Algebra contest

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 $\boxed{2003}$

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

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

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.

Unordered triplets are (1,0,0) and (-1,0,0).

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 $0$.

Is thE Answer 0?

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 : $\displaystyle (\frac{\sqrt{a^2+ab}}{2},\frac{\sqrt{b^2+ab}}{2})$

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

Circumvented? And is this in the topic list at the starting of note?

I got $tanx$=$\sqrt {a/b}$. Idk how to find circumcentre

Is the answer only 1? (f(x) = 0)

I got a weird thing. All of the functions of the type $x^{2n}-1$ are satisfying according to me. Please check.

Two functions f (x)=0 and f (x)=x-1

f(x)=-cos((2n+1)×pi×x/2)

f (x)=sin (n×pi×x)

One more f (x)={x} fractional part. 6 are done

f (x)=[x]-1 (floor function)

There are infinite functions I can make

Is there any mistakes I made?

Post easy, I will solve haha

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.

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

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

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

For those who want the expansion:

$\begin{aligned} & \dfrac {2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right )}{\left (n + \frac {n-1}{2r} \right ) - 2 \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+4k} \right )}\\ & = \dfrac{2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right )}{\frac {k^2 (n-1)}{r(k+r)(2k+r)}}\\ & = \dfrac{r \left ( 2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right ) \right )(k+r)(k+2r)}{k^2 (n-1)}\\ &=\dfrac{r \left (\frac {k^2 n(n-1)}{r(k+r)(2k+r)} \right )(k+r)(k+2r)}{k^2 (n-1)}\\ &= \dfrac {k^2 n (n-1)}{k^2 (n-1)}\\ &= n \end{aligned}$

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

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

Thus, proven.

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!

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

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.

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.

Put y=0. f (f (x))=f (x)

This implies either f (x) is a constant function or f (x)=x.

One more: f (x)=|x|, one more f (x)=-|x|

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.

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

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!

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.

B

B) and D) both can be correct. Check your question again unless it is a multiple options correct question.

And are you guys kidding? Why three contests are going on simultaneously?

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).

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, $\displaystyle r=\frac {1}{\sqrt[3]{2}}$.

is it ? i replaced theta with x $\frac12.\frac{2e^{ix}-\cos x - \cos x e^{2ix}}{\cos^2 x e^{ix} - \cos x - \cos x e^{2ix}+e^{ix}}$

$\boxed{168}$

Here is the solution:

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

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

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

$\begin{aligned} a_n &= 4n^2 + 2 - 2\sqrt{4n^4 + 1}\\ &= (\sqrt{2n^2 + 2n + 1})^2 + (\sqrt{2n^2 - 2n + 1})^2 - 2 \times \sqrt{2n^2 + 2n + 1} \times \sqrt{2n^2 - 2n + 1}\\ &= (\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1})^2\\ \sqrt{a_n} &= \sqrt{2n^2 + 2n + 1} - \sqrt{2n^2-2n +1} \end{aligned}$

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.

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

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

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

$\left (\dfrac {b}{\sqrt{abc}}, \dfrac {a}{\sqrt{abc}}, \dfrac {c}{\sqrt{abc}} \right ), \left (\dfrac {-b}{\sqrt{abc}}, \dfrac {-a}{\sqrt{abc}}, \dfrac {-c}{\sqrt{abc}} \right )$

$\displaystyle \sum_{k=1}^n \dfrac {1}{1-x_k} = \dfrac{\left[\left(1-x_1\right)\left(1-x_2\right)\cdots\left(1-x_{n-1}\right)\right]+\cdots+ \left[\left(1-x_2\right)\left(1-x_3\right)\cdots\left(1-x_{n}\right) \right]}{\left(1-x_1\right)\left(1-x_2\right)\cdots\left(1-x_{n}\right)}$ (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 $\dfrac{\dfrac{n\left(n+1\right)}2}{n+1} = \dfrac n2$.

$Q.E.D$

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

Please post next

Here's the solution:

$\begin{aligned}\dfrac 1a + \dfrac 2b + \dfrac 3c &= \dfrac 1{2a} + \dfrac 1{2b} + \dfrac 3{2b} + \dfrac 9{4c} + \dfrac 3{4c} + \dfrac 1{2a} \\ &\geq 2\sqrt{\dfrac 1{2a} \cdot \dfrac 1{2b}} + 2\sqrt{\dfrac 3{2b} \cdot \dfrac 9{4c}} + 2\sqrt{\dfrac 3{4c} \cdot \dfrac 1{2a}} \\ &= \sqrt{\dfrac 1{ab}} + \sqrt{\dfrac {27}{2bc}} + \sqrt{\dfrac{3}{2ca}} \\ &\geq 3 \cdot \sqrt[3]{\sqrt{\dfrac 1{ab}} \cdot \sqrt{\dfrac {27}{2bc}} \cdot \sqrt{\dfrac{3}{2ca} }} \\ &= 3 \cdot \sqrt{\sqrt[3]{\dfrac 1{ab}} \cdot \sqrt[3]{\dfrac {27}{2bc}} \cdot \sqrt[3]{\dfrac{3}{2ca}}} \\ &\geq 3 \cdot \sqrt{\dfrac{3}{ab + \dfrac{2bc}{27} + \dfrac{2ca}{3} }} \\ &= 3 \cdot \sqrt{\dfrac{81}{27ab + 2bc + 18ca }} \\ &= 3. \ \square \end{aligned}$