### LMAO Revenge

Continuing the tradition of past years, our seniors at the Indian IMO camp(an unofficial one happened this year) once again conducted LMAO, essentially ELMO but Indian. Sadly, only those who were in the unofficial IMOTC conducted by Pranav, Atul, Sunaina, Gunjan and others could participate in that. We all were super excited for the problems but I ended up not really trying the problems because of school things and stuff yet I solved problem 1 or so did I think.

Problem 1: There is a $2022 \times 2022$ grid of real numbers. In a move, you can pick any real number $c$, and any row or column and replace every entry $x$ in it with $c-x$. Is it possible to reach any grid from any other by a finite sequence of such moves?

It turned out that I fakesolved and oh my god I was so disgusted, no way this proof could be false and then when I was asked Atul, it turns out that even my answer was wrong and he didn't even read the proof, this made me even more angry and guess what? I was not alone, Krutarth too fakesolved the problem and a few others did too.

Problem 1: Find all pairs of integers $x$ and $y$ such that they satisfy $$y^{42068(\varphi{X}+\varphi{Y})+1} = x^2+763$$ where $\varphi{K}$ is the number of problems contributed by $K$ in LMAO.

Solution: Hmm, so before we encounter this problem, we must first identify who are $X$ and $Y$. Notice that there are only three people in the story who proposed problems to LMAO namely Atul, Pranjal and Mahavir, so $A$, $X$ and $Y$ is some permutation of them, now notice how $X$ smiled when he saw geometry which didn't even look nice, I mean come on, it was a geometric inequality, so $X$ has to be a super passionate geo main and we know that Mahavir always proposes geo and is a very good geometer. Also notice how Pranav remarks "else who is gonna troll $Y$ about how he solved $69$ generalizations of $PZ$. We know that Atul didn't solve the P4 at IMO which was the only geometry in the contest, and thus we can conclude that $X$ and $Y$ must have been Mahavir and Atul respectively. Now scrolling through Atul's AoPS post, its clear that $\varphi{X}+\varphi{Y}=6$!! As you might have figured out already, this constant we get at the top is absolutely random and all we needed was to make it $1 \pmod{4}$.

We claim that there are no solutions, we get that $\mathbb{Q}(\sqrt{-763})$ has class number $4$. Lets assume $763=d$ and then $y^{4k+1}=(x+\sqrt{-d})(x-\sqrt{-d})$. So the ideals $(x+\sqrt{-d})=I^{4k+1}$ for some ideal $I$ and $(x-\sqrt{d})=J^{4k+1}$ for some ideal $J$. Now since $\mathbb{Q}(\sqrt{-d})$ has class number $4$, we have that $I^4$ is principal and hence so is $I^{4k}$. But $(x+\sqrt{-d})$ is principal too, so $I$ is infact principal. So we get that $$x+\sqrt{-d}=(a+b\sqrt{-d})^{4k+1}$$ and $$x-\sqrt{-d}=(a-b\sqrt{-d})^{4k+1}$$ so subtracting you get that $$\frac{(a+b\sqrt{-d})^{4k+1}-(a-b\sqrt{-d})^{4k+1}}{2 \sqrt{-d}}=1$$.  But then $a=1$. Suppose $$a_n=\frac{(a+b\sqrt{-d})^{n}-(a-b\sqrt{-d})^{n}}{2 \sqrt{-d}}$$ we have the recurrence $$a_{n+2}=2a\times a_{n+1}-(a^2+db^2)a_n$$ so we can show $a_n$ is   increasing thus $4k+1 >2$ and $a_2=1$ since $a_2 \le a_{4k+1}$ but $a_i$ are positive integers as well since $a,b$ are positive integers so now just note that $a_2=ab$, so $a=b=1$ that gives $x^2+d=(1+d)^{4k+1}$, now taking modulo $8$, $8 \mid x^2+3$ which is impossible. $\blacksquare$

Now let's look at problem 4 which was proposed by Krishna!!

Problem 4: In a non-equilateral triangle ABC, let the distance from $A$ to the euler line be $\psi(A)$, prove that $$\psi(A) \leq \sqrt{\frac{(b^2+c^2)^2}{4(a^2+b^2+c^2)}}$$

Sketch: We consider the reflection of $A$ across the euler line, then we have $$AP^2+BP^2+CP^2=AA^2+AB^2+AC^2=AB^2+AC^2$$ because for any $D$, we have that $AD^2+BD^2+CD^2$ is some linear function of $GD^2$ where $G$ is the centroid(this is not hard to prove say with coordinates oops). Then by Ptolemy's theorem, we have that $$AP\times BC =BP \times AC+ CP \times AB$$

I'll leave the finish to the reader which is an application of Cauchy's Inequality!!

Remark: The equality case of this inequality is quite surprising and cool, equality holds if and only if the foot of the perpendicular from $A$ to the euler line is the centroid!!

Now its time to discuss the only not so horrendous looking problem from LMAO Revenge which was proposed by Ananda!!!

Problem 5: If $k < n$ squares are deleted from an $n \times n$ grid, find in terms of $n$ and $k$ the smallest possible size of the largest connected component.

Solution: The following solution and writeup is due to Ananda,

The answer is $n^2-\frac{k(k+1)}{2}$. To achieve this, delete the $k$ squares of a diagonal which has exactly $k$ squares.

Since $k<n$, there is a row $R$ and a column $C$ of the grid which do not have any deleted squares. We claim that the connected component which contains $R$ and $C$ has at least $n^2-\frac{k(k+1)}{2}$ cells. Suppose $m$ is a cell which has not been deleted but is not part of this connected component. Then there is a deleted cell $m_1$ which is in the same column as $m$ and between $m$ and row $R$ and a deleted cell $m_2$ which is in the same row as $m$ and between $m$ and column C. Assign to $m$ the unordered pair $(m_1,m_2)$ (if you have a choice, assign one arbitrarily). If two cells have the same pair assigned to them,  then they must be reflections in the line joining this pair of cells.  It is easy to see that the reflected square cannot be blocked by the same pair of cells. Since no two cells which are not part of this connected component can have the same pair assigned to them, there are at most $\binom{k}{2}$ undeleted cells which are not part of this connected component. $\blacksquare$

And now comes the best problem on the test which had the most submissions but yet had no correct solutions proposed by Anurag!!

Problem 6: What is the maximum number of knights that can be placed on a regular chessboard such that no knight attacks another knight?

Solution: HAHA trivial problem, its 32 right?? Just place all the knights on squares with the same color and proving the upper bound is not hard either right? No.

The answer is $64$, ever seen two white knights attack each other? Right, just place $64$ knights of the same color on the chessboard. $\blacksquare$

Finally, it's time for problem 9 which also as well had $0$ solves!!! Who could have done this to Mahavir? Whose plan was it?

Well, it was none other than Atul, no one else could have done something this morbid to anyone other than Atul, also after all it was LMAO revenge and we make the organizing committee the villains, right? Well that was what Atul thought, it turns out the actual person behind all this was none other than Krutarth, now that you know the answer, can you figure out the details which point out to him? Tell me in the comments or in the OMC discord lounge!! Any other new theories are welcome as well!!! I hope you liked this blogpost!!!

-Debayu

Hii everyone! Today I will be discussing a few geometry problems in which once you "guess" or "claim" the important things, then the problem can easily be finished using not-so-fancy techniques (e.g. angle chasing, power-of-point etc. Sometimes you would want to use inversion or projective geometry but once you have figured out that some particular synthetic property should hold, the finish shouldn't be that non trivial) This post stresses more about intuition rather than being rigorous. When I did these problems myself, I used freehand diagrams (not geogebra or ruler/compass) because I feel that gives a lot more freedom to you. By freedom, I mean, the power to guess. To elaborate on this - Suppose you drew a perfect  diagram on paper using ruler and compass, then you would be too rigid on what is true in the diagram which you drew. But sometimes that might just be a coincidence. e.g. Let's say a question says $D$ is a random point on segment $BC$, so maybe