442. Problems from and for baccalaureate II.

by Virgil Nicula, Apr 14, 2016, 4:15 PM

P1. Find the distance between the lines $\odot\begin{array}{cccc} \nearrow & d_1\ : & \frac{x-1}2=\frac {y+1}3=z & \searrow\\\\ \searrow & d_2\ : & \frac {x+1}3=\frac y4=\frac {z-1}3 & \nearrow\end{array}\odot$ in the analytical space.

Proof 1. $\left\{\begin{array}{cccc} d_1\ : & \frac{x-1}2=\frac {y+1}3=z=m & \implies & M(2m+1,3m-1,m)\\\\ d_2\ ; & \frac {x+1}3=\frac y4=\frac {z-1}3=n & \implies & N(3n-1,4n,3n+1)\end{array}\right\|$ $\implies$ $MN^2=(2m-3n+2)^2+(3m-4n-1)^2+(m-3n-1)^2=2[f(m,n)+3]$ ,

where $f(m,n)=7m^2+17n^2-21mn+n\ .$ I"ll search $\min_{\{m,n\}\subset\mathbb R}f(m,n)=\min_{m\in\mathbb R}\min_{n\in\mathbb R}g_m(n)$ , where $g_m(n)=17\underline n^2-(21m-1)\underline n+7m^2$ . Hence $\boxed{n=\frac {21m-1}{34}}\ (*)$ and

$h(m)=\min_{n\in\mathbb R}g_m(n)=-\frac {\Delta (m)}{4\cdot 17}$ , where $\Delta (m)=(21m-1)^2-68\cdot 7m^2=-35m^2-42m+1\implies$ $\min_{\{m,n\}\subset\mathbb R}f(m,n)=\min_{m\in\mathbb R}h(m)\ ,\ h(m)=\frac 1{68}\cdot\left(35m^2+42m-1\right)$ .

Observe that $\boxed{m=-\frac 35\ \wedge\ n=-\frac 25}\ (1)$ and $\min_{\{m,n\}\subset\mathbb R}f(m,n)=\min_{m\in\mathbb R}h(m)=-\frac 15$ . In conclusion, $\min MN^2=2\left(-\frac 15+3\right)=\frac {28}5$ , i.e. $\boxed{MN=2\cdot\sqrt{\frac 75}}\ (2)$ .

Remark. $\left\{\begin{array}{cccc} m=-\frac 35 & \implies & M\left(-\frac 15,-\frac {14}5,-\frac 35\right)\\\\ n=-\frac 25 & \implies & N\left(-\frac {11}5,-\frac 85,-\frac 15\right)\end{array}\right\|$ $\implies$ the equation of $MN$ is $\frac {x+\frac 15}{-\frac {11}5+\frac 15}=\frac {y+\frac {14}5}{-\frac 85+\frac {14}5}=\frac {z+\frac 35}{-\frac 15+\frac 35}\implies$ $\boxed{\frac {5x+1}{-5}=\frac {5y+14}3=\frac {5z+3}1}\ (3)$ is surjective.

Verify: $MN^2=2^2+\left(\frac 65\right)^2+\left(\frac 25\right)^2=4+\frac {36+4}{25}=4+\frac 85\implies$ $MN^2=\frac {28}5$ and $\left\{\begin{array}{ccc} MN\perp d_1 & \iff & 2\cdot (-5)+3\cdot 3+1\cdot 1=0\\\\ MN\perp d_2 & \iff & 3\cdot (-5)+4\cdot 3+3\cdot 1=0\end{array}\right\|$ .

Proof 2. The equation of the plane $\pi$ for which $d_1\subset\pi\parallel d_2$ is $\left|\begin{array}{ccc} x-1 & y+1 & z\\\ 2 & 3 & 1\\\ 3 & 4 & 3\end{array}\right|\iff$ $\pi (x,y,z)\equiv 5x-3y-z-8=0$ .

Choose the point $P(-1,0,1)\in d_2$ and calculate the distance $\delta_{\pi}(P)=\frac {|\pi (-1,0,1)|}{\sqrt{5^2+3^2+1^2}}=\frac {14}{\sqrt{35}}\implies$ $\boxed{\delta\left(d_1,d_2\right)=2\cdot\sqrt{\frac 75}}\ .$

Proof 3. I"ll use the well-known property $\left\{\begin{array}{ccc} ax+by+cz & = & 0\\\\ a'x+b'y+c'z & = & 0\end{array}\right\|\iff$ $\frac x{\left|\begin{array}{cc} b & c\\\ b' & c'\end{array}\right|}=\frac y{\left|\begin{array}{cc} c & a\\\ c' & a'\end{array}\right|}=\frac z{\left|\begin{array}{cc} a & b\\\ a' & b'\end{array}\right|}$ . If the "slope" of

the required $MN$ is $s(p,q,r)$ , then $\left\{\begin{array}{ccc} MN\perp d_1 & \iff & 2p+3q+r=0\\\\ MN\perp d_2 & \iff & 3p+4q+3r=0\end{array}\right\|\iff$ $\frac p{\left|\begin{array}{cc} 3 & 1\\\ 4 & 3\end{array}\right|}=\frac q{\left|\begin{array}{cc} 1 & 2\\\ 3 & 3\end{array}\right|}=\frac r{\left|\begin{array}{cc} 2 & 3\\\ 3 & 4\end{array}\right|}$ $\iff$

$\frac p5=\frac q{-3}=\frac r{-1}\iff$ the equation of the plane $\pi$ for which $d_1\subset\pi\parallel d_2$ is $5(x-1)-3(y+1)-z=0$ $\iff$ $5x-3y-z-8=0$ a.s.o.

P2. Fie planul $\pi$ care sectioneaza muchiile tetraedrului $ABCD$ in punctele interioare $\left\{\begin{array}{ccccccc} M\in (AB) & , & \frac {MA}{MB}=m & ; & N\in (BC) & , & \frac {NB}{NC}=n\\\\ P\in (CD) & , & \frac {PC}{PD}=p & ; & Q\in (DA) & , & \frac {QD}{QA}=q\end{array}\right\|$ si in punctele exterioare

$\left\{\begin{array}{ccccc} R\in MN\cap AC\cap PQ & \implies & C\in (AR) & \mathrm{si} & \frac {RC}{RA}=r\\\\ S\in NP\cap BD\cap MQ & \implies & D\in (BS) & \mathrm{si} & \frac{SB}{SD}=s\end{array}\right\|$ . Sa se afle relatiile intre $\{m,n,p,q,r,s\}$ si raportul in care planul $\pi$ imparte volumul $v$ al tetraedrului $ABCD$ .

Dem.. Aplicam teorema Menelaus la $:\ \left\{\begin{array}{cc} \overline{MNR}/\triangle ABC\ : & \frac {RC}{RA}\cdot\frac {MA}{MB}\cdot\frac {NB}{NC}=1\implies rmn=1\\\\ \overline{QPR}/\triangle ACD\ : & \frac {RC}{RA}\cdot \frac {QA}{QD}\cdot\frac {PD}{PC}=1\implies r=pq\end{array}\right|\left|\begin{array}{cc} \overline{SQM}/\triangle ABD\ : & \frac {SB}{SD}\cdot\frac {QD}{QA}\cdot\frac {MA}{MB}=1\implies sqm=1\\\\ \overline{SPN}/\triangle BCD\ : & \frac {SB}{SD}\cdot \frac {PD}{PC}\cdot\frac {NC}{NB}=1\implies s=pn\end{array}\right\|$

Asadar, $\boxed{\begin{array}{ccccc} pq & = & r & = & \frac 1{mn}\\\\ pn & = & s & = & \frac 1{mq}\end{array}}\ (*)$ $\implies$ $\boxed{mnpq=1}\ (1)$ $\implies$ $\left\{\begin{array}{c} \frac {v[RAMQ]}{v[ABCD]}=\frac {[AMQ]\cdot\delta_{ABD}(R)}{[ABD]\cdot\delta_{ABD}(C)}=\frac {[AMQ]}{[ABD]}\cdot\frac {\delta_{ABD}(R)}{\delta_{ABD}(C)}=\frac {AM}{AB}\cdot\frac {AQ}{AD}\cdot\frac {RA}{CA}=\frac m{m+1}\cdot \frac 1{q+1}\cdot \frac 1{1-r}\\\\ \frac {v[RCNP]}{v[ABCD]}=\frac {[CNP]\cdot\delta_{BCD}(R)}{[BCD]\cdot\delta_{BCD}(A)}=\frac {[CNP]}{[CBD]}\cdot\frac {\delta_{BCD}(R)}{\delta_{BCD}(A)}=\frac {CN}{CB}\cdot\frac {CP}{CD}\cdot\frac {RC}{CA}=\frac 1{n+1}\cdot \frac p{p+1}\cdot \frac r{1-r}\end{array}\right\|$ . In concluzie,

$\frac {v[ACMNPQ]}{v[ABCD]}=\frac {v[RAMQ]}{v[ABCD]}-\frac {v[RCNP]}{v[ABCD]}=$ $\frac m{m+1}\cdot \frac 1{q+1}\cdot \frac 1{1-r}-$ $\frac 1{n+1}\cdot \frac p{p+1}\cdot \frac r{1-r}=$ $\frac 1{1-r}\cdot \left[\frac m{(m+1)(q+1)}-\frac {pr}{(n+1)(p+1)}\right]=f(m,n,p,q)$ , $r=pq$ .

Caz particular. Daca $m=p=1$ , atunci $r=q$ si $nq=1$ , adica punctele $M$ si $P$ sunt mijloacele muchiilor opuse $[AB]$ si $[CD]$ .Se obtine

usor ca $\frac {v[ACMNPQ]}{v[ABCD]}=\frac 1{1-q}\cdot\left[\frac 1{2(q+1)}-\frac {q^2}{2(q+1)}\right]=\frac {1-q^2}{2(1-q)(1+q)}=\frac 12$ , adica $2\cdot v[ACMNPQ]=v[ABCD]$ .

P3. Study if the system $\left\{\begin{array}{ccc} a+b+c & = & m\\\\ a^2+b^2+c^2 & = & n \end{array}\right\|$ has real solutions, where $\{m,n\}\subset\mathbb R$ .

Proof. $\left(\sum a\right)^2=\sum a^2+2\sum (bc)\iff$ $bc+a(b+c)=\frac {m^2-n}2\iff$ $\ \stackrel{b+c=m-a}{\implies}\ bc=$ $\frac {m^2-n}2-a(m-a)\iff$ $\odot \begin{array}{ccccc} \nearrow & b+c & = & m-a & \searrow\\\\ \searrow & bc & = & a^2-am+\frac {m^2-n}2 & \nearrow\end{array}\odot$ $\implies$

$t^2-(m-a)t+\left(a^2-am+\frac {m^2-n}2\right)=0\begin{array}{cc} \nearrow & b\\\\ \searrow & c\end{array}\ ,$ where $\Delta(a)=(m-a)^2-4a^2+4am-2\left(m^2-n\right)\implies$ $\Delta (a)=-3a^2+2am-m^2+2n$ . In conclusion,

$\Delta\ge 0\iff$ $\boxed{3a^2-2am+\left(m^2-2n\right)\le 0}\ (*)\ ,$ i.e. $m^2\le 3n$ and in this case for $|3a-m|\le \sqrt{2\left(3n-m^2\right)}$ we have $\{b,c\}=\left\{\frac {m-a\pm\sqrt{-3a^2+2am-m^2+2n}}2\right\}\ .$

P4. Solve the system of equations in real numbers $\left\{\begin{array}{ccc} y^2+yz+z^2 & = & a\\\ z^2+zx+x^2 & = & b\\\ x^2+xy+y^2 & = & c\end{array}\right\|$ , where $a+c=2b$ and $x+y+z\ne 0\ .$

Proof. $a+c=2b\iff$ $\left(y^2+yz+z^2\right)+\left(x^2+xy+y^2\right)=2\left(z^2+zx+x^2 \right)\iff$ $2y^2+y(x+z)=x^2+2xz+z^2\iff$ $(x+z)^2-y(x+z)-2y^2=0\iff$

$\left(\frac {x+z}y\right)^2-\frac {x+z}y-2=0\ \stackrel{x+z=ty}{\iff}\ t^2-t-2=0\begin{array}{ccccc} \nearrow & t_1=-1 & \implies & x+y+z=0 & \searrow\\\\ \searrow & t_2=2 & \implies & x+z=2y & \nearrow\end{array}\odot$ Since $x+y+z\ne 0$ obtain that $\boxed{x+z=2y}\ (*)$ . Therefore,

$x^2+xz+z^2=b\iff (x+z)^2-xz=b\iff \boxed{xz=4y^2-b}$ . Hence the equations $\odot\begin{array}{ccccccccc} \nearrow & t^2 & - & 2yt & + & \left(4y^2-b\right) & = & 0 & \searrow\\\\ \searrow & t^2 & + & yt & + & \left(y^2-a\right) & = & 0 & \nearrow\end{array}z\ ($ have at least a common root $z)$ .

So $\left|\begin{array}{cc} 1 & -2y\\\\ 1 & y\end{array}\right|\cdot \left|\begin{array}{cc} -2y & 4y^2-b\\\\ y & y^2-a\end{array}\right|=\left|\begin{array}{cc} 1 & 4y^2-b\\\\ 1 & y^2-a\end{array}\right|^2\iff$ $3y^2\cdot \left|\begin{array}{cc} -2 & 4y^2-b\\\\ 1 & y^2-a\end{array}\right|=\left[3y^2+(a-b)^2\right]^2$ $\iff$ $3y^2\left[6y^2-(2a+b)\right]+9y^4+6(a-b)y^2+(a-b)^2=0$

$\iff$ $\boxed{27y^4-9by^2+(a-b)^2=0}\ (1)$ a.s.o. Particular case. $\left\{\begin{array}{ccc} a & = & 19\\\ b &= & 28\\\ c & = & 37\end{array}\right\|$ . Thus, $(1)$ becomes $27y^4-9\cdot 28y^2+81=0\iff$ $3y^4-28y^2+9=0\begin{array}{cccc} \nearrow & y^2 & = & 9\\\\ \searrow & y^2 & = & \frac 13\end{array}$ .

In conclusion, $y\in\left\{3\ ;\ -3\ ;\ \frac {\sqrt 3}3\ ;\ -\frac {\sqrt 3}3\right\}\implies$ $\left\{\begin{array}{ccc} y=3 & \implies & \odot\begin{array}{cccccc} \nearrow & x^2+3x+9=37 & \implies & x^2+3x-28=0 & \implies & x\in\{-7 ;4\}\\\\ \searrow & z^2+3z+9=19 & \implies & z^2+3z-10=0 & \implies & z\in\{-5;2\}\end{array}\\\\ y=-3 & \implies & \odot\begin{array}{cccccc} \nearrow & x^2-3x+9=37 & \implies & x^2-3x-28=0 & \implies & x\in\{7;-4\}\\\\ \searrow & z^2-3z+9=19 & \implies & z^2-3z-10=0 & \implies & z\in\{5;-2\}\end{array}\end{array}\right\|\ \stackrel{x+z=2y}{\implies}$

$(4,3,2)$ and $(-4,-3,-2)$ a.s.o.

P5. Find the complex numbers $z$ for which $\boxed{\ z^2+2\mathrm i\overline z=0\ }$ .

Proof 1. $\left\{\begin{array}{ccc} z^2+2\mathrm i\overline z & = & 0\\\\ \overline z^2-2\mathrm iz & = & 0\end{array}\right\|\ominus$ $\implies$ $z^2-\overline z^2+2\mathrm i\left(\overline z+z\right)=0\implies$ $\left(z+\overline z\right)\left(z-\overline z+2\mathrm i\right)=0\implies$ $2\Re(z)\cdot\left[2\mathrm i\Im (z) +2\mathrm i\right]=0\implies$ $\Re (z)\cdot\left[\Im (z)+1\right]=0\implies$

$\odot\begin{array}{cccccccccc} \nearrow & \Re (z)=0 & \implies & z=bi & \implies & (b\mathrm i)^2+2\mathrm i(-b\mathrm i)=0 & \implies & -b^2+2b=0 & \implies & \odot\begin{array}{cccc} \nearrow & \Im (z)=0 & \implies & z=0\\\\ \searrow & \Im (z)=2 & \implies & z=2\mathrm i\end{array}\\\\ \searrow & \Im (z)=-1 & \implies & z=a-i & \implies & (a-\mathrm i)^2+2\mathrm i(a+\mathrm i)=0 & \implies & -a^2-3=0 & \implies & \odot\begin{array}{cccc} \nearrow & \Re (z)=\sqrt 3 & \implies & z=\sqrt 3-\mathrm i\\\\ \searrow & \Re (z)=-\sqrt 3 & \implies & z=-\sqrt 3-\mathrm i\end{array}\end{array}$

Proof 2. $z=a+b\mathrm i\implies$ $(a+b\mathrm i )^2+2\mathrm i\left(\overline {a+b\mathrm i}\right)=0$ $\iff$ $a^2+2ab\mathrm i -b^2+2a\mathrm i+2b=0$ $\begin{array}{ccc} \nearrow & 2a(b+1)=0 & \searrow\\\\ \searrow & a^2-b^2+2b=0 & \nearrow\end{array}\odot\implies$ $z\in\left\{0\ ,\ 2\mathrm i\ ,\ \sqrt 3-\mathrm i\ ,\ -\sqrt 3-\mathrm i\right\}$ .

P6. Find the range $\mathrm {Im(f)}$ of the function $f(x)=x+\sqrt {x^4+3}$ , where $x\in\mathbb R\ .$

Proof. $(\forall )\ x\ge 0$ the function $f$ is increasing $(\nearrow )$ and $f(x)\ge \sqrt 3$ . Now suppose that $x<0$ . Thus, $f(x)\ge 1=f(-1)\iff$ $\sqrt{x^4+3}\ge 1-x\iff$ $x^4+3\ge (1-x)^2\iff$

$x^4-x^2+2x+2\ge 0\iff$ $(x+1)^2\left(x^2-2x+2\right)\ge 0$ what is truly. So $\boxed{(\forall )\ x\in\mathbb R\ ,\ f(x)\ge 1=f(-1)\ ,\ \mathrm{i.e.}\ f:\mathbb R\rightarrow [1,\infty)}$ and $\mathrm {Im(f)}=f(\mathbb R)=[1,\infty ).$ See here.

P7. Find the maximum of the expression $\left|2x+y-z\right|$ where $\{x,y,z\}\subset\mathbb R$ and $x^2+3y^2+z^2=2$ .

Proof. $(2x+y-z)^2=\left[2\cdot x+\frac 1{\sqrt 3}\cdot y\sqrt 3+(-1)\cdot z\right]^2\ \stackrel{(C.B.S)}{\le}\ \left[2^2+\left(\frac 1{\sqrt 3}\right)^2+(-1)^2\right]$ $\cdot\left[x^2+\left(y\sqrt 3\right)^2+z^2\right]=$ $\frac {16}3\cdot 2=$ $\frac {32}3\implies$ $|2x+y-z|\le \frac {4\sqrt 6}3\ .$

P8. Prove that for any $\{x,y\}\subset\mathbb R$ there is the chain $4\le (x+y)\left(\frac 1x+\frac 1y\right)\le \left(\frac xy+\frac yx\right)^2\ .$

Proof. The inequalities $(x+y)\left(\frac 1x+\frac 1y\right)\ge 4$ and $\left(\frac xy+\frac yx\right)^2\ge 4$ are well known. Thus, $\left\{\begin{array}{ccc} x^2+y^2 & \ge & 2xy\\\\ 2\left(x^2+y^2\right) & \ge & (x+y)^2\end{array}\right\|\ \bigodot\implies$

$\left(x^2+y^2\right)^2\ge xy(x+y)^2\ \stackrel{:\ \left(x^2y^2\right)}{\implies}\ \left(\frac{x^2+y^2}{xy}\right)^2\ge (x+y)\cdot \frac {x+y}{xy}\implies$ $(x+y)\left(\frac 1x+\frac 1y\right)\le \left(\frac xy+\frac yx\right)^2\ .$

Remark. The chain of the inequalities $\frac 2{\frac 1x+\frac 1y}\le$ $\sqrt {xy}\le \frac {x+y}2\le\sqrt{\frac {x^2+y^2}2}$ $\mathrm{(\ H.M.\ \le\ G.M.\ \le\ A.M.\ \le\ P.M.\ )}$ is well-known.

P9. Prove that $8x(2x^2-1)(8x^4-8x^2+1)=1\iff(2 x-1)(8 x^3-6x-1)(8 x^3+4x^2-4x-1)=0\ (*)$ and solve the equation $(*)\ .$

P10. Let the equation $t^3+mt^2+nt+p=0\implies\odot \begin{array}{ccc} \nearrow & a & \searrow\\\\ \rightarrow & b & \rightarrow\\\\ \searrow & c & \nearrow\end{array}\odot$ Solve the equation $\frac {x-(b+c)}a+\frac {x-(c+a)}b+\frac {x-(a+b)}c=k\ .$

Proof. Observe that (Viete's relations) $:\ \left\{\begin{array}{cccc} a+b+c & = & -m & (1)\\\\ ab+bc+ca & = & n & (2)\\\\ abc & = & -p & (3)\end{array}\right\|\ .$ Hence $\sum \frac {x-(b+c)}a=k\ \stackrel{(1)}{\iff}\ \sum\frac {x+m+a}{a}=k\iff$ $\sum\left(1+\frac {x+m}a\right)=k\iff$

$(x+m)\cdot\sum\frac 1a=k-3\iff$ $(x+m)\cdot \frac {ab+bc+ca}{abc}=k-3\ \stackrel{2\wedge 3}{\iff}\ (x+m)\cdot \frac np=3-k\iff$ $x+m=\frac {p(3-k)}n\iff$ $\boxed{x=\frac {p(3-k)}n-m}\ (*)\ .$

P11. Find $L\equiv \lim_{x\to\infty}x\cdot\left(1-\frac {\ln x}{x}\right)^x\ .$

Proof. $L\equiv\lim_{x\to\infty}\ x\cdot\left(1-\frac {\ln x}{x}\right)^x\ \stackrel{\left(x:=\frac 1x\right)}{\ =\ }\ \lim_{x\to 0}\ \frac {(1+x\ln x)^{\frac 1x}}{x}=e^l\ ,$ where $l=\lim_{x\to 0}\ \frac {\ln (1+x\ln x)-x\ln x}{x}\ \stackrel {(l^{\prime}H)}{\ =\ }$

$\lim_{x\to 0}\ \left[\frac {1+\ln x}{1+x\ln x}-(1+\ln x)\right]=-\lim_{x\to 0}\ \frac {x\ln x(1+\ln x)}{1+x\ln x}=0\ .$ In conclusion, $\lim_{x\to\infty}x\cdot\left(1-\frac {\ln x}{x}\right)^x=1\ .$

Remark. I used an wel known limit $\boxed{\lim_{x\to 0}\ x\ln^{\alpha} x=0\ ,\ \alpha > 0\ }\ .$ Here is an its proof. I used the substitution $x:=x^{\alpha}\ \ :\ \lim_{x\to 0}\ x\ln^{\alpha} x=$

$\lim_{x\to 0}\ x^{\alpha}\left(\ln x^{\alpha}\right)^{\alpha}=\alpha ^{\alpha}\cdot\lim_{x\to 0}\ \left(x\ln x\right)^{\alpha}=0$ because $\lim_{x\to 0}\ x\ln x\ \stackrel{\left(x:=\frac 1x\right)}{\ =\ }\ -\lim_{x\to\infty}\ \frac {\ln x}{x}=-0\ .$ Remain to show $\lim_{x\to\infty}\ \frac {\ln x}{x}=0\ \ldots$

P12. $\{A,B,C\}\subset\mathrm M_3(R)\ \wedge\ \det A=\det B=\det C\ \wedge\ \det (A+iB)=\det (C+iA)$ $\Longrightarrow$ $\det (\mathrm A+\mathrm B)=\det (\mathrm C+\mathrm A)\ .$

Proof. Let $f(x)=\det (\mathrm{A+xB})-\det (\mathrm{C+xA})\ ,\ x\in\mathbb R\ .$ Observe that $f(0)=\underline{\det A-\det C=0}\ \ ,\ \underline{f(i)=0}\ .$ The polynomial $f$ has real coefficients and $\mathrm{gr} f\le 3\ .$ Thus,

$f(x)=kx\left(x^2+1\right)\ .$ Hence $k=\lim_{x\to\infty}\ \frac {f(x)}{x^3}=$ $\lim_{x\to\infty}\ \frac {1}{x^3}\cdot \left[\det (\mathrm{A+xB})-\det (\mathrm{C+xA})\right]=$ $\lim_{x\to\infty}\ \left[\det \left(\mathrm{\frac 1x A+B}\right)-\det \left(\mathrm{\frac 1xC+A}\right)\right]=$ $\underline{\det \mathrm B-\det \mathrm A=0}\ .$

In conclusion, $k=0$ , i.e. $f$ is the null polynomial. In particular, $f(1)=0$ , i.e. $\det (\mathrm A+\mathrm B)=\det (\mathrm C+\mathrm A)\ .$ I used $a\in\mathbb R\ \wedge\ A\in\mathrm M_n(R)$ $\Longrightarrow$ $\det\left(\mathrm{a\cdot A}\right)=a^n\cdot \det \mathrm A\ .$

P13. Gasiti $m\in\mathbb Q$ stiind ca $f(x)\equiv x^4 - 7x^3 + (13 + m)x^2 - (3 + 4m)x + m = 0$ admite radacinile $x_1 = 2 + \sqrt 3$ si $x_3$ , $x_4$ unde $x_3 = 2x_4\ \vee\ x_4 = 2x_3\ .$

Dem. Coeficientii ecuatiei date sunt rationali $\implies$ $x_2=2-\sqrt 3$ $\implies$ $\left\|\begin{array}{c} x_1=2+\sqrt 3\\\\ x_2=2-\sqrt 3\end{array}\ \ \right\|\Longleftrightarrow$ $\left\|\begin{array}{c} S=4\\\\ P=1\end{array}\ \right\|$ $\Longleftrightarrow$ $\boxed{x^2-4x+1=0\ \begin{array}{cc} \nearrow & x_1\\\\ \searrow & x_2\end{array}}\ .$ Relatia $x_3=2x_4\ \ \vee\ \ x_4=2x_3$

inseamna $\left(x_3-2x_4\right)\left(x_4-2x_3\right)=0\ \Longleftrightarrow\ 5P=2S_2\ \Longleftrightarrow\ 2S^2=9P\ \Longleftrightarrow$ exista $n\in\mathbb Q$ astfel incat $\left\|\ \begin{array}{c} S=9n\\\\ P=18n^2\end{array}\ \right\|\ \Longleftrightarrow\ \boxed{x^2-9nx+18n^2=0\ \begin{array}{cc} \nearrow & x_3\\\\ \searrow & x_4\end{array}}\ .$

Deci $X^4 - 7X^3 + (13 + m)X^2 - (3 + 4m)X + m \equiv\left(X^2-4X+1\right)\left(X^2-9nX+18n^2\right)$ $\Longleftrightarrow$ $\left\{\begin{array}{ccccc} \mathrm{coeff.\ of} & x^3\ : & -7=-9n-4 & \iff & n=\frac 13\\\\ \mathrm{coeff.\ of} & x^0\ : & m=18n^2 & \iff & m=2\end{array}\right\|\ .$

In concluzie, ecuatia devine $x^4-7x^3+15x^2-11x+2=\left(x^2-4x+1\right)\left(x^2-3x+2\right)\ \odot \begin{array}{cccc} \nearrow & x_{1,2} & = & 2\pm\sqrt 3\\\\ \rightarrow & x_3 & = & 1\\\\ \searrow & x_4 & = & 2\end{array}\ .$

P14. Sa se determine $m\in\mathbb R$ astfel incat intre radacinile $x_1$ si $x_2$ ale ecuatiei $x^2-(2m+1)x+(m+1)=0$ sa existe relatia $x_1^2=2x_2-3$ sau $x_2^2=2x_1-3\ .$

Metoda 1. Se observa ca $\left\{\begin{array}{ccccc} S & = & x_1+x_2 & = & 2m+1\\\\ P & = & x_1x_2 & = & m+1\end{array}\right\|\ .$ Asadar, $x_1^2=2x_2-3$ sau $x_2^2=2x_1-3\iff$ $\left(x_1^2-2x_2+3\right)\left(x_2^2-2x_1+3\right)=0\iff$

$P^2-2S_3+3S_2+4P-6S+9=0\iff$ $P^2-2\left(S^3-3PS\right)+3\left(S^2-2P\right)+4P-6S+9=0\iff$ $2S^3-6PS-3S^2+6S-P^2+2P-9=0\iff$

$2(2m+1)^3-6(m+1)(2m+1)-3(2m+1)^2+6(2m+1)-(m+1)^2+2(m+1)-9=0\iff$ $16m^3-m^2-6m-9=0\iff$

$(m-1)\left(16m^2+15m+9\right)=0\iff$ $\boxed{\ m=1\ }$ si in acest caz radacinile ecuatiei date sunt $x_1=1$ , $x_2=2\ .$ Se observa ca $x_1^2=2x_2-3$ , adica $1^2=2\cdot 2-3\ .$

Metoda 2. $x_1^2-2x_2+3=0\iff$ $x_1^2-2(S-x_1)+3=0\iff$ $x_1^2+2x_1+(3-2S)=0\ .$ Asadar, ecuatiile $\left\{\begin{array}{ccccccc} x^2 & - & Sx & + & P & = & 0\\\\ x^2 & + & 2x & + & (3-2S) & = & 0\end{array}\right\|$

au cel putin o radacina comuna. In concluzie, $\left|\begin{array}{cc} 1 & P\\\\ 1 & 3-2S\end{array}\right|^2=\left|\begin{array}{cc} 1 & -S\\\\ 1 & 2\end{array}\right|\cdot\left|\begin{array}{cc} -S & P\\\\ 2 & 3-2S\end{array}\right|\iff$ $(2S+P-3)^2=(S+2)\left(2S^2-3S-2P\right)\iff$

$25m^2=(2m+3)\left(8m^2-3\right)\iff$ $16m^3-m^2-6m-9=0\iff$ $(m-1)\left(16m^2+15m+9\right)=0\ \stackrel{(m\in\mathbb R)}{\iff}\ \boxed{\ m=1\ }\ .$

Metoda 3. $x_1^2-Sx_1+P=0\implies$ $x_1^2=Sx_1-P\implies$ $2x_2-3=Sx_1-P\implies$ $\left\{\begin{array}{ccccc} Sx_1 & - & 2x_2 & = & P-3\\\\ x_1 & + & x_2 & = & S\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} (S+2)x_1 & = & 2S+P-3\\\\ (S+2)x_2 & = & S^2-P+3\end{array}\right\|\ \bigodot$ $\implies$

$P(S+2)^2=(2S+P-3)\left(S^2-P+3\right)\implies$ $2S^3-3S^2-6PS+6S-P^2+2P-9=0\implies$ $16m^3-m^2-6m-9=0\iff$ $\boxed{\ m=1\ }\ .$

P15. Aratati ca in orice $\triangle ABC$ are loc relatia $\sin ^2\ \frac{A-B}{2}+\sin \ A\sin \ B+\sin ^2\ \frac{C}{2}=1$ (ON 1980).

Comentariu. Obisnuiam sa spun elevilor mei ca oridecateori intalnesc un produs (in particular puteri !) de tipul $\sin x\sin y$ , $\sin x\cos y$ sau $\cos x\cos y$ sa

inmulteasca cu doi relatia si sa aplice pachetul de formule care exprima asemenea produse printr-o suma de functii trigonometrice de acelasi tip, adica

$\boxed{\ \begin{array}{c} 2\sin x\cos y=\sin (x+y)+\sin (x-y)\\\\ 2\sin x\sin y=\cos (x-y)-\cos (x+y)\\\\ 2\cos x\cos y=\cos (x+y)+\cos (x-y)\end{array}\ }\ \ \blacktriangleleft$ $\blacktriangleright\ \ \boxed{\ \begin{array}{cc} 2\sin^2 x=1-\cos 2x & 2\cos^2x=1+\cos 2x\\\\ 2\sin x\cos x=\sin 2x & \tan^2x=\frac {1-\cos 2x}{1+\cos 2x}\end{array}\ }\ \ \blacktriangleleft$ $\blacktriangleright\ \ \boxed{\begin{array}{c} \tan x=\frac {\sin 2x}{1+\cos 2x}=\frac {1-\cos 2x}{\sin 2x}\\\\ \sin^2x-\sin^2y=\sin (x+y)\sin (x-y)\\\\ \sin^2x+\sin^2y+\cos (x+y)\cos (x-y)=1\end{array}}\ .$

Aplicam recomandarea exercitiului propus. Nu prea imi vine sa cred ca este de la ON - 1980. Nici pe vremea mea 1956-1960 (liceu) nu s-au dat asemenea probleme.

Metoda 1. $\boxed{\begin{array}{c} \sin ^2\ \frac{A-B}{2}+\sin A\sin B+\sin ^2\ \frac{C}{2}=1\\\\ 2\sin ^2\ \frac{A-B}{2}+2\sin A\sin B+2\sin ^2\ \frac{C}{2}=2\\\\ \left[\underline 1-\underline{\underline{\cos (A-B)}}\right]+\left[\underline{\underline{\cos (A-B)}}-\underline{\underline{\underline{\cos (A+B)}}}\right]+\left[\underline 1-\underline{\underline{\underline{\cos C}}}\right]=\underline 2\end{array}}$ deoarece $(A+B)+C=\pi\ \Longrightarrow\ \cos (A+B)=-\cos C\ .$

Metoda 2. $\boxed{\begin{array}{c} \sin ^2\ \frac{A-B}{2}+\sin A\sin B+\sin ^2\ \frac{C}{2}=1\\\\ \sin A\sin B=\sin^2\ \frac {A+B}{2}-\sin^2\ \frac {A-B}{2}\\\\ \sin A\sin B=\sin\ \frac {(A+B)+(A-B)}{2}\ \sin\ \frac {(A+B)-(A-B)}{2}\end{array}}$ deoarece $\sin^2x-\sin^2y=\sin (x+y)\sin (x-y)$ si $\frac {A+B}{2}+\frac C2=\frac {\pi}{2}\ \Longrightarrow$ $\cos\frac C2=\sin\frac {A+B}{2}\ .$

Va propun sa retineti si sa dovediti $:\ \boxed{\begin{array}{c} p(p-a)+(p-b)(p-c)=bc\\\\ p(p-a)-(p-b)(p-c)=bc\cdot \cos A\end{array}}$ $\Longleftrightarrow$ $\boxed{\begin{array}{c} 2p(p-a)=bc(1+\cos A)\\\\ 2(p-b)(p-c)=bc(1-\cos A)\end{array}}\ \stackrel{\pm}{\iff}\ \boxed{\begin{array}{c} \sin\frac A2=\sqrt {\frac {p-b)(p-c)}{bc}}\\\\ \cos\frac A2=\sqrt {\frac {p(p-a)}{bc}}\\\\ \tan\frac A2=\sqrt {\frac {(p-b)(p-c)}{p(p-a)}}\end{array}}\ .$

$\blacktriangleleft\blacktriangleright\ \boxed{\begin{array}{c} a=b\cdot\cos C+c\cdot \cos B\\\\ b=c\cdot\cos A+a\cdot\cos C\\\\ c=a\cdot\cos B+b\cdot\cos A\end{array}}\ \left|\begin{array}{cccc} \odot & (-a) & a & a\\\\ \odot & b & (-b) & b\\\\ \odot & c & c & (-c)\end{array}\right|\ \bigoplus\ \Longrightarrow\ \boxed{\begin{array}{c} -a^2+b^2+c^2=2bc\cdot\cos A\\\\ a^2-b^2+c^2=2ca\cdot\cos B\\\\ a^2+b^2-c^2=2ab\cdot\cos C\end{array}}\ .$

P16. Sa se construiasca media armonica a doua segmente date.

Demonstratie 1. Pe o dreapta $d$ asezam segmentele date astfel: $OA$ si $OB$ astfel incat $A\in (OB)\ .$ Pe o alta dreapta $d'$ care trece prin $O$ si este diferita de $d$ alegem

doua puncte $X$ , $Y$ simetrice fata de $O\ .$ Notam $Z\in AX\cap BY$ si construim punctul $T\in d'$ astfel incat $TZ\parallel d\ .$ Se arata usor ca $\frac 1{OA}+\frac 1{OB}=\frac 2{TZ}\ .$ Intr-adevar,

$TZ\parallel d\implies\odot \begin{array}{cccc} \nearrow & \triangle ZXT\ : & \frac {TZ}{OA}=\frac {XT}{XO} & \searrow\\\\ \searrow & \triangle OYB\ : & \frac {TZ}{OB}=\frac {YT}{YO} & \nearrow\end{array}\bigoplus\implies$ $\frac {TZ}{OA}+\frac {TZ}{OB}=\frac {XT}{XO}+$ $\frac {YT}{YO}\ \stackrel{OX=OY}{\implies}\ TZ\cdot$ $\left(\frac 1{OA}+\frac 1{OB}\right)=\frac {XY}{OX}=2 \implies$ $\boxed{\frac 1{OA}+\frac 1{OB}=\frac 2{TZ}}\ (1)\ .$

Demonstratie 2. Fie $:\ ABCD$ cu $AB\parallel CD\ ,\ I\in AC\cap BD\ ;\ X\in AD$ , $Y\in BC$ astfel incat $I\in XY$ si $XY\parallel AB\ .$ Se arata usor $IX=IY$ si $\boxed{\frac 1{AB}+\frac 1{CD}=\frac 2{XY}}\ (2)$

See and here

P17. Let $\triangle ABC$ with the circumcircle $w\ .$ Denote $P\in AA\cap CC$ and $\{B,D\} = PB\cap w\ .$ Prove that $\frac {DB}{DC} = 2\frac {AB}{AC}\ .$ Denote $XX$ - the tangent line to $w$ at $X\in w\ .$

Proof. I"ll use the Ptolemy's theorem $:\ \boxed{BD\cdot b=AD\cdot a+DC\cdot c}\ (1)\ .$ Observe that $\left\{\begin{array}{cccc} \triangle ADP\sim\triangle BAP & \implies & \frac {AD}c=\frac {AP}{BP} & (2)\\\\ \triangle CDP\sim\triangle BCP & \implies & \frac {CD}a=\frac {CP}{BP} & (3)\end{array}\right\|\ \stackrel{1\wedge 2\wedge 3}{\implies}$

$\frac {b\cdot BD}{ac}=\frac {AD}c+\frac {DC}a=$ $\frac {AP}{BP}+\frac {CP}{BP}=$ $2\cdot \frac {CP}{BP}=$ $2\cdot \frac {CD}a\implies$ $\frac {b\cdot BD}{ac}=2\cdot \frac {CD}a\implies$ $\frac {DB}{DC}= 2\cdot \frac {AB}{AC}\ .$

P18. Se considera paralelogramele $ABCD$ si $XYZT$ (situate in acelasi plan). Sa se arate ca mijloacele segmentelor $[AX]$ , $[BY]$ , $[CZ]$ , $[DT]$ sunt varfurile unui paralelogram.

Demonstratie. Fie mijloacele $M$ , $N$ , $P$ , $R$ ale segmentelor $[AX]$ , $[BY]$ , $[CZ]$ , $[DT]$ respectiv si $U\in AC\cap BD$ , $V\in XZ\cap YT$ . Vom folosi o proprietate cunoscuta sau usor

de dovedit $:$ intr-un patrupunct (nu neaparat in plan sau daca-i in plan, nu neaparat convex !) bimedianele au mijlocul comun. Astfel pentru patrupunctele $ACZX$ , $BDYT$ se observa ca

segmentele $[MP]$ , $[NR]$ , $[UV]$ au acelasi mijloc. In particular segmentele $[MP]$ , $[NR]$ se taie in segmente egale ceea ce inseamna ca patrulaterul $MNPR$ este un paralelogram.

P19. Denote $2s=a+b+c\ .$ Prove that $D\equiv \left|\begin{array}{ccccc} (b-c)(s-a) & & a(s-b) & & a(s-c)\\\\ (c-a)(s-b) & & b(s-c) & & b(s-a)\\\\ (a-b)(s-c) & & c(s-a) & & c(s-b)\end{array}\right|=\frac 14\cdot (a-b)(b-c)(c-a)(a+b+c)\left(a^2+b^2+c^2\right)\ .$

Proof. Inmultim fiecare linie a determinantului cu $2\ ($ implicit impartim $D$ prin $8\ )\ :\ D=\frac 18\ \cdot\left|\begin{array}{ccc} (b-c)(b+c-a) & a(c+a-b) & a(a+b-c) \\\\ (c-a)(c+a-b) & b(a+b-c) & b(b+c-a) \\\\ (a-b)(a+b-c) & c(b+c-a) & c(c+a-b)\end{array}\right|$

$\odot\ \boxed{C_1:=C_1-C_2+C_3}\Longrightarrow$ $D=\frac 18\cdot\left|\begin{array}{ccc} (b-c)(a+b+c) & a(c+a-b) & a(a+b-c) \\\\ (c-a)(a+b+c) & b(a+b-c) & b(b+c-a) \\\\ (a-b)(a+b+c) & c(b+c-a) & c(c+a-b)\end{array}\right|=$ $\frac {a+b+c}8\cdot\left|\begin{array}{ccc} b-c & a(c+a-b) & a(a+b-c) \\\\ c-a & b(a+b-c) & b(b+c-a) \\\\ a-b & c(b+c-a) & c(c+a-b)\end{array}\right|$

$\odot\left\{\begin{array}{cc} (1) & \boxed{C_3:=C_3-C_2} \\\\ (2) & \boxed{L_1:=L_1+L_2+L_3}\end{array}\right\|\ \stackrel{(1)}{\Longrightarrow}\ \ D=$ $\frac {a+b+c}8\ \cdot\left|\begin{array}{ccc} b-c & a(c+a-b) & 2a(b-c) \\\\ c-a & b(a+b-c) & 2b(c-a) \\\\ a-b & c(b+c-a) & 2c(a-b)\end{array}\right|\ \stackrel{(2)}{=}$ $\frac {a+b+c}8\ \cdot\left| \begin{array}{ccc} 0 & a^2+b^2+c^2 & 0 \\\\ c-a & b(a+b-c) & 2b(c-a) \\\\ a-b & c(b+c-a) & 2c(a-b)\end{array}\right|$ $\Longrightarrow$

$D=\frac {a+b+c}8\cdot\left(a^2+b^2+c^2\right)\cdot\left|\begin{array}{cc} a-c & 2b(a-c) \\\\ a-b & 2c(a-b)\end{array}\right|=\boxed{\frac 14\cdot (a-b)(b-c)(c-a)(a+b+c)\left(a^2+b^2+c^2\right)}\ .$ See here another determinants.

P20. Let two fixed $A\ ,$ $B\$ and a circle $w=\mathbb C(O,r)$ where $O\in (AB)\ .$ Find $\max_{P\in w}(PA+PB)\ ,$ where $P\in w\ .$

Proof. Denote $OA=a\ ,$ $OB=b$ and apply the Stewart's relation to the cevian $PO/\triangle APB\ :\ \boxed{b\cdot PA^2+a\cdot PB^2=\left(r^2+ab\right)(a+b)}\ (*)\ .$ Therefore,

$(PA+PB)^2=$ $\left(PA\sqrt b\cdot \frac 1{\sqrt b}+PB\sqrt a\cdot\frac 1{\sqrt a}\right)^2\ \stackrel{\mathrm{C.B.S.}}{\le}\ \left(b\cdot PA^2+a\cdot PB^2\right)\left(\frac 1b+\frac 1a\right)\ \stackrel{(*)}{=}\ \left(r^2+ab\right)(a+b)\cdot\frac {a+b}{ab}$ $\implies$ $\boxed{PA+PB\le (a+b)\cdot\sqrt{1+\frac {r^2}{ab}}}\ .$

We have the equality $\iff$ $\frac {PA\sqrt b}{\frac 1{\sqrt b}}=\frac {PB\sqrt a}{\frac 1{\sqrt a}}\iff$ $b\cdot PA=a\cdot PB\iff$ $\frac {PA}a=\frac {PB}b\ ,$ i.e. the ray $[PO$ is the bisector of the angle $\widehat{APB}\ .$

P21. Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ and its interior point $P$ so that $\left\{\begin{array}{ccc}\widehat{PAB} & \equiv & \widehat{PCA}\\\\ \widehat{PAC} & \equiv & \widehat{PBA}\end{array}\right\|\ .$ Prove that $:\ \left\{\begin{array}{cccc} PA^2 & = & PB\cdot PC & (1)\\\\ \frac {PB}{PC} & = & \left(\frac{AB}{AC}\right)^2 & (2)\\\\ OP & = & R\cdot \frac{|PB-PC|}{BC} & (3)\end{array}\right\|\ .$

Proof. $\triangle ABP\sim\triangle CAP\iff$ $\frac {AB}{CA}=\frac {BP}{AP}=\frac {AP}{CP}\iff$ $\frac {AB}{CA}=\frac {BP}{AP}=\frac {AP}{CP}\implies\odot \begin{array}{cccccc} \nearrow & \frac {BP}{AP}=\frac {AP}{CP} & \implies & PA^2=PB\cdot PC & \implies & (1)\\\\ \searrow & \begin{array}{ccc} \nearrow & \frac {PB}{PA}=\frac {AB}{CA} & \searrow\\\\ \searrow & \frac {PA}{PC}=\frac {AB}{CA} & \nearrow\end{array}\bigodot & \implies & \frac {PB}{PC}=\left(\frac{AB}{AC}\right)^2 & \implies & (2)\end{array}$

Suppose w.l.o.g. that $OM$ separates $P\ ,$ $C\ ,$ where $M$ is the midpoint of $[BC]\ .$ Observe that $m\left(\widehat{BPC}\right)=2A\ ,$ i.e. $\widehat{BPC}\equiv\widehat{BOC}\iff$ $BPOC$

is cyclic. Apply the Ptolemy's relation to $BPOC\ :\ BC\cdot OP+PB\cdot OC=PC\cdot OB\iff$ $BC\cdot OP=R\cdot |PC-PB|\ ,$ i.e. the relation $(3)\ .$

P22. Let $A$ - right $\triangle ABC\ ,$ $D\in (AC)$ so that $m\left(\widehat{ABD}\right)=C$ and $E\in BC$ so that $DE\perp BC\ .$ Prove that $AE=c\ .$

Proof. Observe that $\triangle ADB\sim\triangle ABC\implies$ $\frac {AD}{c}=\frac {DB}{a}=\frac cb$ $\implies$ $\left\|\begin{array}{c} AD=\frac {c^2}{b}\\\\ BD=\frac {ac}{b}\end{array}\right\|$ $\implies$ $CD=AC-AD=b-\frac {c^2}{b}$ $\implies$ $CD=\frac {b^2-c^2}{b}\ .$ $\triangle CDE\sim\triangle CBA$

$\implies$ $\frac {CD}{a}=\frac {DE}{c}=\frac {CE}{b}$ $\implies$ $\left\|\begin{array}{c} DE=\frac ca\cdot CD\\\\ CE=\frac ba\cdot CD\end{array}\right\|$ $\implies$ $\left\|\begin{array}{ccc} DE=\frac ca\cdot \frac {b^2-c^2}{b} & \implies & DE=\frac {c\left(b^2-c^2\right)}{ab}\\\\ CE=\frac ba\cdot \frac {b^2-c^2}{b} & \implies & CE=\frac {b^2-c^2}{a}\end{array}\right\|$ $\implies$ $BE=BC-EC=a-\frac {b^2-c^2}{a}\implies BE=\frac {2c^2}{a}\ .$ Apply

Ptolemy's theorem to $ABED\ :\ AB\cdot DE+$ $AD\cdot BE=BD\cdot AE\iff$ $c\cdot \frac {c\left(b^2-c^2\right)}{ab}+\frac {c^2}{b}\cdot\frac {2c^2}{a}=\frac {ac}{b}\cdot AE\iff$ $c^2\left(b^2-c^2\right)+2c^4=a^2c\cdot AE\iff$ $\boxed{AE=c}$

Remark. $m(\widehat{ABD})=C\implies\Delta ADB\sim\Delta ABC\iff m(\widehat{ADB})=B\ .$ Also, $DE\bot BC\wedge AB\bot AC\implies ABED$ cyclic $\implies$ $\widehat{AEB}\equiv\widehat{ADB}\equiv\widehat{ADB}$ $\implies AE=c\ .$

P23. Dacă $\{a, b, c\}\subset\mathbb R^*\ ,$ $a+b+c=0$ si $a^3+b^3+c^3=a^5+b^5+c^5\ ,$ atunci $a^2+b^2+c^2=\frac{6}{5}\ .$

Demonstratie. Fie ecuatia $x^3+mx+n=0$ ale carei radacini sunt $\{a,b,c\}\ .$ Deci $\left\{\begin{array}{c} s_2=ab+bc+ca=m\\\\ s_3=abc=-n\ne 0\end{array}\right|\ .$ Notam $S_k=a^k+b^k+c^k\ ,$ unde $k\in\mathbb N\ .$

Prin urmare, $S_0=3\ ,$ $S_1=0\ ,$ $S_2=-2m$ si pentru orice $k\in\mathbb N$ avem $S_{k+3}=-mS_{k+1}-nS_k\ .$ Se obtine usor succesiv $S_3=-3n\ ,$ $S_4=2m^2$ si $S_5=5mn\ .$

Relatia din ipoteza $S_3=S_5$ inseamna $-3n=5mn\ ,$ adica $m=-\frac 35$ deoarece $n\ne 0\ .$ In concluzie, $S_2=-2m\ ,$ adica $a^2+b^2+c^2=\frac 65\ .$

P24. Fie $ABCDEF$ un hexagon inscris intr-un cerc. Notam $M\in AC\cap BD\ ,$ $N\in BE\cap CF$ si $P\in AE\cap DF\ .$ Aratati ca $P\in MN\ .$

Proof. Apply the Pascal's theorem to the cyclical hexagon $ACFDBE\ :\ \begin{array}{c}\nearrow\\\\ \rightarrow\\\\ \searrow\end{array}\begin{array}{c} M\in AC\cap DB\\\\ N\in CF\cap BE\\\\ P\in FD\cap EA\end{array}\begin{array}{c}\searrow\\\\ \rightarrow\\\\ \nearrow\end{array}\ \implies\ P\in MN\ .$

P25. Fie $ABCD$ un trapez in care $AB\parallel CD$ . Notam punctul $P$ de intersectie al bisectoarelor exterioare ale unghiurilor $\widehat{ABC}$ , $\widehat{BCD}$ si punctul

$Q$ de intersectie al bisectoarelor exterioare ale unghiurilor $\widehat{ADC}$ , $\widehat{DAB}$ . Sa se arate ca $\boxed{2\cdot PQ=AB+BC+CD+DA}\ .$ (OMN, Mexic).

Dem. Se arata usor ca $\left\{\begin{array}{ccc} \delta_{AB}(G)=\delta_{AD}(G)=\delta_{CD}(G) & \implies & G\in EF\\\\ \delta_{AB}(P)=\delta_{BC}(P)=\delta_{CD}(P) & \implies & P\in EF\end{array}\right\|$ unde am notat $\delta_{XY}(Z)$ - distanta punctului $Z$ la dreapta $XY$ .

Asadar punctele $P$ , $Q$ apartin liniei mijlocii $EF$ a trapezului dat, unde $E\in AD$ si $F\in BC$ . Deoarece $GA\perp GD$ si $PB\perp PC$ obtinem ca

$AB+BC+CD+DA=$ $(AB+CD)+AD+BC=$ $2\cdot EF+2\cdot GE+2\cdot PF=$ $2\cdot (EF+GE+PF)=2\cdot PQ.$

P26. Let $A$ -isosceles $\triangle ABC$ with $AB=AC=a$ and $BC=b.$ Prove that $\boxed{A=140^{\circ}\iff b^3=a^3+3a^2b}\ .$

Proof. Denote the midpoint $M$ of the side $[BC].$ Observe that $A=140^{\circ}\iff B=20^{\circ}\iff$ $\cos 20^{\circ}=\cos B=\frac {AM}{BM}=\frac b{2a}\iff$ $\boxed{\cos 20^{\circ}=\frac b{2a}}\ (1).$

I"ll use the identity $\boxed{\cos 2x=2\cos^2x-1}\ (*)$ for $x=20^{\circ}\ :\ \cos 40^{\circ}=2\cdot \left(\frac b{2a}\right)^2-1=\frac {b^2}{2a^2}-1=\frac {b^2-2a^2}{2a^2}\iff$ $\boxed{\cos 40^{\circ}=\frac {b^2-2a^2}{2a^2}}\ (2).$

I"ll apply the identity $\boxed{2\cos x\cos y=\cos (x+y)\cos (x-y)}\ (**)$ for $x=20^{\circ}$ and $y=\cos 40^{\circ}\ :\ 2\cos 20^{\circ}\cdot \cos 40^{\circ}=\cos 60^{\circ}+\cos 20^{\circ}\iff$

$\frac ba\cdot \frac {b^2-2a^2}{\cancel 2a^2}=\frac 1{\cancel 2}+\frac b{\cancel 2a}\iff$ $\frac {b\left(b^2-2a^2\right)}{\cancel a\cdot a^2}=\frac {a+b}{\cancel a}\iff$ $b\left(b^2-2a^2\right)=a^2(a+b)\iff$ $b^3=a^3+3a^2b.$

P27. Let an acute $\triangle ABC$ with the Euler's circle $w$ and $\{X,Y,Z\}\subset w$ so that $A\in XX,$ $B\in YY$ and $C\in ZZ.$ Prove that $AX^2+BY^2+CZ^2=\frac 14\cdot\left(a^2+b^2+c^2\right).$

Proof. Let the midpoints $M,$ $N,$ $P$ of the sides $[BC],$ $[CA],$ $[AB]$ and the projections $D,$ $E,$ $F$ of $A,$ $B,$ $C$ on the same sides respectively. Is well known that $\{M,N,P;D,E,F\}\subset w.$

I"ll use the power $p_w(X)$ of $X$ w.r.t. a given circle $w.$ Thus, $AX^2=p_w(A)=$ $AP\cdot AF=\frac c2\cdot b\cos A=$ $\frac {2bc\cos A}4=$ $\frac 14\cdot\left(b^2+c^2-a^2\right)\implies$ $\boxed{4\cdot AX^2=b^2+c^2-a^2}\ .$

Obtain analogously $4\cdot BY^2=c^2+a^2-b^2$ and $4\cdot CZ^2=a^2+b^2-c^2.$ In conclusion, $AX^2+BY^2+CZ^2=\frac 14\cdot\left(a^2+b^2+c^2\right).$

P28 (Ruben Auqui). Let a square $ABCD$ with $AB=r$ and the circle $w=\mathbb C(A,r).$ For an interior $P$ of the square and which belongs to $w$ consider

$R\in BP\cap CD$ and $m\left(\widehat{CDP}\right)=\alpha,$ $m\left(\widehat{CPD}\right)=\beta .$ Prove that $\cot\alpha +\cot\beta=2$ and $PC=2\cdot PR\iff$ $R$ is the midpoint of $[CD].$

Proof. Prove easily that $\left\{\begin{array}{ccc} m\left(\widehat{BPC}\right) & = & 225^{\circ}-\beta\\\ m\left(\widehat{CBP}\right) & = & 45^{\circ}-\alpha\end{array}\right\|\ .$ Apply the theorem of Sines in $:\ \left\{\begin{array}{cccc} \triangle CPD\ : & \frac {PC}{\sin \widehat{CDP}}=\frac {CD}{\sin \widehat{CPD}} & \iff & \frac {PC}{\sin\alpha}=\frac r{\sin\beta}\\\\ \triangle CPB\ : & \frac {PC}{\sin \widehat{CBP}}=\frac {BC}{\sin \widehat{BPC}} & \iff & \frac {PC}{\sin\left(45^{\circ}-\alpha\right)}=\frac r{\sin\left(\beta -45^{\circ}\right)} \end{array}\right\|\implies$

$\frac {\sin\alpha}{\sin\left(45^{\circ}-\alpha\right)}=\frac {\sin\beta}{\sin\left(\beta -45^{\circ}\right)}\iff$ $\frac{\tan\alpha}{1-\tan\alpha}=\frac {\tan\beta}{\tan\beta -1}\iff$ $\tan\alpha +\tan\beta =2\tan\alpha\tan\beta\iff$ $\boxed{\cot\alpha +\cot\beta =2}\iff$ $\sin\left(\alpha +\beta\right)=2\sin\alpha\sin\beta\ .$

P29. Let $\triangle ABC$ with $A=70^{\circ}$ and $B=30^{\circ}.$ Prove that $\boxed{a^3=b^3+3ab^2}\ .$

Proof 1 (trigonometric). Denote $\boxed{t=\frac ab}\ (*)$ Therefore, $t=\frac {\sin A}{\sin B},$ i.e. $\boxed{t=2\cos 20^{\circ}}\ (1)$ and $a^3=b^3+3ab^2\iff$ $t^3=3t+1\ \stackrel{(1)}{\iff}\ 8\cos ^320^{\circ}=6\cos 20^{\circ}+1\iff$

$4\cos 20^{\circ}(1+\cos 40^{\circ})=6\cos 20^{\circ}+1\iff$ $4\cos 20\cos 40^{\circ}=1+2\cos 20^{\circ}\iff$ $2(\cos 60^{\circ}+\cos 20^{\circ})=1+2\cos 20^{\circ},$ what is true. See and proposed upperly P26.

Proof 2 ("slicing").

P30. Let $z=a+bi\in\mathbb C$ , where $\{a,b\}\subset\mathbb N^*$ . Find the smallest possible value of $a+b$ for which $z+z^2+z^3\in \mathbb R$ .

Proof. $z\not\in\mathbb R$ , i.e. $z\ne \overline z$ . Thus, $z+\overline z=2a$ , $z\overline z=a^2+b^2$ , $z^2+\overline z^2=(z+\overline z)^2-2z\overline z=$ $2\left(a^2-b^2\right).$ Thus, $z+z^2+z^3\in \mathbb R\iff$ $z+z^2+z^3=\overline z+\overline z^2+\overline z^3\iff$

$1+\left(z+\overline z\right)+\left(z^2+z\overline z+\overline z^2\right)=0\iff$ $1+2a+2\left(a^2-b^2\right)+\left(a^2+b^2\right)=0\iff$ $\boxed{3a^2+2a+1=b^2}\ ,$ i.e. the smallest positive integer $a$ for which $3a^2+2a+1$

is a positive integer. Found $a=6$ , i.e. $a=6$ and $b=11$ . In conclusion, the required sum $a+b$ is equal to $17$ .

This post has been edited 470 times. Last edited by Virgil Nicula, Feb 19, 2018, 8:29 PM

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August 2018

462. Diverse probleme de geometrie.

January 2018

461. Problems for the relaxation in ... weekend (II).

October 2017

460. Working page VI.

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459.Working page V.

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458. Working page IV.

457. Relatii metrice si inegalitati geometrice.

456. Probleme de Algebra.

July 2017

456* Integrals.

455. Problems for the relaxation in ... weekend!

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454. Mathematical note: "trig. ineg."- GMB 6/1992, page 201.

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453. Working page III.

December 2016

452. Working page II.

November 2016

451. ABC with C(O,R), C(I,r) and N-Nagel point ==> ON=R-2r.

450. Nota matematica: "Asupra unei inegalitati remarcabile".

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449. Working page I.

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448. Extensions or generalizations.

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447. Locuri geometrice remarcabile.

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446. Mathematics "instant" for the middle school.

445. Mathematics "instant" for the high school.

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444. Remarkable points in a triangle.

443. Problems of the type "instant" (continuare).

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442. Problems from and for baccalaureate II.

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438. Tangential quadrilateral. Zaslavsky's problem.

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435. Geometry problems in space.

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433. Some properties of remarkable lines in a triangle II.

432. Geometry problems with perpendicularities II.

431. Trigonometry in geometry II.

430. Trigonometrical identities.

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429. Bretschneider's_formula

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428. Some metrical problems from the contests II.

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427. Selected nice or well-known geometry problems.

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426. Problems from and for baccalaureate.

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425. Geometrical extremity II.

424. Some nice problems of E. E. Q. Rodriguez and others.

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423. Geometry problems.

422. Geometrical problems.

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419. Some geometry problems for hight school I

418. Some nice problems with polynomials/equations III

417 <bis> 418 (educatie: "14.27.28.32.57"/\"127.248").

417. The incircle of ABC and other two incircles.

416. Cateva solutii ale unei probleme de tip slicing.

415. Geometry 8.

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411. Geometry 4.

410. Geometry 3.

409. Geometry 2.

408. Geometry 1.

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407. Cyclical quadrilaterals.

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400. Some nice geometry problems for the high school I.

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399. Algebra (I).

398. Geometry problems (II).

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396. Own inegalities stronger than the Panaitopol's.

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395. Metrical relations in a quadrilateral.

394. A cevian in a triangle and two incircles.

393. Some problems with circles.

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392. Art of Problem Solving (geometry).

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391. CRUX Mathematicorum.

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390. Probleme de analiza matematica.

389. Some geometric properties in a triangle I.

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388. Some interesting "slicing" problems.

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387. Algebra (II).

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386. Barycentrical coordinates.

385. Some nice geometry problems.

384. Geometry problems (I).

383. Trigonometry in geometry (middle or high school) I.

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382. Peruan "jewels" (Israel Diaz, M. O. Sanchez a.s.o.)

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379. Geometrical extremity I.

378. Some metrical relations.

377. Some remarkable properties of the symmedian.

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376. Trigonometrical identities.

374. Some proofs of the Law of Cosines.

373. Extension of Chebyshev's inequality.

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372. The distancies : OI , OI_a , ON a.s.o.

371. Some problems from NMO (final stage) - 2013, Romania.

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370. Some metrical relations in a triangle.

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369. Mixtilinear circles of ABC.

364. Some interesting inequalities I

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362. Fermat's problem and other metrical problems.

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361. Arhimedes theorem of the broken chord in a circle.

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358. Some properties of the remarkable lines in a triangle.

September 2012

357. Morley theorem.

356. Order relation between s & (2R+r) in an acute triangle.

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355. Some problems with polynomials or equations I

354. Gradul II - Univ. B.B. Cluj, aug. 1998 Prof I sau II.

353. Some problems of the analytical geometry.

July 2012

352. Some definite integrals.

351. Problems with areas.

350. Another (easy) integrals.

349. Another interesting limits.

348. Subiectele de la BAC 2012 /Mate-Info.

June 2012

347. Two isogonal lines in a triangle.

346. Some easy integrals.

345. Characteristic function of a set.

344. Some trigonometry problems.

343. Problems with collinear points and concurrent lines.

May 2012

342. Exponential or logarithmic equations/inequations.

April 2012

341. Some nice and interesting "slicing" problems.

340. Geometry problems for middle school.

339. Derivatives. Rolle's function.

338. ROU National Math Olympiad 2012.

February 2012

337. Nice metrical problems.

January 2012

336. Some problems of the analytical geometry.

335. Some difficult geometric/trigonometric inequalities.

334. 2011 Japan Mathematical Olympiad Finals Problem 1.

333. Problems with the common tangent for two circles.

332. Some metrical problems from the contests I.

December 2011

331. Some nice problems with polynomials/equations I.

November 2011

330. A triangle and an its transversal through a point.

329. Some problems with complex numbers.

328. A general identities with areas in a triangle ABC.

October 2011

327. Some geometrical/algebraic extremum problems.

326. Some problems from trigonometry.

325. Without the l'Hospital's rules.

324. Some determinants.

323. Easy, nice and interesting problems for high school.

322. Incenter, Gergonne, Nagel a.s.o. of ABC.

321. Harmonical quadrilateral.

320. Some problems with metrical relations.

September 2011

319. Some problems with induction.

318. IRAN - 2011 (geometrie).

317. Some interesting geometry problems for middle school.

316. Some properties of isogonal rays in an angle of ABC.

August 2011

315. Indian Team Selection Test 2010 ST1 P1 and extension.

314. USAMTS 2009 Round 2 #5.

313. For middle school - some nice problems.

312. Very nice (famous trigonometrical problems) !

311. Some problems from USAMO and other contests.

310. Some properties of the tangential quadrilateral.

309. Beautiful problem ! IRAN Selection Test for IMO, 2004.

308. Some nice and easy properties in a triangle.

July 2011

307. Very nice proofs !

306. Some easy and nice "slicing" problems.

305. Position of the roots w.r.t. a given real number.

304. Integrals ("brothers") III.

303. Integrals II.

302. Some problems with sequencies of the "roots" .

301. Some trigonometrico-geometrical inequalities.

300. Lagrange's multiplier. Examples and problems.

299. A class of the recurrent sequencies.

298. Some nice problems from Crux Mathematicorum.

297. Nice inequalities with concrete numbers.

296. Some nice Vittasko's problems.

295. M1 3th subject, 2c - School Graduate, ROMANIA.

294. The Octav Halunga's inequality in a triangle.

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293. A difficult trigonometric equation (third point !).

292. Length of the angled bisector and some means.

291. Some integrals.

290. Characterize "OP perpendicular on OA" in ABC a.s.o.

289. Nice identities in an algebra/ geometry/trigonometry.

288. An inequality with two A-cevians in a triangle ABC.

287. Some metrical problems with or without trigonometry.

286. ABC is right triangle iff s=2R+r and other similar pp.

285. Nice and easy inequalities. For example, IMAC - 2009.

284. Ellipse (definition).

283. Some nice and easy problems.

282. Factorize of polynomials.

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281. Identity with R in an acute triangle (ILL - 1974).

280. Common roots of two polynomial equations.

279. O.M. Holland , O.M. Ireland 1988 and TST USA 2000, etc.

278. In memoria profesorului Alexandru Lupas.

277. A nice and difficult relation with Lemoine's axis.

276. Chinese TST 2009 and Second Round 2006.

275. Some interesting algebraic equations and systems.

274. An inequality in a triangle for its interior point.

273. The area of the triangle IOH. When I belongs to OH ?

272. Outside of triangle. Extension of Fermat's point.

271. Relations concerning Fermat points.

April 2011

270. Morrocan M.O. 2010 and Croatia TST 2009.

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.

266. Rectangles erected on sides of a triangle. Concurrence.

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.

260. Some nice, simple identities and their applications.

259. ABC equilateral triangle ==> DEF equilateral triangle.

258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

November 2010

179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

October 2010

170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

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

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

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

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

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

132. Asupra unei inegalitati intr-un triunghi.

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.

126. Routh's theorem.

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

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

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

120. Two equal neighbouring angles/sides in a quadrilateral.

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 !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

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

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

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).

105. A problem with three circles.

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

103. Metrical usual relations in triangle. Applications.

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).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

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

72. Colinearity in a triangle - H , P , M .

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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442. Problems from and for baccalaureate II.

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