Get inspired by the success stories of our students in IIT JAM 2021. Learn More

For Students who are passionate for Mathematics and want to pursue it for higher studies in India and abroad.

This is a beautiful sample problem from ISI MStat PSB 2010 Problem 2. It's all about how many isosceles triangle with sides of integers lengths one can construct under certain conditions. We provide a detailed solution with prerequisites mentioned explicitly.

Find the total number of isosceles triangles such that the length of each side is a positive integer less than or equal to 40.

(Here equilateral triangles are also counted as isosceles triangle.)

- Basic counting principles.
- Triangle inequality.

Let the sides of the isosceles triangle are k, k and l units, (length of the equal sides are k units)

So, our problem is all about finding triplets of form (k,k,l), where \(l, k \le 40\) . And how many such triplets can be found !

also since l,k and k are also sides of an (isosceles) triangle, so by triangle inequality,

\(l < 2k\) . Since l is an integer so, \(l \le 2k-1\)

So, we have \(l\le 40\) and \(l \le 2k-1\). combining we have \(l \le min(40, 2k-1)\) .

now let us define, \( l_k \) : possible number of l's for a given k =1,2,....,40

here we must consider two cases.

**Case-1 **: when \(2k-1 \le 40 \Rightarrow k \le \lceil \frac{40+1}{2} \rceil = 20 \)

So, l can be any integer between 1 and 2k-1, since \( l\ \le 2k-1\) when \( k \le 20\)

so,there can be 2k-1 choices for the value of each k, when k=1,...,20.

Or, \( l_k = 2k-1 \) for k=1,2,....,20.

now, **Case-2** : when 2k-1>40 \( \Rightarrow k > \lceil \frac{40+1}{2} \rceil = 20 \)

so, here l can take any integer between 1 to 40. So, there 40 choices for l for a given k, when k=21,22,...,40.

Or, \( l_k = 40 \) for k=21,22,....,40.

Combining** Case-1 **and **Case-2,** we have,

\(l_k = \begin{cases} 2k-1 & k=1,2,...,20 \\ 40 & k=21,22,...,40 \end{cases}\)

So, finally, let all possible number of triplets of form (k,k,l) are = \(T_40\)

So, \( T_{40} \) = \( \sum_{k=1}^{40}{l_k} \) = \( \sum_{k=1}^{20}{(2k-1)}\) + \( \sum_{k=21}^{40}{40} \) =\( (20 )^2 + 40 \times (40-20)=400+800= 1200. \)

So, \(T_{40}\) = 1200. hence we can find 1200 such isosceles triangles with sides of integer lengths such that the length of each sides are less than or equal to 40.

Can you generalize this problem? i.e. can you find How many isosceles triangle one can construct, with sides of integer lengths, such that the sides are less than or equal to any integer N ? Can you find an elegant formula to express \(T_N \), for any integer N?

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIAL
For general $N \in \Bbb N$ we have $$T_N = \left ( \left \lfloor \frac {N+1} {2} \right \rfloor \right )^2 + N^2 - N \left \lfloor \frac {N+1} {2} \right \rfloor.$$

yes you are correct.