\]This is Jensen's Functional Inequality, which we know has solution $f(x)=kx+c$, but only if $d\neq0$. So if $d=0$, then $a=x=y=z=t$, but this contradicts the fact that $x<y<z<t$, so $d$ can't be $0$, so we're done. $\square$
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Ritwin
157 posts
#105 • 1 Y
Y by LLL2019
Wrong

:oops_sign: my solution was wrong :blush:
This post has been edited 1 time. Last edited by Ritwin, May 13, 2022, 3:02 AM
Reason: spoilered the wrong solution :blush:
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megarnie
5610 posts
#106
Y by
Rewritten solution:

Let $P(x,y,z,t)$ denote the given assertion and $d$ be some nonzero rational number.

$P(x,x+d,x+2d,x+3d): f(x)+f(x+3d)=f(x+d)+f(x+2d)$.

$P(x-d,x,x+d,x+2d): f(x-d)+f(x+2d)=f(x)+f(x+d)$.

Adding the equations gives $f(x-d)+f(x)+f(x+2d)+f(x+3d)=2f(x+d)+f(x)+f(x+2d)\implies f(x-d)+f(x+3d)=2f(x+d)$.

Since $x$ and $d$ can be any rational numbers, this simplifies to $f(x)+f(y)=2f\left(\frac{x+y}{2}\right)$. This is Jensen's FE, but I will solve it here.

Let $g(x)=f(x)-f(0)$. Then $g(x)+g(y)+2f(0)=2g\left(\frac{x+y}{2}\right)+2f(0)\implies g(x)+g(y)=2g\left(\frac{x+y}{2}\right)$ with the additional constraint that $g(0)=0$. Let $Q(x,y)$ be this new assertion.

$Q(x,0): g(x)=2g\left(\frac{x}{2}\right)\implies g\left(\frac{x}{2}\right)=\frac{g(x)}{2}$.

Thus, $g(x)+g(y)=g(x+y)$, so $g$ is Cauchy, which is well-known to have solutions $g(x)=kx$ for some constant $k$ over the rationals.

So $f(x)=kx+f(0)$, $\boxed{f(x)=kx+c}$ for rational constants $k$ and $c$. To show that this works, we have $kx+c+kt+c=ky+c+kz+c\iff kx+kt=ky+kz$. Note if $k=0$, then constant solution obviously works. If not, then $kx+kt=ky+kz\iff x+t=y+z$, which is true since they form an arithmetic progression.
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ZETA_in_olympiad
2211 posts
#107
Y by
Answer
Proof
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brainfertilzer
1831 posts
#108
Y by
The only solutions are $\boxed{f(x) = ax + b}$ for all $x$, where $a,b\in\mathbb{Q}$. These work since
\[ (ax + b) + (at + b) = (ay + b) + (az + b)\iff a(x + t - y -z) = 0,\]which is true since $x + t = y + z$. It remains to show that these solutions are the only ones.

Take $\{x,y,z,t\} = \{a, a + d/2, a + d, a + 3d/2\}$ and $\{x,y,z,t\} = \{a + d/2, a + d, a + 3d/2, a + 3d\}$ separately to get the equations
\[ f(a) + f(a + 3d/2) = f(a + d/2) + f(a + d)\qquad (1),\]\[ f(a + d) + f( a + 3d/2) = f(a +d/2) + f(2d)\qquad (2).\]Subtracting $(1)$ from $(2)$ gives $f(a + d) - f(a) = f(a + 2d) - f(a + d)\implies f(a + d) = \frac{f(a) + f(a + 2d)}{2}$. Using the substitution $(x,y) = (a, a+ 2d)$, we have $f\left(\frac{x + y}{2}\right) = \frac{f(x)+ f(y)}{2}$ for all $x,y\in\mathbb{Q}$. Hence, $f$ satisfies Jensen's FE over the rationals, so $f$ is linear.
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rayfish
1121 posts
#109
Y by
The only solutions are linear $f$, or $f(x)=mx+n$ for rational $m$ and $n$, which works. Now, we show these are the only solutions.

Let rational numbers $a<b<c<d<e$ be in arithmetic progression. Then
\begin{align*}
f(b) + f(e) &= f(c) + f(d)\\
f(b) + f(c) &= f(a) + f(d)\\
\end{align*}Subtracting, we get $f(e)-f(c) = f(c)-f(a)$, so $f$ is linear over all arithmetic progressions in $\mathbb{Q}$. In particular, $f$ is linear for all integer inputs. Let $g$ be a linear function over $\mathbb{Q}$ such that $g(x)=f(x)$ for all integral $x$. We will show that $g=f$.

Let $r=\frac p q$ be a nonzero rational number, where $p$ and $q$ are nonzero integers. Since $f$ is linear over $\{ kr \mid k\in\mathbb{Z} \}$, and $f(0)=g(0)$ and $f(p) = g(p)$, we know that $f=g$ over that set, so $f(r)=g(r)$. Hence, $f=g$, as desired.
This post has been edited 5 times. Last edited by rayfish, Dec 5, 2022, 2:45 AM
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RedFireTruck
4226 posts
#110 • 1 Y
Y by BluePoliceCar
Let the average be $a$ and $d=z-a$. Then, $f(a-3d)+f(a+3d)=f(a-d)+f(a+d)$. We can plug in $a+2d$ for $a$ and we get $f(a+5d)-f(a+3d)=f(a+d)-f(a-d)$. We add these equation to get $f(a+5d)+f(a-3d)=2f(a+d)$. We can plug in $a-d$ for $a$ to get $f(a+4d)+f(a-4d)=2f(a)$. We can plug in $4d$ for $d$ to get $f(a+d)+f(a-d)=2f(a)$ for all rationals $a$ and $d$. It is easy to see by induction that $f(cx)=cf(x)-(c-1)f(0)$ for integers $c\ge 1$. This means that $f(p)=qf(\frac{p}{q})-(q-1)f(0)$ so $f(\frac{p}{q})=\frac{f(p)+(q-1)f(0)}{q}=\frac{pf(1)+(q-p)f(0)}{q}=\frac{p}{q}f(1)+(1-\frac{p}{q})f(0)=\frac{p}{q}(f(1)-f(0))+f(0)$ so $f(x)=x(f(1)-f(0))+f(0)=mx+b$ for all rational $x$. Plugging this back in, all $f(x)=mx+b$ work.
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asdf334
7585 posts
#111 • 1 Y
Y by BluePoliceCar
WLOG assume that $f(0)=0$. We may now consider the quantities $f(kd)$ for some integer $k$ and positive rational $d$; by induction we find that the values $f(2kd)$ actually form an arithmetic sequence. This immediately solves the problem over $\mathbb{Z}$ and subsequently $\mathbb{Q}$. $\blacksquare$

(This is essentially the same as @2above.)

Motivation
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HamstPan38825
8867 posts
#112
Y by
Notice that for all $a$ and $d>0$,
\begin{align*}
f(a+d) + f(a+2d) &= f(a) + f(a+3d) \\
f(a) + f(a+d) &= f(a-d) + f(a+2d).
\end{align*}Thus we have $$f(a+3d) + f(a-d) = 2f(a+d),$$so $f$ satisfies Jensen's functional equation. As a result, $f(x) = x+c$ for constants $c$ are the only solutions.
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Infinity_Integral
306 posts
#113
Y by
We claim only all $f(x)=mx+c, m,c\in\mathbb{Q}$ work.
Note that these functions work by checking.
For any $a,d$ we have
$$f(a+d)+f(a+4d)=f(a+2d)+f(a+3d)$$$$f(a+2d)+f(a+5d)=f(a+3d)+f(a+4d)$$So we get $f(a+d)+f(a+5d)=2f(a+3d)$, equivalent to
$$f(x-y)+f(x+y)=2f(x)$$Doing induction similar to the Cauchy Functional Equation over $\mathbb{Q}$ gives the solution set.
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OlympusHero
17020 posts
#114
Y by
Same solution as many others, but I'll post here anyway.

writeup
This post has been edited 1 time. Last edited by OlympusHero, Oct 12, 2023, 1:19 AM
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NamelyOrange
514 posts
#115
Y by
Question: would Jensen's functional equation work here?
This post has been edited 1 time. Last edited by NamelyOrange, Oct 14, 2023, 4:22 PM
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anudeep
199 posts
#116
Y by
We claim $f$ is of the form $\alpha x+\beta$ for some $\alpha, \beta\in\mathbb{Q}$, easy to check why they work and we shall show that these are the only solutions to $f$.
Consider five terms of arithmetic progression namely $a,a+d, \ldots, a+4d$, with $a,d\in\mathbb{Q}$ and $d>0$. Plug these terms in the following manner,
\begin{align*}
f(a+d)-f(a)&=f(a+3d)-f(a+a+2d)\\
f(a+2d)-f(a+d)&=f(a+4d)-f(a+3d).
\end{align*}Adding two of the above equation we deduce to the following,
$$f(a+2d)=\cfrac{f(a+4d)+f(a)}{2}.$$Aha! this resembles the ubiquitous Jensen's equation and turns out it is precisely that as one may substitute $u=a$ and $v=a+4d$ ( this is because $(u,v)$ can generate the whole of $\mathbb{Q}\times\mathbb{Q}$). $\square$
This post has been edited 2 times. Last edited by anudeep, Apr 27, 2025, 2:44 PM
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