Waste less time on Facebook — follow Brilliant.
×

Algebra contest

This contest has ended on 3rd september

TOPICS

Relation and function, matrix and determinants, vector, system of equations, sequence and series, permutation and combination, binomial theorem, complex numbers, polynomials, inequalities .

Rules

  1. I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

  2. A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

  3. Please make a substantial comment.

  4. Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 24 hours. Then, you must post the solution and you have the right to post a new problem.

  5. If the one who solves the last problem does not post a new problem in 2 hours, the creator of the previous problem has the right to post another problem.

  6. It is NOT compulsory to post original problems. But make sure it has not been posted on Brilliant.

  7. If a diagram is involved in your problem please make sure it is drawn by a computer program.

  8. Format your solution in \(\LaTeX\), picture solution will be accepted but only picture will not be, i.e. you can use picture for diagram or something, but not for the complete solution. Also make sure your solution is detailed and make sure to proof all claims.

  9. Do not post problems on same topic frequently like 3 continuous inequality problem are not allowed. 2 are sufficient.

  10. For those who are on slack Please post in #general that new problem is up along with problem number and the link to contest (https://brilliant.org/discussions/thread/algebra-contest/?sort=new)

  11. Handwritten picture solution are accepted only when they are easily understandable!!

Note by Prince Loomba
1 month, 2 weeks ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

PROBLEM 1

Find the number of ordered pairs of real numbers (a,b) such that

\((a+bi)^{2002}=a-bi\)

Here \(i\)=\(\sqrt {-1}\) Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba SOLUTION TO PROBLEM 1 let \[a+bi=e^{ix}\] then \[e^{2002ix}=e^{-ix}\] and \[e^{2003ix}=1\] so all te possible \(a+bi\) are 2003th roots of unity . we know from here that \(a+bi\) has 2003 diffent solutions and for each solution we get a different ordered pair for \((a,b)\) and hence the number of ordered pairs \((a,b)\) is \[\boxed{2003}\] Aareyan Manzoor · 1 month, 2 weeks ago

Log in to reply

@Aareyan Manzoor Right. Post next Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba No, answer should be 2004 as (0,0) is also a solution. Archit Agrawal · 1 month, 1 week ago

Log in to reply

Problem 22

Passed on from Jesse Nieminen, since he didn't wish to post.

Let \(a, b, c \in \mathbb{R}\) be real numbers such that \(a+b+c>0\). Prove the inequality

\[a^3+b^3+c^3 \leq (a^2+b^2+c^2)^{\frac32} + 3abc\] Sharky Kesa · 1 month ago

Log in to reply

@Sharky Kesa \(\begin{align} \left(a^3 + b^3 + c^3 - 3abc\right)^2 &= \left[\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right]^2 \\ &= \left(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\right)\left(a^2 + b^2 + c^2 - ab - bc - ca\right)^2 \\ &\leq \left(a^2 + b^2 + c^2\right)^3 \\ &\implies a^3 + b^3 + c^3 \leq \left(a^2 + b^2 + c^2\right)^{\frac 32} + 3abc. \square \end{align}\)

\(\left(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\right)\left(a^2 + b^2 + c^2 - ab - bc - ca\right)^2 \leq \left(a^2 + b^2 + c^2\right)^3\)
is true because if we substitute \(x = a^2 + b^2 + c^2, y = ab + bc + ca\), we get \(\left(x+2y\right)\left(x-y\right)^2 \leq x^3 \Leftrightarrow 2y \leq 3x \Leftrightarrow 2ab + 2bc + 2ca \leq 3a^2 + 3b^2 + 3c^2 \Leftrightarrow a^2 + b^2 + c^2 + \left(a-b\right)^2 + \left(b-c\right)^2 + \left(c-a\right)^2 \geq 0\) Jesse Nieminen · 1 month ago

Log in to reply

Problem 17

Find real numbers \(a\), \(b\) and \(c\) such that

\[|ax + by + cz| + |bx + cy + az| + |cx + ay + bz| = |x| + |y| + |z|\]

for all real numbers \(x\), \(y\) and \(z\). Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Unordered triplets are (1,0,0) and (-1,0,0). Archit Agrawal · 1 month, 1 week ago

Log in to reply

@Archit Agrawal Post solution too Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Put x,y=0, z=1 from this we get |a|+|b|+|c|=1. Then put x=y=z from this we get |a+b+c|=1. So a,b,c must be if same sign or 0. 1st case none of them is 0. Take x a very large negative number and x,y be very small positive number, the LHS mod will be opened with negative sign. The equation must be true for all x,y,z which is not happening. Do same in case of 1 of them being 0, you will see it is also not possible. So 2 of them must be 0 and other must be 1 or -1. Archit Agrawal · 1 month, 1 week ago

Log in to reply

@Archit Agrawal Yes! You have got the answer! Sharky Kesa · 1 month, 1 week ago

Log in to reply

PROBLEM 9

Let \(a>0\) be a real number and \(f (x)\) a real function defined on all R satisfying for all x belong to R

\(f (x+a)=\frac {1}{2}+\sqrt {f (x)-{f (x)}^{2}}\)

Prove that \(f (x)\) is periodic and give an example for \(f (x)\) when \(a=1\). Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Replace x by x+a, 2 times you will find that f(x+3a)=f(x+a). Hence it is periodic with period 2a. Eg. f(x)=(2+√6)/4. Archit Agrawal · 1 month, 1 week ago

Log in to reply

@Archit Agrawal Absolutely right. Noww post next Prince Loomba · 1 month, 1 week ago

Log in to reply

\(\text{Problem 5:}\)

Find all functions \(f:{\rm Z}\to{\rm Z}\) such that \(\displaystyle f(m+n)+f(m)f(n)=f(mn+1)\) where \(m,n\) are integers. Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma I got a weird thing. All of the functions of the type \(x^{2n}-1\) are satisfying according to me. Please check. Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba hmm ri8 but there are some complicated functions involving mod also. anyways post the solution otherways Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma Mod function: (Alternatively the trig function) \(f(x) = 0\) if \(x\equiv 1, 3 \pmod{4}\), \(f(x) = 1\) if \(x \equiv 2 \pmod{4}\), \(f(x) = -1\) if \(x \equiv 0 \pmod{4}\). Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

@Sharky Kesa ya those are the solutions exactly. you may post the solution Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma I am just saying this. I did not say these are all. Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma Is the answer only 1? (f(x) = 0) Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava NO, \(f(x)=0\) is trivial, And more functions are there. Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava x-1 is also Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Yes one of them, you need to show it other than trial and error. Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma Is there any mistakes I made? Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma

Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba yes it will work , i wll ppst other functions later . for now you may post the next question Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma There are infinite functions I can make Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma f (x)=[x]-1 (floor function) Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma One more f (x)={x} fractional part. 6 are done Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma f(x)=-cos((2n+1)×pi×x/2)

f (x)=sin (n×pi×x) Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Firstly the floor is meaningless here since it's a function from integers to integers. And moreover you are not supposed to guess functions, you have to show the derivation. Otherwise it's not a solution Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma Two functions f (x)=0 and f (x)=x-1 Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba There are 6, excluding the zero function Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma \(x^{2}-1\) Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba see, dont check your answers then the problem is meaningless . i would suggest that if you hve gor the functions post your solution and i wll cmmnt there Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma |x|-1 also satisfies. I am just checking by graph which can hold and verifying my answer Prince Loomba · 1 month, 2 weeks ago

Log in to reply

Problem 4

Let \(a,b\) lie in first quadrant and let \(x\) be an acute angle.

Given that (\(acosx)^2 +(bsinx)^2 = [(sin2x)(\dfrac{a+b}{2} )]^2 \)

Find the value of tanx. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava If we denote the three vertices by \(A(a,b),B(a\csc x,0),C(0,b\sec x)\) then observe that \(\displaystyle |AB|^2=a^2\cot^2 x+b^2,|BC|^2=a^2+b^2\tan^2 x,|CA|^2=a^2\csc^2 x +b^2\sec^2 x\)

which in turn shows \(\displaystyle |AB|^2+|BC|^2=|CA|^2\) and so \(\displaystyle B=\frac{\pi}{2}\).

Now, the circumcentre of a right angled triangle lies at the mid-point of it's hypotenuse and so the co-ordinates of the circumcenter is \(\displaystyle (M,N)=(\frac{a\csc x}{2},\frac{b\sec x}{2}) \)

From the relation we have , \(\displaystyle a^2\cos^2 x+b^2\sin^2 x=\sin^2 x\cos^2 x(a+b)^2\)

\(\displaystyle \implies (a\cos^2 x)^2+(b\sin^2 x)^2-2ab\sin^2 x\cos^2 x =0 \implies a\cos^2 x=b\sin^2 x\)

So we have , \(\displaystyle \csc x = \sqrt{\frac{a+b}{a}} , \sec x=\sqrt{\frac{a+b}{b}}\)

So the circumcenter is : \(\displaystyle (\frac{\sqrt{a^2+ab}}{2},\frac{\sqrt{b^2+ab}}{2}) \) Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma Correct @Aditya Sharma but sorry I rephrased the problem so @Prince Loomba will decide who will post the next problem. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma You post the next problem. The original was this only Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava Wait @Harsh Shrivastava , you can't change the problem for the reason someone can't do or have problems with that, the solution is done. Let the problem be back to it's original state Aditya Sharma · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma You post the next prob. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Aditya Sharma Post the next Prince Loomba · 1 month, 2 weeks ago

Log in to reply

Comment deleted 1 month ago

Log in to reply

@Harsh Shrivastava Meaning of circumvented? Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba I meant circumvented damn autocorrect. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava I got \(tanx\)=\(\sqrt {a/b}\). Idk how to find circumcentre Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Done post the solution. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava

Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Lemme rephrase the problem. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava Circumvented? And is this in the topic list at the starting of note? Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Corrections done.

If I tell the name of the topic, the problem will be trivial :) Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

Problem 3

Let \(a,b,x,f (x)\) be positive integers such that when \(a>b\), \(f (a)>f (b)\).

If \(f (f (x))=x^{2}+2\), find the value of \(f (3)\) Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba

Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava Post next soon Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava Its right. Post next Prince Loomba · 1 month, 2 weeks ago

Log in to reply

Comment deleted 1 month ago

Log in to reply

@Harsh Shrivastava I want solution. Prince Loomba · 1 month, 2 weeks ago

Log in to reply

Comment deleted 1 month ago

Log in to reply

@Harsh Shrivastava Yeah absolutely correct Prince Loomba · 1 month, 2 weeks ago

Log in to reply

Comment deleted 1 month ago

Log in to reply

@Harsh Shrivastava Are you posting? Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava Read point 11 in the note Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Harsh Shrivastava Handwritten will work. The only condition is it should be understandable! Prince Loomba · 1 month, 2 weeks ago

Log in to reply

problem 2 given the roots of \[y^8-y^7+y^6-1=0\] are \(y_1,y_2,...,y_7,y_8\) find the value of \[\dfrac{1}{y_1^9}+\dfrac{1}{y_2^9}+...+\dfrac{1}{y_7^9}+\dfrac{1}{y_8^9}\] Aareyan Manzoor · 1 month, 2 weeks ago

Log in to reply

@Aareyan Manzoor Solution to Problem 2

\[y^8 - y^7 + y^6 - 1 = 0\]

Now we have to form a polynomial whose roots are \(\frac{1}{y_{1}}, .....\)

Now, let \(x = \frac{1}{y}\)

then \(y = \frac{1}{x}\)

Putting it in the above equation, we get

\[x^8 + x^2 - x + 1 = 0\]

\[x^8 = x - x^2 - 1\]

\[x^9 = x^2 - x^3 - x ...(1)\]

Putting \(x = \frac{1}{y_{1}},..., \frac{1}{y_{8}}\) repeatedly in the equation and add them all and using bit Newton Sum (\(S_{1} = S_{2} = S_{3} = 0\))

And answer is \(0\). Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba you are right. Aareyan Manzoor · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Post the next question. Harsh Shrivastava · 1 month, 2 weeks ago

Log in to reply

@Aareyan Manzoor Is thE Answer 0? Kaustubh Miglani · 1 month, 2 weeks ago

Log in to reply

Problem 26

Can someone please post a solution to my problems!!!

Prove for positive \(x\), \(y\) and \(z\)

\[\sqrt{x^2+y^2-xy}+\sqrt{y^2+z^2-yz} \geq \sqrt{z^2+x^2+zx}\] Sharky Kesa · 3 weeks, 3 days ago

Log in to reply

@Sharky Kesa Take a triangle ABC and construct a point P inside the triangle such that angleAPB = angleBPC = angleCPA= 120°.

Now by law of cosines, we can let AP = x,BP = y and CP = z.

The terms in the inequality represents side lengths of triangle ABC and thus the inequality is true by triangle inequality. Harsh Shrivastava · 3 weeks, 3 days ago

Log in to reply

@Sharky Kesa Post easy, I will solve haha Prince Loomba · 3 weeks, 3 days ago

Log in to reply

@Prince Loomba This is very, very simple. You just have to think outside the box, and you will get it. Sharky Kesa · 3 weeks, 3 days ago

Log in to reply

Problem 25

Prove \((a_1+b_1)(a_2+b_2)\ldots (a_n+b_n) \leq (a_1+b_{\sigma(1)})(a_2+b_{\sigma(2)})\ldots (a_n+b_{\sigma(n)})\) where \(\{a\}\) and \(\{b\}\) are similarly ordered sequences and \(\sigma\) denotes a permutation. Sharky Kesa · 3 weeks, 5 days ago

Log in to reply

@Sharky Kesa Time is up bro.. Prince Loomba · 3 weeks, 3 days ago

Log in to reply

@Sharky Kesa Permutations are always greater than or equal to 1 so each bracket of lhs is smaller than corresponding bracket of rhs Prince Loomba · 3 weeks, 5 days ago

Log in to reply

@Prince Loomba No, what is meant is that the permutation is a permutation of the set \({1,2,3,4,...,n}\). e.g. if \(n=3\), we can assume a permutation of it to be \({2,1,3}\), so \(\sigma(1)=2\), \(\sigma(2)=1\) and \(\sigma(3)=3\). Note that the above must hold true for any permutation.


Furthermore, even if permutation is defined as what you think it was, you're wrong. What if \(b\) was negative? Sharky Kesa · 3 weeks, 4 days ago

Log in to reply

Problem 24

Prove \(2^{1/x} < \dfrac {x+1}{x}\) for \(x>1\). Sharky Kesa · 3 weeks, 6 days ago

Log in to reply

@Sharky Kesa Why did no one give the quick and simple solution?!

\[\begin{align} 2^{1/x} &< \dfrac {x+1}{x}\\ 2x^x &< (x+1)^x\\ 2x^x &< x^x + x \cdot x^{x-1} + \{\text{ Stuff }\}\\ 0 &< \{\text{Stuff}\} \end{align}\]

Since the value of \(\text{Stuff}\geq 1\), the final inequality must be true, so the original inequality is true. Thus, proven. Sharky Kesa · 3 weeks, 5 days ago

Log in to reply

@Sharky Kesa I was waiting for someone else to post a solution. I solved this in first 5 minutes. Jesse Nieminen · 3 weeks, 5 days ago

Log in to reply

@Sharky Kesa Counter-example: \(x=50\). Jesse Nieminen · 3 weeks, 6 days ago

Log in to reply

@Jesse Nieminen LOL, sorry, the other way! Sharky Kesa · 3 weeks, 6 days ago

Log in to reply

@Sharky Kesa Time elapsed post next Prince Loomba · 3 weeks, 5 days ago

Log in to reply

Problem 23

Since Jesse Nieminen didn't post, my turn again!

Let \(a,b,c,d\in\mathbb{R}^{+}\) and \(a+b+c+d+abcd=5\). Prove

\[\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {1}{c} + \dfrac {1}{d} \geq 4\] Sharky Kesa · 4 weeks, 1 day ago

Log in to reply

@Sharky Kesa Post next. Time elapsed Prince Loomba · 3 weeks, 6 days ago

Log in to reply

@Prince Loomba I have a solution using Lagrange Multipliers :P Jesse Nieminen · 3 weeks, 6 days ago

Log in to reply

@Jesse Nieminen Just write it. My solution is Lagrange Multipliers as well Sharky Kesa · 3 weeks, 6 days ago

Log in to reply

@Sharky Kesa I found a solution using AM-GM inequality. I will post it in few minutes if my battery doesn't run out. Jesse Nieminen · 3 weeks, 6 days ago

Log in to reply

@Prince Loomba This competition is now me solving or not solving problems by @Sharky Kesa :D Jesse Nieminen · 3 weeks, 6 days ago

Log in to reply

Problem 21

Let \(a,b,c\) be positive real numbers such that \(27ab + 2bc + 18ca = 81\). Prove that

\[\dfrac1a + \dfrac2b + \dfrac3c \geq 3\] Jesse Nieminen · 1 month ago

Log in to reply

@Jesse Nieminen According to Mathematica, the inequality is stricter than the one you have provided us with. Mathematica states that the expression is greater than or equal to \( \frac{7\sqrt{7}}{3\sqrt{3}} \), which is approximately 3.56 Jack Lam · 1 month ago

Log in to reply

@Jack Lam You can post next Prince Loomba · 1 month ago

Log in to reply

@Jesse Nieminen Here's the solution:

\(\begin{align}\dfrac 1a + \dfrac 2b + \dfrac 3c &= \dfrac 1{2a} + \dfrac 1{2b} + \dfrac 3{2b} + \dfrac 9{4c} + \dfrac 3{4c} + \dfrac 1{2a} \\ &\geq 2\sqrt{\dfrac 1{2a} \cdot \dfrac 1{2b}} + 2\sqrt{\dfrac 3{2b} \cdot \dfrac 9{4c}} + 2\sqrt{\dfrac 3{4c} \cdot \dfrac 1{2a}} \\ &= \sqrt{\dfrac 1{ab}} + \sqrt{\dfrac {27}{2bc}} + \sqrt{\dfrac{3}{2ca}} \\ &\geq 3 \cdot \sqrt[3]{\sqrt{\dfrac 1{ab}} \cdot \sqrt{\dfrac {27}{2bc}} \cdot \sqrt{\dfrac{3}{2ca} }} \\ &= 3 \cdot \sqrt{\sqrt[3]{\dfrac 1{ab}} \cdot \sqrt[3]{\dfrac {27}{2bc}} \cdot \sqrt[3]{\dfrac{3}{2ca}}} \\ &\geq 3 \cdot \sqrt{\dfrac{3}{ab + \dfrac{2bc}{27} + \dfrac{2ca}{3} }} \\ &= 3 \cdot \sqrt{\dfrac{81}{27ab + 2bc + 18ca }} \\ &= 3. \ \square \end{align}\) Jesse Nieminen · 1 month ago

Log in to reply

@Jesse Nieminen I did not see that approach at all! Great job! Sharky Kesa · 1 month ago

Log in to reply

@Jesse Nieminen Please post next Prince Loomba · 1 month ago

Log in to reply

@Prince Loomba I give my right to post to everyone. Jesse Nieminen · 1 month ago

Log in to reply

Problem 20

Let \(x_1, x_2, \ldots, x_n\) be the roots of the polynomial \(X^n + X^{n-1} + \ldots + X + 1\). Prove that

\[\displaystyle \sum_{k=1}^n \dfrac {1}{1-x_k} = \dfrac {n}{2}\] Sharky Kesa · 1 month ago

Log in to reply

@Sharky Kesa \(\displaystyle \sum_{k=1}^n \dfrac {1}{1-x_k} = \dfrac{\left[\left(1-x_1\right)\left(1-x_2\right)\cdots\left(1-x_{n-1}\right)\right]+\cdots+ \left[\left(1-x_2\right)\left(1-x_3\right)\cdots\left(1-x_{n}\right) \right]}{\left(1-x_1\right)\left(1-x_2\right)\cdots\left(1-x_{n}\right)}\) (Numerator has total of \(n\) addends, each missing different \(1 - x_k\))

After multiplying out the products we get \(\displaystyle \dfrac{n \times 1 - \left(n-1\right)\times\left(x_1 + \cdots + x_n\right) + \cdots - \left(-1\right)^n \times1\times\left(x_1x_2\cdots x_{n-1} + \cdots + x_2\cdots x_n\right)}{1 - \left(x_1+\cdots+x_n\right) + ... + \left(-1\right)^n \times \left(x_1x_2\cdots x_n\right)}\)

After using Vieta's formulas we get \(\dfrac{\dfrac{n\left(n+1\right)}2}{n+1} = \dfrac n2\).

\(Q.E.D\) Jesse Nieminen · 1 month ago

Log in to reply

@Jesse Nieminen Good solution. You can post the next problem, but see if you can get a simpler, non-expansive solution. Sharky Kesa · 1 month ago

Log in to reply

@Jesse Nieminen Nice solution.

Post the next problem. Harsh Shrivastava · 1 month ago

Log in to reply

Problem 19

Let \(a, b, c > 0\) be real parameters. Solve the system of equations:

\[\begin{align} ax - by + \frac {1}{xy} &= c\\ bz - cx + \frac {1}{zx} &= a\\ cy - az + \frac {1}{yz} &= b \end{align}\] Sharky Kesa · 1 month ago

Log in to reply

@Sharky Kesa This solution is a bit messy, but is quite quick as opposed to the other possible solutions (I think).


Consider \(a\), \(b\) and \(c\) as the unknowns instead, so we get a set of linear equations. If we multiply the first equation by \(x\), and then \(y\) and substitute \(cx\) in the second question and \(cy\) in the last one, we obtain

\[\begin{align} a(x^2+1)-b(xy+z)&=\frac{1}{xz}-\frac{1}{y} \quad (i)\\ a(xy-z)-b(y^2+1)&=- \frac {1}{x} - \frac {1}{yz} \quad (ii) \end{align}\]

Multiply equation \((i)\) by \(y^2+1\) and equation \((ii)\) by \(-(xy+z)\), and add them up. It follows that

\[a(x^2+y^2+z^2+1)=\dfrac{x^2+y^2+z^2+1}{xz}\]

so \(a=\frac {1}{xz}\). Similarly, \(b=\frac{1}{yz}\) and \(c=\frac{1}{zx}\). Thus, \(abc=\frac{1}{(xyz)^2}\), so \(xyz=\pm \frac{1}{\sqrt{abc}}\). From this, we can attain the solutions of the system are

\[\left (\dfrac {b}{\sqrt{abc}}, \dfrac {a}{\sqrt{abc}}, \dfrac {c}{\sqrt{abc}} \right ), \left (\dfrac {-b}{\sqrt{abc}}, \dfrac {-a}{\sqrt{abc}}, \dfrac {-c}{\sqrt{abc}} \right ) \] Sharky Kesa · 1 month ago

Log in to reply

@Sharky Kesa I made a silly mistake while calculating by matrices... Now I got it haha Prince Loomba · 1 month ago

Log in to reply

Problem 18

Since @Archit Agrawal's time ran out, I'll post next question.

Consider the sequence

\[a_n = 2 - \dfrac {1}{n^2 + \sqrt{n^4 + \frac {1}{4}}}, \quad n \geq 1\]

Prove that \(\sqrt{a_1}+\sqrt{a_2}+\ldots+\sqrt{a_{119}}\) is an integer and find the value of it. Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Here is the solution:

We will take the conjugate of the expression to get as follows:

\[\begin{align} a_n &= 2 - \dfrac {1}{n^2 + \sqrt{n^4+\frac {1}{4}}}\\ &= 2 - \dfrac{n^2 - \sqrt{n^4 + \frac {1}{4}}}{n^4 - \left ( n^4 + \frac {1}{4} \right )}\\ &= 2 + 4n^2 - 2 \sqrt{4n^4 + 1} \end{align}\]

We can factorise \(4n^4 + 1\) as \((2n^2 - 2n + 1)(2n^2 + 2n + 1)\). Note that \((2n^2 - 2n + 1) + (2n^2 + 2n + 1) = 4n^2 +2\). Thus, this expression is the perfect square identity. We have

\[\begin{align} a_n &= 4n^2 + 2 - 2\sqrt{4n^4 + 1}\\ &= (\sqrt{2n^2 + 2n + 1})^2 + (\sqrt{2n^2 - 2n + 1})^2 - 2 \times \sqrt{2n^2 + 2n + 1} \times \sqrt{2n^2 - 2n + 1}\\ &= (\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1})^2\\ \sqrt{a_n} &= \sqrt{2n^2 + 2n + 1} - \sqrt{2n^2-2n +1} \end{align}\]

Note that this is a telescoping series so almost all terms cancel out in the, so we are left with \(\sqrt{2\times 119^2 + 2 \times 119 + 1} - \sqrt{2\times 1^2 - 2 \times 1 + 1} = 169 - 1 = 168\), which is an integer. Thus proven. Sharky Kesa · 1 month ago

Log in to reply

@Sharky Kesa \[\boxed{168}\] Aman Rajput · 1 month, 1 week ago

Log in to reply

@Aman Rajput Can you prove it? That was the question asked. Sharky Kesa · 1 month, 1 week ago

Log in to reply

Problem 16

Evaluate: \( \displaystyle \sum_{n=0}^{\infty} \cos^n{\theta} \cos{n \theta} \)

Assume θ is such that the sum converges. Jack Lam · 1 month, 1 week ago

Log in to reply

@Jack Lam is it ? i replaced theta with x \[\frac12.\frac{2e^{ix}-\cos x - \cos x e^{2ix}}{\cos^2 x e^{ix} - \cos x - \cos x e^{2ix}+e^{ix}}\] Aman Rajput · 1 month, 1 week ago

Log in to reply

@Aman Rajput I will prove the fraction simplifies to 2. We are required to prove:

\[\begin{align} \dfrac{2e^{ix}-\cos x - \cos x e^{2ix}}{\cos^2 x e^{ix} - \cos x - \cos x e^{2ix}+e^{ix}}&= 2\\ 2e^{ix} - \cos x - \cos x e^{2ix} &= 2\cos^2 x e^{ix} - 2\cos x - 2\cos x e^{2ix} + 2e^{ix}\\ \cos x e^{2ix} + \cos x &= 2 \cos^2 x e^{ix}\\ \cos x (2 \cos^2 x + 2i \sin x \cos x) &= 2 \cos^2 x (\cos x + i \sin x)\\ 2 \cos^2 x (\cos x + i \sin x) &= 2 \cos^2 x (\cos x + i \sin x) \end{align}\]

which is clearly true. Thus, the original fraction must be equal to 2, so the value of the summation is 1. Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Aman Rajput It is assumed that I want the answer in the simplest possible form... Jack Lam · 1 month, 1 week ago

Log in to reply

Problem 15

Find the least real number \(r\) such that for each triangle with side lengths \(a\), \(b\), \(c\),

\[\dfrac {\max(a, b, c)}{\sqrt[3]{a^3 + b^3 + c^3 + 3abc}} < r\] Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Sharky Kesa WLOG, it can be assumed that \( \displaystyle a \ge b \ge c\)

Using the triangle inequality, and some trivially true statements:

\[b + c - a > 0\]

\[b^2 + c^2 + a^2 + ab + ac - bc > 0\]

\[b^3 + c^3 - a^3 + 3abc > 0\]

\[a^3 + b^3 + c^3 + 3abc > 2a^3\]

The rest should follow after rearranging...

\[\dfrac {a^3}{a^3+b^3+c^3+3abc} < \dfrac {1}{2}\]

\[\dfrac {a}{\sqrt[3]{a^3+b^3+c^3+3abc}} < \dfrac {1}{\sqrt[3]{2}}\]

Thus, \(\displaystyle r=\frac {1}{\sqrt[3]{2}}\). Jack Lam · 1 month, 1 week ago

Log in to reply

PROBLEM 14

Which of the following sets are incomparable

A). Set of all living children, set of all living humans

B). Set of regular polygons, Set of irregular polygons

C). Set of all the triangles in a plane, set of all the squares in a plane

D). Set of all the things in this universe,set of all the chairs.

Question

You have to explain why all the others are wrong and this is right (Your answer) Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Firstly, in A), we have the the set of all living children is a subset of the set of all living humans, so they can be compared.

Secondly, in C), we have the set of all triangles, which is an uncountable infinite, versus the set of all squares, which is also an uncountable infinite, and we cannot compare these two sets.

Thirdly, in D), we has the set of all chairs is a subset of the set of all things in the universe, so they can be compared.

Finally, in B), we have the set of regular polygons is disjoint from the set of irregular polygons. However, note that the probability an arbitrary polygon is regular is 0 (not impossible, 0), so it is almost guaranteed that an arbitrary polygon is irregular. Thus, we must have the set of all regular polygons to be less than the set of irregular polygons.

My answer is C). Sharky Kesa · 1 month, 1 week ago

Log in to reply

Problem 13

In problem 12, if c approaches \(-\infty\), such that when x tends to \(\infty\), \(x^{2}+bx+c\) tends to 0.

Which of the following can be the value of \(|-x^{2}+bx+c|\)?(\(b\neq 0)\)

A) 0

B) \(\infty\)

C) \(-\infty\)

D)\(\sqrt{2016^{2015}}\) Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba And are you guys kidding? Why three contests are going on simultaneously? Aman Rajput · 1 month, 1 week ago

Log in to reply

@Aman Rajput Your time limit to create a question is up. Read point 5. Prince Loomba · 1 month, 1 week ago

Log in to reply

@Aman Rajput Geometry one is stucked, JEE algebra one I told Rajdeep not to start but he started! Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba B) and D) both can be correct. Check your question again unless it is a multiple options correct question. Aman Rajput · 1 month, 1 week ago

Log in to reply

@Aman Rajput Aman you are right. Post the next. I created this question as multicorrect only! Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba B Archit Agrawal · 1 month, 1 week ago

Log in to reply

@Archit Agrawal Wrong. Think again Prince Loomba · 1 month, 1 week ago

Log in to reply

PROBLEM 12

For the equation \(x^{2}+bx+c\), as x approches \(\infty\), it also approaches \(\infty\).

For the equation \(|-x^{2}+bx+c|\), as x approaches \(\infty\), it also approaches \(\infty\)

Can we compare the 2 infinities? If yes, which is greater? Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Yes, we can compare the two infinities. If b is +ve then x^2+bx+c is greater and if b is -ve then |-x^2+bx+c|is greater. Archit Agrawal · 1 month, 1 week ago

Log in to reply

@Archit Agrawal Right post next Prince Loomba · 1 month, 1 week ago

Log in to reply

Problem 11

Let \(f\) be a function \(f:\mathbb{Z} \to \mathbb{Z}\) that satisfies below:

\[f(y+f(x))=f(x-y)+y(f(y+1)-f(y-1))\]

Find all functions that satisfy. Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Your time is up. 24 hours were given to solve the problem and 2 hours to post the next one for you. 26 hours are over, so by point number 5 since I am creator of problem 10, I can post problem 12! Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Sorry, I was busy, I'll post solution ASAP. Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Ok the final answer

As I have wrote f (f (x))=f (x)

The following satisfy

Constant function

f is odd when x is odd and f is even when x is even

f (x)=x Prince Loomba · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Here are the solutions:

1) \(f(x)=a\) if \(x\) is even and \(a\) is even, and \(f(x)=b\) if \(x\) is odd and \(b\) is odd.

2) \(f (x)=c \forall c\)

3)\(f (x)=x \forall x\)

Sharky discussed some time back this with me so it would be unfair for me to post the solution. Zoha Asif · 1 month, 1 week ago

Log in to reply

@Zoha Asif Well I found these solutions but couldn't find a way to prove that these were the only ones, so I gave up. I'm just waiting for the solution now. (is it some kind of induction?) Wen Z · 1 month, 1 week ago

Log in to reply

@Sharky Kesa One more: f (x)=|x|, one more f (x)=-|x| Prince Loomba · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Put y=0. f (f (x))=f (x)

This implies either f (x) is a constant function or f (x)=x. Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Not necessarily. Sharky Kesa · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Then you tell. Idk the solutions of f (f(x))=f (x) Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba And cannot be f(x)=x Kaustubh Miglani · 1 month, 1 week ago

Log in to reply

@Kaustubh Miglani Bro LHS becomes y+f (x)=y+x

RHS becomes x-y+y (y+1-y+1)=x+y

Clearly f (x)=x satisfies

Alternatively take inverse of my equation, you will get f (x)=x Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Sorry calc error Kaustubh Miglani · 1 month, 1 week ago

Log in to reply

@Kaustubh Miglani No problem Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Its a constant function Kaustubh Miglani · 1 month, 1 week ago

Log in to reply

Problem 10

Find the number of polynomials of degree 5 with distinct coefficient from the set

{\(1,2,...9\)}

that are divisible by \(x^{2}-x+1\) Prince Loomba · 1 month, 1 week ago

Log in to reply

@Prince Loomba Let \(P(x)=ax^5 + bx^4 + cx^3 + dx^2 + ex + f\). Note that the roots of \(x^2 - x + 1\) are \(e^{\frac{i\pi}{3}}\) and \(e^{\frac {-i\pi}{3}}\). Thus, \(P(x)\) is divisible by \(x^2 - x + 1\) if \(P(e^{\frac{i\pi}{3}})=0\) and \(P(e^{\frac{-i\pi}{3}}) = 0\). From this, we gather that \(a+b=d+e\) and \(\frac {a}{2} - \frac {b}{2} - c - \frac {d}{2} + \frac {e}{2} + f = 0\), which implies \(e + 2f + a = b + 2c + d\). Also, since \(a+b=d+e\), \(a-d=e-b=c-f\).

We now consider \(P(x)= (k+l)x^5 + 9x^4 + (m+l)x^3 + kx^2 + (9+l)x + m\), with \(l > 0\) and \(k \leq 9 \leq m\). For a given \(l\), there are \(\dbinom{9-k}{3}\) ways of choosing \(k\), \(m\) and \(n\) such that \(l+m \leq 9\). Now since the coefficients are distinct, we must subtract any combination with the difference in \(k\), \(m\) and \(n\) being \(l\).

There are \(9-2l\) ways to select two numbers differing by \(l\) and \(7-l\) ways to select the remaining numbers. This implies that for a given \(l\), there are \(\dbinom{9-l}{3} - (9-2l)(7-l)+9-3l\) polynomials satisfying, which adds up to \(53\). However, we take into account the rearrangements of the coefficients (there are 12 for each) so there are \(53 \times 12 = 636\) polynomials satisfying. Sharky Kesa · 1 month, 1 week ago

Log in to reply

Problem 8

Let \(n\) be a positive integer. Prove \((x^2+x)^{2^n} + 1\) is irreducible. Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

@Sharky Kesa look at the polynomial mod 2

its equivalent to \((x^{2}+x+1)^{2^{n}}\)

so if \((x^{2}+x)^{2^{n}+1}\) \(\equiv\) f(x) g(x) mod 2

\(f(x)=(x^{2}+x+1)^{m}\) mod 2

\(g(x) = (x^{2}+x+1)^{(2^{n}-m)}\) mod 2

\(f(x)=(x^{2}+x+1)^{m} + 2P(x)\)

\(g(x)=(x^{2}+x+1)^{(2^{n}-m)} + 2Q(x)\)

\(f(x)*g(x)= ((x^{2}+x+1)^m + 2P(x))((x^{2}+x+1)^{(2^n-m)} + 2Q(x))\)

Let x=w (complex cube root of unity)

\((w^{2}+w)^{2^{n}} + 1 = 2P(w)*2Q(w)\)

2 = 4P(w)Q(w)

We got P (w)Q (w) as rational numbers but they were integer polynomials.

Hence a contradiction! Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Sharky Kesa It sufficies to consider the equivalent rational polynomial

\( (x^2 -1)^{2^n} + 4^{2^n} \)

It is known that every non-trivial binomial coefficient in the rows of Pascal's Triangle which are powers of 2 are all even.

Thus, in the expasion of the above polynomial, only the leading co-efficient and the constant are not even.

The leading coefficient is clearly not divisible by 2.

The constant term is clearly not divisible by 4.

Therefore, Eisenstein's Criterion holds and the polynomial is irreducible.

Hence, the original polynomial is also irreducible. Jack Lam · 1 month, 1 week ago

Log in to reply

@Sharky Kesa Let \(a\in I\) be a root of it. \((a^{2}+a)^{2^{n}}=-1\). Clearly for n>=1, LHS is a perfect square with range >=0, so inequality cant hold

Thus no integer can be its root.

Hence proved Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Not quite. Your method is incorrect since irreducible just means that it cannot be factored. A property of irreducible polynomials (degree > 1) is that they have no integral roots, but the converse isn't true. Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Ya same way About to post it But u did it first Kaustubh Miglani · 1 month, 2 weeks ago

Log in to reply

@Kaustubh Miglani But it is incorrect. Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

@Sharky Kesa Yeah U are right.Thanks for explaining Kaustubh Miglani · 1 month, 2 weeks ago

Log in to reply

Problem 7

Let \(H_n = \frac{1}{1} + \frac {1}{2} + \ldots + \frac {1}{n}\). Prove that for \(n \geq 2\)

\[n(n+1)^{\frac {1}{n}} < n + H_n\] Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

@Sharky Kesa Note that \(n+H_n = (1+\frac {1}{1}) + (1+\frac{1}{2}) + \ldots + (1+\frac {1}{n})\) (These brackets are all distinct terms). Now, by AM-GM, we have

\[\begin{align} \dfrac {\displaystyle \sum_{i=1}^n \left ( 1+\frac{1}{i} \right ) }{n} &> \sqrt[n]{\displaystyle \prod_{i=1}^n \left ( 1+\frac{1}{i} \right )}\\ \dfrac {H_n + n}{n} &> \sqrt[n]{n+1}\\ H_n+n &> n(n+1)^{\frac{1}{n}} \end{align}\]

Thus, proven. Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

Problem 6

Let \(S_{1}, S_{2}, S_{3}\) denote the sum of n terms of 3 arithmetic progressions with first terms unity and their common differences are in harmonic progression. Prove that

\[\large n=\frac {2S_{3}S_{1}-S_{1}S_{2}-S_{2}S_{3}}{S_{1}-2S_{2}+S_{3}}.\] Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba I would like to cut the algebraic bash out of my solution, which I have, but it is quite a standard expansion.

\[\begin{align} S_1 &= \frac {n}{2} (2a+(n-1)d_1)\\ S_2 &= \frac {n}{2} (2a+(n-1)d_2)\\ S_3 &= \frac {n}{2} (2a+(n-1)d_3)\\ d_1 &= \frac{1}{x}\\ d_2 &= \frac {1}{x+k}\\ d_3 &= \frac {1}{x+2k}\\ a &= 1\\ \implies S_1 &= n + \frac{n-1}{2r}\\ S_2 &= n + \frac{n-1}{2r+2k}\\ S_3 &= n + \frac{n-1}{2r+4k} \end{align}\]

Substituting back into the original expression, after much painful expansion, you get the fraction to simplify to \(n\).


For those who want the expansion:

\[\begin{align} & \dfrac {2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right )}{\left (n + \frac {n-1}{2r} \right ) - 2 \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+4k} \right )}\\ & = \dfrac{2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right )}{\frac {k^2 (n-1)}{r(k+r)(2k+r)}}\\ & = \dfrac{r \left ( 2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right ) \right )(k+r)(k+2r)}{k^2 (n-1)}\\ &=\dfrac{r \left (\frac {k^2 n(n-1)}{r(k+r)(2k+r)} \right )(k+r)(k+2r)}{k^2 (n-1)}\\ &= \dfrac {k^2 n (n-1)}{k^2 (n-1)}\\ &= n \end{align}\] Sharky Kesa · 1 month, 2 weeks ago

Log in to reply

@Sharky Kesa This is not the solution. The solution is 10 times easier. I will consider it as solution either you solve full or take out the shortcut that I have Prince Loomba · 1 month, 2 weeks ago

Log in to reply

@Prince Loomba Can you please post the easy solution. @Prince Loomba Harsh Shrivastava · 1 month, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...