Proposals

Let

S

be a set of positive integers, possibly infinite, such that no positive integer greater than 1 divides all elements of

S

. Determine all non-periodic infinite sequences

a_1,a_2,a_3,\dots

of positive integers such that, for all positive integers

n

,

$a_n\leqslant |a_{n+\ell}-\ell|$ for all $\ell$ in $S$ , and
$a_n=|a_{n+\ell}-\ell|$ for at least one $\ell$ in $S$ .

Made in: August 2024
Personal niceness rating: 7/10
Created forwards

The starting point for this proposal is SYMO 2021 Q3 (see the proposer remarks):

I thought that was cool, so I tried to make a similar sequence. Very quickly, I had a problem statement:

(V1) Consider a sequence $a_1,a_2,\dots$ defined by $a_{n+1}=\left\lfloor\frac{a_na_{n-1}}{a_{n-1}-1}\right\rfloor$ for $n\ge 2$ . Show that for sufficiently large positive integers $n$ , $a_{n+1}=a_n+1$ .

This was quite boring, so I tried to generalise it:

(V2) Consider a sequence $a_1,a_2,\dots$ defined by $a_{n+1}=\left\lfloor\frac{a_n(a_{n-1}+k)}{a_{n-1}}\right\rfloor$ for $n\ge 2$ . Show that for sufficiently large positive integers $n$ , $a_{n+1}-a_n$ is constant.

Still, this was relatively trivial by taking out $a_n$ from the floor function. In order to prevent this, I wanted to get the current term to depend on future terms. (I also changed it into FE form.)

(V3) Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for any $n$ , $f(n+1)=\min \left\lfloor\frac{f(n)f(n+m)}{f(m)}\right\rfloor.$

I tried to solve this a while but I realised it was extremely hard (and also not nice). I wanted to focus on the interesting idea of dependence on future terms, which is what makes the sequence/function hard to control. Making it as simple as possible, I got:

(V4) Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for any $n\in \mathbb{N}$ , $f(n)=\min_{m\in\mathbb{N}} |f(n+m)-m|.$

Unfortunately, this went back to being trivial (show $f$ is strictly increasing). Hence, I tried two different adjustments to the problem:

(V5) Let $k$ be a positive integer. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for any $n\in \mathbb{N}$ , $f(n)=\min_{m\in \mathbb{N}}|f(n+km)-km|.$

(V6) Let $k$ be a positive integer. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for any $n\in \mathbb{N}$ , $f(n)=\min_{m\in \mathbb{N}}|f(n+m)-km|.$

After a while, I realised that the correct adjustment to make the problem harder was to change the range of $m$ , and not whatever I was doing. Here is my first attempt:

(V7) Let $k$ be a positive integer. Find all functions $f:\mathbb{N}\rightarrow \mathbb{N}$ such that for any $n\in \mathbb{N}$ , $f(n)=\min_{m\ge k}|f(n+m)-km|.$

How general can we go? What about:

(V8) Let $A$ be a (possibly infinite) set of positive integers. Find all $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for any positive integer $n$ ,
$f(n)=\min_{m\in A}|f(n+m)-m|.$

This turned out to be a mathematically rich structure to analyse. I spent many hours trying to classify the many pathological solutions but eventually gave up, and reluctantly added two conditions to make the question solvable: a $gcd$ condition and non-periodicity.

I submitted the problem as an FE, and the PSC made it a sequence instead. I think the sequence formulation looks more natural but it is much easier to do the algebra in FE form.

Let

I

be the incenter of

\triangle ABC

and

\omega

be its incircle. Let

D

be the intersection of

\omega

with

BC

. Suppose

X,Y

are points on

\omega

such that lines

IX,IY

and

AD

are tangent to the circumcircle of

\triangle AXY

. Prove that line

XY

bisects segment

BC

.

Made in: December 2024
Personal niceness rating: 8/10
Created backwards

I wrote a post about this one. Parts of it are copied here below:

One day I was looking at a geom solution that involves radius zero circles, and I thought it would be fun to try and construct a problem with the idea.

So what I wanted was a midpoint on a nice collinearity. Why?

If say $M$ is the midpoint of $AA'$ and also lies on line $BB'$ , then we can find a circle tangent to $AA'$ at $A$ for which $M-B-B'$ is the radical axis of this circle and the radius zero circle $A'$ . To make the problem work, $M$ has to be a less well-known point so I can conceal it.

With that in mind, I started scrolling through this handout until I saw this config:

In particular, the line $N_2 - I - M$ . This is true because it is the midline of $A-D'-X$ , a classic collinearity.

As discussed above, there exists a circle tangent to $AD$ at $A$ such that $IM$ is the radical axis of this circle and $(D)$ . What exactly is this circle? Currently, I only had one condition: that it was tangent to $AD$ at $A$ . So, I needed to find one more point on the circle.

Since $I$ is on the radical axis, the power from $I$ to the circle is $ID^2$ , and so $IX,IY$ are tangent to the circle. (This is the $X,Y$ in the original question.)

This means the two circles are orthogonal. Hence, if I marked the point $P$ where $DO$ intersects the incircle, $OA^2=OP\times OD$ , and hence $\angle APD=90^\circ$ .

But we know many properties of $P$ – for example, it passes through the point diametrically opposite $D$ on the incircle, and also passes through the extouch.

If we look at this classic problem, we are motivated to construct the circle through $M$ with radius $MD$ . So I put all of that in a diagram:

At this point, I had three mutually orthogonal circles. This gave me new ways to define $(AXY)$ . For example $X,Y,M$ are collinear and $K',K,I$ are also collinear.

I knew there was a proposal somewhere in this diagram already, and I just had to find what i wanted to ask. There were a lot of possible questions, and I tried to find the one that defines the least number of points.

I forgot exactly what I tried here, but I eventually ended up with the following which I was satisfied with:

Let $I$ be the incenter of $\triangle ABC$ and $\omega$ be its incircle. Let $D$ be the intersection of $\omega$ with $BC$ . Suppose $X,Y$ are points on $\omega$ such that $XY$ bisects $BC$ and $AD$ is tangent to the circumcircle of $\triangle AXY$ . Prove that $IX$ and $IY$ are also tangent to the circumcircle of $\triangle AXY$ .

While lying in bed, I realised that I could combine all the tangencies into one condition and ask to prove $XY$ bisects $BC$ instead, and that’s how I ended up with the actual question. As a bonus, this version had a zero radius circle solution too.

Points

B,A,C,D,E

lie on a circle in that order such that

AB\times CD=AC\times BD \text{ and }AB+CE=AC+BE.

Show that the line through

E

parallel to

BC

and the line through

D

perpendicular to

AD

intersect on the angle bisector of

\angle BAC

.

Made in: April 2026
Personal niceness rating: 10/10
Created backwards

(For this writeup, I will use the same diagrams I sent to a friend while we were making it, so the diagrams may not look nice.)

One day, we were talking about inequality questions that are actually geometry in disguise. I brought up the following question from Deux SL:

Let $a,b,c,d$ be positive reals such that
$a + c = b + d$ and $ab = 3cd$ . Determine the best constant $k$ such that
$k \cdot \frac{ac + bd}{a + b – c – d} \le \sqrt[4]{(a^2 – ab + b^2)(c^2 + cd + d^2)}.$

Rather incredibly, the setup to solve this is to introduce a bicentric quadrilateral with lengths $a,b,c,d$ . The relation $ab=3cd$ makes one of the angles $60^\circ$ by cosine rule on the diagonal, and the rest can be done using some Ptolemy.

That. Is really cool. Let’s generalise.

Consider bicentric quadrilateral $ABCD$ with $AB=a,BC=b,CD=c,DA=d$ . By Pitot, we have $a+c=d+b$ . By cosine rule, $d^2+a^2-2ad\cos(\angle A)=b^2+c^2+2bd\cos(\angle A)$ .

Combining the two equations together, we get $\frac{bc}{ad}=\frac{1-\cos(\angle A)}{1+\cos(\angle A)}$ . Classically, this equals $\tan^2(\frac{\angle A}{2})$ . At the same time, we can use (area) sine rule on $\frac{bc}{ad}$ , which is the ratio $S_{\triangle ADB}/S_{\triangle CDB}$ . This equals $d(A,DB)/d(C,DB)$ .

While the above seems like a trig bash, it is one of the coolest uses of trig I have seen. Let’s summarise what we have done above.

Starting with a triangle $ABC$ , we can construct a point $X$ such that $ABXC$ is bicentric by finding some point $D$ on $BC$ , extending $AD$ to a point such that $AD/DE=\tan^2(\frac{\angle A}{2})$ , and drawing the line through $E$ parallel to $BC$ . $X$ is then one of the intersections between this line and $(ABC)$ .

Constructing a bicentric quadrilateral is not an easy task, so there was something going on in here. Let’s remove the trig by choosing a nice point $D$ . Time for a diagram:

I realised that if $D$ is the foot of the angle bisector, and $E,F$ lie on $AB,AC$ such that $DEF\perp AD$ , then $(AEF)\cap AD$ gives a point with the desired ratio $\tan^2(\frac{\angle A}{2})$ .

I also realised that there were two intersections, $X_1,X_2$ . After a while, I discovered that $ABX_2C$ was an ex-bicentric quadrilateral, which was a cool find.

After this, I played a while with the two circles (the incircle and excircle) to see if I could get any results from there. Here was one of my images from the discussion:

(There are some non-trivial collinearities going on.)

After a while, I realised that if I let $(AEF)\cap (ABC)$ at $T$ , then $ABTC$ is harmonic. This is obvious if you know your configs, but I was a bit rusty. This allows the removal of points $E,F$ , since $G$ can be defined as the intersection of the perpendicular from $T$ to $AT$ and $AD$ .

I sent everything I had back to my friend and he came up with this version of a problem with only 6 points:

(V1) Let $ABTC$ be a cyclic harmonic quadrilateral ( $AB\le AC$ ). The line through $T$ perpendicular to $AT$ intersects the angle bisector of $\angle BAC$ at $G$ . Let $X$ be a point on the minor arc $BT$ of $(ABTC)$ . Prove that $ABXC$ has an incircle if and only if $XG\parallel BC$ .

The rest of it can be summarised in two screenshots:

It took me embarrassingly long (measured in weeks) to find a simple synthetic solution to this that didn’t go back through the trig bash route, but I got there eventually. (The solution is beautiful too.)

I love this question. Would engrave the statement on my tombstone.

Let

\triangle ABC

be an acute triangle with altitudes

AD,BE,CF

, orthocenter

H

and circumcircle

\Gamma

. Let

W (\neq A)

be the intersection of

AH

and

\Gamma

, and

T

be the intersection of

EF

and

BC

. The circumcircle of

\triangle TDW

meets

\Gamma

again at

X

and the circumcircle of

\triangle BHC

at

Y

and

Z

.

Prove that the largest of $\angle EWF$ , $\angle EXF$ , $\angle EYF$ and $\angle EZF$ equals the sum of the other three.

Made in: November 2024
Personal niceness rating: 6/10
Created backwards

One day, I noticed the following: if we take points such that $AB^2=AC^2=AD\times AE$ ( $A,D,E$ are collinear), then because of the two pairs of similar triangles, we can angle chase to get $\angle BEC+\angle BAC=\angle BDC$ . In other words, if I define $f(X)=\angle BXC$ , then $f(E)+f(A)=f(D)$ .

I realised that this fit very well into the orthocenter config, since we get $ME^2=MF^2=MB^2=MC^2$ where $BE,CF$ are heights. I thus define $f(X)=\angle EXF$ . How do I fit the config above into the diagram? Realising that $MB^2$ is the power of $M$ to $(ABC)$ , I want to draw a line through $M$ which intersects the circumcircle at two points, and then reflect one of these points over $M$ .

Still, there isn’t a proposal in here. The next step is to disguise the config. I thought it would be cool to completely remove the point $M$ by using the config twice to get two equations and removing $f(M)$ . I also realised that the reflection of $(ABC)$ across $M$ is conveniently the circle $(BHC)$ , so I drew that in.

Using the points in the final diagram above, this means that for any line $Y'W'MZ'X'$ , I get $f(X')+f(M)=f(Z')$ and $f(Y')+f(W')=f(M)$ . Removing $f(M)$ , this means $f(Z')=f(W')+f(X')+f(Y')$ .

This was interesting, but I did not want to make it any arbitrary line, else it will give away the configuration. I realised I could transfer the angles by doing an inversion at $T$ , which removes the line altogether.

Finally, to disguise the fact that any circle through $T$ and $D$ works, I decided to fix $W$ as well. On hindsight, I’m not sure if this was the correct thing to do.

I figured this problem should never appear on any contest for the sanity of the contestants, so I used it for VT. I am quite happy with the phrasing of the problem, it looks goofy and a bit unhinged.

Let

n

be a positive integer and

\{d_1,d_2,\dots,d_k\}

be the set of all positive divisors of

n

. Prove that for any positive integer

m

for which

\gcd(m,k)=1

,

d_1+d_2+\dots+d_k\mid d_1^m+d_2^m+\dots+d_k^m.

Made in: September 2024
Personal niceness rating: 7/10
Created backwards

The starting point for this proposal is this problem from Iran MO 2024:

For a given positive integer number $n$ find all subsets $\{r_0,r_1,\cdots,r_n\}\subset \mathbb{N}$ such that $n^n+n^{n-1}+\cdots+1 \mid n^{r_n}+\cdots+ n^{r_0}.$
For this question, the key is to realise that you can replace $r_i$ with $r_i\pmod{n+1}$ , as $n^{n+1}-1$ is a multiple of the LHS. Hence, if your $r_i$ s are a CRS modulo $n+1$ , the divisibility holds.

What happens if we force $r_i=i\cdot k$ ? Then, the statement holds when $\gcd(k,n+1)=1$ . This gives an interesting fact: $\gcd(k,n+1)=1$ implies $\frac{m^{n+1}-1}{m-1}\text{ is divisible by } \frac{m^{k(n+1)}-1}{m^k-1}.$ This is reminiscent of the sum of divisor formula.

In fact, taking $m$ as primes and multiplying these divisibilities together, we get the original statement. The individual $gcd$ conditions nicely multiply together as well.

At this point, the problem is still pretty trivial. I tried two ways to make it harder.

First, the divisibility and the gcd condition is not if and only if. There are exceptions where the divisibility can hold and the gcd is not 1. I tried to study these exceptions but they were very hard to control (because primes).

Second, I tried to make it into an analytic question. For example, what is the smallest $k$ that makes the divisibility true? This is upper bounded by the smallest $k$ satisfying $\gcd(m,k)=1$ , which is around $\ln(m)=\ln(d(n))$ . While this is definitely harder as it hides the gcd condition, it also looks ugly.

Hence, when we were finding easy Q1s for CWMIST, I submitted the original version of the question. I firmly believe it is way harder to make good easy questions than good hard questions. Put another way, it is trivial to make easy and trivial problems but non-trivial to make easy and non-trivial problems.

Consider a polynomial

P(x)

with integer coefficients. For any positive integer

n

, define

f(n)

to be the number of coefficients in

P(x)^n

that are not multiples of

67

. Show that for all sufficiently large

n

, at least

n^{0.99}

of the numbers

f(1),f(2),\dots,f(n)

are perfect

2026

th powers.

Made in: February 2026
Personal niceness rating: 9/10
Created backwards

I was scrolling through the IMO Longlist for reasons I cannot remember, and I stumbled upon the following problem from the 1988 ILL:

Let $n$ be a positive integer. Find the number of odd coefficients of the polynomial $u_n(x) = (x^2 + x + 1)^n.$

The solution uses the fact that $(x^2+x+1)^{2^k}\equiv x^{2^{k+1}}+x^{2^k}+1\pmod{2}$ by Fermat Little Theorem. Hence, we can write $n$ in binary and expand it (similar to fast exponentiation algorithm).

I tried to do this for arbitrary primes and polynomials. If we let $p$ be a prime and $k=a_tp^t+a_{t-1}p^{t-1}+\dots+a_1p+a_0$ (base $p$ representation), then $P(x)^k\equiv \prod_{i=0}^tP(x)^{a_ip^i}\equiv \prod_{i=0}^tP\left(x^{p^i}\right)^{a_i}\pmod{p}.$ It is not very easy to simplify this further, because when you fully the expand the RHS, some terms have the same degree. But what if I force them to not do that?

The easiest way to make $a_i$ small compared to $p$ . How small before no two terms in the expansion have the same degree? We need $a_i\le\frac{p}{deg(P)}$ .

At this point, I already had a problem somewhere, I just needed to figure out how to ask it. Firstly, what do I do after the products do not interact? The simplest way is to talk about perfect squares, since each digit just needs to be even.

After that, my priority was to hide the key fact (importance of size of $a_i$ ). I thought the best way to do it is to count the number of such $k$ (such that all base $p$ digits is $\le\frac{p}{deg(P)}$ ). To my slight surprise, there were many such numbers, larger than any $\mathcal{O}(n^{1-\varepsilon})$ , as long as $n$ and $p$ are large enough.

I sent this to a friend to testsolve, and he found out that if $p$ is too small, the same argument can be done with a sufficiently large power of $p$ . After a bit of cosmetic adjustment, this leads to the final problem statement. My intention is that a solver who first looks at the problem will think “wow that’s a lot of 2026th powers”.