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I.S.I. and C.M.I. Entrance

An isosceles triangle,on Trigonometry, I.S.I Entrance 2016, Solution to Subjective problem no. 6

Understand the problem

Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. If \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\), then prove that the triangle is isosceles.

Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 6 from 2016 
Topic
Trigonometry

Difficulty Level
7 out of 10

Suggested Book
 Plane Trigonometry by S.L. Loney

Start with hints

Do you really need a hint? Try it first!

Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. Given \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\)    

\(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})\) \(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\)  

\(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\) \(\Rightarrow  b\sin A\cos B=a\cos A \sin B\) \(\Rightarrow  (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \)

 

\(\Rightarrow  (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}\)  [since \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)] \(\Rightarrow \frac{\cos B}{\cos A}=1\) \(\cos B=\cos A\) \(A=B\).

\(\Rightarrow \Delta ABC\) is isosceles  (Proved).

Connected Program at Cheenta

I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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I.S.I. and C.M.I. Entrance

Four Points on a Circle, ISI Entrance 2017, Subjective Problem no 2

Understand the problem

 Consider a circle of radius 6 as given in the diagram below. Let \(B,C,D\) and \(E\) be points on the circle such that \(BD\) and \(CE\), when extended, intersect at \(A\). If \(AD\) and \(AE\) have length 5 and 4 respectively, and \(DBC\) is a right angle, then show that the length  of \(BC\) is \(\frac{12+\sqrt{15}}{5}\).

Diagram : click here.

Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017
Topic
Plane Geometry

Difficulty Level
6 out of 10

Start with hints

Do you really need a hint? Try it first!

Given, \(AD=5, AE=4\) and \(\angle DBC=90^\circ \).

As \(D,B,C\) are points on the circle having radius 6.

Therefore \(DC\) is the diameter of the circle

\(\Rightarrow DC=6×2=12\).

Now  DC is diameter of the circle \(\Rightarrow \angle DEC=90^\circ \).

Therefore \(\angle DEA\) is also right angle.

The length of \(DE=\sqrt{5^2-4^2}=3\). 

And the length of \(EC=\sqrt{12^2-3^2}=3\sqrt{15}\).

Therefore \(AC=AE+EC=4+3\sqrt{15}\).

From \(∆AED\) and \(∆ABC\) we have,

\(\frac{AD}{DE}=\frac{AC}{BC}  \Rightarrow  BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}\).

Therefore the length of \(BC\) is \(\frac{12+9\sqrt{15}}{5}\).   (Ans.)

Watch the video ( Coming Soon … )

Connected Program at Cheenta

I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Product of Digits, ISI Entrance 2017


Understand the problem

  Let \(g : \mathbb{N} \to \mathbb{N} \) with \( g(n) \) being the product of digits of \(n\).

       (a) Prove that \( g(n)\le n\) for all \( n \in \mathbb{N} \) .

       (b) Find all \(n \in \mathbb{N} \) , for which \( n^2-12n+36=g(n) \).

Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 5 from 2017
Topic
Inequality

Difficulty Level
9 out of 10

Suggested Book

Challenge and Thrill of Pre-College Mathematics by V.Krishnamuthy , C.R.Pranesachar, ect.

Start with hints

Do you really need a hint? Try it first!

Let \(n\) be a \(k\) digit(s) number number , then \(n\) can be written as

                            \(n=a_0+10a_1+10^2a_2+\cdots+10^{k-1}a_{k-1}\)

Where ,\(a_0,a_1,…,a_{k-1}\) are digits of \(n\).

\(a_0,a_1,…,a_{k-1} \in [1,9] \) as the range of the function \(g\) is \(\mathbb{N}\)

\(\Rightarrow a_0,a_1,…,a_{k-1}\neq 0\) .

Now \(g(n)=a_0a_1a_2\cdots a_{k-1}\le \underbrace{10\cdot10\cdot10\cdots 10}_{(k-1) times}  \cdot a_{k-1}\)   [Since \(a_0,a_1,…,a_{k-1} \le 10 \)] 

Equality holds when \(k=1\) .

 

\(\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}\)

\(\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}+\cdots+10^2a_2+10a_1+a_0\)    [Since \(a_0,a_1,…,a_{k-1} >0\)] 

\(\Rightarrow g(n)\le n\)        (Proved) .

 

\(n^2-12n+36=g(n)\)

\(\Rightarrow n^2-12n+36 \le n\)     [Since \(g(n) \le n\) ]

\(\Rightarrow n^2-13n+36 \le 0\)

\(\Rightarrow (n-9)(n-4) \le 0\)

\(\Rightarrow 4\le n\le 9\)

\(\Rightarrow n={4,5,6,7,8,9}\) (Ans.)  . 

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Three Primes, ISI Subjective Entrance 2017

Understand the problem

Let \(p_1,p_2,p_3\)  be primes with \(p_2\neq p_3\), such that \(4+p_1p_2\) and \(4+p_1p_3\) are perfect squares. Find all possible values of \(p_1,p_2,p_3\). 

 

Start with hints

Do you really need a hint? Try it first!

Let \(4+p_1p_2=m^2\) and \(4+p_1p_3=n^2\), where \( m,n \in \mathbb{N}\).

\(\Rightarrow p_1p_2=(m-2)(m+2)\) and \(p_1p_3=(n+2)(n-2)\).

Since \(p_1,p_2,p_3\) are primes with \(p_2\neq p_3\) \(\Rightarrow m\neq n\).

 Case –I: \(p_2<p_3   \Rightarrow m<n\).

Clearly, \(p_1=m+2=n-2,    \Rightarrow n=m+4\)

             \(p_2=m-2\)    and 

              \(p_3=n+2=m+6\).

Therefore, \((m+2),(m-2),(m+6)\) are all prime numbers.

              We see that \(m=5\) satisfy the above condition. And then,\( p_1=7,p_2=3,p_3=11\) .

 

 

Case-II: \(p_2>p_3    \Rightarrow   m>n\).

\(\Rightarrow   p_1=m-2=n+2     \Rightarrow     m=n+4\)

                       \( p_2=m+2=n+6\)   and 

                        \(p_3=n-2\).

Now \((n+2),(n+6), (n-2)\) all are primes. Again , \(n=5\) satisfy this condition. Hence \(p_1=7,p_2=11,p_3=3\).

 

Thus all possible values of \(p_1,p_2,p_3\) are ( 7,3,11)  and (7,11,3).

Now need to conclude that there does not exist any more triple of prime numbers satisfying the given condition.

Consider these numbers:

           \(p_1=m+2,p_2=m-2,p_3=m+6\) , now the gaps between \(p_1,p_2,p_3\) are given by:

                \(p_1-p_2=4,   p_3-p_2=8  \)  and  \(p_3-p_1=4\).

We see that for \(m>9\) these three gaps cannot be 4,8 and 4 simultaneously . That is at least one of these three gaps is greater than 4 for \(m>9\) . And between 1 to 9 only \(m=5\) satisfy the given condition. Hence there does not exist any more triples. 

 

Connected Program at Cheenta

Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2017. Subjective Problem no. 6.

Topic
Number Theory

Difficulty Level

8.5 out of 10

I.S.I. & C.M.I. Entrance Program

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System of n equations of Real Analysis , I.S.I Entrance 2008, Solution to Subjective Problem No. 9.

Understand the problem

 For \(n\ge3 \), determine all real solutions of the system of \(n\) equations :

                                               \(x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\)

                                                                 ………………………….

                      \(x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n-1}+x_n=\frac{1}{x_i}\)

                                                                 …………………………..

                                               \(x_2+x_3+\cdots+x_{n-1}+x_n=\frac{1}{x_1}\).

Source of the problem

I.S.I. (Indian Statistical Institute) B.Math.(Hons.) Entrance Examination 2008. Subjective Problem no. 9.

Topic

System of \(n\) Equations

Difficulty Level

9 out of 10

Suggested Book

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

Start with hints

Do you really need a hint? Try it first!

 Let \( s = x_1+x_2+\cdots+x_i+\cdots +x_{n-1}+x_n\).

Then the system of \(n\) equations are equivalent to \(x_i^2-sx_i+1=0\) for \(i=1,2,….,n\).

It follows that \(x_1,x_2,…..,x_{n-1},x_n\) are solutions to the quadratic equation: \(u^2-su+1=0\).                                                   \(\cdots\cdots\cdots\cdots (i)\)

Now we have two possible cases.

Case – I :  Two roots of the equation (i) are equal, then , all \(x_i\) are equal. 

i.e. , \(x_1=x_2=\cdots=x_{n-1}=x_n=u\).

\(\Rightarrow s=nu\).

Putting this value of \(s\) in equation (i) we get, \((n-1)u^2=1\).

\(\Rightarrow u=\frac{1}{\pm \sqrt{n-1}}\).

Case – II : Two roots of equation (i) are not equal. Then let these two roots are \(u_1 \) and \(u_2\). Also let among {\(x_1,x_2,……,x_{n-1},x_n\)}   \( k\)  of them equal to \(u_1\) and \((n-k) \) of them equal to \(u_2\), where \(0<k<n\).

      In this case, we have , (a) \(u_1+u_2=s\).    [Sum of roots of equation (i)]

                                              (b) \(u_1\cdot u_2=1\).    [ Product of roots of equation (i)].

Now,.   \(s=x_1+x_2+\cdots+x_{n-1}+x_n=k\cdot u_1+(n-k)\cdot u_2\)

        \(\Rightarrow s=(u_1+u_2)+(k-1)\cdot u_1+(n-k-1)\cdot u_2\) 

        \(\Rightarrow (k-1)\cdot u_1+(n-k-1)\cdot u_2=0\)   [using (a)]

        \(\Rightarrow  (k-1)\cdot u_1^2=(k+1-n)\cdot u_1\cdot u_2=k+1-n\le 0\) [using (b) and since ,\( k<n\Rightarrow k+1\le n\)].

      \(\Rightarrow u_1^2\le 0\)

But \(u_1\neq 0\) as the coefficient of \(u^0\) is 1(\(\neq 0\)) in equation (i).

        \(\Rightarrow u_1^2<0\)

\(\Rightarrow u_1\) is not real. So we have no real solution to the system of \(n\) equations in this case.

 

 

 

Thus the total no. of real solutions to the system of \(n\) equations is two and these two solutions are:

                        \(x_1=x_2=\cdots=x_{n-1}=x_n=\frac{1}{\sqrt{n-1}}\)      and

                        \(x_1=x_2=\cdots=x_{n-1}=x_n=-\frac{1}{\sqrt{n-1}}\).(Ans.)

Connected Program at Cheenta

I.S.I. & C.M.I. Entrance Program

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Trigonometric Substitution, ISI Entrance 2019 – Problem 6

Understand the problem

 For all natural numbers\(n\), let 

        \(A_n=\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}\)           (\( n\) many radicals)

(a) Show that for \(n\ge 2,  A_n=2\sin \frac{π}{2^{n+1}}\).

(b) Hence, or otherwise, evaluate the limit

                             \(\displaystyle\lim _{n\to \infty} 2^n A_n\) .

Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 6.
Topic
Trigonometric Substitution 

Difficulty Level
7.5 out of 10

Start with hints

Do you really need a hint? Try it first!

For \(n\ge 2\),

\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}\)       (\( n\) many radicals)

=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+0}}}\)

=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+2\cos \frac{π}{2}}}}\). 

 

=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2(1+\cos \frac{π}{2})}}}\)

=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{4\cos^2 \frac{π}{2^2}}}}\)

=\(\sqrt{2+\sqrt{2+\cdots +2\cos \frac{π}{2^2}}}\). (\(n-1\) many radicals)

     ……..    ………..     ……….      …   …….  …

     …….     ………     …………    .….     ………

=\(2\cos \frac{π}{2^n}\)        \([n\ge 2]\).

 

Now \(A_n=\sqrt{2-2\cos \frac{π}{2^n}}\)

\(\Rightarrow  A_n= \sqrt{2(1-\cos \frac{π}{2^n})}\)

\(\Rightarrow  A_n= \sqrt{4\sin^2 \frac{π}{2^{n+1}}}\)

 

 \(\Rightarrow  A_n= 2\sin \frac{π}{2^{n+1}}\). 

Now \(\displaystyle\lim_{n\to \infty} 2^nA_n=\displaystyle\lim_{n\to \infty} 2^{n+1}\cdot \sin \frac{π}{2^{n+1}}\).

Since , \(n \to \infty \)

\(\Rightarrow 2^{n+1}\to \infty \)

\(\Rightarrow \frac{π}{2^{n+1}}\to 0\)

Let \(\frac{π}{2^{n+1}}=z \Rightarrow z \to 0\), when \(n \to \infty\).

Therefore, \(\displaystyle\lim_{n\to \infty} 2^n A_n \)=\(\displaystyle\lim_{z \to 0}\frac{\sin z}{z}\cdot π =1\times π=π\). (Ans.)         

 

 

 

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Two Similar Triangles, ISI Entrance Subjective 2018


Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 2.
Topic
Difficulty Level

5.5 out of 10

Suggested Book

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Start with hints

Do you really need a hint? Try it first!

\(PQ\) and \( RS\) are two chords of the circle \(C\) , intersecting at the point \(O\). See figure: click here.

Given \(PO=3\) cm

\(SO=4\) cm

\([\triangle POR]= 7 cm^2\).

 

From the triangles \(POS\) and \(QOS\) we have,

                \(\angle POR=\angle SOQ\) [Opposite angles]

                \(\angle SRP=\angle SQO \) [Angle on the same semi-circle \(STP\)]

                \(\angle QSO= \angle OPR\) [Angle on the same semi-circle \(ST’P\)] 

Therefore the \(\triangle POR\) and \(\triangle SOQ\) are similar triangles .

 

\(\frac{[\triangle POR]}{OP^2}=\frac{[\triangle SOQ]}{SO^2}.\)

\(\Rightarrow [\triangle SOQ]=\frac{SO^2}{PO^2}\cdot [\triangle POR]\)=\(\frac{4^2}{3^2}\cdot 7=12\frac{4}{9}\).(Ans.)

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Pythagorean Triple, ISI Subjective Entrance 2018, Problem 7

Understand the problem

 Let \(a,b,c \in \mathbb{N}\) be such that
\(a^2+b^2=c^2\) and \(c-b=1\).
Prove that
(i) \(a\) is odd,
(ii) \(b\) is divisible by 4,
(iii) \(a^b+b^a\) is divisible by \(c\).

 

Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 7.
Topic
Number Theory

Difficulty Level
8 out of 10

Suggested Book

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Start with hints

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\(a^2+b^2=c^2=(b+1)^2=b^2+2b+1\)

\(\Rightarrow a^2=2b+1\) Which implies \(a^2\) is odd integer.

\(\Rightarrow a\) is also an odd integer =\(2k+1\) (say).

\((2k+1)^2+b^2=c^2\)

\(\Rightarrow 4k^2+4k+1+b^2=b^2+2b+1\)

\(\Rightarrow b=2k(k+1)\).

Now \(k(k+1) \) is always even =\(2l\) (say).

Therefore, \(b=4l\), i.e. \(b\) is divisible by 4.

 

\(a^b+b^a=a^{4l}+b^{2k+1}=(a^2)^{2l}+(c-1)^{2k+1}\)

=\((2b+1)^{2l}+(c-1)^{2k+1}\)

=\((2c-1)^{2l}+(c-1)^{2k+1}.\)

Now \(2l\) is even , therefore , \((2c-1)^{2l} \) is of the form : \( 2cp+1\) where \(p \in \mathbb{N}\).

And \(2k+1\) is odd , therefore \((c-1)^{2k+1} \) is of the form : \(cq-1\), where \(q \in \mathbb{N}\).

Therefore \(a^b+b^a=(2cp+1)+(cq-1)=c\cdot (2p+q)\).

\(\Rightarrow c|a^b+b^a\).

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Powers of 2 – I.S.I. Entrance 2019 Subjective Problem 1


Understand the problem

 Prove that the positive integers \(n\) that cannot be written as a sum of \(r\) consecutive positive integers, with \(r>1\) ,are of the form \(n=2^l\) for some \(l\ge 0\).

 

Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 1.

Topic
Number Theory

Difficulty Level

8.5 out of 10

Start with hints

Do you really need a hint? Try it first!

Claim: Any positive integer \(n\) can be written as \(n=2^k\cdot m\) , where \(k\ge0\) and \(m\) is an odd positive integer.

To prove this claim use the fact :  \(n=2^k\cdot p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}\), where \(k\) and all \(k_i\) are non-negetive integers and all \(p_i\) are odd primes.

The sum of (any) \(r\) consecutive positive integers is given by,

\((q+1)+(q+2)+(q+3)+\cdots+(q+r)\)

=   \(q\cdot r+(1+2+3+\cdots+r)\)

=   \(q\cdot r+\frac{r(r+1)}{2}\).

 

Equating this sum to \(n\) we get,

 

        \(2^k\cdot m = q\cdot r+\frac{r(r+1)}{2}\)

 

Or,     \(2^{k+1}\cdot m = 2q\cdot r+r(r+1)\)

 

Or,     \(2^l\cdot m = r(2q+r+1)\) , where \(l=k+1\ge1\).

In particular if we take \(n=2^l\) then \(m\) is equal to 1.

Since both \(r\) and \((2q+r+1)\) are greater than 1, so they can’t be equal to \(m\) in this case. Again one of \(r, (2q+r+1)\) is odd integer which implies the product \(r(2q+r+1)\) can’t be equal to \(2^l\). 

\(\Rightarrow 2^l \neq r(2q+r+1) \)

\(\Rightarrow 2^l\) can’t be expressed as the sum of \(r\) consecutive positive integers with \(r>1\) and \(l\ge 1\).

Now, \(n=2^0=1 \) also can’t be written in the same manner when \(l=0\). Therefore the positive integers \(n\) that cannot be written as a sum of \(r\) consecutive positive integers, with \(r>1\) , are of the form \(n=2^l\) for some \(l\ge 0\).

To show that any number other than of the form \(2^l\) is sum of consecutive integers:
Observe that for \(n=2^l.m\), where \(m\) is an odd number greater than 1, there are two cases:
1. \( m < 2^l \) 
Select \(r\) and \(q\) such that \(r = m\) and \(2q+r+1 = 2^l \).

2. \( m > 2^l \)
Select \(r\) and \(q\) such that \( 2q+r+1 = m\) and \( r = 2^l \).

(QED).

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