An arc is defined as a homeomorphism from an interval to $X$ (and its range).
Let $x, y \in X$ be any two points in an arc $J$. Then $[x, y]$ denotes the subset of the arc from $x$ to $y$.
We say $J$ is a quasi-arc if there is a number $L \geq 1$ such that $ diam ([x, y]) < L \cdot d(x, y) $ for all $x, y \in J$. This is a classical concept from geometry of metric spaces.
We propose to generalize this notion into that of coarse-arc as follows. $J$ is a $f-$ coarse arc if there exists a function $f : \mathbb{R} \to \mathbb{R} $ such that $ diam ([x, y]) < f (d(x, y)) $ for all $x, y \in J$
Moreover we say a metric space $X$ has an intrinsic curvature $f$ if any arc $J$ in $X$ can be approximated by a $f$-coarse arc and cannot be approximated by any $f’$-coarse arc such that $f’ < f$.
I sometimes create small problems (mostly at pre-college level), to have fun. Here are three recent ones. Readers may give it a try or point out issues with problem-statement or indicate that it is trivial.
Please note that some version of these problems may exist somewhere as there is a tendency of familiar ideas to echo through the brain. May be I have read something like this somewhere in the past. At any rate, this is just some musing for fun.
Problem 1
A connected, directed graph is called a ‘Calcutta Graph’ if degree of every vertex is 3, each vertex has exactly two incoming edges and one outgoing edge.
Suppose G is a Calcutta Graph with n vertices. Then at most how many directed circuits can G have?
Problem 2
A $n \times n$ grid is made up of $n^2$ squares. In each square you are allowed to draw one diagonal. If $k$ diagonals line-up, end-to-end, then they make a path of length $k$. The figure below shows a $4 \times 4$ grid with a path of length $6$.
Find the maximum number of paths of length $2n$ in a $n \times n$ grid.
Problem 3
Suppose $n$ circles and $n$ (infinite) lines are drawn in the plane such that
No three lines pass through a single point
No three circles pass through a single point
Every pair of circles intersect each other at two distinct points
Every pair of lines intersect each other at one point
Every line cuts every circle at two distinct points.
The graduate-school days are zooming away quickly from my life. It seems that the space of human memory is hyperbolic in nature. Things get thin and small at an exponential rate. I defended my thesis in July 2020 and reached India in August of the same year. The pandemic was in full swing. It was a conscious choice to return to my aging family who needed support.
Almost all of 2021 was spent on the paper that Chris and I were working on. It is an extension of the results in my doctoral thesis. We showed that connected boundary of a relatively hyperbolic group is locally connected. This removed some of the tameness restrictions that Bowditch’s theorem has on the peripheral subgroups. At the end of 2021, my advisor suggested that I should work on some projects alone.
For a few days, I felt like a radarless ship in the ocean of mathematics. Since I was not associated with any university at the time, research had to be a solo adventure. I decided to build a small research group at Cheenta. It is the organisation that I developed from scratch since 2010.
Cheenta was conceived as a training school for math olympiads for school students. Subsequently we have also accepted college students for university level programs. We already have a strong alumni and student base spread all around the world. I could easily get a few people who became curious about geometric group theory.
We started meeting weekly. In order to keep a psychological leverage, I put the meeting time on Tuesdays at 10:30 PM IST or 11 AM CST. In graduate school, that was the time when I met my advisor weekly. My brain-clock responded to this procedure and a group of 7 students was assembled for weekly adventures in geometric group theory.
2022 was also productive. I managed to prove a small theorem related to Dehn fillings and connectedness of Bowditch boundary. The entire team participated in a translation project of the famous green book by Ghys and Harpe from French to English. I also started collaborating with Arka Banerjee on another problem related to embedding of hyperbolic plane in relatively hyperbolic groups.
I hope 2023 will be productive. I want to understand how small cancellation theory, and splittings of relatively hyperbolic groups interact. It could be a powerful source of examples in group theory. Another area that interests me is the theory of hierarchically hyperbolic groups and spaces.
There are a few obstacles for my research activities. Books and journals are not easily available outside the university system. Access to conferences is hard. I was invited to speak at a conference in Ohio (to be held in April). However due to VISA and funding issues I was forced to decline the offer. There are few positive ends as well. My work at Cheenta allows me to have flexible work-hours and financial security. It also helps me to stay with my family at home.
Lets hope 2023 will be productive with what I have.
This is an ongoing survey of olympiad problems. Source material is INMO, USAMO and IMO. The goal is to indicate key ideas involved in the proof.
USAMO 2010 Problem 1
Let $A X Y Z B$ be a convex pentagon inscribed in a semicircle of diameter $A B$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $A X, B X$, $A Z, B Z$, respectively. Prove that the acute angle formed by lines $P Q$ and $R S$ is half the size of $\angle X O Z$, where $O$ is the midpoint of segment $A B$.
Key Concepts
Angle at centre is twice the angle at circumference
External angle in a cyclic quadrilateral is equal to the interior opposite angle
Simson Line
Cyclic Pentagon
Idea of Proof
If we use Simson line (feet of perpendiculars from a point in the circumference, in this case $Y$, to three sides of a triangle (in this case $\Delta A X B$ and $\Delta A Z B$) are collinear), then we immediately conclude that $PQ$ and $RS$ meet at a point $M$ on the diameter $AB$. Here is part of the picture.
Next notice that we have a cyclic pentagon $A P Y R M$. Now angle chasing shows $\angle Z M Y = \angle Z A Y$ and so on. This proof is suggested in Evan Chen’s notes.
Alternatively we may do angle chasing. Here the key idea is to show $M C Y D$ is a cyclic quadrilateral.
INMO 2014 Problem 5
In an acute-angled triangle $A B C$, a point $D$ lies on the segment $B C$. Let $O_1, O_2$ denote the circumcentres of triangles $A B D$ and $A C D$, respectively. Prove that the line joining the circumcentre of triangle $A B C$ and the orthocentre of triangle $O_1 O_2 D$ is parallel to $B C$.
Key Concepts
Similarity of triangles
Cyclic pentagon
Angles at orthocenter
Angle at the centre of a circle is twice the angle at circumference
Idea of Proof
Notice the $H$ and $O$ are orthocenter and circumcenter of two different triangles $\Delta O_1 O_2 D$ and $\Delta ABC$. One recurring theme in geometry is to bind special points in different triangles via cyclic pentagons. A major step in this problem is to identify $A O_1 H O O_2$ as a cyclic pentagon. In fact notice that $\Delta A O_1 O_2$ is similar to $\Delta A B C$ is similar to $O_1 O_2 D$.
INMO 2016 Problem 1
Let \( ABC \) be triangle in which \( AB = AC \). Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio \( \frac{AB}{BC} \).
Idea of proof
Suppose \( H \) is the orthocenter and \( I \) is the incenter. Suppose \(AD\) is the perpendicular cum angle bisector passing through \( H \) and \( I\) meeting \( BC \) at \( D \).
Then \( \tan \frac{B}{2} = \frac{r}{BD} \), the inradius.
We can also show \( \tan \frac{A}{2} = \frac{2r}{BD} \).
Since \( A = \pi – 2B \), doing some trigonometry we obtain the value of \( \cos B \). The final answer is equivalent to \( \frac{1}{2 \cos B } \) which turns out to be \( \frac {3}{4} \).
INMO 2014 Problem 1
In a triangle \( ABC \), let \(D\) be a point on the segment \(BC\) such that \(AB + BD = AC + CD\). Suppose that the points \(B, C\) and the centroids of triangles \(ABD\) and \(ACD\) lie on a circle. Prove that \(AB = AC\).
Key Concepts
Parallel lines divide sides in equal ratios
Apollonious Theorem
Triangular Inequality
Centroid divides median in \( \frac{2}{1} \) ratio.
Proof Idea
Suppose \( T \) is the midpoint of \(AD\). If \(G_1\) and \( G_2 \) are the two medians, we show that \(G_1 G_2\) is parallel to \( BC \).
Then \(B G_1 G_2 C\) is a cyclic trapezium. This implies \(BT = CT\)
Now apply Apollonious theorem to deduce that \(AB^2 + BD^2 = AC^2 + CD^2\)
Finally notice that \(AB + BD = AC + CD\) from given hypothesis. Transposing we get \(AB^2 – AC^2 = CD^2 – BD^2\). Hence either \(CD – BD\) cancels out with \(AB – AC \) provided both the non-zero, or we have \(AB + AC = BD + CD\). The second case is impossible as it violates triangular inequality. Hence the first case holds.
