Post your solution then.zadid xcalibured wrote:I have a solution for problem 18.So let's move on to the next one.Someone post problems.
Secondary and Higher Secondary Marathon
- Tahmid Hasan
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- zadid xcalibured
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Re: Secondary and Higher Secondary Marathon
Solution $18$:let $BE$ and $AC$ intersect at $Q$.As $\angle{PDB}=\angle{DEB}=\angle{BQP}$ ,$PBQD$ is cyclic.And $PB=PD$ implies $\angle{BQP}=\angle{DQP}$ which implies $\angle{DQP}=\angle{CQE}$ which eventually implies $\triangle{ADQ}$ and $\triangle{CQE}$ are congruent as $AC$ and $DE$ are parallel.
Re: Secondary and Higher Secondary Marathon
$BB' || AC$
rest of it should be clear form the picture.
rest of it should be clear form the picture.
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Re: Secondary and Higher Secondary Marathon
Problem- $\boxed {18}$
Let $B'\in \omega$ so that $BB'\parallel AC$. Then $\angle EBC=\angle EAC=\angle DCA=\angle DBA$ since $ACDE$ is a cyclic trapezoid. Again $ACBB'$ is also a cyclic trapezoid. So $\angle BCA=\angle B'AC$. Now between $\triangle BCF$ and $\triangle B'AF$, $\angle FBC=\angle FB'C, \angle FCB=\angle FAB', BC=B'A$. Thus $\triangle BCF$ and $\triangle B'AF$ are equivalent. Thus $AF=CF$.
Let $B'\in \omega$ so that $BB'\parallel AC$. Then $\angle EBC=\angle EAC=\angle DCA=\angle DBA$ since $ACDE$ is a cyclic trapezoid. Again $ACBB'$ is also a cyclic trapezoid. So $\angle BCA=\angle B'AC$. Now between $\triangle BCF$ and $\triangle B'AF$, $\angle FBC=\angle FB'C, \angle FCB=\angle FAB', BC=B'A$. Thus $\triangle BCF$ and $\triangle B'AF$ are equivalent. Thus $AF=CF$.
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- zadid xcalibured
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Re: Secondary and Higher Secondary Marathon
Problem $19$:Let $f (x) = a_0x^n + a_1x^{n−1} + ⋯ + a_n$ be a polynomial with real coefficients such that $a_0 ≠ 0$ and for all real $x$,$f (x) f (2x^2) = f (2x^3+x)$.Prove that $f(x)$ has no real root.
Last edited by zadid xcalibured on Wed Dec 19, 2012 11:11 pm, edited 1 time in total.
Re: Secondary and Higher Secondary Marathon
Shouldn't it be $x^i$ ?zadid xcalibured wrote:Problem $19$:Let $f (x) = a_0x_n + a_1x_{n−1} + ⋯ + a_n$ be a polynomial with real coefficients such that $a_0 ≠ 0$ and for all real $x$,$f (x) f (2x^2) = f (2x^3+x)$.Prove that $f(x)$ has no real root.
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- zadid xcalibured
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Re: Secondary and Higher Secondary Marathon
Oops.Edited.
- Phlembac Adib Hasan
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Re: Secondary and Higher Secondary Marathon
Let $\exists q:q\neq 0$ and $f(q)=0$. Then the equation will imply $f(q^3+q)=0$ and so on. Since $x^3+x=x$ is possible only for $x=0$, so this will imply $f(x)$ has infinitely many roots in $\mathbb {R}$ which is impossible.zadid xcalibured wrote:Problem $19$:Let $f (x) = a_0x^n + a_1x^{n−1} + ⋯ + a_n$ be a polynomial with real coefficients such that $a_0 ≠ 0$ and for all real $x$,$f (x) f (2x^2) = f (2x^3+x)$.Prove that $f(x)$ has no real root.
If $q=0$, then $a_n=0$. So $f(x)=x^kP(x)$. where $P(0)\neq 0$. Now in the given equation we find $x^{3k}P(x)P(2x^2)=x^k(2x^2+1)^kP(2x^3+x)$. Since we assumed $P(x)$ has no zero root, so RHS has free $x$ to the power $k$ where LHS has $x^{3k}$. So a contradiction and so $f(0)\neq 0$. Therefore $f(x)$ has no real root.
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- Fahim Shahriar
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Re: Secondary and Higher Secondary Marathon
I'm going to post problem no. 20.
$\boxed {20}$
A series is formed in the following manner:
$A(1) = 1$;
$A(n) = f(m)$ numbers of $f(m)$ followed by $f(m)$ numbers of $0$;
$m$ is the number of digits in $A(n-1)$
Find $A(30)$. Here $f(m)$ is the remainder when $m$ is divided by $9$.
Source:BD National Math Olympiad 2010, Category: Secondary/Higher Secondary
$\boxed {20}$
A series is formed in the following manner:
$A(1) = 1$;
$A(n) = f(m)$ numbers of $f(m)$ followed by $f(m)$ numbers of $0$;
$m$ is the number of digits in $A(n-1)$
Find $A(30)$. Here $f(m)$ is the remainder when $m$ is divided by $9$.
Source:BD National Math Olympiad 2010, Category: Secondary/Higher Secondary
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
- zadid xcalibured
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Re: Secondary and Higher Secondary Marathon
This sequence must be periodic with period at most $9$.Actually this is periodic with period $6$.So $A(30)=A(6)=77777770000000$