441. LaTeX (common simbols).

by Virgil Nicula, Apr 7, 2016, 12:27 AM

LaTeX <= click => AoPS cyclic quadrilateral $\bf\color{red}\bigstar\ L\ a\ T\ e\ X\ \bigstar$ $\bigstar\ \mathrm{L\ a\ T\ e\ X}\ \bigstar$
$\blacktriangleright\ \mathrm{\underline{Operators}.}$

$\begin{array}{ccccccccccccccccccccccccccccccccccccccc} \pm & \mp & \times & \div & \cdot & \dots & \cdots & \ddots & \iddots & \ast & \star & \dagger & \ddagger & \amalg & \cap & \cup & \uplus & \sqcap & \sqcup & \vee & \wedge & \oplus & \ominus & \otimes\\\\ \circ & \bullet & \diamond & \lhd & \rhd & \unlhd & \unrhd &\oslash & \odot & \bigcirc & \triangleleft & \Diamond & \bigtriangleup & \bigtriangledown & \Box & \triangleright & \setminus & \wr & \sqrt{x} & x^{\circ} & \triangledown & \sqrt[n]{x} & a^x & a^{xyz} \end{array}$

$\blacktriangleright\ \mathrm{\underline{Relations}.}$

$\begin{array}{ccccccccccc cccccccccc cccccccccc} \le & \ge & \neq & \sim & \ll & \gg & \doteq & \simeq & \subset & \supset & \approx & \asymp & \subseteq & \supseteq & \cong & \smile & \sqsubset & \sqsupset & \equiv & \frown & \sqsubseteq & \sqsupseteq & \propto &\bowtie & \in & \ni & \prec & \succ & \vdash & \dashv\\\\ \preceq & \succeq & \models & \perp & \parallel & \gg & \mid & \bumpeq & \nmid & \nleq & \ngeq & \nsim & \ncong & \nparallel & \not< & \not> & \not= & \not\le & \not\ge & \not\sim & \not\approx & \not\cong & \not\equiv & \not\parallel & \nless & \ngtr & \lneq & \gneq & \lnsim & \lneqq\gneqq\end{array}$

$\blacktriangleright\ \mathrm{\underline{Greek Letters}.}$

$\begin{array}{cccccccccccccccccccccccccccccccccccccccc} \alpha & \beta & \gamma & \delta & \epsilon & \varepsilon & \zeta & \eta & \theta &\vartheta & \iota & \kappa & \lambda & \mu & \nu & \xi & \pi & \varpi & \rho & \varrho & \sigma & \varsigma & \tau & \upsilon & \phi & \varphi & \chi & \psi & \omega & \Gamma & \Delta & \Theta & \Lambda & \Xi & \Pi & \Sigma & \Upsilon &\Phi & \Psi & \Omega\end{array}$

$\blacktriangleright\ \mathrm{\underline{Arrows}.}$

$\begin{array}{cccccccccccccccccccccccccccccccccc} \gets & \to & \leftarrow & \Leftarrow & \rightarrow & \Rightarrow & \leftrightarrow & \Leftrightarrow & \mapsto & \hookleftarrow & \leftharpoonup & \leftharpoondown & \rightleftharpoons & \longleftarrow & \Longleftarrow & \longrightarrow & \longleftrightarrow & \Longleftrightarrow & \longmapsto & \hookrightarrow & \rightharpoonup & \rightharpoondown & \leadsto & \uparrow & \Uparrow & \downarrow & \Downarrow & \updownarrow & \Updownarrow & \nearrow & \searrow & \swarrow & \nwarrow\end{array}$

$\blacktriangleright\ \mathrm{\underline{Accents}\ and\ \underline{Others}.}$

$\begin{array}{cccccccccccccccccccccccccccccc} \hat{x} & \check{x} & \dot{x} & \breve{x} & \acute{x} & \ddot{x} & \grave{x} & \tilde{x} & \mathring{x} & \bar{x} & \vec{x} & \vec{\jmath} & \tilde{\imath} & \widehat{3+x} & \widetilde{abc} & \infty & \triangle & \angle & \aleph & \hbar & \imath & \jmath & \ell & \wp & \Re & \Im & \mho & \prime & \emptyset & \nabla\\\\ \surd & \partial & \top & \bot & \vdash & \dashv & \forall & \exists & \neg & \flat & \natural & \sharp & \backslash & \Box & \Diamond & \clubsuit & \diamondsuit & \heartsuit & \spadesuit & \Join & \blacksquare & \S & \P & \copyright & \pounds & \overarc{ABC} & \underarc{XYZ} & \bigstar & \square & \mathbb{R}\checkmark\end{array}$

$\blacktriangleright\ \mathrm{\underline{Command\ Symbols}.\ \underline{European\ Language\ Symbols}.\ \underline{Bracketing\ Symbols}.\ \underline{Multi-Size\ Symbols}.}$

$\begin{array}{cccccccccccccccccccccccccccccccccccccccccccc} \textdollar & \& & \% & \# & \_ & \{ & \} & \backslash & \oe & \ae & \o & \OE & \AE & \AA & \O & \l & \ss & !` & \L & \SS & \{ & \} & \| & \backslash & \lfloor & \rfloor & \lceil & \rceil & \langle & \rangle & \sum & \int & \oint & \prod & \coprod & \bigcap & \bigcup & \bigsqcup & \bigvee & \bigwedge & \bigodot & \bigotimes & \bigoplus & \biguplus\end{array}$

$\blacktriangleright\ \sum\textstyle\sum$ $\int \textstyle\int$ $\oint \textstyle\oint$ $\prod \textstyle\prod$ $\coprod \textstyle\coprod$ $\bigcap \textstyle\bigcap$ $\bigcup \textstyle\bigcup$ $\bigsqcup \textstyle\bigsqcup$ $\bigvee \textstyle\bigvee$ $\bigwedge \textstyle\bigwedge$ $\bigodot \textstyle\bigodot$ $\bigotimes \textstyle\bigotimes$ $\bigoplus \textstyle\bigoplus$ $\biguplus \textstyle\biguplus$ $\ominus$ $\oplus$ $\otimes\ ;\ \underbrace {f(f(\ldots f(x))\ldots)}_{n \textrm{ times}}\ ;\ \cancel{1024}\ ;\ \textrm{cis } x = \cos x + \textrm{i}\sin x=e^{\text{i}x}\ ;\ k\in\overline{1,n}\ .$

Examples. $\overline{XYZT}\ ;\ \underline{UVW}\ ;\ \underleftrightarrow{\overbrace{XYZT}}\ ;\ \overbrace{\underleftarrow{ABCD}}\ ;\ \underbrace{\overbrace{XYZT}}$ $;$ $\overleftrightarrow{\underbrace{ABCD}}\ ;\ \displaystyle\lim_{x\to a}$ $;$ $\underleftarrow{ABCD}$ $;$ $\overleftarrow{ABCD}$ $;$ $\overrightarrow{ABCD}\ ;\ \underleftarrow{\overleftrightarrow{ABCD}}$ $;$ $E_n=\frac{\lim\limits_{x\searrow 0}\frac {\sin x}x+2^{n-1}\cdot\left(1+\prod\limits_{k=1}^na_k\right)}{\prod\limits_{k=1}^n\left(1+a_k\right)}\ ,\ n\in\mathbb{N}^*\ \ ;$

$\binom{\widehat{\odot \triangledown \odot}}{\wr \wr}$ $;$ $\bf\color{red}Nice\ proof\ !\ \bf\color{black}\ ;\ \underbrace{\text{sequence of letters}}_{\text{stuff going underneath}}$ $;$ $\overbrace{\text{sequence of letters}}^{\text{stuff going above}}$ $;$ $\overarc{ABC}$ $;$ $\overbrace {\underbrace {\overline {\underline {\left| \underline {\overline {\left| NOBODY ?\right| }}\right| }}}}\ ;\ E_n=\frac{\lim_{x\searrow 0}\frac {\sin x}x+2^{n-1}\cdot\left(1+\prod_{k=1}^na_k\right)}{\prod_{k=1}^n\left(1+a_k\right)}\ ,\ n\in\mathbb{N}^*\ ;\ \cancel{\bigcirc}$

$\boxed{\begin{array}{ccccc} \mathrm{C}\ \mathrm{R}\ \mathrm{Q}\ \mathrm{Z}\ \mathrm{rm} & \mathbb{C}\ \mathbb{R}\ \mathbb{Q}\ \mathbb{Z}\ \mathrm{bb} & \mathbf{C}\ \mathbf{R}\ \mathbf{Q}\ \mathbf{Z}\ \mathrm{bf} & \mathcal{C}\ \mathcal{R}\ \mathcal{Q}\ \mathcal{Z}\ \mathrm{cal} & \mathfrak{C}\ \mathfrak{R}\ \mathfrak{Q}\ \mathfrak{Z}\ \mathrm{frak}\end{array}}\ ;\ \bf\color{red}Here\ are\ some\ links\ :$ Casey's theorem $\blacktriangleleft\blacktriangleright$

All the best !

$\mathrm{END}$
P1. $\left\{\ z_1\ ,\ z_2\ ,\ z_3\ \right\}\ \subset\ \mathbb{C}\ \ \wedge\ \ \left\|\ \begin{array}{c} |z_1|=|z_2|=|z_3|=1\\\\ z_1^3+z_2^3+z_3^3+z_1z_2z_3=0\end{array}\ \right\|\ \Longrightarrow\ |z_1+z_2+z_3|\ \in\ \{\ 1\ ,\ 2\ \}\ .$

Proof. $\ \ \frac {z_1}{u}=\frac {z_2}{v}=\frac {z_3}{1}=\frac {z_1+z_2+z_3}{u+v+1}\ \Longrightarrow\ \left\|\ \begin{array}{c} |u|=|v|=1\\\\ u^3+v^3+1+uv=0\ (1)\\\\ \left|z_1+z_2+z_3\right|=|u+v+1|\end{array}\ \right\|\ .$ Thus, $(1)\ \ \wedge\ \ \overline u=\frac 1u\ \ \wedge\ \ \overline v=\frac 1v\ \Longrightarrow\ u^3+v^3+u^3v^3+u^2v^2=0\ (2)\ .$

$(1)\ \ \wedge\ \ (2)\ \Longrightarrow\ 1+uv=u^2v^2(1+uv)=-\left(u^3+v^3\right)\ \Longrightarrow$ $(1+uv)^2(1-uv)=0\ \ \wedge\ \ \left(u^3+v^3\right)+(1+uv)=0\ .$

$\blacktriangleright\ \ uv=1\ \wedge\ \left(u^3+v^3\right)+2=0\ \Longrightarrow\ (u+v)^3-3(u+v)+2=0\ \Longrightarrow\ u+v\in\{1,-2\}\ \Longrightarrow\ |u+v+1|=\left|z_1+z_2+z_3\right|\in\{1,2\}\ .$

$\blacktriangleright\ \ uv=-1\ \wedge\ u^3+v^3=0\ \Longrightarrow\ (u+v)^3+3(u+v)=0\ \Longrightarrow\ u+v\in\left\{0,\pm i\sqrt 3\right\}\ \Longrightarrow$ $|u+v+1|=\left|z_1+z_2+z_3\right|\in\{1,2\}\ .$

$\left\|\ \begin{array}{c} |u|=|v|=1\\\\\\ u^3+v^3+1+uv=0\end{array}\ \right\|\ \Longrightarrow\ u+v\ \in\ \left\{\ -2\ ,\ 1\ ,\ 0\ ,\ \pm i\sqrt 3\ \right\}\ .$

$\left|\begin{array}{c} u+v=1\\\\ uv=1\end{array}\right|\ \ \vee\ \ \left|\ \begin{array}{c} u+v=-2\\\\ uv=1\end{array}\right|\ \ \vee\ \ \left|\ \begin{array}{c} u+v=0\\\\ uv=-1\end{array}\right|\ \ \vee\ \ \left|\ \begin{array}{c} u+v=i\sqrt 3\\\\ uv=-1\end{array}\right|\ \ \vee\ \ \left|\ \begin{array}{c} u+v=-i\sqrt 3\\\\ uv=-1\end{array}\right|\ \ .$

OFF-topic. In general in viata este mai usor sa gresesti decat sa intelegi unde si de ce ai gresit. Si in matematica se poate intampla sa gresesti, insa
astfel apare o noua problema - descoperirea greselii - care te proiecteaza spre alte metode sau chiar extinderea teoriei. Cu alte cuvinte, greseala in
matematica uneori este benefica. Insa aici in noua problema aparuta ai "dezavantajul" ca trebuie sa te descurci singur, cum de altfel s-a si intamplat.

P2. Fie triunghiul $ABC$ cu $m\left(\widehat{BAC}\right)=120^{\circ}\ .$ Punctele $M$ , $N$ si $P$ sunt picioarele bisectoarelor din $A\ ,$ $B\ ,$ $C$ pe $BC\ ,$ $CA\ ,$ $AB$ respectiv. Sa se arate ca $m\left(\widehat{BNM}\right)=30^{\circ}\ .$

Demonstratie. Notam incentrele $I\ ,$ $R$ ale triunghiurilor $ABC\ ,$ $ABM$ respectiv. Se observa ca $AR\perp AN\ ,$ adica $[AN$ este bisectoarea exterioara din varful $A$

in $\triangle ABI$ ceea ce inseamna $\frac {RB}{RI}=\frac {NB}{NI}\ .$ Deci si $[MR$ , $[MN$ sunt bisectoarele (interioara si exterioara) din varful $M$ in $\triangle BMI\ .$ asadar, $MR\perp MN$ ,

adica patrulaterul $ARMN$ este inscris in cercul de diametru $[RN]\ \Longrightarrow$ $\widehat{BNM}\equiv\widehat{RNM}\equiv\widehat{RAM}\ \Longrightarrow\ m\left(\widehat{BNM}\right)=30^{\circ}$ - concluzia problemei.

Generalizare. Fie $ABC$ un triunghi pentru care $m\left(\widehat {BAC}\right)\ >\ 90^{\circ}\ .$ Consideram bisectoarea interioara $[BN$ din varful $B$

unde $N\in (AC)$ si punctul $M\in (BC)$ pentru care $m\left(\widehat {MAC}\right)\ =\ 180^{\circ}-A\ .$ Sa se arate ca $m\left(\widehat {BNM}\right)=A-90^{\circ}\ .$

P3. Fie $\triangle\ ABC$ si $D\in (AB)$ , $E\in (AC)$ ca $\overrightarrow{DB}+\overrightarrow{EC}=p\cdot \left(\overrightarrow{DA}+\overrightarrow{EA}\right)\ .$ Notam $T\in BE\ \cap\ CD\ .$ Sa se afle $\alpha =\alpha (p) \in\mathbb R$ astfel incat $\overrightarrow{TB}+\overrightarrow{TC}=\alpha\cdot \overrightarrow{TA}\ .$

Demonstratie. Alegem originea sistemului in $A\ ,$ adica $\overrightarrow A=\vec 0$ si pentru orice $X\ ,$ $Y$ avem $\overrightarrow{AX}=X\ \ ,\ \ \overrightarrow{XY}=Y-X\ .$ Deci exista $\{m,n\}\subset (0,1)$ ca $\boxed{D=mB\ \ \wedge\ \ E=nC}\ .$

$\blacktriangleright\ \overrightarrow{DB}+\overrightarrow{EC}=p\cdot \left(\overrightarrow{DA}+\overrightarrow{EA}\right)$ $\Longleftrightarrow$ $(B+C)-(D+E)=-p(D+E)$ $\Longleftrightarrow(1-p)(D+E)=$

$B+C\Longleftrightarrow$ $(1-p)(mB+nC)=B+C$ $\Longleftrightarrow$ $\boxed{m=n=\frac {1}{1-p}}\ \ (\ DE\ \parallel\ BC\ )\ .$

$\blacktriangleright\ T\in BE\ \cap\ CD$ $\Longleftrightarrow$ exista $u$ , $v$ reale ca $T=uD+(1-u)C=vE+(1-v)B$ $\Longleftrightarrow$ $T=muB+(1-u)C=(1-v)B+nvC$ $\Longleftrightarrow$ $\left\{\begin{array}{c} mu=1-v\\\\ nv=1-u\end{array}\right\|$ $\Longleftrightarrow$

$\left\{\begin{array}{c} mu+v=1\\\\ u+nv=1\end{array}\right\|$ $\Longleftrightarrow$ $\left\{\begin{array}{c} u=\frac {n-1}{mn-1}\\\\ v=\frac {m-1}{mn-1}\end{array}\right\|$ si in acest caz $\boxed{T=\frac {m(n-1)}{mn-1}\cdot B+\frac {n(m-1)}{mn-1}\cdot C}$ (prelucrare in general, fara a tine seama de relatia din ipoteza !).

$\blacktriangleright\ m=n=\frac {1}{1-p}$ $\implies$ $\boxed{\ T=\frac {1}{2-p}\cdot (B+C)\ }$ si $\overrightarrow{TB}+\overrightarrow{TC}=$ $\alpha \cdot \overrightarrow{TA}$ $\Longleftrightarrow$ $B+C=(2-\alpha )\cdot T$ $\Longleftrightarrow$ $(B+C)=\frac {2-\alpha}{2-p}\cdot (B+C)$ $\Longleftrightarrow\boxed{\ \alpha =p\ }\ .$

P4. Sa se arate ca un triunghi $ABC$ este dreptunghic $\Longleftrightarrow \ \cos\frac A2\cos\frac B2\cos\frac C2-\sin\frac A2\sin\frac B2\sin\frac C2=\frac 12\ \iff\ p=2R+r\ .$

Demonstratie 1. Folosim identitatile $:\ \left\|\ \begin{array}{ccc} \cos\frac A2\cos\frac B2\cos\frac C2 & = & \frac{p}{4R}\\\\ \sin\frac A2\sin\frac B2\sin\frac C2 & = & \frac{r}{4R}\ \end{array}\right\|\ .$ Astfel egalitatea devine $:\ p=2R+r\ \Longleftrightarrow\ p^{2}=4R^{2}+4Rr+r^{2}\ \Longleftrightarrow\ 2p^{2}-8Rr-2r^{2}=$

$8R^{2}\Longleftrightarrow$ $a^{2}+b^{2}+c^{2}=8R^{2}\ \Longleftrightarrow\ \sin^{2}A+\sin^{2}B+\sin^{2}C=2\ \Longleftrightarrow\ \cos 2A+\cos 2B+\cos 2C+1=0$ $\Longleftrightarrow\ 2\cos(A-B)\cos (A+B)+2cos^{2}C=0\ \Longleftrightarrow$

$\cos C=0$ sau $\cos C=\cos(A-B)$ $\Longleftrightarrow\ C=90^{\circ}$ sau $A=B+C$ sau $B=A+C$ $\Longleftrightarrow\ 90^{\circ}\in\{A,\ B,\ C\}\ .$

Demonstratie 2. Se arata usor prin relatiile Viete ca ecuatia $f(x)=p\cdot x^3-(4R+r)\cdot x^2+p\cdot x-r=0$ admite radacinile $\left\{\tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\right\}\ .$

Se observa ca $f(1)=2[p-(2R+r)]\ .$ Asadar, $p=2R+r\ \Longleftrightarrow\ f(1)=0\ \Longleftrightarrow\ 1\in \left\{\tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\right\}\ \Longleftrightarrow\ 90^{\circ}\in\{A,B,C\}\ .$

Demonstratie 3. $p=2R+r\iff$ $(p-a)=(2R-a)+(p-a)\tan\frac A2\iff$ $(p-a)\left(1-\tan\frac A2\right)=2R(1-\sin A)\ \stackrel{\tan\frac A2=t}{\iff}\ (p-a)(1-t)=$ $2R\cdot\left(1-\frac {2t}{1+t^2}\right)$

$\iff$ $(p-a)(1-t)=2R\cdot\frac {(1-t)^2}{1+t^2}\iff$ $t=1\ \vee\ (p-a)\left(1+t^2\right)=2R(1-t)\ .$ Deci $t=1\iff \underline{\underline{A=90}}^{\circ}\ .$ Presupunem f.r.g. $t\ne 1$ , adica $A\ne 90^{\circ}\ .$ In acest caz

$p-a=2R\cos\frac A2\left(\cos\frac A2-\sin\frac A2\right)\iff$ $p-a=R\left(2\cos^2\frac A2-\sin A\right)\iff$ $b+c-a=4R\cos^2\frac A2-a\iff$ $b+c=4R\cos^2\frac A2\iff$ $\sin B+\sin C=2\cos^2\frac A2$

$\iff 2\cos\frac A2\cos\frac {B-C}2=2\cos^2\frac A2\iff$ $\cos\frac {B-C}2=\cos\frac A2\iff$ $B=A+C$ sau $C=A+B\iff \underline{\underline{B=90^{\circ}\ \vee\ C=90}}^{\circ}\ .$ In concluzie, $\boxed{90^{\circ}\in\{A,B,C\}}\ .$

P5 (inegalitate cu medii). Daca $a,b>0$ sa se arate ca $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}\ .$

Demonstratie. $\left(\sqrt {ab}+\sqrt {\frac {a^2+b^2}{2}}\right)^2\le 2\cdot \left(ab+\frac {a^2+b^2}{2}\right)=2ab+a^2+b^2=(a+b)^2\ \Longrightarrow$ $\boxed{\ \sqrt {ab}+\sqrt {\frac {a^2+b^2}{2}}\le a+b\ }\ (*)\ .$

Se observa ca $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}\ \Longleftrightarrow\ \frac {a+b}{2}-\frac {2ab}{a+b}\le\sqrt {\frac {a^2+b^2}{2}}-\sqrt {ab}\ \Longleftrightarrow$

$\left(\frac {a+b}{2}-\frac {2ab}{a+b}\right)\cdot\left(\sqrt {\frac {a^2+b^2}{2}}+\sqrt {ab}\right)\le\frac {a^2+b^2}{2}-ab\ \Longleftrightarrow\ \frac {(a-b)^2}{2(a+b)}\cdot\left(\sqrt {\frac {a^2+b^2}{2}}+\sqrt {ab}\right)\le\frac {(a-b)^2}{2}\ \Longleftrightarrow\ (*)\ .$

P6 (O.J.M). Notam in $\triangle ABC$ mijlocul $M$ al laturii $[AB]$ si piciorul $D\in (AC)$ al bisectoarei din $B\ .$ Aratati ca $DM \perp DB\ \Longleftrightarrow\ AB=3\cdot BC\ .$

Demonstratie 1. Fie $N\in (AB)$ pentru care $ND\parallel BC\ .$ Astfel, $\widehat {NDB}\equiv\widehat {CBD}\equiv \widehat {NBD}\ ,$ adica $NB=ND\ .$ Din $DN\parallel BC$ obtinem

$\frac {AN}{AB}=\frac {DN}{BC}$ $\Longleftrightarrow$ $\frac {AN}{AB}=\frac {NB}{BC}\ (*)$ In concluzie, $DM\perp DB\ \Longleftrightarrow\ NB=NM=DN=\frac c4\stackrel {(*)}{\ \Longleftrightarrow\ }\frac 34=\frac {\frac c4}{a}\ \Longleftrightarrow\ c=3a\ .$

Demonstratie 2. Se arata usor ca $m\left(\widehat{ADM}\right)=\frac {C-A}2\ .$ Voi folosi o relatia cunoscuta $\frac {MA}{MB}=\frac {DA}{DB}\cdot\frac {\sin\widehat{MDA}}{\sin\widehat{MDB}}\iff$ $DA\cdot\sin\frac {C-A}2=DB\iff$

$\frac {bc}{a+c}\sin\frac {C-A}2=\frac {2ac}{a+c}\cdot\cos\frac B2\iff$ $\boxed{\begin{array}{ccc} b\cdot\sin\frac {C-A}2 & = & 2a\cdot\cos\frac B2\\\\ 2\cos \frac {C+A}2 & = & 2\sin\frac B2\end{array}}\ \bigodot\iff$ $b(\sin C-\sin A)=2a\sin B\iff$ $b(c-a)=2ab\iff c=3a\ .$

Extindere. Notam in $\triangle ABC$ piciorul $D\in (AC)$ al bisectoarei din $B$ si consideram un punct $M\in (AB)$ pentru care $DM\perp DB\ .$ Aratati ca $\frac {AM}{AB}=\frac {c-a}{c+a}\ .$

Generalizare. Fie $\triangle ABC$ si $\left\|\begin{array}{ccc} M\in (AB) & , & \frac {MA}{MB}=m\\\\ N\in (CA) & , & \frac {NA}{NC}=n\end{array}\right\|\ .$ Sa se arate ca $NM\perp NB\ \ \Longrightarrow\ \ n(2m+1-n)\cdot b^2+(n+1)(n-2m)\cdot c^2=n(n+1)(2m+1)\cdot a^2\ .$

Proof. Denote $\left\{\begin{array}{ccc} m\left(\widehat{ANM}\right) & = & u\\\\ m\left(\widehat{BCN}\right) & = & v\end{array}\right\|$ and apply the theorem of Sines in the triangles $:$

$\blacktriangleright\ \triangle AMN\ :\ \frac {AM}{\sin \widehat{ANM}}=\frac {AN}{\sin\widehat{AMN}}\iff$ $\frac {mc}{(m+1)\sin u}=\frac {nb}{(n+1)\sin (A+u)}\iff$ $cm(n+1)(\sin A+\cos A\tan u)=bn(m+1)\tan u\iff$

$\tan u=\frac {cm(n+1)\sin A}{bn(m+1)-cm(n+1)\cos A}\iff$ $\tan u=\frac {m(n+1)\cdot 2bc\sin A}{2b^2n(m+1)-m(n+1)\cdot 2bc\cos A}\iff$ $\tan u=\frac {4m(n+1)S}{2b^2n(m+1)-m(n+1)\left(b^2+c^2-a^2\right)}\iff$

$\boxed{\tan u=\frac {4m(n+1)S}{m(n+1)\left(a^2-c^2\right)+(mn+2n-m)b^2}}\ (1)\ .$

$\blacktriangleright\ \triangle BCN\ :\ \frac {BC}{\sin \widehat{BNC}}=\frac {NC}{\sin\widehat{CBN}}\iff$ $\frac {a}{\sin v}=\frac {b}{}(n+1)\sin (C+v)\iff$ $a(n+1)(\sin C+\cos C\tan v)=b\tan v\iff$ $\tan v=\frac {a(n+1)\sin C}{b--a(n+1)\cos C}\iff$

$\tan v=\frac {(n+1)\cdot 2ab\sin C}{2b^2-(n+1)\cdot 2ab\cos C}\iff$ $\tan v=\frac {4(n+1)S}{2b^2-(n+1)\left(a^2+b^2-c^2\right)}\iff$ $\boxed{\tan v=\frac {4(n+1)S}{-(n+1)\left(a^2-c^2\right)+(1-n)b^2}}\ (2)\ .$ Therefore, $NM\perp NB\iff$

$u+v=90^{\circ}\iff$ $\tan u\tan v=1\ \stackrel{1\wedge 2}{\iff}\ m(n+1)^2\cdot 16S^2=$ $\left[m(n+1)\left(a^2-c^2\right)+(mn+2n-m)b^2\right]\cdot\left[-(n+1)\left(a^2-c^2\right)+(1-n)b^2\right]\iff$

$m(n+1)^2\cdot \left[2\cdot\sum \left(b^2c^2\right)-\sum a^4+\left(a^2-c^2\right)^2\right]=(1-n)(mn+2n-m)b^4+\left[m\left(1-n^2\right)-(1+n)(mn+2n-m)\right]b^2\left(a^2-c^2\right)\iff$

$m(n+1)^2\cdot \left[2\left(a^2+c^2\right)-b^2\right]=$ $(1-n)(mn+2n-m)b^2+\left[m\left(1-n^2\right)-(1+n)(mn+2n-m)\right]\left(a^2-c^2\right)\iff$

$n(2m+1-n)\cdot b^2+(n+1)(n-2m)\cdot c^2=n(n+1)(2m+1)\cdot a^2\ .$

Caz particular. If $n=2m\ ,$ then $NM\perp NB\iff b=(n+1)a$ (urmeaza sa va "jucati" particularizand "m" si "n" pentru a obtine o relatie frumoasa intre a , b , c !

P7. Fiind dat $\triangle ABC$ notam piciorul $D\in (BC)$ al bisectoarei interioare din varful $A$ si pentru un punct $P\in (AD)$

notam $\left\{\begin{array}{ccc} E\in BP\cap AC & ; & F\in CP\cap AB\\\\ X\in BE\cap DF & ; & Y\in CF\cap DE\end{array}\right\|\ .$ Sa se arate ca $[AD$ este bisectoarea unghiului $\widehat {XAY}\ .$

Generalizare. Fie $\triangle ABC\ ,$ un interior $P$ si $\left\|\begin{array}{c} D\in AP\cap BC\\\\ E\in BP\cap AC\\\\ F\in CP\cap AB\end{array}\right\|\ ,$ $\left\|\begin{array}{c} X\in BE\cap DF\\\\ Y\in CF\cap DE\end{array}\right\|$ si $\left\|\begin{array}{ccc} m(\angle BAX)=x & ; & m(\angle CAY)=y\\\\ m(\angle PAX)=u & ; & m(\angle PAY)=v\end{array}\right\|\ .$ Sa se arate $\frac {\sin (x+u)}{\sin (y+v)}=\frac {\sin x\sin v}{\sin y\sin u}\ .$

Demonstratie : Aplicam teorema lui Ceva sub forma trigonometrica pentru punctul/triunghiul $:$

$\left\| \begin{array}{cccccccccccccccccc} X/\triangle ABD & : & \frac{\sin x}{\sin u} & \cdot & \frac{\sin(\angle ADF)}{\sin(\angle BDF)} & \cdot & \frac{\sin(\angle DBP)}{\sin(\angle PBA)} & = & 1 & \Longleftrightarrow & \frac{\sin x}{\sin u} & = & \frac{\sin(\angle BDF)}{\sin(\angle ADF)} & \cdot & \frac{\sin(\angle ABE)}{\sin(\angle CBE)} & & (1) & \\\\ Y/\triangle ADC & : & \frac{\sin v}{\sin y} & \cdot & \frac{\sin(\angle ACP)}{\sin(\angle DCP)} & \cdot & \frac{\sin(\angle CDE)}{\sin(\angle EDA)} & = & 1 & \Longleftrightarrow & \frac{\sin v}{\sin y} & = & \frac{\sin(\angle BCF)}{\sin(\angle ACF)} & \cdot & \frac{\sin(\angle ADE)}{\sin(\angle CDE)} & & (2) & \\\\ P/\triangle ABC & : & \frac{\sin(\angle ABE)}{\sin(\angle CBE)} & \cdot & \frac{\sin(\angle BCF)}{\sin(\angle FCA)} & \cdot & \frac{\sin(\angle CAD)}{\sin(\angle BAD)} & = & 1 & \Longleftrightarrow & \frac{\sin(x+u)}{\sin(v+y)}& = & \frac{\sin(\angle ABE)}{\sin(\angle CBE)} & \cdot & \frac{\sin(\angle BCF)}{\sin(\angle FCA)} & & (3) & \ \end{array}\right\|$

$\stackrel{1\wedge 2}{\iff}\ \begin{array}{ccccccccccc} \frac{\sin x}{\sin u} & \cdot & \frac{\sin v}{\sin y} & = & \frac{\sin(\angle ABE)}{\sin(\angle CBE)} & \cdot & \frac{\sin(\angle BCF)}{\sin(\angle ACF)} & \cdot & \frac{\sin(\angle BDF)}{\sin(\angle ADF)} & \cdot & \frac{\sin(\angle ADE)}{\sin(\angle CDE)}\end{array}\ .$

$\stackrel{3}{\iff}\ \begin{array}{ccccccccc} \frac{\sin x}{\sin u} & \cdot & \frac{\sin v}{\sin y} & = & \frac{\sin(x+u)}{\sin(v+y)} & \cdot & \frac{\sin(\angle BDF)}{\sin(\angle ADF)} & \cdot & \frac{\sin(\angle ADE)}{\sin(\angle CDE)} \end{array}\ \ \ (4)\ .$ Pe de alta parte avem : $\left\|\ \begin{array}{ccccc} \frac{BF}{FA} & = & \frac{BD}{AD} & \cdot & \frac{\sin(\angle BDF)}{\sin(\angle ADF)} \\\\ \frac{AE}{EC} & = & \frac{AD}{DC} & \cdot & \frac{\sin(\angle ADE)}{\sin(\angle CDE)} \end{array}\right|\ \bigodot\ \Longrightarrow$

$\frac{BF}{FA}\cdot\frac{AE}{EC}=$ $\frac{BD}{DC}\cdot\frac{\sin(\angle BDF)}{\sin(\angle ADF)}\cdot$ $\frac{\sin(\angle ADE)}{\sin(\angle CDE)}\ \stackrel{\mathrm{Ceva}}{\iff}\ \frac{\sin(\angle BDF)}{\sin(\angle ADF)}\cdot\frac{\sin(\angle ADE)}{\sin(\angle CDE)}=1\ (5)\ .$ Asadar, din relatiile $4$ si $5\ \Longrightarrow$ $\boxed{\ \frac{\sin x}{\sin u}\cdot\frac{\sin v}{\sin y}=\frac{\sin(x+u)}{\sin(v+y)}\ }\ .$

P8. Fie $\triangle ABC$ cu cercul inscris $w=\mathbb C(I,r)$ si punctele $D\in AI\cap BC\ ,$ $E\in BI\cap CA\ ,$ $F\in CI\cap AB\ .$

1. Notam $X\in DE$ si $Y\in DF$ pentru care $I\in XY\parallel BC\ .$ Sa se arate ca $IX=IY=\frac {abc}{(b+c)(a+b+c)}\ .$

2. Notam punctul $M$ unde bisectoarea $[AI$ taie a doua oara cercul circumscris $w$ al $\triangle ABC$ si punctele $U\in XY\cap AC\ ,$

$V\in XY\cap AB\ ,$ $N\in MU\cap BI\ ,$ $P\in MV\cap CI\ .$ Sa se arate ca $\{N,P\}\subset w\ .$

Demonstratie.

1. Notam $R\in AX\cap BC\ .$ Din asemanarea $\triangle AIX\sim\triangle ADR\ \Longrightarrow\ \frac{IX}{DR}=\frac{AI}{AD}\ \Longleftrightarrow\ \boxed{IX=\frac{AI}{AD}\cdot DR}\ (1)\ .$ In $\triangle ABD$ aplicam teorema bisectoarei $\Longrightarrow$

$\frac{AI}{ID}=\frac{AB}{BD}=\frac{c}{\frac{ac}{b+c}}\iff$ $\frac{AI}{ID}=\frac{b+c}{a}\iff$ $\boxed{\frac{AI}{AD}=\frac{b+c}{a+b+c}}\ (2)$ si din $IX\parallel DR\ \Longrightarrow$ $\frac{AX}{XR}=\frac{b+c}{a}\ (3)\ .$ Aplicam teorema lui Menelaus transversalei

$\overline{EXD}$ si $\triangle ACR$ $\Longrightarrow$ $\frac{AE}{EC}\ \cdot\ \frac{CD}{DR}\ \cdot\ \frac{RX}{XA}=1$ $\Longleftrightarrow$ $DR=\frac ca\ \cdot\ \frac{ab}{b+c}\cdot$ $\frac{RX}{XA}\ \stackrel{3}{\iff}\ DR=\frac ca\cdot\frac{ab}{b+c}\cdot\frac{a}{b+c}\iff$ $\boxed{DR=\frac{abc}{(b+c)^2}}\ (4)\ .$

Acum, revenind in relatia $\ (1)\ \stackrel{2\wedge 4}{\iff}\ \boxed{IX=\frac{abc}{(b+c)(a+b+c)}}\ .$ Analog se calculeaza si segmentul $[IY]\ .$

2. Pentru a arata ca $P$ se afla pe cercul circumscris $\triangle ABC$ este suficient sa aratam ca $ACMP$ este inscriptibil. Astfel ne ramane sa demonstram ca $\angle MPC=\angle MAC=\frac{\angle A}{2}\ .$ Din $\angle AIE$ este exterior $\triangle AIB$ $\Longrightarrow\ \angle AIE=\frac{\angle A+\angle B}{2}\ \iff$ $\angle BIM=\angle AIE=\frac{\angle A+\angle B}{2}\ .$ Dar $ABMC$ este inscriptibil $\iff\angle MBC=\angle MAC=\frac{\angle A}{2}\iff$ $m(\angle MBI)=\frac{\angle A+\angle B}{2}=\angle BIM\ .$ Deci $\triangle BIM$ este isoscel cu $BM=BI\ (*)\ .$ Acum din $UV\parallel BC\iff$ $\angle IUE=\angle C$ si deoarece $\angle IEU=180^{\circ}-\angle C-\frac{\angle B}{2}\iff$ $\angle BIV=\angle EIU=\frac{\angle B}2\ \triangle BIV$ este isoscel cu $BV=VI\ (**)$ $\stackrel{(*)\wedge(**)}{\iff}\ VM$ este mediatoarea lui $[BI]$ $\Longleftrightarrow\ PM$ este mediatoarea lui $[BI]\ .$ Notam $\{O\}=PM\cap BI\ .$ Atunci $\triangle POI$ este dreptunghic in $O$ si cum $\angle PIB$ este exterior $\triangle BIC\ \Longrightarrow\ \angle PIO=\frac{\angle B+\angle C}{2}\ .$ Asadar, $\angle MPC=\angle OPI=$ $90^{\circ}-\frac{\angle B+\angle C}{2}=$ $\frac{\angle A}{2}$ c.c.t.d. Se arata analog ca si $N$ se afla pe cercul circumscris $\triangle ABC\ .$

P9 (Ruben Dario). Fie punctele coliniare $(A,C,E,O,D,B)\subset d$ in aceasta ordine astfel incat $O$ este mijlocul lui $[AB]$ si $EA=a\ ,$ $EB=b\ ,$ $AC=2c\ ,$ $DB=2d\ .$ Consideram $:$

semicercul $w$ cu diametrul $[AB]\ ;$ semicercul $w_1$ cu centrul $J$ si diametrul $[AC]\ ;$ semicercul $w_2$ cu centrul $K$ si diametrul $[BD]\ ;$ cercul $\Omega =\mathbb C(I,x)$ care este tangent exterior cercurilor

$w_1\ ,$ $w_2\ ,$ tangent interior semicercului $w$ si tangent dreptei $d$ in punctual $E\ .$ Intregul desen este situat in semiplanul $(d,I)\ .$ Sa se arate ca $\frac 1a+\frac 1b=\frac 1x$ si $\frac 1c-\frac 1d=4\left(\frac 1a-\frac 1b\right)\ .$

Demonstratie. Se arata usor ca $:\ \left\{\begin{array}{ccccccc} \triangle EIO\ : & IO^2=EI^2+EO^2 & \implies & \left(\frac {a+b}2-x\right)^2=x^2+\left(\frac {a-b}2\right)^2 & \implies & \frac 1a+\frac 1b=\frac 1x & (*)\\\\ \triangle EIJ\ : & IJ^2=EJ^2+EI^2 & \implies & (x+c)^2=(a-c)^2+x^2 & \implies & 2c=\frac {a^2}{x+a} & (1)\\\\ \triangle EIK\ : & IK^2=EK^2+EI^2 & \implies & (x+d)^2=(b-d)^2+x^2 & \implies & 2d=\frac {b^2}{x+b} & (2)\end{array}\right\|\ .$ Din diferenta relatiilor $(1)$ si $(2)$

obtinem $2(d-c)=\frac {b^2}{x+b}-\frac {a^2}{x+a}=$ $\frac {x\left(b^2-a^2\right)+ab(b-a)}{(x+a)(x+b)}\implies$ $\frac {2(d-c)}{b-a}=$ $\frac {x(a+b)+ab}{(x+a)(x+b)}\ \stackrel{(*)}{=}\ \frac {2ab}{(x+a)(x+b)}$ $\implies$ $\boxed{\frac {d-c}{b-a}=\frac {ab}{(x+a)(x+b)}}\ (3)\ .$

Din produsul relatiilor $(1)$ si $(2)$ obtinem $\boxed{4cd=\frac {a^2b^2}{(x+a)(x+b)}}\ (4)\ .$ Din raportul relatiilor $(3)$ si $(4)$ obtinem $\frac {d-c}{b-a}\cdot \frac 1{4cd}=\frac {ab}{(x+a)(x+b)}\cdot \frac {(x+a)(x+b)}{a^2b^2}\iff$

$\frac {d-c}{b-a}=\frac {4cd}{ab}\iff$ $\frac {d-c}{cd}=4\cdot \frac {b-a}{ab}\iff$ $\frac 1c-\frac 1d=4\left(\frac 1a-\frac 1b\right)\ .$ Nice problem!

P10 (M. O. Sanchez). Let $\triangle ABC$ with the midpoint $M$ of $[AC]\ ,$ the bisector $[CN\ ,$ where $N\in (AB)$

and the Lemoine point $K$ of $\triangle ABC\ .$ Prove that $K\in MN\iff$ $ab+c^2=a^2\iff A=B+2C\ .$

Proof. Denote $L\in BC\cap AK\ .$ Is well known that $\frac {LB}{c^2}=\frac {LC}{b^2}=\frac a{b^2+c^2}$ and $\left\{\begin{array}{cccc} \frac {NB}{NA} & = & \frac ab\\\\ \frac {MC}{MA} & = & 1\\\\ \frac {KL}{KA} & = & \frac {a^2}{b^2+c^2}\end{array}\right\|\ .$ Apply the Cristea's theorem $:\ \frac {NB}{NA}\cdot LC+\frac {MC}{MA}\cdot LB=\frac {KL}{KA}\cdot BC\iff$ $\frac ab\cdot \frac {ab^2}{b^2+c^2}+1\cdot\frac {ac^2}{b^2+c^2}=\frac {a^2}{b^2+c^2}\cdot a\iff$ $\boxed{ab+c^2=a^2}\ (*)\ \iff$ $ab=a^2-c^2\iff$ $\sin A\sin B=\sin^2A-\sin^2C=$ $\sin (A-C)\sin (A+C)=$ $\sin (A-C)\sin B$

$\iff$ $\sin A=\sin (A-C)\iff$ $A+(A-C)=180^{\circ}\iff$ $\boxed{2A=180^{\circ}+C}\ (1)\ .$ Otherwise. Let $D\in (BC)$ so that $CA=CD\ .$ Observe that $BD=a-b$ and

$ab+c^2=a^2\iff$ $c^2=a(a-b)\iff$ $BA^2=BD\cdot BC\iff$ $\triangle BAD\sim \triangle BCA\iff$ $\widehat{BAD}\equiv\widehat {BCA}\iff$ $\boxed{A=B+2C}\iff$ $2A=180^{\circ}+C\ ,$ i.e. the relation $(1)\ .$

Particular case. $\frac A4=\frac B2=\frac C1=\frac {\pi}7\implies$ $A=B+2C\implies$ $K\in MN\ .$

P11. Sa se arate ca in orice $\triangle ABC$ exista inegalitatea $:\ \frac 1{b+c}+\frac 1{c+a}+\frac 1{a+b}\ \le\ \frac{3(R+r)}{4S}\ .$

Proof. $\frac {9}{4p}\stackrel{(CBS)}{\le}\sum\frac {1}{b+c}\le\sum\frac {b+c}{4bc}=$ $\sum\frac {a(b+c)}{4abc}=\frac {\sum bc}{2abc}=$ $\frac {p^2+r^2+4Rr}{8RS}\ \stackrel{(\mathrm{Gerretsen})}{\le}\ \frac {\left(4R^2+4Rr+3r^2\right)+r^2+4Rr}{8RS}=$ $\frac {(R+r)\cdot (R+r)}{2RS}\le$ $\frac {\frac {3R}{2}\cdot (R+r)}{2RS}=$

$\frac {3(R+r)}{4S}$ $\Longleftrightarrow$ $\boxed{\frac {9}{4p}\ \le\ \sum\frac {1}{b+c}\ \le\ \frac {3(R+r)}{4S}}\ .$ Retineti $\boxed{4r(5R-r)\ \le\ ab+bc+ca\ \le\ 4(R+r)^2}\ \stackrel{\mathrm{Gerretsen}}{\iff}\ \boxed{16Rr-5r^2\ \le\ p^2\ \le\ 4R^2+4Rr+3r^2}\ .$

Aplicatie. Sa se arate ca $\left(\frac {p}{R+r}\right)^2\ \le\$ $\sum\frac {a^2}{bc}\ \le\ \frac {2R-r}{r}$ si $\sum \frac {1}{a(b+c)}\ \ge$ $\frac {9}{8(R+r)^2}\ .$

Demonstratie. $\left(\frac{p}{R+r}\right)^2=$ $\frac{4p^2}{(4R^2+4Rr+3r^2)+r^2+4Rr}\ \stackrel{\mathrm{Gerretsen}}{\le}\ \frac{4p^2}{p^2+r^2+4Rr}=$ $\frac{(a+b+c)^2}{ab+bc+ca}\ \stackrel{\mbox{(C.B.S)}}{\le}\ \sum\frac{a^2}{bc}\ .$

$\sum\frac{a^2}{bc}=\frac{a^3+b^3+c^3}{abc}=\frac{2p(p^2-6Rr-3r^2)}{4RS}=\frac{p^2-6Rr-3r^2}{2Rr}\ \stackrel{\mathrm{Gerretsen}}{\le}\ \frac{4R^2+4Rr+3r^2-6Rr-3r^2}{2Rr}=\frac{2R-r}{r}\ .$

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$\sum\frac{1}{a(b+c)}\ \stackrel{\mathrm{C.B.S.}}{\ge}\ \frac{9}{2(ab+bc+ca)}=\frac{9}{2\underline{\underline{p^2}}+2r^2+8Rr}\ \stackrel{\mathrm{Gerretsen}}{\ge}\ \frac{9}{2(\underline{\underline{4R^2+4Rr+3r^2}})+2r^2+8Rr}=\frac{9}{8(R+r)^2}\ .$

P12. Sa se arate ca daca $\triangle ABC$ este ascutitunghic atunci exista inegalitatea $p^2\ge 2R^2+8Rr+3r^2$ (Walker).

Demonstratie 1. Inegalitatea este echivalenta cu $\boxed{a^2+b^2+c^2 \geq 4(R+r)^2}$ a carei demonstratie se gaseste aici.

Demonstratie 2. Identitati cunoscute $:\ \left\{\begin{array}{cc} a^2+b^2+c^2=2\left(p^2-r^2-4Rr\right) & (1)\\\\ 4S=\left(b^2+c^2-a^2\right)\tan A & (2)\\\\ \sin 2A+\sin 2B+\sin 2C=\frac {2S}{R^2} & (3)\\\\ \cos A+\cos B+\cos C=1+\frac rR & (4)\end{array}\right\|$ $\implies$ $a^2+b^2+c^2=$ $\sum\left(b^2+c^2-a^2\right)\ \stackrel {(2)}{=}$ $4S\sum \frac {\cos A}{\sin A}=$ $8S\sum \frac {\cos^2A}{\sin 2A}\ \stackrel{C.B.S.}{\ge}$

$8S\frac {\left(\sum \cos A\right)^2}{\sum\sin 2A}\ \stackrel{(3)\wedge (4)}{=}\ 8S\frac {\left(1+\frac rR\right)^2}{\frac {2S}{R^2}}=$ $4(R+r)^2\Longrightarrow$ $\boxed{a^2+b^2+c^2\ge 4(R+r)^2}\ \stackrel{(1)}{\Longrightarrow}\ 2\left(p^2-r^2-4Rr\right)\ge 4(R+r)^2\ \Longrightarrow\ \boxed{p^2\ge 2R^2+8Rr+3r^2}\ .$

P13. Fie doua numere pozitive $a$ si $b\ .$ Se arata usor ca $(a+b)^4\le \left(a^2+3b^2\right)\left(b^2+3a^2\right)\ .$ Intr-adevar, $(a^2+b^2+b^2+b^2)(a^2+a^2+a^2+b^2)\ge$

$(aa+ba+ba+bb)^2=(a+b)^4\ .$ Sa se arate o "intarire" a acesteia $:\ (a+b)^4\le\boxed{(a+b)^3\sqrt {2\left(a^2+b^2\right)}\le \left(a^2+3b^2\right)\left(b^2+3a^2\right)}\ (*)\ .$

Demonstratie. Notam $\frac{a}{b}=x>0$ si $(*)\ \iff\ (x^2+3)^2(1+3x^2)^2\ge 2(x+1)^6(x^2+1)$ $\iff$ $(x-1)^2(7x^6+2x^5+25x^4-4x^3+25x^2+2x+7)\ge 0\ .$

P14. Sa se arate ca daca $ABC$ este un triunghi ascutitunghic, atunci $\sum\ \frac {b+c}{a}\ \ge\ \frac {(R+r)(R+2r)}{Rr}\ \ge\ 3\left(1+\sqrt[3]{\frac {R}{2r}}\right)\ .$

Demonstratie. Obtinem succesiv $:\ \sum \frac{b+c}{a}= \frac{ \sum a \sum bc}{abc}-3=\frac{p^2+r^2-2Rr}{2Rr}\ .$ Folosind inegalitatea Walker $p^2\ge 2R^2+8Rr+3r^2$ care se gaseste aici

vom obtine $\sum \frac{b+c}{a}\geq \frac{R^2+2r^2+3rR}{Rr}=\frac {(R+r)(R+2r)}{Rr}\ .$ Pentru partea a doua, notand cu $x^3=\frac{R}{2r}$ inegalitatea $\frac {(R+r)(R+2r)}{Rr}\ \ge\ 3\left(1+\sqrt[3]{\frac {R}{2r}}\right)$

se scrie echivalent $(1+2x^3)(1+\frac{1}{x^3})\ \ge\ 3(1+x)$ echivalent cu $x^3+x^3+\frac{1}{x^3} \geq 3x$ care reiese usor din inegalitatea mediilor.

P15. Sa se arate ca in orice $\triangle ABC$ are loc relatia $\boxed{p-\sqrt 3 \cdot(R+r)=8R\cdot\sin\left(\frac{\pi}{6}-\frac A2\right)\cdot\sin\left(\frac{\pi}{6}-\frac B2\right)\cdot\sin\left(\frac{\pi}{6}-\frac C2\right)}\ .$

Demonstratie. $x+y+z=0\Longrightarrow \prod\sin x=-\frac 14\cdot \sum \sin 2x$ (se arata usor de la dreapta spre stanga !). Deoarece $\sum\left(\frac{\pi}{6}-\frac A2\right)=0$

putem aplica lema de mai sus. Asadar $8R\cdot\prod \sin\left(\frac{\pi}{6}-\frac A2\right)=$ $-2R\cdot \sum\sin\left(\frac{\pi}{3}-A\right)=$ $-2R\cdot\sum\left(\frac {\sqrt 3}{2}\cdot\cos A-\frac 12\cdot\sin A\right)=$

$-2R\cdot\left(\frac {\sqrt 3}{2}\cdot\sum\cos A-\frac 12\cdot\sum\sin A\right)=$ $-2R\cdot\left[\frac {\sqrt 3}{2}\cdot\left(1+\frac rR\right)-\frac 12\cdot\frac pR\right]=p-\sqrt 3 \cdot(R+r)\ .$

P16. Fie un triunghi $ABC$ cu centrul de greutate $G$ si centrul cercului inscris $I\ .$ Sa se arate ca exista inegalitatea $\boxed{\frac {AG}{AI}+\frac {BG}{BI}+\frac {CG}{CI}\ \ge\ \frac 13\cdot (a+b+c)\cdot\left(\frac 1a+\frac 1b+\frac 1c\right)}\ .$

Demonstratie. Se observa ca $:\ \mathrm{LHS}=\sum\frac{\frac 23\cdot m_a}{\frac{b+c}{a+b+c}\cdot l_a}=$ $\frac{2(a+b+c)}{3}\ \cdot\ \sum\frac{m_a}{l_a(b+c)}\ .$ Prin urmare va fi suficient sa aratam ca $\sum\frac{m_a}{l_a(b+c)}\ \ge\ \frac 12\left(\frac 1a+\frac 1b+\frac 1c\right)\ .$ Dar in orice

triunghi este adevarata inegalitatea $:\ \boxed{m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2}\ .$ Intr-adevar, $m_a=\frac{\sqrt{2(b^2+c^2)-a^2}}{2}=$ $\frac{\sqrt{(b+c)^2-2bc+2bc\cos A}}{2}=$ $\frac{\sqrt{(b+c)^2-2bc(1-\cos A)}}{2}\ge$

$\frac{\sqrt{(b+c)^2-\frac{(b+c)^2}{2}(1-\cos A)}}{2}$ $\iff$ $m_a\ge\frac{b+c}{2}\sqrt{1-\frac{1-\cos A}{2}}=$ $\frac{b+c}{2}\sqrt{\frac{1+\cos A}{2}}=$ $\frac{b+c}{2}\ \cdot\ \cos\frac A2\implies$ $\frac{m_a}{l_a(b+c)}\ge$ $\frac{\frac{b+c}{2}\cos\frac A2}{\frac{2bc}{b+c}\cos\frac A2\ \cdot\ (b+c)}=$ $\frac{b+c}{4bc}\ .$ Deci

$\sum\frac{m_a}{l_a(b+c)}\ge$ $\sum\frac{b+c}{4bc}=$ $\frac 12\left(\frac 1a+\frac 1b+\frac 1c\right)\ .$ Remarca. In rezolvarea problemei esentiala a fost inegalitatea $m_a\ \ge\ \frac{b+c}{2}\cdot\cos\frac A2\ ,$ "mai tare" decat $m_a\ge\sqrt{p(p-a)}\ .$

P17 (M.O. Sanchez). Let a right trapezoid $ABCD$ where $AD\parallel BC$ and $AB\perp AD\ .$ Suppose that exists $E\in (AB)$ so that $\triangle CDE$ is

equilateral. Denote $BC=n$ and $AD=m\ ,$ where $n<m\ .$ Prove that the area $\boxed{[CDE]=\frac {2\left(m^3+n^3\right)}{(m+n)^3}\cdot S}\ (*)\ ,$ where $S$ is the area of $ABCD\ .$

Proof. Denote $CD=l\ ,$ the midpoint $M$ of $[CD]$ and $EA=u\ ,$ $EB=v\ .$ Observe that $AEMD$ and $BEMC$ are cyclic quadrilaterals

and $ABM$ is an equilateral triangle, i.e. $MA=MB=u+v\ .$ Apply the Ptolemy's theorem to the following quadrilaterals $:$

$\left\{\begin{array}{cccccccc} AEMD\ : & m\cdot EM+u\cdot MD=ED\cdot AM & \implies & m\cdot \frac {EM}{ED}+u\cdot\frac {MD}{ED}=AM & \implies & \frac {m\sqrt 3}2+\frac u2=u+v & \implies & u+2v=m\sqrt 3\\\\ BEMC\ : & n\cdot EM+v\cdot MC=EC\cdot BM & \implies & n\cdot \frac {EM}{EC}+v\cdot\frac {MC}{EC}=BM & \implies & \frac {n\sqrt 3}2+\frac v2=u+v & \implies & 2u+v=n\sqrt 3\end{array}\right\|\bigoplus\implies$

$\boxed{m+n=(u+v)\sqrt 3}\ (1)\ .$ Let $P\in AD$ so that $CP\perp AD\ .$ Thus, $PC^2+PD^2=CD^2\iff$ $(u+v)^2+(m-n)^2=CD^2\ \stackrel{(1)}{\iff}\ \frac {(m+n)^2}3+(m-n)^2=l^2\iff$

$\boxed{4\left(m^2-mn+n^2\right)=3l^2}\ (2)\ .$ Therefore, $[CDE]=\frac {l^2\sqrt 3}4\ \stackrel{(2)}{=}\ \frac {m^2-mn+n^2}{\sqrt 3}=\frac{m^3+n^3}{(m+n)\sqrt 3}\implies$ $\boxed{[CDE]=\frac {m^3+n^3}{(m+n)\sqrt 3}}\ (3)\ .$ Observe analogously that

$[ABCD]=\frac {m+n}2\cdot (u+v)\ \stackrel{(1)}{=}\ \frac {m+n}2\cdot \frac {m+n}{\sqrt 3}=\frac {(m+n)^2}{2\sqrt 3}\implies$ $\boxed{[ABCD]=\frac {(m+n)^2}{2\sqrt 3}}\ (4)\ .$ In conclusion, $\frac {[CDE]}{[ABCD]}=\frac {m^3+n^3}{(m+n)\sqrt 3}\cdot\frac {2\sqrt 3}{(m+n)^2}\implies\ (*)\ .$

Extension. Let a right trapezoid $ABCD$ where $AD\parallel BC$ and $AB\perp AD\ .$ Suppose that exists $E\in (AB)$ so that $\triangle CDE$ is $E$ -isosceles

with $m\left(\widehat{ECD}\right)=\phi\ .$ Denote $BC=n$ and $AD=m\ ,$ where $n<m\ .$ Prove that the area $[CDE]=\frac {\sqrt 3\left(m^2+2mn\cos 2\phi +n^2\right)}{(m+n)^2\sin2\phi}\cdot S\ (*)$

Proof. Denote $CD=l\ ,$ the midpoint $M$ of $[CD]$ and $EA=u\ ,$ $EB=v\ .$ Observe that $AEMD$ and $BEMC$ are cyclic

and $ABM$ is a $M$ -isosceles triangle, i.e. $MA=MB=\frac {u+v}{2\cos\phi}\ .$ Apply the Ptolemy's theorem to the following quadrilaterals $:$

$\left\{\begin{array}{cccccccc} AEMD\ : & m\cdot EM+u\cdot MD=ED\cdot AM & \implies & m\cdot \frac {EM}{ED}+u\cdot\frac {MD}{ED}=AM & \implies & m\sin\phi +u\cos\phi =\frac {u+v}{2\cos\phi} & \implies & v-u\cos 2\phi =m\sin 2\phi\\\\ BEMC\ : & n\cdot EM+v\cdot MC=EC\cdot BM & \implies & n\cdot \frac {EM}{EC}+v\cdot\frac {MC}{EC}=BM & \implies & n\sin\phi +v\cos\phi =\frac {u+v}{2\cos\phi} & \implies & u-v\cos 2\phi =n\sin 2\phi\end{array}\right\|\bigoplus\implies$

$\boxed{m+n=(u+v)\tan\phi }\ (1)\ .$ Let $P\in AD$ so that $CP\perp AD\ .$ Thus, $PC^2+PD^2=CD^2\iff$ $(u+v)^2+(m-n)^2=l^2\ \stackrel{(1)}{\iff}\ \frac {(m+n)^2}{\tan^2\phi}+(m-n)^2=$ $l^2\iff$ $(m+n)^2\cos^2\phi +(m-n)^2\sin^2\phi =$ $l^2\sin^2\phi\iff$ $\boxed{m^2+2mn\cos 2\phi +n^2=l^2\sin^2\phi}\ (2)\ .$ Therefore, $\boxed{[CDE]=\frac {\sqrt 3\left(m^2+2mn\cos 2\phi +n^2\right)}{4\sin^2\phi}}\ (3)\ .$ Observe

analogously that $[ABCD]=\frac {m+n}2\cdot (u+v)\ \stackrel{(1)}{=}\ \frac {m+n}2\cdot \frac {m+n}{\tan\phi}=\frac {(m+n)^2}{2\tan\phi}\implies$ $\boxed{[ABCD]=\frac {(m+n)^2}{2\tan\phi}}\ (4)\ .$ In conclusion,

$\frac {[CDE]}{[ABCD]}=\frac {\sqrt 3\left(m^2+2mn\cos 2\phi +n^2\right)}{4\sin^2\phi}\cdot\frac {2\tan\phi}{(m+n)^2}=\frac {\sqrt 3\left(m^2+2mn\cos 2\phi +n^2\right)}{(m+n)^2\sin2\phi}\implies\ (*)\ .$

$\mathrm{END}$

This post has been edited 339 times. Last edited by Virgil Nicula, Aug 31, 2016, 12:18 PM

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462. Diverse probleme de geometrie.

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456* Integrals.

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

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

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449. Working page I.

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444. Remarkable points in a triangle.

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396. Own inegalities stronger than the Panaitopol's.

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392. Art of Problem Solving (geometry).

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391. CRUX Mathematicorum.

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387. Algebra (II).

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357. Morley theorem.

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

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352. Some definite integrals.

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

314. USAMTS 2009 Round 2 #5.

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312. Very nice (famous trigonometrical problems) !

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

308. Some nice and easy properties in a triangle.

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307. Very nice proofs !

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300. Lagrange's multiplier. Examples and problems.

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268. Saraghin contest, 2011.

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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

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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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