355. Some problems with polynomials or equations I

by Virgil Nicula, Aug 28, 2012, 1:21 PM

PP1. Prove easily that $\boxed{g(x)=(x+y+z)^3-\left(x^3+y^3+z^3\right)=3(x+y)(y+z)(x+z)}$ . Prove similarly that there is

$p(x,y,z)$ such that $h(x,y,z)\equiv (x+y+z)^5-\left(x^5+y^5+z^5\right)=(x+y)(y+z)(x+z)p(x,y,z)$ . Find $p(x,y,z)$ .

Proof 1. Since $g(-y,y,z)=g(-z,y,z)=g(x,-z,z)=0$ what means exists $k\in\mathbb R^*$ such that $g(x,y,z)=k(x+y)(x+z)(y+z)$ for any $\{x,y,z\}\subset\mathbb R$ .

Thus, $g(1,1,0)=2k\implies 2^3-2=2k\implies k=3$ . In conclusion, $g(x)=3(x+y)(y+z)(x+z)$ . Observe that $h(-y,y,z)=h(-z,y,z)=h(x,-z,z)=0$

what means that exists $\{m,n\}\subset\mathbb R$ so that $h(x,y,z)=(x+y)(x+z)(y+z)\left[m(x+y+z)^2+n(xy+yz+zx)\right]$ for any $\{x,y,z\}\subset\mathbb R$ .

Thus, $\left\{\begin{array}{ccc} h(1,1,0)=30=2(4m+n) & \implies & 4m+n=15\\\\ h(1,1,1)=240=24(3m+n) & \implies & 3m+n=10\end{array}\right\|\implies$ $\left\{\begin{array}{c} m=5\\\ n=-5\end{array}\right\|\implies$

$\boxed{\ \left(x+y+z\right)^5=x^5+y^5+z^5+5(x+y)(y+z)(z+x)\left(x^2+y^2+z^2+xy+yz+zx\right)\ }$ .

Verify : $x=1\ ,\ y=2\ ,\ z=2\ \implies\ 5^5=65+5\cdot 3\cdot 3\cdot 4\cdot (9+8)\iff$ $612=36\cdot 17$ .

Remark. Prove easily that $3\cdot \left[(x+y+z)^5-\left(x^5+y^5+z^5\right)\right]=$ $5\cdot\left[(x+y+z)^3-\left(x^3+y^3+z^3\right)\right]\left(x^2+y^2+z^2+xy+yz+zx\right)$ .

Proof 2 (with fundamental symmetrical forms). Denote the equation $(t-x)(t-y)(t-z)=t^3-s_1t^2+s_2t-s_3=0$ with the roots $\{x,y,z\}$ , where $\left\{\begin{array}{c} x+y+z=s_1\\\\ xy+yz+zx=s_2\\\\ xyz=s_3\end{array}\right\|$

and $(\forall )\ n\in\mathbb N$ , $S_n=x^n+y^n+z^n$ . Prove easily that $S_0=3$ , $S_1=s_1$ , $S_2=s_1^2-2s_2$ and $S_{n+3}=s_1\cdot S_{n+2}-s_2\cdot S_{n+1}+s_3\cdot S_n$ . Therefore,

$\left\{\begin{array}{ccccc} S_3 & = & x^3+y^3+z^3 & = & s_1^3-3s_1s_2+3s_3\\\\ S_4 & = & x^4+y^4+z^4 & = & s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2\\\\ S_5 & = & x^5+y^5+z^5 & = & s_1^5-5s_1^3s_2+5s_1^2s_3-5s_2s_3+5s_1s_2^2\end{array}\right\|\implies$ $s_1^5=S_5+5\left(s_1^2-s_2\right)\left(s_1s_2-s_3\right)$ . Prove easily that $(a+b)(b+c)(c+a)=s_1s_2-s_3$

and $a^2+b^2+c^2+ab+bc+ca=s_1^2-s_2$ . Thus, $\left(x+y+z\right)^5=x^5+y^5+z^5+5(x+y)(y+z)(z+x)\left(x^2+y^2+z^2+xy+yz+zx\right)$ .

Remark. $(a+b+c)^5=\sum (b+c-a)^5+80abc\left(a^2+b^2+c^2\right)$ and $(a+b+c)^5\ge 81abc\left(a^2+b^2+c^2\right)$ .

PP2. Let $z_k\ ,\ k\in\overline{1,4}$ be the roots of the equation $(z - i)^4=(2z - i)^4$ . Determine the value of the product $P\equiv \prod_{k=1}^4\left(z_k^2 + 1\right)$ .

Proof. Let $f(z)=(2z - i)^4-(z - i)^4=15\cdot\prod_{k=1}^4\left(z-z_k\right)$ . Thus, $P=\prod_{k=1}^4\left[\left(i-z_k\right)\cdot\left(-i-z_k\right)\right]=$ $\prod_{k=1}^4\left(i-z_k\right)\cdot \prod_{k=1}^4\left(-i-z_k\right)=$ $\frac {f(i)}{15}\cdot \frac {f(-i)}{15}=\frac {65}{225}\implies$ $\boxed{\ P=\frac {13}{45}\ }$

PP3. Let a cubic polynomial $P$ . You are also give $\left\{\begin{array}{c} Q_1(X) = X^2 + (k-29)X - k\\\\ Q_2(X) = 2X^2 + (2k-43)X + k\end{array}\right|$ . If both $Q_1$ and $Q_2$ are factors of the polynomial $P$ , then compute largest possible value of $k$

Proof. $Q_1$ , $Q_2$ are factors of $P$ iff the equations $Q_1(x)=0$ and $Q_2(x)=0$ have at least one common root iff $\left|\begin{array}{cc} 1 & -k\\\\ 2 & k\end{array}\right|^2=\left|\begin{array}{cc} 1 & k-29\\\\ 2 & 2k-43\end{array}\right|\cdot\left|\begin{array}{cc} k-29 & -k\\\\ 2k-43 & k\end{array}\right|\iff$

$9k^2=15k(3k-72)\iff$ $k\in \{0,30\}$ . In conclusion, $k=30$ and in this case $P(x)=0\implies$ $x\in \left\{-6,-\frac 52,5\right\}$ .

Remark (see PP12). I used the remarkable property: the equations $\left\{\begin{array}{c} a_1x^2+b_1x+c_1=0\\\\ a_2x^2+b_2x+c_2=0\end{array}\right|$ , where $a_1a_2\ne 0$ have at least one common root $\iff$ $\left|\begin{array}{cc} a_1 & c_1\\\\ a_2 & c_2\end{array}\right|^2=\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|\cdot\left|\begin{array}{cc} b_1 & c_1\\\\ b_2 & c_2\end{array}\right|$ .

PP4. Let $f = aX^3 + bX^2 + cX + d$ . Given that $f(2)= 5$ , $f(3)= 14$ and $f(8) - f(7) = 64$ . Find the value of $f(12) - f(11)$ .

Proof. Denote $g(X)=f(X)-f(X-1)-X^2$ . Observe that $\mathrm{gr}(g)\le 2$ , $g(X)=(3a-1)X^2+(2b-3a)X+a-b+c$

and $\left\{\begin{array}{ccc} x:=3 & \implies & g(3)=f(3)-f(2)-3^2=14-5-9=0\\\\ x:=8 & \implies & g(8)=f(8)-f(7)-8^2=64-6=0\end{array}\right|\implies$ $G(X)=(3a-1)(X-3)(X-8)$ .

In conclusion, $f(12)-f(11)=12^2+(3a-1)(12-3)(12-8)=$ $144+36(3a-1)\implies$ $\boxed{f(12)-f(11)=108(a+1)}$ .

PP5. Let the equation $x^{3} +ax +b=0\odot \begin{array}{ccc} \nearrow & x_1 &\searrow\\\\ \rightarrow & x_2 & \rightarrow\\\\ \searrow & x_3 & \nearrow\end{array}\odot$ . Find the equation with the roots $\left\{\begin{array}{c} y_1=\frac {x_2}{x_3}+\frac {x_3}{x_2}\\\\ y_2=\frac {x_3}{x_1}+\frac {x_1}{x_3}\\\\ y_3=\frac {x_1}{x_2}+\frac {x_2}{x_1}\end{array}\right|$ .

Proof. Denote $S_2=x_1^2+x_2^2+x_3^2$ and $\left\{\begin{array}{ccc} s_1 & = & x_1+x_2+x_3\\\\ s_2 & = & x_1x_2+x_2x_3+x_3x_1\end{array}\right|$ . Observe that $\left\{\begin{array}{c} s_1=0\\\\ s_2=a\end{array}\right|$ and $S_2=s_1^2-2s_2$ $\implies$ $S_2=-2a$ . Thus, $y_1=\frac {x_2}{x_3}+\frac {x_3}{x_2}=$ $\frac {x_2^2+x_3^2}{x_2x_3}=$

$\frac {S_2-x_1^2}{s_2-x_1\left(s_1-x_1\right)}\implies$ $\boxed{y_1=-\frac {x_1^2+2a}{x_1^2+a}}$ , i.e. $y_k=-\frac {x_k^2+2a}{x_k^2+a}\ ,\ k\in\overline{1,3}$ . I"ll eliminate $x$ between $\left\{\begin{array}{c} x^3+ax+b=0\\\\ y+\frac {x^2+2a}{x^2+a}=0\end{array}\right|$ $\iff$ $\left\{\begin{array}{c} x^3+ax+b=0\\\\ x^2=\frac {-a(y+2)}{y+1}\end{array}\right|$

$\iff$ $\left\{\begin{array}{c} x\cdot \frac {-a(y+2)}{y+1}+ax+b=0\\\\ x^2=\frac {-a(y+2)}{y+1}\end{array}\right|$ $\iff$ $\left\{\begin{array}{c} x=\frac {b(y+1)}{a}\\\\ x^2=\frac {-a(y+2)}{y+1}\end{array}\right|$ $\iff$ $\left[\frac {b(y+1)}{a}\right]^2+\frac {a(y+2)}{y+1}=0\iff$ $b^2(y+1)^3+a^3(y+2)=0$ .

In conclusion, the required equation is $\boxed{b^2y^3+3b^2y^2+\left(a^3+3b^2\right)y+\left(2a^3+b^2\right)=0}$ .

PP6. Solve the following system of equations $\left\{\begin{array}{cc} f(x,y)\equiv x^2+3y^2+4xy-18x-22y+31=0\\\\ g(x,y)\equiv 2x^2+4y^2+2xy+6x-46y+175=0\end{array}\right\|$ .

Proof 1. The idea is to let $x=u+a, y=v+b$ and find $a,b$ so that the coefficients of degree 1 become zero. Basically what we have to do is to take derivatives

of the two LHS in terms of $x$ and $y$ . Then let all 4 derivatives equal zero and solve that system of equations to obtain $a,b$ . Therefore, $\left\{\begin{array}{c} f'_x=2x+4y-18=0\\\\ f'_y=6y+4x-22=0\\\\ g'_x=4x+2y+6=0\\\\ g'_y=8y+2x-46=0\end{array}\right|\implies$

$x=-5$ and $y=7$ . Let $x=u-5, y=v+7$ we have the system $\left\{\begin{array}{cc} u^2+3v^2+4uv=1\\\\ 2u^2+4v^2+2uv=1\end{array}\right\|$ . Substract (3) from (4) we have $(u-v)^2=0 \Leftrightarrow u=v$ a.s.o.

Proof 2. $\left\{\begin{array}{c} x^2+3y^2+4xy-18x-22y+31=0\\\\ 2x^2+4y^2+2xy+6x-46y+175=0\end{array}\right\|\iff$ $\odot\begin{array}{ccc} \nearrow & x^2+2(2y-9)x+\left(3y^2-22y+31\right)=0 & \searrow\\\\ \searrow & 2x^2+2(y+3)x+\left(4y^2-46y+175\right)=0 & \nearrow\end{array}\odot x_{\mathrm{com}}\iff$

$\left|\begin{array}{cc} 1 & 3y^2-22y+31\\\\ 1 & 4y^2-46y+175\end{array}\right|^2=$ $\left|\begin{array}{cc} 1 & 2(2y-9)\\\\ 1 & 2(y+3)\end{array}\right|\cdot$ $\left|\begin{array}{cc} 2(2y-9) & 3y^2-22y+31\\\\ 2(y+3) & 4y^2-46y+175\end{array}\right|\iff$ $\left(y^2-24y+144\right)^2=$

$4(12-y)\left[(2y-9)\left(4y^2-46y+175\right)-(y+3)\left(3y^2-22y+31\right)\right]\iff$

$(y-12)^4+4(y-12)\left(5y^3-115y^2+799y-1668\right)=0\iff$

$(y-12)^4+4(y-12)^2\left(5y^2-55y+139\right)=0\iff$ $(y-12)^2\left(21y^2-244y+700\right)=0\ .$ a.s.o.

PP7. Prove that for any real number $x$ there is the inequality $x^8-x^7+x^2-x+15>0$ .

Proof. Denote $f(x)=x^8-x^7+x^2-x+15\ ,\ x\in\mathbb R$ . Therefore, $\left\{\begin{array}{ccccccc} x\not\in(0,1) & \implies & f(x) & = & 15+x(x-1)\left(x^6+1\right) & \ge & 15\\\\ x\in (0,1) & \implies & f(x) & = & 14+x^2+(x-1)^2\sum\limits_{k=0}^6x^k & > & 14\end{array}\right\|$ $\implies (\forall )\ x\in\mathbb R\ ,\ f(x)>0$ .

PP8. Let $(a,b,c,d)$ be a solution of the system $\left\{\begin{array}{c} a+b=15\\\\ ab+c+d=78\\\\ ad+bc=160\\\\ cd=96.\end{array}\right|$ . Find the greatest possible value of $a^2+b^2+c^2+d^2$ .

Proof. Let $P(x)=$ $(x^2-ax+b)(x^2-cx+d)$ . Expanding gives $P(x)=x^4-(a+b)x^3+(ab+c+d)x^2-(ad+bc)+cd=$ $x^4-15x^3+78x^2-160x+96=$

$(x-1)(x-4)^2(x-6)$ . Suppose w.l.o.g. that $x-1$ is a factor of $x^2-ax+b$ . The other factor is either $x-4$ or $x-6$ . Thus, $(a, b, c, d)=(5, 4, 10, 24)$ with sum of squares $717$

or $(a, b, c, d)=(7, 6, 8, 16)$ with sum of squares $405$ . In conclusion, $717$ is the greatest.

PP9. Let $\{x,y,z\}\subset\mathbb C$ such that $(y+z)(x-y)(x-z)=(z+x)(y-z)(y-x)=(x+y)(z-x)(z-y)=1$ . Find all possible values of $(y+z)(x+y)(z+x)$ .

Proof. So $x\ne y\ne z\ne x$ . Let $\left\{\begin{array}{c} y-z=a\\\ z-x=b\\\ x-y=c\end{array}\right|$ , where $a+b+c=0$ and $abc\ne 0$ . Thus, $bc(y+z)=ca(z+x)=ab(x+y)=-1\iff$ $\frac {y+z}a=\frac {z+x}b=$ $\frac {x+y}c=$

$\frac {2(x+y+z)}{a+b+c}=$ $-\frac 1{abc}\implies$ $\boxed{x+y+z=0}$ . Thus, $(y+z)(x-y)(x-z)=1\stackrel{(y+z=-x)}{\iff}$ $x(x-y)(x-z)+1=0\iff$ $x^3-x^2(y+z)+xyz+1=0$ a.s.o. $\stackrel{(y+z=-x)}{\iff}$

$\left\{\begin{array}{c} 2x^3=-(xyz+1)\\\\ 2y^3=-(xyz+1)\\\\ 2z^3=-(xyz+1)\end{array}\right|\bigodot\implies$ $(2xyz)^3+(xyz+1)^3=0\iff$ $2xyz+xyz+1=0\iff$ $xyz=-\frac 13\iff$ $\prod (y+z)=\frac 13$ .

PP10. Factorize over $\mathbb R$ (with the details and no extra assist) the polynomial $x^8+98x^4+1$ .

Proof 1. I"ll use the standard substitution $\boxed{x+\frac 1x=t\iff xt=x^2+1}$ . Therefore, our polynomial becomes :

$\boxed{\ \begin{array}{c} \\ x^8+98x^4+1\ \iff\ x^4\left[\left(x^4+\frac 1{x^4}\right)+98\right]\ \iff\ x^4\left[\left(x^2+\frac 1{x^2}\right)^2+96\right]\\\\ x^4\left\{\left[\left(x+\frac 1x\right)^2-2\right]^2+96\right\}\ \iff\ x^4\left[\left(t^2-2\right)^2+96\right]\ \iff\ x^4\left(t^4-4t^2+100\right)\\\\ x^4\left[\left(t^2+10\right)^2-24t^2\right]\ \iff\ x^4\left(t^2-2\sqrt 6\cdot t+10\right)\left(t^2+2\sqrt 6\cdot t+10\right)\\\\ \left[(xt)^2-2\sqrt 6\cdot x(xt)+10x^2\right]\left[(xt)^2+2\sqrt 6\cdot x(xt)+10x^2\right]\\\\ \left[\left(x^2+1\right)^2-2\sqrt 6\cdot x\left(x^2+1\right)+10x^2\right]\left[\left(x^2+1\right)^2+2\sqrt 6\cdot x\left(x^2+1\right)+10x^2\right]\\\\ \boxed{\left(x^4-2\sqrt 6\cdot x^3+12x^2-2\sqrt 6\cdot x+1\right)\left(x^4+2\sqrt 6\cdot x^3+12x^2+2\sqrt 6\cdot x+1\right)}\\\\ \left(x^4+12x^2+1\right)^2-\left[2\sqrt 6x\left(x^2+1\right)\right]^2\\\ \end{array}\ }$
Proof 2. I"ll use the standard substitution $\boxed{x-\frac 1x=t\iff xt=x^2-1}$ . Therefore, our polynomial becomes :

$\boxed{\ \begin{array}{c} \\ x^8+98x^4+1\ \iff\ x^4\left[\left(x^4+\frac 1{x^4}\right)+98\right]\ \iff\ x^4\left[\left(x^2+\frac 1{x^2}\right)^2+96\right]\\\\ x^4\left\{\left[\left(x-\frac 1x\right)^2+2\right]^2+96\right\}\ \iff\ x^4\left[\left(t^2+2\right)^2+96\right]\ \iff\ x^4\left(t^4+4t^2+100\right)\\\\ x^4\left[\left(t^2+10\right)^2-16t^2\right]\ \iff\ x^4\left(t^2-4t+10\right)\left(t^2+4t+10\right)\\\\ \left[(xt)^2-4x(xt)+10x^2\right]\left[(xt)^2+4x(xt)+10x^2\right]\\\\ \left[\left(x^2-1\right)^2-4x\left(x^2-1\right)+10x^2\right]\left[\left(x^2-1\right)^2+4x\left(x^2-1\right)+10x^2\right]\\\\ \boxed{\left(x^4-4x^3+8x^2+4x+1\right)\left(x^4+4x^3+8x^2-4x+1\right)}\\\\ (x^4 + 8x^2 + 1)^2 - (4x^3 - 4x)^2\\\ \end{array}\ }$
Proof 3. $x^8 + 98x^4 + 1 = (x^4 + 8x^2 + 1)^2 - (4x^3 - 4x)^2 \implies$ $x^8 + 98x^4 + 1 =(x^4 + 4x^3 + 8x^2 - 4x + 1)(x^4 - 4x^3 + 8x^2 + 4x + 1)$ .

Remark. The last two proofs present the factorize over $\mathbb Z$ .

Proof 4. $x^8+98x^4+1=\left(x^4+49\right)^2+1-49^2=$ $\left(x^4+49\right)^2-48\cdot 50=$ $\left(x^4+49\right)^2-\left(20\sqrt 6\right)^2\implies$ $x^8+98x^4+1=\left(x^4+49+20\sqrt 6\right)\left(x^4+49-20\sqrt 6\right)$ .

PP11. Let $\{a,b,c\}\subset\mathbb R$ such that $abc=-1$ , $a+b+c=4$ and $\sum\frac{a}{a^2-3a-1}=\frac{4}{9}$ . Find the value of $a^2+b^2+c^2$ .

Proof. $ab+bc+ca=m$ , $x^3-4x^2+mx+1=0$ $\odot\begin{array}{ccc} \nearrow & a & \searrow\\\ \rightarrow & b & \rightarrow\\\ \searrow & c & \nearrow\end{array}\odot$ . Thus, $x^3-4x^2+mx+1=0\stackrel{(*)}{\iff}$ $\frac {x}{x^2-3x-1}=\frac {x-1}{2-m}$ . So $\sum\frac{a}{a^2-3a-1}=\frac{4}{9}$ $\iff$

$\sum\frac {a-1}{2-m}=\frac 49\iff$ $9\left(-3+\sum a\right)=4(2-m)\iff$ $9=4(2-m)\iff$ $\boxed{m=-\frac 14}$ . In conclusion, $a^2+b^2+c^2=$ $(a+b+c)^2-2(ab+bc+ca)=$ $16+\frac 12$ $\implies$

$a^2+b^2+c^2=$ $\frac {33}{2}$ . $*\blacktriangleright$ Remark. $0=x^3-4x^2+mx+1=(x^2-3x-1)(x-1)+(m-2)x\implies$ $\frac {x}{x^2-3x-1}=\frac {x-1}{2-m}$ .

PP12. The equations $\left\{\begin{array}{c} a_1x^2+b_1x+c_1=0\\\\ a_2x^2+b_2x+c_2=0\end{array}\right\|$ have at least one common root $\iff$ $\left|\begin{array}{cc} a_1 & c_1\\\\ a_2 & c_2\end{array}\right|^2=\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|\cdot\left|\begin{array}{cc} b_1 & c_1\\\\ b_2 & c_2\end{array}\right|$ and the common root is $x_{\mathrm{com}}=-\frac {\left|\begin{array}{cc} a_1 & c_1\\\\ a_2 & c_2\end{array}\right|}{\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|}$ .

Proof. Using the Cramer's rules solve the system $\left\{\begin{array}{c} a_1\cdot \left(x^2\right)+b_1\cdot (x)=-c_1\\\\ a_2\cdot\left(x^2\right)+b_2\cdot (x)=-c_2\end{array}\right\|\implies$ $\left(x^2\right)=\frac {\left|\begin{array}{cc} b_1 & c_1\\\\ b_2 & c_2\end{array}\right|}{\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|}\ \ \wedge\ \ (x)=-\frac {\left|\begin{array}{cc} a_1 & c_1\\\\ a_2 & c_2\end{array}\right|}{\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|}$ .

In conclusion, $(x)^2=\left(x^2\right)\implies$ $\left|\begin{array}{cc} a_1 & c_1\\\\ a_2 & c_2\end{array}\right|^2=\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|\cdot\left|\begin{array}{cc} b_1 & c_1\\\\ b_2 & c_2\end{array}\right|$ and in this case the common root is $x_{\mathrm{com}}=-\frac {\left|\begin{array}{cc} a_1 & c_1\\\\ a_2 & c_2\end{array}\right|}{\left|\begin{array}{cc} a_1 & b_1\\\\ a_2 & b_2\end{array}\right|}$

PP13. Prove that the roots of the equation $\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = \frac{1}{x}$ , where $0<c<b<a$ , are real.

Proof 1 (without derivatives). $\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = \frac{1}{x}\iff$ $f(x)\equiv x[(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)]-(x-a)(x-b)(x-c)=0$ .

Thus, $f(x)\equiv 2x^3-(a+b+c)x^2+abc=0\odot\begin{array}{ccc} \nearrow & x_1 & \searrow\\\\ \rightarrow & x_2 & \rightarrow\\\\ \searrow & x_3 & \nearrow\end{array}\odot$ and prove easily that $\left\{\begin{array}{cccc} f(-\infty ) & = & -\infty & -\\\\ f(0) & = & abc & +\\\\ f(c) & = & c(c-a)(c-b) & +\\\\ f(b) & = & b(b-a)(b-c) & -\\\\ f(a) & = & c(c-a)(c-b) & +\\\\ f(\infty ) & = & \infty & +\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccccc} f(-\infty ) & < & 0 & < & f(0)\\\\ f(c) & > & 0 & > & f(b) \\\\ f(b) & < & 0 & < & f(a)\end{array}\right\|$ .

In conclusion, the equation $f(x)=0$ has three real roots so that $x_1<0<c<x_2<b<x_3<a$ .

Proof 2 (with derivatives). Denote $\frac {a+b+c}3=m$ and prove easily that $f(x)=x\sum (x-b)(x-c)-(x-a)(x-b)(x-c)=0\iff$

$f(x)=2x^3-(a+b+c)x^2+abc=0$ . Since $f^{\prime}(x)=6x^2-2(a+b+c)x=0\begin{array}{ccc} \nearrow & 0 & \searrow\\\\ \searrow & \frac {a+b+c}3 & \nearrow\end{array}\odot$ obtain that

$\left\{\begin{array}{cccc} -\infty\ <\ x\ <\ 0 & \implies & f\ \nearrow & 0\ >\ -\infty \ =\ f(-\infty)\ <\ f(x)\ <\ f(0)=abc\ >\ 0\\\\\ 0\ <\ x\ <\ m & \implies & f\ \searrow & 0\ <\ abc\ =\ f(0)\ >\ f(x)\ >\ f(m)=-m^3+27abc\ <\ 0\\\\ m\ <\ x\ <\ \infty & \implies & f\ \nearrow & 0\ >\ -m^3+27abc\ =\ f(m)\ <\ f(x)\ <\ f(\infty )\ =\ \infty \ >\ 0\end{array}\right\|$ . Thus, $0<c<m<a$ and $f(-\infty)<0<f(c)=$

$(c-a)(c-b)>0>f(m)<0<f(a)=(a-b)(a-c)$ . In conclusion, $(\exists )\ \left\{x_1,x_2,x_3\right\}\subset\mathbb R$ so that $x_1<0<c<x_1<m<x_3<a$ .

PP14. Find $a\in\mathbb R$ so that exist $\{x,y\}\subset\mathbb R^*$ for which $\left\{\begin{array}{cc} x^2-2xy-3y^2=8\\\\ 2x^2+4xy+5y^2=m\end{array}\right\|$ , where $m=a^4-4a^3+4a^2-12+\sqrt{105}$ .

Proof. Let $\boxed{\ t=\frac xy\ }$ . Thus, $m=2x^2+4xy+5y^2=$ $y^2\left(2t^2+4t+5\right)>0$ $\implies$ $\boxed{ m > 0\ }\ (*)$ . Therefore, $\frac { x^2-2xy-3y^2}{2x^2+4xy+5y^2}=$ $\frac 8m\iff$ $\frac { t^2-2t-3}{2t^2+4t+5}=$ $\frac 8m\iff$

$(m-16)t^2-2(m+16)t-(3m+40)=0$ . Since $t\in\mathbb R$ must $\Delta^{\prime}(m)=$ $(m+16)^2+(m-16)(3m+40)\ge 0$ $\iff$ $4m^2+24m-384\ge 0$ $\iff$

$m^2+6m-96\ge 0\stackrel{(*)}{\iff}$ $\boxed{\ m\ge -3+\sqrt{105}\ }$ . In conclusion, $a^4-4a^3+4a^2-12+\sqrt{105}\ge -3+\sqrt{105}$ $\iff$ $a^4-4a^3+4a^2-9\ge 0\iff$

$(a+1)(a-3)(a^2-2a+3)\ge 0$ $\iff$ $\boxed{\ a\in (-\infty,-1]\cup[3,\infty)}$ .

PP15. Find the remainder when the polynomial $P(x)\equiv (x+2)^{2011}-(x+1)^{2011}$ is divided by $x^{2}+x+1$ .

Proof. Let $x^2+x+1=0\begin{array}{ccc} \nearrow & w & \searrow\\\\ \searrow & \overline w & \nearrow\end{array}\odot$ where $\left\{\begin{array}{ccc} w & = & \cos\frac {2\pi}3+i\sin\frac {2\pi}3\\\\ w^3=1 & ; & w^2+w+1=0\end{array}\right\|$ . Exist $\{a,b\}\subset\mathbb R$ so that $P(x)=\left(x^2+x+1\right)\cdot Q(x)+ax+b$ .

$\left\{\begin{array}{ccc} (w+2)^{2011}=(w+2)^{2\cdot 1005+1}=\left(w^2+4w+4\right)^{1005}(w+2)=\left(-3w^2\right)^{1005}(w+2) & \implies & (w+2)^{2011}=-3^{1005}(w+2)\\\\ (w+1)^{2011}=(w+1)^{2\cdot 1005+1}=\left(w^2+2w+1\right)^{1005}(w+1)=w^{1005}(w+1) & \implies & (w+1)^{2011}=w+1\end{array}\right\|$ . Therefore,

$P(w)=(w+2)^{2011}-(w+1)^{2011}\implies$ $P(w)=-3^{1005}(w+2)-(w+1)$ . In conclusion, $\boxed{P(w)=-\left(3^{1005}+1\right)\cdot w-\left(2\cdot 3^{1005}+1\right)}\ (*)$ .

Since $\{a,b\}\subset\mathbb R$ obtain that $P(w)=aw+b\ \stackrel{(*)}{\implies}$ $\left\{\begin{array}{ccc} a & = & -\left(3^{1005}+1\right)\\\\ b & = & -\left(2\cdot 3^{1005}+1\right)\end{array}\right\|$ .

PP16. Let $x^3-x^2+m(9x-1)=0$ be an equation with the roots $\{x_1,x_2,x_3\}\subset \mathbb R^*_+$ , where $m\in\mathbb R$ . Prove that $x_1=x_2=x_3$ .

Proof. Observe that $\left\{\begin{array}{ccccc} s_1 & = & x_1+x_2+x_3 & = & 1\\\\ s_2 & = & x_1x_2+x_2x_3+x_3x_1 & = & 9m\\\\ s_3 & = & x_1x_2x_3 & = & m\end{array}\right\|\implies$ $\frac 1{x_1}+\frac 1{x_2}+\frac 1{x_3}=\frac {s_2}{s_3}=9\implies$ $9=\left(x_1+x_2+x_3\right)\left(\frac 1{x_1}+\frac 1{x_2}+\frac 1{x_3}\right)\ge 9\implies$ $x_1=x_2=x_3$ .

This post has been edited 91 times. Last edited by Virgil Nicula, Nov 17, 2015, 7:04 AM

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140. A synthetical property <==> linear/angled relation.

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136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

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133. Some problems with projections/perpendiculars (1).

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131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

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125. I like very much this Darij Grinberg's message.

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122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

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119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

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112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

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108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

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104. Concursul Revistei de Matematica din Galati (RMG).

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102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

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August 2010

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94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

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70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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by OlympusHero, Sep 14, 2022, 4:44 AM
blog
by tigerzhang, Aug 1, 2021, 12:02 AM
Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
by GoogleNebula, Apr 14, 2021, 5:25 AM
Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

355. Some problems with polynomials or equations I

by Virgil Nicula, Aug 28, 2012, 1:21 PM

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