Cheenta Reading Room

Outstanding problems, discussion and more

Leibniz Rule, ISI 2018 Problem 4

The Problem Let \(f:(0,\infty)\to\mathbb{R}\) be a continuous function such that for all \(x\in(0,\infty)\), $$f(2x)=f(x)$$Show that the function \(g\) defined by the equation $$g(x)=\int_{x}^{2x} f(t)\frac{dt}{t}~~\text{for}~x>0$$is a constant function. Key Ideas One...

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Functional Equation – ISI 2018 Problem 3

The Problem Let \(f:\mathbb{R}\to\mathbb{R}\) be a continuous function such that for all \(x\in\mathbb{R}\) and for all \(t\geq 0\), $$f(x)=f(e^tx)$$Show that \(f\) is a constant function. Key Ideas Set \( \frac{x_2}{x_1} = t \) for all \( x_1, x_2 > 0 \). Do the same...

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Power of a Point – ISI 2018 Problem 2

The Problem Suppose that \(PQ\) and \(RS\) are two chords of a circle intersecting at a point \(O\). It is given that \(PO=3 \text{cm}\) and \(SO=4 \text{cm}\). Moreover, the area of the triangle \(POR\) is \(7 \text{cm}^2\). Find the area of the triangle \(QOS\). Key...

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Geometry of Algebra

It is interesting to represent Algebraic Identities using Geometric objects. One easy example is presented in this video. https://youtu.be/ZCRV-cvTqIc Think! Try to draw the following algebraic identities using geometric objects \(a^2 - b^2 = (a+b)(a-b) \) \( (a+b)^3...

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Injection Principle – Combinatorics

The central goal of Combinatorics is to count things. Usually, there is a set of stuff that you would want to count. It could be number of permutations, number of seating arrangements, number of primes from 1 to 1 million and so on. Counting number of elements in a...

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লিনিয়ার বীজগণিত (বাংলায় আধুনিক গণিত)

লিনিয়ার বীজগণিত নিয়ে আমরা একটি ভিডিও সিরিজ তৈরী করছি। 'চিন্তা'-র কলেজ গণিত প্রোগ্রামে যদিও প্রধানত ইংলিশে আলোচনা হয়, আমরা চেষ্টা করি বিভিন্ন আঞ্চলিক ভাষা গুলোতে কিছু আলোচনা করতে। https://youtu.be/W7eAu8LCbik   পরবর্তী আলোচনা গুলো খুব আসছে এই...

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Orthocenter and equal circles

Orthocenter (or the intersection point of altitudes) has an interesting construction. Take three equal circles, and make them pass through one point H. Their other point of intersection creates a triangle ABC. Turns out, H is the orthocenter of ABC. In this process,...

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