405. Art of problem solving (algebra).

by Virgil Nicula, Nov 19, 2014, 9:49 AM

E0. Prove that $(\forall )\ \left\{a,b,c\right\}\subset\mathbb C$ exists the chain $:\ ABC$ is equilateral $\iff ABC\sim BCA\iff$ $\left|\begin{array}{ccc}
1 & 1 & 1\\\\
a & b & c\\\\
b & c & a\end{array}\right|\iff$ $a^2+b^2+c^2=ab+bc+ca$ .

Proof 1. Let $w=\cos\frac {\pi}{3}+i\sin\frac {\pi}{3}$ , where $\left\{\begin{array}{c}
w^2+1=w\\\\
\overline w=\frac 1w=-w^2\end{array}\right\|$ . Thus, $ABC$ is equilateral $\iff b-a=w(c-a)\ \vee\ b-a=\overline w(c-a)\iff$

$\frac {b-a}{c-a}\in\{w,\overline w\}\iff$ $\left(\frac {b-a}{c-a}\right)^2+1=\frac {b-a}{c-a}\iff$ $\boxed{(b-a)^2+(c-a)^2=(b-a)(c-a)}\iff$ $a^2+b^2+c^2=ab+bc+ca$ .

Proof 2. Let the midpoint $M(m)$ of $[BC]$ , where $2m=b+c$ . Thus, $\left\{\begin{array}{c}
AM\perp BC\\\\
2\cdot AM=\sqrt 3\cdot BC\end{array}\right\|$ . Thus, $M\left(\frac {b+c}2\right)$ and $2AM=BC\sqrt 3\implies 2\cdot |m-a|=|b-c|\sqrt 3$ .

$ABC$ is equilateral $\iff (m-a)|b-c|=\pm i(b-c)|m-a|\iff$ $2\left(\frac {b+c}{2}-a\right)=\pm i(b-c)\sqrt 3\iff$ $\boxed{(b+c-2a)^2+3(b-c)^2=0}\iff$ $\sum a^2=ab+bc+ca$ .



E1. Let $\{A(a),B(b),C(c)\}\subset\mathbb C$ . Prove that $A\in BC\iff$ $a\cdot \overline b+b\cdot \overline c+c\cdot \overline a\in\mathbb R\iff$ $\left|\begin{array}{ccc}
1 & 1 & 1\\\
a & b & c\\\
\overline a & \overline b & \overline c\end{array}\right|=0$ .

Proof. $A\in BC\iff$ $(\exists ) \lambda \in \mathbb R$ so that $a-b=\lambda \cdot (c-b)\iff$ $(a-b)\cdot (\overline c-\overline b)=(c-b)\cdot (\overline a-\overline b)\iff$

$a\overline c-a\overline b-b\overline c=c\overline a -c\overline b-b\overline a\iff$ $z=a\overline b+b\overline c+c\overline a=a\overline c+c\overline b+b\overline a=\overline z\iff$ $z=a\overline b+b\overline c+c\overline a\in\mathbb R\iff$ $\left|\begin{array}{ccc}
1 & 1 & 1\\\
a & b & c\\\
\overline a & \overline b & \overline c\end{array}\right|=0$ .



E2. Find the roots of 2th equation with complex coefficients.

Proof. Let $az^2+bz+c=0$ be a second degree polynomial equation with $\{a,b,c\}\subset\mathbb C$ and $a\ne 0$ . Suppose w.l.o.g. that exists at least a nonreal

number in the set $\{a,b,c\}$ and $\Delta=b^2-4ac=u+iv$ , where $\{u,v\}\subset\mathbb R$ and $v\ne 0$ . Thus, $az^2+bz+c=0\iff (2az+b)^2=\Delta\ (*)$ . Solve

$y^2=u+iv$ . Prove easily that $y_{1,2}=\pm w$ , where $w=\sqrt{\frac {r+u}{2}}+i\cdot\mathrm{sgn}(v)\cdot\sqrt {\frac {r-v}{2}}$ and $r=|\Delta|$ . From $(*)$ obtain that $z_{1,2}=\frac {-b\pm w}{2a}$ .



E3. Let $z_{1,2}\in\mathbb C$ be the roots of the equation with complex coefficients $az^2+bz+c=0\ ,\ a\ne 0$ . Prove that $\boxed{\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}}$ .

Proof. $\left\{\begin{array}{cccc}
\left|z_1-z_2\right|^2=\left|\left(z_1+z_2\right)^2-4z_1z_2\right| & \implies & \left|z_1-z_2\right|^2=\left|\frac {b^2-4ac}{a^2}\right| &  (1)\\\\
\boxed{\left|z_1+z_2\right |^2+\left|z_1-z_2\right|^2=2\left(\left|z_1\right|^2+\left|z_2\right|^2\right)} & \stackrel{(1)}{\implies} & \left|z_1\right|^2+\left|z_2\right|^2=\frac 12\cdot \left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{|a|^2}\right) &  (2)\\\\
\left(\left|z_1\right|+\left|z_2\right|\right)^2=\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1z_2\right| & \stackrel{(2)}{\implies} & \left(\left|z_1\right|+\left|z_2\right|\right)^2=\frac 12\cdot\left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{2|a|^2}\right)+2\cdot\left|\frac ca\right| & (3)\end{array}\right\|$ $\stackrel{(3)}{\implies}$ $\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}$ .

Example. Let $\left\{z_1,z_2\right\}\subset\mathbb C$ be the roots of $z^2-2iz+m=0$ . Find the values of $m$ for which $\left|z_1\right|+\left|z_2\right|=2$ .

Proof. For $a:=1\ ,\ b:=-2i$ and $c:=m\in\mathbb C$ obtain that $\left|z_1\right|+\left|z_2\right|=2\iff$ $\sqrt {\frac {4+4|1+m|+4|m|}{2}}=2\iff$ $|m|+|m+1|=1$ . Let

$M(m)\ ,\ A(0)\ ,\ B(-1)$ . Thus, $|m|+|m+1|=1\iff$ $MA+MB=AB\iff$ $M\in[AB]\iff m\in\mathbb R\ ,\ m\in [-1,0]$ . Nice exercise !



E4. Find the minimum value of $E(x,y)=x^2+y^2$ , where $2x+3y=6$ and $\{x,y\}\subset\mathbb R^*_+$ .

Proof 1. I"ll use the Lagrange's identiy $(ax+by)^2+(bx-ay)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)$ . Thus, for $a:=2$ and $b:=3$ obtain that

$x^2+y^2=\frac 1{13}\cdot\left[36+(3x-2y)^2\right]\ge \frac {36}{13}\implies$ $\boxed{x^2+y^2\ge\frac {36}{13}}$ . We"ll have the equality iff $3x-2y=0$ , i.e. $3x=2y\iff$ $\frac x2=\frac y3=\frac 6{13}$ .

Proof 2. $2x+3y=6$ is the equation of $AB$ , where $\left\{\begin{array}{c}
A(3,0)\subset Ox\\\\
B(0,2)\subset Oy\end{array}\right\|$ and $E(x,y)=x^2+y^2=MO^2$ and $M(x,y)\in [AB]$ is a mobile point.

Thus, $E(x,y)$ is $\min\iff\ M$ becomes the projection $P$ of the origin $O$ on $AB$ , i.e. $OP\cdot AB=OA\cdot OB\iff$ $OP=\frac {2\cdot 3}{\sqrt{13}}$ and in this case $E(x,y)\ge OP^2=\frac {36}{13}$ .



E5. Prove that $x\in\mathbb R\ \implies\ a\sin x+b\cos x\le \sqrt{a^2+b^2}$ and $x\in\left(0,\frac {\pi}2\right)\implies 1<\sin x+\cos x\le \sqrt 2$ .

Proof. $a\sin x+b\cos x\le \sqrt{a^2+b^2}\iff$ $(a\sin x+b\cos x)^2\le a^2+b^2\iff$ $2ab\sin x\cos x\le a^2\left(1-\sin^2x\right)+b^2\left(1-\cos^2x\right)\iff$

$2ab\sin x\cos x\le (a\cos x)^2+(b\sin x)^2\iff$ $(a\cos x-b\sin x)^2\ge 0$ , what is truly. We'll have the equality iff $a\cos x=b\sin x\iff$ $\tan x=\frac ab$ .

Analogously $1<\sin x+\cos x\iff$ $1<(\sin x+\cos x)^2\iff$ $0<2\sin x\cos x\iff$ $0<\sin 2x$ , what is truly. We"ll have the equality iff $x\in\left\{0,\frac {\pi}2\right\}$ .



E6. Find the minimum value of $E(x,y,z)=x^2+y^2+z^2$ , where $ax+by+cz=1$ and $\{a,b,c\}\subset\mathbb R^*$ .

Proof. I"ll use the Lagrange's identity $(ax+by+cz)^2+(ay-bx)^2+(bz-cy)^2+(cx-az)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)$ .

Thus, $\boxed{\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\ge (ax+by+cz)^2}$ (C.B.S. inequality). We"ll have the equality iff $ay=bx\ ,\ bz=cy\ ,\ cx=az$ ,

i.e. $\frac ax=\frac by=\frac cz$ . Thus, $x^2+y^2+z^2\ge\frac 1{a^2+b^2+c^2}$ with equality iff $\frac ax=\frac by=\frac cz=a^2+b^2+c^2$ .



E7. Find the minimum value of an expression $E(x)=x(x+1)(x+2)(x+3)$ , where $x\in\mathbb R$ .

Proof. Let $x(x+3)=y\ge -\frac 94$ . Thus, $E(x)$ is $\min \iff$ $f(y)=y(y+2)$ is $\min\iff$ $\min_{4y+9\ge 0}\left(y^2+2y\right)=-1$ and is touched in $y=-1$ , i.e. $x_{\min}\in\left\{\frac {-3\pm\sqrt 5}2\right\}$ . .


E8. Ascertain the minimum value of $E(a,b )=a^2+b^2$ for which $E(x)=x^4+ax^3+bx^2+ax+1=0$ has at least one real root.

Proof. Since $x\ne 0$ can denote $x+\frac 1x=t$ , where $|t|\ge 2$ . Obtain that $at+b+t^2-2=0$ which for fixed $t$ is the equation of a line in the analytical plane $xOy$ of the points $(a,b)$ .

The distance from the origin to this line is $\boxed{d(t)=\frac {\left|t^2-2\right|}{\sqrt{t^2+1}}}\ (*)$ . The required value is the minimum of $d^2(t)$ . Obtain that $t_{\min}\in\{\pm 2\}$ and $\min\left(a^2+b^2\right)=d^2\left(\pm 2\right)=\frac 45$ .



E9. Is there an easy way to find two real numbers given their sum and product ? For example, the sum is $21$ and their product is $104$ .

Proof. Prove easily or it's well-known that $(\forall )\ \{x,y\}\subset\mathrm R^2$ exist the identities $\left\{\begin{array}{cc}
2\cdot \mathrm{max}\{x,y\}=(x+y)+|x-y| & (1)\\\\
2\cdot \mathrm{min}\{x,y\}=(x+y)-|x-y| & (2)\end{array}\right|$ .

Thus, $\left|\begin{array}{ccc}
x+y & = & 21\\\
xy & = & 104\end{array}\right|\implies$ $(x-y)^2=x^2+y^2-2xy=$ $(x+y)^2-4xy=21^2-4\cdot 104=441-416=25\implies$

$|x-y|=5\stackrel{(1)}{\implies}$ $2\mathrm{max}\{x,y\}=(x+y)+|x-y|=21+5=26\implies$ $\left|\begin{array}{ccc}
\mathrm{max}\{x,y\} & = & 13\\\\
x+y & = & 21\end{array}\right|\implies$ $\{x,y\}=\{8,13\}$ .

Remark. Prove that $\left\{\begin{array}{ccc}
x+y & = & S\\\
xy & = & P\end{array}\right|$ , where $S^2\ge 4P\ \implies\ \{x,y\}=$ $\left\{\frac {S\pm\sqrt{S^2-4P}}{2}\right\}$ .



E10. Find $\{a,b,c\}\subset\mathbb R$ for which $a^2+b^2+c^2=$ $a^3+b^3+c^3=a^4+b^4+c^4$ .

Proof. $\sum a^2=\sum a^3=\sum a^4\implies$ $\sum\left(a^2-a\right)^2=\sum a^4-2\cdot\sum a^3+\sum a^2=0\implies$ $\sum\left(a^2-a\right)^2=0\implies$ $\left(a^2=a\right)\ \wedge\ \left(b^2=b\right)\ \wedge\ \left(c^2=c\right)\implies$

$\{a,b,c\}\subset\{0,1\}$ .We"ll have $1+3+3+1=8$ solutions, i.e. $(a,b,c)\in\{(0,0,0)\ ;\ (0,0,1)\ ;\ (0,1,0)\ ;\ (0,1,1)\ ;\ (1,$ $0,0)\ ;\ (1,0,1)\ ;\ (1,1,0)\ ;\ (1,1,1)\}$ .



E11. Prove that $(\forall )\ x\in\mathbb R\ ,\ x\ne \frac 23$ exists the equivalence $1<\frac {2x+1}{3x-2} <3\iff 1<x<3$ .

Proof. I"ll use the evident property $a<X<b\iff (a-X)(b-X)<0$ . Thus, $1<\frac {2x+1}{3x-2}<3\iff$

$\left(1-\frac {2x+1}{3x-2}\right)\left(3-\frac {2x+1}{3x-2}\right)<0\iff$ $(x-3)(x-1)<0\iff 1<x<3$ .



E12. Prove that $ x=\frac{2(ab+bc+ca)}{a+b+c}\ \ \vee\ \ x=$ $0\ \iff\ \frac{x+a}{x-a}+\frac{x+b}{x-b}+\frac{x+c}{x-c}=\frac{6abc}{(x-a)(x-b)(x-c)}+3 $ .

Proof. $\sum\frac {x+a}{x-a}=\frac{6abc}{(x-a)(x-b)(x-c)}+3\iff$ $\sum\left(\frac {x+a}{x-a}-1\right)=\prod\frac {2a}{x-a}$ $\iff$ $\sum\frac {a}{x-a}=3\cdot\prod\frac {a}{x-a}$ .

Thus, $\sum a(x-b)(x-c)=3abc\iff$ $x^2\sum a-2x\sum bc=0\iff$ $ x=\frac{2(ab+bc+ca)}{a+b+c}\ \ \vee\ \ x=0$ .



E13. Calculate the sum $1^2-3^2+5^2-7^2+\ldots +97^2-99^2$ .

Proof. Denote $S_n=\left(1^2-3^2\right)+\left(5^2-7^2\right)+\ldots+\left[(4n-3)^2-(4n-1)^2\right]$ , where $n\in\mathbb N^*$ . We must calculate $S_{50}$ . Observe that $S_1=-8$ and is well-known that

$\sum_{k=1}^n(2k-1)=n^2$ . Thus, $S_k-S_{k-1}=$ $(4k-3)^2-(4k-1)^2=-8(2k-1)\ ,\ k>1$ $\implies$ $\sum_{k=2}^n\left[S_k-S_{k-1}\right]=-8\sum_{k=2}^n(2k-1)=$ $-8\left(n^2-1\right)\implies$ $S_n-S_1=$

$-8\left(n^2-1\right)\implies$ $S_n=-8-8\left(n^2-1\right)=-8n^2\implies$ $\boxed{S_n=-8n^2\ ,\ (\forall )\ n\in\mathbb N^*}$ . In the particular case $n:=50$ obtain that $S_{50}=-8\cdot 50^2\implies$ $S_{50}=-20000$ .



E14. Let $a_i\in\mathbb N^*\ ,\ i\in\overline{1,m}$ and $b_j\in\mathbb N^*\ ,\ j\in\overline{1,n}$ . Denote $X=\prod_{i=1}^ma_i$ and $Y=\prod_{j=1}^nb_j$ . Prove that $\prod_{\begin{array}{c}
1\le i\le m\\\
1\le j\le n\end{array}}\frac {a_i}{b_j}=\frac {X^n}{Y^m}$ .

Proof. $a_i=b_j\cdot \frac {a_i}{b_j}$ , where $i\in \overline{1,m}$ and $j\in\overline{1,n}\implies$ $X=\prod_{i=1}^ma_i=\prod_{i=1}^m\left(b_j\cdot \frac {a_i}{b_j}\right)=$ $\prod_{i=1}^mb_j\cdot \prod_{i=1}^m\frac {a_i}{b_j}=$ $b_j^m\cdot \prod_{i=1}^m\frac {a_i}{b_j}\implies$

$X^n=\prod_{j=1}^nX=\prod_{j=1}^n\left(b_j^m\cdot \prod_{i=1}^m\frac {a_i}{b_j}\right)=$ $\left(\prod_{j=1}^nb_j\right)^m\cdot \left(\prod_{j=1}^n\prod_{i=1}^m\frac {a_i}{b_j}\right)=$ $Y^m\cdot \prod_{\begin{array}{c}
1\le i\le m\\\
1\le j\le n\end{array}}\frac {a_i}{b_j}\implies$ $\prod_{\begin{array}{c}
1\le i\le m\\\
1\le j\le n\end{array}}\frac {a_i}{b_j}=\frac {X^n}{Y^m}$ .



PP15 (England - NMO - 2008). Solve over $\mathbb R$ the following system$\ :\ \left\{\begin{array}{ccc}
xy(z+1) & = & 4\\\\
yz(x+1) & = & 12\\\\
zx(y+1) & = & 4\end{array}\right\|$

Proof. Denote $\boxed{\ xyz=t\ }\ (*)$ , where $\{x,y,z\}\subset\mathbb R^*$ . System becomes $\left\{\begin{array}{ccc}
xy & = & 4-t\\\\
yz & = & 12-t\\\\
zx & = & 4-t\end{array}\right\|\ \bigodot\ \stackrel{(*)}{\implies}\ t^2=(4-t)(12-t)(4-t)\iff$

$(t-4)^2(t-12)+t^2=0\iff$ $t^3-19t^2+112t-192=0\iff$ $(t-3)(t-8)^2=0\implies\odot\begin{array}{ccccccc}
\nearrow & t=3 & \implies & x=\frac 13 & ; & y=z=3 & \searrow\\\\
\searrow & t=8 & \implies & x=2 & ; & y=z=-2 & \nearrow\end{array}\odot$



PP16. Let the "triangle" of all odd natural numbers $\begin{array}{ccc}
L_1 & : &  1\\\
L_2 & : & 3\ \ 5\\\
L_3 & : & 7\ \ 9\ 11\\\
L_4 & : & 13\ 15\ 17\ 19\\\
\ldots & : & \ldots\ \ldots\ \ldots\ \ldots\end{array}$ . Ascertain $:$

$\blacktriangleright$ The first term and the last term of the $\mathrm{n^{th}}$ line, where $n\in \mathbb N^*\ ;$

$\blacktriangleright$ The sum of the terms from the $\mathrm{n^{th}}$ line $\ ;$

$\blacktriangleright$ Prove that $1^3+2^3+\ \ldots\ +(n-1)^3+n^3=\left[1+2+3+\ \ldots\ +(n-1)+n\right]^2\ ;$

$\blacktriangleright$ Find the "coordinates" of the number $1001$ , i.e. find the line that belongs to and where is number the $1001$ in this line.



PP17. A polynomial $f$ with rational coefficients leaves remainder $15$ when divided by $(X-3)$ and remainder

$(2X+1)$ when divided by $(X-1)^2$ .Find remainder when divided polynomial $f$ by $(X-3)(X-1)^2$ .


Proof. The polynomial $f$ leaves remainder $(2X+1)$ when divided by $(X-1)^2\iff$ exists $g\in\mathbb Q[X]$ so that $\boxed{f=(X-1)^2\cdot g+(2X+1)}\ (*)$ . The polynomial $f$ leaves

remainder $15$ when divided by $(X-3)\iff f(3)=15\ \stackrel{(*)}{\iff}$ $4g(3)+7=15\iff$ $\boxed{g(3)=2}\iff$ the polynomial $g$ leaves remainder $2$ when divided by $(X-3)\iff$

exists $h\in\mathbb Q[X]$ so that $\boxed{g=(X-3)\cdot h+2}$ . In conclusion, the required polynomials $f$ with given properties are $f\ \stackrel{(*)}{=}\ (X-1)^2\cdot \left[(X-3)\cdot h+2\right]+(2X+1)=$

$(X-3)(X-1)^2\cdot h+2(X-1)^2+2X+1\iff$ $\boxed{f=(X-3)(X-1)^2\cdot h+\left(2X^2-2X+3\right)}\ ,$ where $h\in\mathbb Q[X]$ .



PP18. Solve over $\mathbb R$ the system $:\ \left\{\begin{array}{ccc}
a^3-2ab^2-4a-8b & = & 0\\\\
2b^3-a^2b-4a+16b & = & 0\end{array}\right\|$ .

Proof. Eliminate $a\ :\ \left\{\begin{array}{cccccc}
 a^3 & * & -2(b^2+2)a & -8b & = & 0\\\\
* & ba^2 & 4a & -2b(b^2+8) & = & 0\end{array}\right|\left|\begin{array}{cc}
\odot & (-b)\\\\
\odot & a\end{array}\right\|\bigoplus$ $\implies$ $4a^2-12ba+8b^2=0\implies$ $\boxed{a^2-3ab+2b^2=0}\odot\begin{array}{cc}
\nearrow & a=b\\\\
\searrow & a=2b\end{array}$ .

For $\underline{a=b}$ in the upper second line obtain that $b^3+4b-2b(b^2+8)=0$ , i.e. $b^3+12b=0\implies \underline {b=0}$ . For $\underline{\underline{a=2b}}$ in the same equation obtain

that $4b^3+8b-2b(b^2+8)=0\implies$ $2b^3-8b=0\implies$ $b^2=4\implies$ $\underline{\underline{b\in\{-2,2\}}}$ . In conclusion, the solutions are: $(0,0)\ ;\ (4,2)\ ;\ (-4,-2)$ .

Remark. With the general method there are yet two steps: $\left\{\begin{array}{cccccc}
b\cdot a^2 & 4\cdot a & -2b(b^2+8) & = & 0\\\\
1\cdot a^2 & -3b\cdot a & 2b^2 & = & 0\end{array}\right\|\implies$ $\left|\begin{array}{cc}
b & -2b(b^2+8)\\\\
1 & 2b^2\end{array}\right|^2=$ $\left|\begin{array}{cc}
b & 4\\\\
1 & -3b\end{array}\right|\cdot\left|\begin{array}{cc}
4 & -2b(b^2+8)\\\\
-3b & 2b^2\end{array}\right|\iff$

$8b(b^2+4)^2=(3b^2+4)(3b^3+20b)\iff$ $\boxed{b=0}$ or $8(b^2+4)^2=(3b^2+4)(3b^2+20)$ . Denote $b^2=t\ge 0$ . Thus, $8(t+4)^2=(3t+4)(3t+20)=0\iff$

$t^2+8t-48=0\iff t\in \{4,-12\}$ . Since $t\ge 0$ the final solutions are $b^2\in \{0,4\}$ , i.e. $b\in\{0,2,-2\}$ a.s.o.



PP19. Let $r$ be a root of the cubic function $f(x)=x^3+x^2-4x+1$ . Show that $r^2+r-3$ is also a root of $f(x)$ .

Proof. Let $m=r^2+r-3$ , where $r\ne 0$ . Since $f(x)=x\left(x^2+x-3\right)-(x-1)$ get $0=f(r)=rm-(r-1)\implies$ $\boxed{m=\frac {r-1}{r}}$ .

$\blacktriangleright\ f(m)=f\left(\frac {r-1}{r}\right)=$ $\frac {(r-1)^3+r(r-1)^2-4r^2(r-1)+r^3}{r^3}=-\frac {r^3+r^2-4r+1}{r^3}=-\frac {f(r)}{r^3}=0\implies$ $f(m)=0$ .



PP20 ("instant"). Prove without derivatives that $:\ \left\{\begin{array}{ccccccccc}
f(x) & = & x^2(36-x) & \le & f(24) & = & 6912 & \mathrm{for\ any} &  x\in (0,36]\\\\
g(x) & = & x^2(6-x) & \le & g(4) & = & 32 & \mathrm{for\ any} &  x\in (0,6]\end{array}\right\|$

Proof. $f$ is $\max$ $\iff$ $\frac x2\cdot\frac x2\cdot (36-x)$ is $\max .$ Observe that $\frac x2+\frac x2+(36-x)=36$ (constant). Hence $\frac x2=36-x=\frac {36}3=12\iff$ $\frac {3x}2=36\iff$ $x=24\in (0,36].$

$g$ is $\max$ $\iff$ $\frac x2\cdot\frac x2\cdot (6-x)$ is $\max .$ Observe that $\frac x2+\frac x2+(6-x)=6$ (constant). Hence $\frac x2=6-x=\frac 63=2\iff$ $\frac {3x}2=6$ and $x=4\in (0,6],$ what is compatibly.

Remark.

$f(x)\le f(24)\iff$ $x^2(36-x)\le 6912\iff$ $x^3-36x^2+6912\ge 0\iff$ $(x+12)(x-24)^2\ge 0,$ what is true for any $x\in (0,36].$

$g(x)\le g(4)\iff$ $x^2(6-x)\le 32\iff$ $x^3-6x^2+32\ge 0\iff$ $(x+2)(x-4)^2\ge 0,$ what is true for any $x\in (0,36].$



PP21. Sa se rezolve peste $R$ sistemul format din ecuatiile $\odot\begin{array}{ccccccc}
\nearrow & x^3 & - & y^3 & = & 74 & \searrow\\\\
\searrow & x^2 & + & y^2 & = & 26 & \nearrow\end{array}\odot$

Proof.
This post has been edited 61 times. Last edited by Virgil Nicula, Feb 8, 2018, 9:09 AM

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    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: 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 :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404396
  • Total comments: 37
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