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

by Virgil Nicula, Jul 9, 2017, 2:12 PM

$\boxed{\mathrm{LEMMA\ :}\ (\forall )\ x>0\ ,\ x^3+1\ge x\sqrt{2\left(x^2+1\right)}\ .\ \mathrm{APPLICATION\ :}\ (\forall )\ \triangle\ ABC\ ,\ a^3+b^3+c^3\ \ge\ \left(2+\sqrt 2\right)abc}$

P1. Prove that $\sum_{\mathrm{cyc}} a^2(s-a)=4rs(R+r)$ and $\sum_{\mathrm{cyc}}\frac {a^2}{(s-b)(s-c)}\ge 12\ ,$ where $2s=a+b+c\ ($ standard notations for $\triangle ABC)\ .$

Proof.

$\blacktriangleright$ Denote $\left\{\begin{array}{ccc} x & := & s-a\\\ y & := & s-b\\\ z & := & s-c\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} a & = & y+z\\\ b & = & x+z\\\ c & = & x+y\end{array}\right\|\ .$ Hence $\left\{\begin{array}{ccc} x+y+z & = & s\\\\ a+b+c & = & 2s\end{array}\right\|\ .$ Therefore, $\sum_{\mathrm{cyc}} a^2(s-a)=$ $\sum x(y+z)^2=$ $\sum \left[2xyz+x\left(y^2+z^2\right)\right]=$

$2\cdot \sum (xyz)+\sum x\left(y^2+z^2\right)=$ $6xyz+\sum yz(y+z)=$ $6xyz+\sum yz[(x+y+z)-x]=6xyz+\sum yz\cdot \sum x-\sum (xyz)=$ $3xyz+\sum (s-b)(s-c)\cdot\sum (s-a)=$

$3(s-a)(s-b)(s-c)+s\cdot \sum [bc-s(s-a)]=$ $3sr^2+s\left(\cancel{s^2}+r^2+4Rr\right)-\cancel{s^3}=4sr^2+4Rsr=4sr(R+r)\implies$ $\boxed{\sum_{\mathrm{cyc}} a^2(s-a)=4sr(R+r)}\ .$ We have equality iff $P=1\ ,$ i.e. $yz=1\ .$

$\blacktriangleright\ \sum_{\mathrm{cyc}}\frac {a^2}{(s-b)(s-c)}=\sum_{\mathrm{cyc}}\frac {(y+z)^2}{yz}=$ $\sum_{\mathrm{cyc}}\left(\frac yz+\frac zy+2\right)=6+\sum\left(\frac yz+\frac zy\right)\ge 12\implies$ $\boxed{\sum_{\mathrm{cyc}}\frac {a^2}{(s-b)(s-c)}\ge 12}\ .$ I used the identities $\left\|\begin{array}{c} s(s-a)+(s-b)(s-c)=bc\\\\ ab+bc+ca=s^2+r^2+4Rr\\\\ (s-a)(s-b)(s-c)=sr^2\end{array}\right\|\ .$

Remark. $\sum\frac {a^2}{(s-b)(s-c)}=\frac 1{(s-a)(s-b)(s-c)}\cdot\sum a^2(s-a)=$ $\frac {4sr(R+r)}{sr^2}=\frac {4(R+r)}r\ge \frac {4(2r+r)}r=12\implies$ $\sum\frac {a^2}{(s-b)(s-c)}\ge 12\ .$

P2. For an $A$ -rightangled $\triangle ABC$ denote $:\ M\ -$ the midpoint of $[AB]\ ;\ D\ -$ the projection of $A$ on $BC\ .$ Prove that the equivalence $\boxed{m\left(\widehat{MCA}\right)=2\cdot m\left(\widehat{MCB}\right)\iff\frac {DB}{DC}=\frac {1+\sqrt {17}}2}\ .$

Proof. Let $m\left(\widehat{MCB}\right)=\phi$ and $\boxed{t:=\frac {c^2}{b^2}}\ .$ $\implies$ $\frac {DB}{DC}=\frac {a\cdot DB}{a\cdot DC}=\frac {c^2}{b^2}=t\implies$ $\boxed{\frac{DB}{DC}=t}\ (*)\ .$ Thus, $1=\frac {MA}{MB}=\frac {CA}{CB}\cdot\frac {\sin\widehat{MCA}}{\sin\widehat{MCB}}=$ $\frac ba\cdot \frac {\sin 2\phi}{\sin\phi}\implies$ $1=\frac {2b\cos\phi}a \implies$ $\cos\phi =\frac a{2b}\ .$

Hence $\cos^2\phi =\frac {a^2}{4b^2}=\frac {b^2+c^2}{4b^2}=\frac {t+1}4\implies$ $2\cos^2\phi -1=\frac {t+1}2-1=\frac {t-1}2\implies$ $\boxed{2\cos^2\phi -1=\frac {t-1}2}\ (1)\ .$ Thus, $\cos 2\phi =\frac {CA}{CM}=\frac {CA}{\sqrt {AC^2+AM^2}}=\frac b{\sqrt{b^2+\frac {c^2}4}}=$ $\frac {2b}{\sqrt{4b^2+c^2}}=\frac 2{\sqrt{t+4}}\implies$ $\boxed{\cos 2\phi =\frac 2{\sqrt{t+4}}}\ (2)\ .$ Thus, $\cos2\phi=2\cos^2\phi -1\ \stackrel{(1\wedge 2)}{\iff}\ \frac 2{\sqrt {t+4}}=\frac {t-1}2\iff$ $(t-1)\sqrt{t+4}=4\iff$ $t>1$ and $t^3+2t^2-7t-12=0\iff$

$(t+3)(t^2-t-4)=0\iff t^2-t-4=0\iff t=\frac {1+\sqrt {17}}2\ \stackrel{(*)}{\iff}\ \boxed{\frac {DB}{DC}=\frac {1+\sqrt {17}}2}\ .$

P3 (Kadir Altintas). Let $\triangle ABC$ with $b\ne c,$ the incenter $I,$ the orthocenter $H,$ the Nagel's point $N$ and the midpoint $M$ of $[IH].$ Prove that $NM\perp BC\iff b+c=5a.$

Proof. Denote $\left\{\begin{array}{cccc} D\in AN\cap BC\ : & DB=s-c\ ; & DC=s-b\\\\ E\in BN\cap BC\ : & EC=s-a\ ; & EA=s-c\\\\ F\in CN\cap BC\ : & FA=s-b\ ; & FB=s-a\end{array}\right\|\ .$ From the Aubel's relation obtain that $\frac {NA}{ND}=\frac {FA}{FB}+\frac {EA}{EC}=\frac {s-c}{s-a}+\frac {s-b}{s-a}=\frac a{s-a}$ $\implies$ $\boxed{\frac {NA}{ND}=\frac a{s-a}}\ (*)$ $\implies$

$\frac {NA}a=\frac {ND}{s-a}=\frac {AD}s\ .$ Apply the Stewart's relation to the ray $[AD$ in $\triangle ABC\ :\ \boxed{a\cdot AD^2+a(s-b)(s-c)=c^2(s-b)+b^2(s-c)}\ .$ Apply the theorem of median to the following triangles $:$

$\left\{\begin{array}{cc} BM/\triangle BIH\ : & 4\cdot MB^2=2\cdot\left(BI^2+BH^2\right)-IH^2\\\\ CM/\triangle CIH\ : & 4\cdot MC^2=2\cdot\left(CI^2+CH^2\right)-IH^2\end{array}\right\|\ .$ Product of the substruction between the previous relations becomes $4\cdot\left(MB^2-MC^2\right)=$ $2\left(BI^2-CI^2\right)+2\cdot\left(BH^2-CH^2\right)=$

$2\cdot\left[\frac {ac(s-b)}s-\frac {ab(s-c)}s\right]+2\cdot 4R^2\left(\cos^2B-\cos^2C\right)=2\cdot [(ac-4Rr)-(ab-4Rr)]+2\cdot 4R^2\left(\sin^2C-\sin^2B\right)=$ $2a(c-b)+2\left(c^2-b^2\right)=$ $2(c-b)(a+b+c)=$

$4s(c-b)\implies$ $4\cdot\left(MB^2-MC^2\right)=$ $4s(c-b)\iff$ $\boxed{MB^2-MC^2=s(c-b)}\ (1)\ .$ Apply the Stewart's relation to the ray $[NB/\triangle ABD\ :\ NB^2\cdot AD+AD\cdot NA\cdot ND=$

$c^2\cdot ND+(s-c)^2\cdot NA\iff$ $NB^2+\frac {a(s-a)}{s^2}\cdot AD^2=$ $\frac {c^2(s-a)}s+\frac {a(s-c)^2}s\iff$ $\boxed{s^2\cdot NB^2+a(s-a)\cdot AD^2=s\cdot\left[c^2(s-a)+a(s-c)^2\right]}\ .$ Get analogously the similar relation

$\boxed{s^2\cdot NC^2+a(s-a)\cdot AD^2=s\cdot \left[b^2(s-a)+a(s-b)^2\right]}\ .$ From the substruction between the previous relations get $s\cancel{^2}\cdot\left(NB^2-NC^2\right)=\cancel s(s-a)(c-b)(c+b)-\cancel sa^2(c-b)\iff$

$s\cdot\left(NB^2-NC^2\right)=(s-a)(c-b)(c+b)-a^2(c-b)=$ $(c-b)\left[(s-a)(b+c)-a^2\right]=(c-b)[s(b+c)-a(a+b+c)]=s[(c-b)[(b+c)-2a]\iff$

$\boxed{NB^2-NC^2=(c-b)(b+c-2a)}\ (2)\ .$ In conclusion, $NM\perp BC\iff$ $NB^2-NC^2=MB^2-MC^2\ \stackrel{1\wedge 2}{\iff}\ b+c-2a=s\iff 2(b+c)-4a=a+(b+c)\iff$ $b+c=5a\ .$

Remark 1. If $E$ is the Euler's nine-point center, then prove easily that $\boxed{EB^2-EC^2=\frac {c^2-b^2}2}\ .$ From PP3 we have $\boxed{NB^2-NC^2=(c-b)(b+c-2a)}\ .$ Therefore, $NE\perp BC\iff$

$NB^2-NC^2=EB^2-EC^2\iff$ $(\cancel{c-b})[(b+c)-2a]=\frac {(\cancel{c-b})(c+b)}2\iff$ $2(b+c)-4a=c+b\iff$ $b+c=4a.$ In conclusion, $\boxed{NE\perp BC\iff b+c=4a}\ .$

Remark 2. $I -$ the incenter of $\triangle ABC\implies$ $IB^2-IC^2=\frac {ac(s-b)}s-\frac {ab(s-c)}s=$ $ac-ab=a(c-b),$ i.e. $\boxed{IB^2-IC^2=a(c-b)}\ .$ From the previous remark obtain that

$\boxed{EB^2-EC^2=\frac {c^2-b^2}2}\ .$ Hence $IE\perp BC\iff$ $IB^2-IC^2=EB^2-EC^2\iff$ $a(c-b)=\frac {(c-b)(c+b)}2\iff$ $a=\frac{c+b}2\iff$ $b+c=2a,$ i.e. $\boxed{IE\perp BC\iff b+c=2a}$

P4 (Kadir Altintas). Let $\triangle ABC$ with $b\ne c,$ the incenter $I,$ the centroid $G,$ the Euler's nine-point center $E$ and the Lemoine's point $K\ .$ Prove that $\left\{\begin{array}{ccc} KG\perp BC & \iff & b^2+c^2=5a^2\\\\ KE\perp BC & \iff & b^2+c^2=3a^2\\\\ KI\perp BC & \iff & a=b+c-\sqrt {2bc}\end{array}\right\|\ .$

Proof. I"ll use the relations $\boxed{s_a=\frac {2bc}{b^2+c^2}\cdot m_a}\ (*)$ and $\boxed{4m_a^2=2\left(b^2+c^2\right)-a^2}\ (1)\ ,$ where $m_a$ is the length of the $A$ -median and $s_a$ is the length of the $A$ -symmedian. Let $Y\in BK\cap AC$ and

$Z\in CK\cap AB\ .$ Prove easily that $\left\{\begin{array}{ccc} \frac {KB}{a^2+c^2}=\frac {BY}{a^2+b^2+c^2}=\frac {2acm_b}{\left(a^2+c^2\right)\left(a^2+b^2+c^2\right)} & \implies & KB=\frac {2acm_b}{a^2+b^2+c^2}\\\\ \frac {KC}{a^2+b^2}=\frac {CZ}{a^2+b^2+c^2}=\frac {2acm_c}{\left(a^2+b^2\right)\left(a^2+b^2+c^2\right)} & \implies & KB=\frac {2acm_b}{a^2+b^2+c^2}\end{array}\right\|\ .$ Therefore, $KB^2-KC^2=\left(\frac {2acm_b}{a^2+b^2+c^2}\right)^2-\left(\frac {2abm_c}{a^2+b^2+c^2}\right)^2=$

$\frac {4a^2\left(c^2m_b^2-b^2m_c^2\right)}{\left(a^2+b^2+c^2\right)^2}=$ $\frac {a^2}{\left(a^2+b^2+c^2\right)^2}\cdot\left[c^2\left(2a^2+2c^2-\cancel{b^2}\right)-b^2\left(2a^2+2b^2-\cancel{c^2}\right)\right]=$ $\frac {a^2\left[2a^2\left(c^2-b^2\right)+2\left(c^4-b^4\right)\right]}{\left(a^2+b^2+c^2\right)^2}=$ $\frac {2a^2\left(c^2-b^2\right)\left(a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)^2}=$ $\frac {2a^2\left(c^2-b^2\right)}{a^2+b^2+c^2}\implies$

$\boxed{KB^2-KC^2=\frac {2a^2\left(c^2-b^2\right)}{a^2+b^2+c^2}}\ (2)\ .$ Prove easily that $\boxed{GB^2-GC^2=\frac {c^2-b^2}3}\ (3)\ .$ Hence $KG\perp BC\iff$ $KB^2-KC^2=GB^2-GC^2\stackrel{2\wedge 3}{\iff}\ \frac {2a^2\left(c^2-b^2\right)}{a^2+b^2+c^2}=\frac {c^2-b^2}3\iff$

$a^2+b^2+c^2=6a^2\iff b^2+c^2=5a^2\ .$ In conclusion, $\boxed{KG\perp BC \iff b^2+c^2=5a^2}\ .$ In the previous remark 2 proved that $EB^2-EC^2=\frac {c^2-b^2}2\ .$ Hence $KE\perp BC\iff$

$KB^2-KC^2=EB^2-EC^2\iff$ $\frac {2a^2\left(c^2-b^2\right)}{a^2+b^2+c^2}=\frac {c^2-b^2}2\iff$ $a^2+b^2+c^2=4a^2\iff$ $b^2+c^2=3a^2\ .$ In conclusion, $\boxed{KE\perp BC\iff b^2+c^2=3a^2}\ .$ From the previous

remark 2 obtain $IB^2-IC^2=a(c-b)\ .$ Hence $KI\perp BC\iff$ $KB^2-KC^2=IB^2-IC^2\iff$ $\frac {2a^2\left(c^2-b^2\right)}{a^2+b^2+c^2}=a(c-b)\iff$ $2a(b+c)=a^2+b^2+c^2\iff$

$a^2-2a(b+c)+(b+c)^2-2bc=0\iff$ $(b+c-a)^2=2bc\iff$ $b+c-a=\sqrt{2bc}\iff$ $a=b+c-\sqrt {2bc}\ .$ In conclusion, $\boxed{KI\perp BC\iff a=b+c-\sqrt {2bc}}\ .$

An easy extension. Let $M(x,y,z)$ be the point with barrycentrical coordinates w.r.t. $\triangle ABC,$ where $x+y+z=1.$ Prove that $\boxed{MB^2-MC^2=(z-y)\cdot a^2+x\left(c^2-b^2\right)}$ (standard notations).

Proof. If $M_k\left(x_k,y_k,z_k\right)$ where $k\in\overline{1,2},$ then $\boxed{M_1M_2^2=-\left[\left(y_1-y_2\right)\left(z_1-z_2\right)\cdot a^2+\left(z_1-z_2\right)\left(x_1-x_2\right)\cdot b^2+\left(x_1-x_2\right)\left(y_1-y_2\right)\cdot c^2\right]}\ .$ For example, if $M(x,y,z)$ and $A(1,0,0)$ then:

$MA^2=-\left[yz\cdot a^2+z(x-1)\cdot b^2+(x-1)y\cdot c^2\right]\implies\odot$ $\begin{array}{ccc} \nearrow & MB^2=-\left[zx\cdot b^2+x(y-1)\cdot c^2+z(y-1)\cdot a^2\right] & \searrow\\\\ \searrow & MC^2=-\left[xy\cdot c^2+y(z-1)\cdot a^2+x(z-1)\cdot b^2\right] & \nearrow\end{array}\odot\implies$ $MB^2-MC^2=(z-y)\cdot a^2+x\left(c^2-b^2\right)\ .$

Exemple de coordonate baricentrice pentru cateva puncte remarcabile ale unui triunghi $ABC,$ unde $A(1,0,0)\ ;\ B(0,1,0)\ ;\ C(0,0,1)$ sunt varfurile triunghiului $ABC\ .$

$\blacktriangleright\ D\left(0,\frac 12,\frac 12\right)$ - mijlocul lui $[BC]\ ;\ A_1\left(-1,1,1\right)$ - simetricul lui $A$ fata de $D\ ;\ A_2\left(2,-\frac 12,-\frac 12\right)$ - simetricul lui $D$ fata de $A\ ;\ G\left(\frac 13,\frac 13,\frac 13\right)$ - centru de greutate al triunghiului $ABC\ .$

$\blacktriangleright\ I\left(\frac a{2s},\frac b{2s},\frac c{2s}\right)$ - centrul cercului inscris $w=\mathbb (I,r)\ ,\ a+b+c=2s\ ;\ O\left(\frac {R^2\sin 2A}{2S},\frac {R^2\sin 2A}{2S},\frac {R^2\sin 2A}{2S}\right)$ - centrul cercului circumscris $\Omega =\mathbb C\left(O,R\right)\ ,\ \sum \sin 2A=\frac {2S}{R^2}\ .$

$\blacktriangleright\ H\left(\cot B\cot C,\cot C\cot A,\cot A\cot B\right)$ - ortocentrul $,\ \sum\cot B\cot C=1\ ;\ I_a\left(\frac {-a}{2(s-a)},\frac b{2(s-a)},\frac c{2(s-a)}\right)$ - centrul cercului exinscris $w_a=\mathbb C\left(I_a,r_a\right)\ ,\ -a+b+c=2(s-a).$

$\blacktriangleright\ N\left(\frac {s-a}s,\frac {s-b}s,\frac {s-c}s\right)$ - punctul lui Nagel $,\ \sum (s-a)=s\ ;\ \Gamma\left(\frac {(s-b)(s-c)}{r(4R+r)},\frac {(s-c)(s-a)}{r(4R+r)},\frac {(s-a)(s-b)}{r(4R+r)}\right)$ - punctul lui Gergonne $,\ \sum (s-b)(s-c)=r(4R+r)\ .$

$\blacktriangleright\ L\left(\frac {a^2}{\sum a^2},\frac {b^2}{\sum a^2},\frac {a^2}{\sum a^2}\right)$ - punctul lui Lemoine (centrul simedian). Daca $s_a-$ lungimea $A$ -simedianei si $m_a-$ lungimea $A$ -medianei, atunci $\boxed{s_a=\frac {2bc}{b^2+c^2}\cdot m_a}$ si $\boxed{4m_a^2+a^2=2\left(b^2+c^2\right)}\ .$

Applications. $\left\{\begin{array}{ccccccc} GA^2=\frac {4m_a^2}9 & \implies & GB^2-GC^2=\frac 13\cdot \left(c^2-b^2\right) & ; & IA^2=bc-4Rr & \implies & IB^2-IC^2=a(c-b)\\\\ \begin{array}{cccc} I_aB^2 & = & r_a^2+(s-c)^2 & \searrow\\\\ I_aC^2 & = & r_a^2+(s-b)^2 & \nearrow\end{array} & \implies & I_aB^2-I_aC^2=a(b-c) & ; & I_aA^2=r_a^2+s^2 & \implies & \begin{array}{ccccc} \nearrow & I_aA^2-I_aB^2=c(a+b)\\\\ \searrow & I_aA^2-I_aC^2=b(a+c)\end{array}\\\\ HA^2=4R^2-a^2 & \implies & HB^2-HC^2=c^2-b^2 & ; & LA=\frac {2bcm_a}{a^2+b^2+c^2} & \implies & LB^2-LC^2=\frac {2a^2\left(c^2-b^2\right)}{a^2+b^2+c^2}\end{array}\right\|\ \mathrm{a.s.o.\ !}$

$\boxed{an_a^2+a(s-b)(s-c)=c^2(s-b)+b^2(s-c)\implies \odot\begin{array}{ccccc} \nearrow & s^2\cdot NB^2+a(s-a)n_a^2 & = & s\cdot \left[c^2(s-a)+a^2(s-c)^2\right] & \searrow\\\\ \searrow & s^2\cdot NC^2+a(s-a)n_a^2 & = & s\cdot \left[b^2(s-a)+a^2(s-b)^2\right] & \nearrow \end{array} \odot \implies NB^2 - NC^2 = (c-b)(b+c-2a)}$

P5:

Proof. $f(\theta )=\frac 1{2+\cos\theta}+\frac 1{2+\sin\theta}=$ $\frac {(\sin\theta +\cos\theta)+4}{\sin\theta\cos\theta +2(\sin\theta +\cos\theta)+4},$ where $\theta\in\mathbb R\ .$ I"ll use the substitution $\boxed{t=h(\theta )=\sin\theta +\cos\theta }\ ,$ where $|t|\le \sqrt 2$ and $\boxed{\sin\theta\cos\theta =\frac {t^2-1}2}\ (*)\ .$

Hence $f(\theta )=(g\circ h)(\theta),$ where $g(t)=\frac {2(t+4)}{t^2+4t+7}$ and $|t|\le\sqrt 2\ .$ Prove easily that $g'(t)\ s.s\ -(t^2+8t+9)\ s.s.\ -\left(t+4-\sqrt 7\right),$ i.e. $g$ is increasing $(\nearrow )$ on $\left[-\sqrt 2,\sqrt 7-4\right]$ and decreasing

$(\searrow )$ on $\left[\sqrt 7-4,\sqrt 2\right]\ .$ Thus, $g(-4+\sqrt 7)=\boxed{\frac {2+\sqrt 7}3\ge g(t)\ge \frac {2(4-\sqrt 2)}7}=g\left(\sqrt 2\right)\ .$ For $\{a,b\}\subset\mathbb R$ defined the equivalence $:\ a\ s.s\ b\iff a=b=0$ or $ab>0\iff\mathrm{sign} a=\mathrm{sign} b.$

$\boxed{\begin{array}{ccccccc} t & \| & -\sqrt 2 & & \sqrt 7-4 & & \sqrt 2\\\ == & == & === & == & === & == & ===\\\ g^{\prime}(t) & \| & + & + & 0 & - & -\\\ == & == & === & == & === & == & ===\\\ g(t) & \| & \frac {2\left(4+\sqrt 2\right)}7 & \nearrow & \frac {2+\sqrt 7}3 & \searrow & \frac {2\left(4-\sqrt 2\right)}7\end{array}}\iff\left\|\begin{array}{c} \min\limits_{t\in\left[-\sqrt 2,\sqrt 2\right]}g(t)=\min\limits_{x\in\mathbb R}f\left(\theta\right)=\frac {2\left(4-\sqrt 2\right)}7\\\\ \max\limits_{t\in \left[-\sqrt 2,\sqrt 2\right]}g(t)=\max\limits_{x\in\mathbb R}f\left(\theta\right)=\frac {2+\sqrt 7}3\end{array}\right\|\iff$ $f\left(I\right)=\left[\frac {2\left(4-\sqrt 2\right)}7,\frac {2+\sqrt 7}3\right],$ where $I=\left[-\sqrt 2,\sqrt 2\right]$

P6 (Carlos Hugo Olivera Díaz). Let an acute $\Delta\ ABC$ with the excircle $w_a$ for what denote $\left\{\begin{array}{ccc} M\in AB\cap w_a\ & ; & X\in BC\ ,\ MX\perp BC\ ,\ MX=m\\\\ N\in AC\cap w_a\ & ; & Y\in BC\ ,\ NY\perp BC\ ,\ NY=n\\\\ D\in BC & ; & AD\perp BC \ ,\ AD=h_a\end{array}\right\|\ .$ Prove that $h_a=\frac {2mn}{1-\cos A}\ .$

Proof. $\left\{\begin{array}{ccccc} MX\parallel AD & \iff & \frac {AD}{MX}=\frac {AB}{BM} & \iff & \frac {h_a}m=\frac c{s-c}\\\\ NY\parallel AD & \iff & \frac {AD}{NY}=\frac {AC}{CN} & \iff & \frac {h_a}n=\frac b{s-b}\end{array}\right\|$ $\bigodot$ $\implies$ $\frac {h_a^2}{mn}=\frac {bc}{(s-b)(s-c)}=\csc^2\frac A2=\frac 1{\sin^2\frac A2}=\frac 2{1-\cos A}\implies$ $\boxed {h_a^2=\frac {2mn}{1-\cos A}}\ .$

P7 (clasa a IV - a). Sa se determine perimetrul figurii mentionate aici (<= click).

Proof. The closed polygonal line $ABCDEFGHA$ has $:\ AB=315\ ,\ BC=300\ ,\ EF=55\ ;\ AB\parallel GH\parallel EF\parallel CD\ ;\ BC\parallel ED\parallel FG\parallel AH\ ;\ K\in AH\cap CD\ ,\ L\in HG\cap DE\ .$

The perimeter $p(CDEFGHA)=CD+DE+EF+FG+GL+LH+HA=$ $(CD+LH)+(LG+EF)+(DE+FG+HA)=$ $AB+2\cdot EF+BC\ .$ In conclusion,

$p(CDEFGHA)=$ $AB+2\cdot EF+BC$ and the required perimeter $p(ABCDEFGHA)=$ $AB+BC+p(CDEFGHA)=$ $AB+BC+(AB+2\cdot EF+BC)=$

$2\cdot (AB+BC+EF)=$ $2(315+300+55)=$ $2\cdot 670=1340$ $\implies$ $\boxed{p(ABCDEFGHA)=1340}\ .$

P8 (Miguel Ochoa Sanchez). Let $ABCD$ be a square with the circumcircle $w$ and choose $\left|\begin{array}{c} M\in \overarc{BC}\ ;\ N\in \overarc{CD}\\\\ m\left(\widehat{MAN}\right)=45^{\circ}\end{array}\right|$ for which denote $F\in BC\cap w,$ $E\in CD\cap w\ .$ Prove that $\frac {MC}{AF}+\frac {NC}{AE}=1\ .$

Proof. Let $\left\{\begin{array}{c} m\left(\widehat{CAM}\right)=m\left(\widehat{DAN}\right)=\alpha\\\\ m\left(\widehat{CAN}\right)=m\left(\widehat{BAM}\right)=\beta\end{array}\right\|$ where $\alpha +\beta =45^{\circ}\ .$ Thus, $\left\{\begin{array}{ccccc} \frac {MC}{AF}=\frac {AC}{AB}\cdot\frac {MC}{AC}\cdot\frac {AB}{AF}=\sqrt 2\cdot\sin \alpha \cdot\cos\beta\\\\ \frac {NC}{AE}=\frac {AC}{AD}\cdot\frac {NC}{AC}\cdot\frac {AD}{AE}=\sqrt 2\cdot\sin\beta\cdot\cos\alpha\end{array}\right\|\bigoplus\implies$ $\frac {MC}{AF}+\frac {NC}{AE}=\sqrt 2\cdot (\sin\alpha\cos\beta +\sin\beta\cos\alpha )=$

$\sqrt 2\cdot\sin (\alpha +\beta )=\sqrt 2\cdot\frac {\sqrt 2}2=1\implies$ $\frac {MC}{AF}+\frac {NC}{AE}=1\ .$ Observe that $:\ BMCN,$ $CNDM,$ $ABMN,$ $ADNM$ are isosceles trapezoids and $MN=AB\ ;\ \triangle BMC\equiv\triangle CND\ ;$

the point $H\in BN\cap DM$ is the orthocenter of $\triangle AMN\ ;$ a.s.o. Remark. If $\alpha +\beta =\phi <90^{\circ},$ then $\frac {MC}{AF}+\frac {NC}{AE}=1+\sqrt 2\cdot\sin\left(\phi -45^{\circ}\right)\cdot\cos\left(\alpha -\beta \right)\ .$

Lemma. Prove that for any $x>0$ there is the inequality $x^3+1\ge x\sqrt {2\left(x^2+1\right)}\ .$

Proof 1. $x^3+1\ge \frac 12\cdot (x+1)\left(x^2+1\right)=\frac 12\cdot (x+1)\cdot \sqrt{x^2+1}\cdot\sqrt{x^2+1}\ge \frac 12\cdot 2\sqrt x\cdot \sqrt{2x}\cdot\sqrt{x^2+1}=x\sqrt{2\left(x^2+1\right)}\implies$ $\boxed{x^3+1\ge x\sqrt{2\left(x^2+1\right)}}\ .$

Proof 2. $\left\{\begin{array}{ccc} x^2+1 & \ge & 2x\\\\ x^4-x^2+1 & \ge & x\left(x^2-x+1\right)\end{array}\right\|\bigodot\implies$ $x^6+1\ge 2x^2\left(x^2-x+1\right)\iff$ $x^6+2x^3+1\ge 2x^2\left(x^2+1\right)\iff$ $\boxed{x^3+1\ge x\sqrt{2\left(x^2+1\right)}}\ .$

Proof 3. Observe that $x^3+1\ge\frac {3x^2+1}2\iff$ $2x^3-3x^2+1\ge 0\iff$ $(2x+1)(x-1)^2\ge 0$ what is truly. Hence $\boxed{x^3+1\ge \frac {3x^2+1}2}\ (*)\ .$

Therefore, $x\sqrt {2\left(x^2+1\right)}=\sqrt{2x^2\cdot \left(x^2+1\right)}\le$ $\frac {2x^2+\left(x^2+1\right)}2=$ $\frac {3x^2+1}2\ \stackrel{(*)}{\le}\ x^3+1$ $\implies$ $\boxed{x^3+1\ge x\sqrt {2\left(x^2+1\right)}}\ .$

P9 (Van Khea - Cambodgia). Prove that for any $A$ -right-angled $\triangle ABC$ there is the inequality $a^3+b^3+c^3\ge \left(2+\sqrt 2\right)abc\ .$ (standard notations).

Proof 1. Suppose w.l.o.g. there is $x\in\left(0,\frac {\pi}2\right)$ so that $a=1\ ,\ b=\sin x$ and $c=\cos x\ .$ I"ll use the substitution $\boxed{t:=\sin x+\cos x}\ ,$ where $t\in\left(1,\sqrt 2\right]$ and $\boxed{\sin x\cos x=\frac {t^2-1}2}\ .$ The our inequality

becomes $1+(\sin x+\cos x)(1-\sin x\cos x)\ge \left(2+\sqrt 2\right)\sin x\cos x\iff$ $1+t\cdot\left(1-\frac{t^2-1}2\right)\ge \left(2+\sqrt 2\right)\cdot \frac{t^2-1}2\iff$ $2(t+1)-t\left(t^2-1\right)\ge$ $\left(2+\sqrt 2\right)\left(t^2-1\right)\ \stackrel{t+1>0}{\iff}$

$2-t(t-1)\ge \left(2+\sqrt 2\right)(t-1)\iff$ $t^2+\left(1+\sqrt 2\right)t-\left(4+\sqrt 2\right)\le 0\iff$ $\left(t-\sqrt 2\right)\left(t+1+2\sqrt 2\right)\le 0\iff$ $t\in\left[-1-2\sqrt 2,\sqrt 2\right]\cap \left(1,\sqrt 2\right]\iff$ $t\in \left(1,\sqrt 2\right]$ what is truly.

Proof 2. from $a^3=a\cdot a^2=$ $a\cdot \left(b^2+c^2\right)\ge 2abc$ obtain that $\boxed{a^3\ge 2abc}\ (1)\ .$ I"ll use the previous lemma in the particular case $x=\frac bc\ :$

$b^3+c^3\ge bc\sqrt{2\left(b^2+c^2\right)}\iff\boxed{b^3+c^3\ge abc\sqrt 2}\ (2)\ .$ From the sum of the relations $(1)$ and $(2)$ get the required inequality $\boxed{a^3+b^3+c^3\ge \left(2+\sqrt 2\right)abc}\ .$

P10 (Kunihiko Chikaya). Prove that $\int_0^{\frac {\pi}2}\sqrt{1-2\sin 2x+3\cos^2x}\ \mathrm{dx}=\int_0^{\frac {\pi}2}|2\cos x-\sin x|\ \mathrm{dx}=2\sqrt 5-3\ .$

Proof. $1-2\sin 2x+3\cos^2x=$ $\left(1-\cos^2x\right)-4\sin x\cos x+4\cos^2x=$ $\sin^2x-4\sin x\cos x+4\cos^2x=(2\cos x-\sin x)^2\implies$

$\int_0^{\frac {\pi}2}\sqrt{1-2\sin 2x+3\cos^2x}\ \mathrm{dx}=\int_0^{\frac {\pi}2}|f(x)|\ \mathrm{dx}\ ,\$ where $f:I\rightarrow \mathbb R\ ,\ f(x)=2\cos x-\sin x$ and $I=\left[0,\frac {\pi}2\right]\ .$ Prove easily that $f$ is strict

decreasing $(\searrow)$ on $I$ and $2=f(0)\ \searrow\ f(\arctan 2)=0\ \searrow\ -1=f\left(\frac {\pi}2\right)\ .$ Let $F(x)=\int_I f(x)\ \mathrm{dx}=2\sin x+\cos x+\mathbb C\ .$ Therefore,

$\int_0^{\frac {\pi}2}|f(x)|\ \mathrm{dx}=$ $\int_0^{arctan 2} f(x)\ \mathrm{dx}-\int_{\arctan 2}^{\frac {\pi}2} f(x)\ \mathrm{dx}=$ $2\cdot F(\arctan 2 )-F(0)-F\left(\frac {\pi}2\right)=2\cdot \left(2\cdot\frac 2{\sqrt 5}+\frac 1{\sqrt 5}\right)-3=2\sqrt 5-3\ .$

P11 (Kunihiko Chikaya). Let $\{x,y\}\subset\mathbb R$ such that $2\sin x\sin y+3\cos y+6\cos x\sin y=7\ .$ Find the value of $\tan^2x+2\tan^2y\ .$

Proof. I"ll use the welll-known inequality $\boxed{\left|a\sin\phi +b\cos\phi\right|\le \sqrt{a^2+b^2}}\ (*)\ .$ For example $\boxed{|\sin x+3\cos x|\le\sqrt {10}}\ (1)$ and $2\sin x\sin y+3\cos y+6\cos x\sin y=7\iff$

$2(\sin x+3\cos x)\cdot \underline{\sin y}+3\cdot\underline{\cos y}=7\implies$ $4(\sin x+3\cos x)^2+9\ge 49\implies$ $(\sin x+3\cos x)^2\ge 10\ \stackrel{(1)}{\implies}\ \sqrt {10}\ge |\sin x+3\cos x|\ge \sqrt{10}\implies\boxed{|\sin x+3\cos x|=\sqrt {10}}\ (2)$

$\iff$ $(\sin x+3\cos x)^2=10\left(\sin^2+\cos^2x\right)\iff$ $\left(\tan x+3\right)^2=10\left(\tan^2x+1\right)\iff$ $9\tan^2x-6\tan x+1=0\iff$ $(3\tan x-1)^2=0\implies$ $\boxed{\tan^2 x=\frac 19}\ (3)\ .$ Observe that

$7-3\cos y=2\sin y(\sin x+3\cos x)\ \stackrel{(2)}{\implies}\ |7-3\cos y|=2\sqrt{10}|\sin y|\iff$ $(7-3\cos y)^2=40\sin^2y\iff$ $49+9\cos^2y-42\cos y=40\sin^2y\iff$ $(7\cos y-3)^2=0\iff$

$\cos y=\frac 37\iff$ $\tan^2y=\frac {1-\cos^2y}{\cos^2y}=\frac {40}{9}\implies$ $\boxed{\tan^2y=\frac {40}9}\ (4)\ .$ In conclusion, from the sum of $(3)$ and $(4)$ obtain that $\boxed{\tan^2x+2\tan^2y=9}\ .$ Very nice problem! Thank you.

P12 (Kunihiko Chikaya). Let the square $ABOC$ with $AB=1$ and the circle $w=\mathbb C(O,1)\ .$ For a mobile point $P\in w$ so that $BC$

separates $P$ and $O$ denote $\{P,Q\}=AP\cap w\ .$ Prove that the area $S$ of the triangle $POQ$ is $\max\ \iff\ m\left(\widehat{PAB}\right)=\frac {\pi}{12}\ .$

Proof. Denote the midpoint $M$ of $[PQ]$ and $m\left(\widehat{PAB}\right)=\phi\ .$ Suppose w.l.o.g. that $\boxed{0<\phi <\frac {\pi}4}\ ,$ i.e. $\tan\phi =a\in (0,1)\ .$ Observe that $:\ OA=\sqrt 2$ and $m\left(\widehat{OAM}\right)=\frac {\pi}4-\phi\implies$

$\left\{\begin{array}{cccc} AM=OA\cos\widehat{OAM}=\sqrt 2\cdot \cos\left(\frac {\pi}4-\phi\right) & \implies & \boxed{AM=\cos\phi +\sin\phi} & (1)\\\\ OM=OA\sin\widehat{OAM}=\sqrt 2\cdot \sin\left(\frac {\pi}4-\phi\right) & \implies & \boxed{OM=\cos\phi-\sin\phi} & (2)\end{array}\right\|\ .$ Therefore, $AP\cdot AQ=AB^2\ ,$ i.e. $\boxed{AP\cdot AQ=1}$ and $AP+AQ=2\cdot AM\ \stackrel{(1)}{=}\ 2(\cos\phi+\sin\phi )$

$\implies\boxed{AP+AQ=2(\cos\phi +\sin\phi )}\ (3)\ .$ Hence $(AP-AQ)^2=(AP+AQ)^2-4\cdot AP\cdot AQ=4(1+\sin 2\phi)-4=4\sin2\phi\implies$ $\boxed{PQ=2\sqrt {\sin2\phi}}\ (4)\ .$ Denote $\boxed{\cos \phi -\sin\phi =t}\ .$

Thus, $\boxed{\sin2\phi =1-t^2}\ ,$ where $t\in (0,1)\ .$ In conclusion, $\cancel 2S=PQ\cdot OM=\cancel 2(\cos\phi -\sin\phi)\sqrt {\sin 2\phi}=\cancel 2t\sqrt{1-t^2}\implies$ $\boxed{S^2=t^2\left(1-t^2\right)}\ (5)\ .$ Thus, $S$ is $\max\iff$ $t^2\left(1-t^2\right)$ is $\max$ with

$t^2+\left(1-t^2\right)=1$ (constant) $\implies$ $t^2=1-t^2=\frac 12\iff$ $\boxed{t=\frac {\sqrt 2}2}\iff$ $\sin 2\phi =1-t^2=1-\frac 12=\frac 12\iff \sin 2\phi =\frac 12\iff 2\phi =\frac {\pi}6\iff \boxed{\phi =\frac {\pi}{12}}\ ,\ a=\tan\phi =2-\sqrt 3\ ,\ S=\frac 12$

P13 (Mehmet Şahin, Turkey). Let $ABCD$ be a square with the circumcircle $w$ and a point $P\in \overarc{AD}\ .$ Prove that $2\cdot [ABCP]=AB^2\ .$

Proof 1. Denote $I\in BP\cap AC$ and $\left\{\begin{array}{ccc} m\left(\widehat{PBA}\right) & = &\alpha\\\ m\left(\widehat{PBD}\right) & = & \beta\\\ m\left(\widehat{BIC}\right) & = & \phi\end{array}\right\|\ ,$ where $\alpha +\beta =\frac {\pi}4$ and $\sin\phi =\cos \beta\ .$ Hence $PB=BD\cos\beta =AC\sin\phi\implies$

$\boxed{AC\sin\phi =PB}\ (*)\ \implies$ $2\cdot [ABCP]=PB\cdot (AC\sin\phi )\ \stackrel{(*)}{=}\ PB\cdot PB=PB^2\implies$ $[ABCP]=\frac {PB^2}2\ .$

Proof 2 . $\left\{\begin{array}{c} E\in PB\ ;\ AE\perp BP\\\\ F\in PB\ ;\ CF\perp BP\end{array}\right\|\implies$ $\triangle ABE\equiv\triangle BCF\implies$ $BE=CF,$ $AE=BF\implies$ $E$ -right-angled $\triangle AEP,$ $F$ -right-angled $CFP$ are isosceles: $EA=EP,$ $FC=FP\implies$

$\left\{\begin{array}{c} BE=CF=FP\\\\ AE=BF=EP\end{array}\right\|\implies$ $AE+CF=BF+FP=BP$ $\implies$ $\underline{PB^2}=PB\cdot PB=$ $PB\cdot (AE+CF)=$ $PB\cdot AE+PB\cdot CF=$ $2\cdot [ABP]+2\cdot [CBP]=$ $\underline{2\cdot [ABCP]}$ (S. Fulger).

Proof 3.

P14 (Carlos Hugo Olivera Diaz, Peru). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and the circumcircle $\Omega =\mathbb (O,R)\ .$

Denote $\left\{\begin{array}{ccccc} Q\in BI\cap \Omega & ; & P\in CI\cap \Omega\\\\ M\in AB\cap w & ; & N\in AC\cap w\end{array}\right\|\ .$ Prove that $\boxed{M\in PQ\iff A=90^{\circ}\iff N\in PQ}\ .$

Proof. $M\in PQ\iff$ $\frac {MA}{MB}=\frac {QA}{QB}\iff$ $\frac {s-a}{s-b}=\frac {\sin\frac B2}{\sin\left( A+\frac B2\right)}\iff$ $\frac {s-a}{s-b}=\frac {2\cos\frac B2\sin\frac B2}{2\cos\frac B2\sin \left(A+\frac B2\right)}\iff$ $\frac {s-a}{s-b}=\frac {\sin B}{\sin A+\sin C}\iff\frac {s-a}{s-b}=\frac b{a+c}\ .$ In conclusion,

$\boxed{M\in PQ\iff\frac {s-a}{s-b}=\frac b{a+c}}\iff$ $\frac {s-a}{(s-a)+(s-b)}=\frac b{a+b+c}\iff$ $2s(s-a)=bc\iff$ $2s(s-a)=s(s-a)+(s-b)(s-c)\iff$ $s(s-a)=(s-b)(s-c)\iff$

$\tan^2\frac A2=\frac {(s-b)(s-c)}{s(s-a)}=1\iff A=90^{\circ}\ .$ Prove analogously that $N\in PQ\iff$ $\frac {NA}{NC}=\frac {PA}{PC}\iff$ $\frac {s-a}{s-c}=\frac {\sin \frac C2}{\sin\left(A+\frac C2\right)}=\frac c{a+b}\iff$ $\boxed{N\in PQ\iff \frac {s-a}{s-c}=\frac c{a+b}}$ a.s.o.

P15 (Leo Giugiuc & Gh. Duca). Let a regular polygon $A_1A_2\ldots A_{n-1}A_n$ with the circumcircle $w=\mathbb C(O,R)\ .$ Prove that for any $P\in w$ there is the identity $\sum_{k=1}^n PA_k^2=2nR^2\ .$

Proof. Can use the wellknown implication $\{M,N\}\subset w\implies MN=2R\sin\frac{\widehat{MON}}2$ and the trigonometrical identity $:\ \sum_{k=1}^n\cos x_k=\frac {\cos\frac {x_1+x_n}2\sin\frac {nr}2}{\sin\frac {nr}2}\ ,$ where $x_k=x_1+(k-1)r\ ,\ \forall\ k\in\overline{1,n}\ .$

P16 (Adil Abdullayev, Baku). Prove $(\forall )\ \triangle ABC$ there is the following inequality $:\ \frac {r_a}{\cos^2\frac A2}+\frac {r_b}{\cos^2\frac B2}+\frac {r_c}{\cos^2\frac C2}\ge 6R$ (standard notations).

Proof. $\frac {r_a}{\cos^2\frac A2}=$ $\frac {2r_a\sin\frac A2}{\cos\frac A2\sin A}=$ $\frac {2r_a\tan\frac A2}{\sin A}=$ $\frac {2r_a^2}{s\cdot \sin A}\implies$ $\sum\frac {r_a}{\cos^2\frac A2}=\frac 2s\cdot\sum\frac {r_a^2}{\sin A}\ \stackrel{C.B.S}{\ge}\ \cdot\frac {2\left(\sum r_a\right)^2}{s\sum \sin A}=$ $\frac {2R(4R+r)^2}{s^2}\ge 6R$ because is wellknown the inequality $\boxed{s\sqrt 3\le 4R+r}\ .$

P17 (Sefket Arslanagic, Berlin). Prove that the inequality $\sqrt {a-1}+\sqrt {b-1}+\sqrt {c-1}\le \sqrt {abc+\min \{a,b,c\}}$ for any $\{a,b,c\}\subset (1,\infty )\ .$

Proof. Suppose w.l.o.g. $a\le b\le c$ and denote $\left\{\begin{array}{ccccc} \sqrt {a-1}=x>0 & \implies & a=x^2+1>1\\\\ \sqrt {b-1}=y>0 & \implies & b=y^2+1>1\\\\ \sqrt {c-1}=z>0 & \implies & c=z^2+1>1\end{array}\right\|\ .$ Thus, our inequality becomes $\sqrt {a-1}+\sqrt {b-1}+\sqrt {c-1}\le \sqrt {abc+\min \{a,b,c\}}$ for any

$\{a,b,c\}\subset (1,\infty )\ ,$ i.e. $x+y+z\le \sqrt {\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)+x^2+1}\ (*)\ .$ Denote $\boxed{y+z=S\ \mathrm{and}\ yz=P}\ (1)\ .$ In conclusion, the inequality becomes $[x+(y+z)]^2\le$

$x^2+1+\left(y^2+1\right)\left(z^2+1\right)\cdot x^2+\left(y^2+1\right)\left(z^2+1\right)\ \stackrel{(1)}{\iff}$ $\cancel {x^2}+2Sx+\cancel{S^2}\le \cancel{x^2}+1+\left[S^2+(P-1)^2\right]\cdot x^2+$ $\cancel{S^2}+(P-1)^2\implies$ $\left[S^2+(P-1)^2\right]\cdot \underline x^2-2S\cdot \underline x+$

$\left[(P-1)^2+1\right]\ge 0$ what is true because $\Delta^{\prime}=$ $S^2-\left[S^2+(P-1)^2\right]\cdot\left[(P-1)^2+1\right]=-(P-1)^2\left[\left(S^2+1\right)+(P-1)^2\right]\le 0\ .$ We have equality iff $P=1\ ,$ i.e. $yz=1\ .$

click => P18 (Nelson Cayo Deza Velasquez, Lima-PERU).

Proof. Suppose w.l.o.g. that $:$ the big circle is $\alpha =\mathbb C(O,1)$ and touches $AD$ at $P\ ;$ the semicircle is $\beta =\mathbb C(I,r)\ ,$ touches $\alpha$ at $T\ ,$ where $r\in (0,1)\ .$ Denote $PE=y$ and apply the Pythagoras' theorem

in $\triangle OPI\ :\ OI^2=PO^2+PI^2\iff$ $(r+1)^2=1^2+(y+r)^2\iff$ $\cancel{r^2}+2r+\cancel 1=\cancel 1+y^2+2ry+\cancel{r^2}\iff$ $2r=y^2+2ry\iff$ $2r(1-y)=y^2\iff$ $\boxed{\ 2r=\frac {y^2}{1-y}\ }\ (*)\ .$ Observe

that $:\ ED=PD-PE=1-y\implies$ $\boxed{\ ED=1-y\ }\ (1)\ ;$ $\boxed{\ EF=2r\ }\ (2)\ ;$ $AF=AE+EF=1+y+2x\ \stackrel{(*)}{=}\ 1+y+\frac {y^2}{1-y}=$ $\frac {(1+y)(1-y)+y^2}{1-y}=\frac {1-\cancel{y^2}+\cancel{y^2}}{1-y}=\frac 1{1-y}\implies$

$\boxed{\ AF=\frac 1{1-y}\ }\ (3)\ .$ In conclusion, $\sqrt {ED}+\sqrt {EF}\ \stackrel {1\wedge 2}{=}\ \sqrt {1-y}+\sqrt{2r}\ \stackrel{(*)}{=}\ \sqrt {1-y}+\frac y{\sqrt {1-y}}=\frac {(1-y)+y}{\sqrt {1-y}}=\frac 1{\sqrt {1-y}}\ \stackrel{(3)}{=}\ \sqrt {AF}\implies$ $\boxed{\sqrt {ED}+\sqrt {EF}=\sqrt{AF}}\ .$ Nice problem!

P19 (Kadir Altintas, Turkey). Let $\triangle ABC$ with the centroid $G$ and the orthocenter $H\ .$ Prove that $\boxed{\ GH\perp GA\iff b^2+c^2=2a^2\ }\ .$ (standard notations).

Proof 1. Denote the midpoint $M$ of the side $[BC]$ and the projection $D$ of the vertex $A$ on $BC\ .$ Observe that $GH\perp GA\iff$ $HDMG$ is cyclic, i.e. $AH\cdot AD=AG\cdot AM\iff$

$2R\cos A\cdot h_a=\frac {2m_a}3\cdot m_a\iff$ $2Rh_a\cdot \cos A=\frac {4m_a^2}6\ .$ I"ll use the well-known identities $\boxed{2Rh_a=bc}\ (*)\ ,\ b^2+c^2-a^2=2bc\cdot \cos A$ and $4m_a^2+a^2=2\left(b^2+c^2\right)\ .$

Hence $GH\perp GA\iff$ $6bc\cdot \cos A=4m_a^2\iff$ $3\left(b^2+c^2-a^2\right)=2\left(b^2+c^2\right)-a^2\iff$ $3\left(b^2+c^2\right)-3a^2=2\left(b^2+c^2\right)-a^2\iff$ $b^2+c^2=2a^2\ .$

Proof 2. I'll apply identities $\boxed{\ HG=2\cdot OG\ }$ and $\boxed{\ OG^2=R^2-\frac {a^2+b^2+c^2}9\ }\ .$ From the Pythagoras' theorem obtain that $\ :\ GH\perp GA\iff$ $GH^2+GA^2=AH^2\iff$

$4\cdot\left(\cancel{R^2}-\frac {a^2+b^2+c^2}9\right)+\frac {2\left(b^2+c^2\right)-a^2}9=\cancel{4R^2}-a^2\iff$ $4\left(a^2+b^2+c^2\right)-2\left(b^2+c^2\right)+a^2=9a^2\iff$ $4a^2+4\left(b^2+c^2\right) -2\left(b^2+c^2\right)+a^2=9a^2\iff$ $b^2+c^2=2a^2\ .$

P20 (Kadir Altintas, Turkey). Let $\triangle ABC$ with the centroid $G$ and the incenter $I$ for what denote $\left\{\begin{array}{ccc} E & \in & IG\cap (AC)\\\\ F & \in & IG\cap (AB)\end{array}\right\|\ .$ Prove that $\boxed{IG\perp IA\iff \frac 1b+\frac 1c=\frac 6{a+b+c}}\ .$

Proof. Denote the midpoint $M$ of $[BC]$ and $D\in AI\cap BC\ .$ Observe that $IA\perp IG\iff AE=AF=m\ ,$ i.e. $\left\{\begin{array}{ccc} FB & = & c-m\\\\ EC & = & b-m\end{array}\right\|\ .$ Apply the well-known Cristea's theorem to the points $:$

$\odot \begin{array}{cccccccccc} \nearrow\ G & : & \frac {FB}{FA}\cdot MC+\frac {EC}{EA}\cdot MB=\frac {GM}{GA}\cdot BC & \iff & \frac {c-m}m+\frac {b-m}m=1 & \iff & (c-m)+(b-m)=m & \iff & \boxed{3m=b+c} & \searrow\\\\ \searrow\ I & : & \frac {FB}{FA}\cdot DC+\frac {EC}{EA}\cdot DB=\frac {ID}{IA}\cdot BC & \iff & \frac {c-m}m\cdot b+\frac {b-m}m\cdot c= a & \iff & b(c-m)+c(m-b)=am & \iff & \boxed{2bc=m(a+b+c)} & \nearrow\end{array}\bigodot\implies$

$3\cancel m\cdot 2bc=(b+c)\cdot \cancel m(a+b+c)\iff 6bc=(b+c)(a+b+c)\iff$ $\frac {b+c}{bc}=\frac 6{a+b+c}\iff$ $\frac 1b+\frac 1c=\frac 6{a+b+c}\ .$ Done. Nice problem !

An easy extension. Let an acute $\triangle ABC$ with the incenter $I$ and its interior point $P$ with the barycentrical coordinates $(x,y,z)\ ,$

$x+y+z=1$ w.r.t. $\triangle ABC$ for what denote $\left\{\begin{array}{ccc} E & \in & IP\cap (AC)\\\\ F & \in & IP\cap (AB)\end{array}\right\|\ .$ Prove that $\boxed{IP\perp IA\iff \frac yb+\frac zc=\frac 2{a+b+c}}\ .$

P21 (for Mathskidd - elementary proof). Let $ABC$ be a triangle with the orthocentre $H$ and the circumcircle $w=\mathbb C(O,R)\ .$ Prove the identity $OH^2=9R^2-\left(a^2+b^2+c^2\right)$ (standard notations).

Proof 1. Let the midpoint $M$ of $[BC]$ and the diameter $[AS]$ of $w\ .$ $\left\{\begin{array}{cccc} BH\perp AC\ ;\ SC\perp AC & \implies & SC\parallel BH\\\\ CH\perp AB\ ;\ SB\perp AB & \implies & SB\parallel CH\end{array}\right\|$ $\implies$ $BHCS$ is a parallelogram $\implies M\in HS$ and $HB=CS$ $\implies$

$HB^2=CS^2=AS^2-AC^2=4R^2-b^2\implies$ $HB^2=4R^2-b^2$ a.s.o. $\implies$ $\boxed{HA^2+a^2=HB^2+b^2=HC^2+c^2=4R^2}\ (1)\ .$ Median $HM/\triangle BHC\ :\ HS^2=4\cdot HM^2=$

$2\left(HB^2+HC^2\right)-BC^2\ \stackrel{(1)}{=}\ 2\left[\left(4R^2-b^2\right)+\left(4R^2-c^2\right)\right]-a^2$ $\implies$ $\boxed{HS^2=16R^2-2\left(b^2+c^2\right)-a^2}\ (2)\ .$ Median $HO/\triangle AHS\ :\ 4\cdot HO^2+AS^2=$ $2\left(AH^2+HS^2\right)\ \stackrel{(1\wedge 2)}{\implies}$

$4\cdot HO^2=$ $2\left[\left(4R^2-a^2\right)+\left(16R^2-2b^2-2c^2-a^2\right)\right]-4R^2=$ $36R^2-4\left(a^2+b^2+c^2\right)$ $\implies$ $4\cdot HO^2=36R^2-4\left(a^2+b^2+c^2\right)\implies$ $OH^2=9R^2-\left(a^2+b^2+c^2\right)\ .$

Proof 2. Observe that $m\left(\widehat{OAH}\right)=|B-C|\ .$ Hence can apply the generalized Pythagoras' theorem to $\triangle AOH\ :\ HO^2=AH^2+AO^2-2\cdot AH\cdot AO\cdot\cos\widehat{OAH}=\left(4R^2-a^2\right)+R^2-$

$4R^2\cos A\cos (B-C)=$ $5R^2-a^2+4R^2\cos (B+C)\cos (B-C)=$ $5R^2-a^2+2R^2\cos 2B\cos 2C=$ $5R^2-a^2+2R^2\left[\left(1-2\sin^2B\right)+\left(1-2\sin^2C\right)\right]=$ $5R^2-a^2+$

$4R^2\left(1-\sin^2B-\sin^2C\right)=$ $5R^2-a^2+4R^2-b^2-c^2\implies$ $HO^2=9R^2-\left(a^2+b^2+c^2\right)\ .$ I used the identities $\cos 2x=1-2\sin^2x$ and $\cos x+\cos y=2\cdot\cos\frac {x+y}2\cdot \cos\frac {x-y}2\ .$

Remark.

$\blacktriangleright\ 2R^2\cdot\sum\cos 2A=$ $2R^2\cdot\sum\left(1-2\sin^2A\right)=$ $6R^2-4R^2\cdot\sum\sin^2A=$ $6R^2-\sum (2R\sin A)^2=$ $6R^2-\sum a^2$ $\implies$ $\boxed{\sum a^2=6R^2-2R^2\cdot\sum\cos 2A}\ (1)\ .$

$\blacktriangleright\ \sum\cos 2A=$ $\cos 2A+(\cos 2B+\cos 2C)=\cos 2A+2\cos (B+C)\cos (B-C)=2\cos^2A-1-2\cos A\cos (B-C)=$

$-1-2\cos A[\cos (B-C)-\cos A]=-1-2\cos A[\cos (B-C)+\cos (B+C)]=-1-4\cos A\cos B\cos C\implies$ $\boxed{\sum\cos 2A=-1-4\prod\cos A}\ (2)\ .$

$\blacktriangleright\ HO^2=9R^2-\sum a^2\ \stackrel{(1)}{=}\ 9R^2-\left(6R^2-2R^2\cdot\sum\cos 2A\right)=$ $3R^2+2R^2\cdot\sum\cos 2A\ \stackrel{(2)}{=}\ 3R^2+2R^2\left(-1-4\prod\cos A\right)=$

$R^2-8R^2\cdot\prod\cos A=$ $R^2\left(1-8\cdot\prod\cos A\right)\implies$ $\boxed{HO^2=R^2\cdot\left(1-8\prod\cos A\right)}\ (3)\ .$ The power of $H$ w.r.t. $w$ is $p_w(H)=OH^2-R^2\ ,$ i.e. $\boxed{p_w(H)=-8R^2\prod\cos A}\ (4)\ .$

P21 (Boris COLAKOVIC). Prove that $(\forall )\ \triangle\ ABC\ \mathrm{there\ is\ the\ following\ inequality\ :}\ \frac 1a+\frac 1b+\frac 1c\le\frac {\sqrt 3}{2r}$ .

Proof. I"ll use only the remarkable inequality $\boxed{\ s\le\frac {3R\sqrt 3}2\ }\ (*)$ and the well-known inequality $\boxed{\ 3(bc+ca+ab)\le (a+b+c)^2\ }\ (1)\ .$

Therefore, $\sum\frac 1a=$ $\frac {ab+bc+ca}{abc}\ \stackrel{(1)}{\le}\ \frac {(a+b+c)^2}{3\cdot abc}=$ $\frac {4s^2}{3\cdot 4Rrs}=$ $\frac s{3Rr}\ \stackrel{(*)}{\le}\ \frac 1{\cancel{3R}r}\cdot \frac {\cancel{3R}\sqrt 3}2=\frac {\sqrt 3}{2r}\implies$ $\boxed{\ \frac 1a+\frac 1b+\frac 1c\le \frac {\sqrt 3}{2r}\ }\ .$

Remark. This inequality is equivalent with $\underline{\underline{\boxed{\ h_a+h_b+h_c\le s\sqrt 3\ }}}\ .$ Indeed, $\sum h_a\le s\sqrt 3\iff$ $\sum\frac {bc}{2R}\le s\sqrt 3\iff$ $\sum bc\le 2Rs\sqrt 3\iff$ $\frac {\sum bc}{abc}\le\frac {2Rs\sqrt 3}{4Rsr}\iff$

$\underline{\underline{\boxed{\ \frac 1a+\frac 1b+\frac 1c\le \frac {\sqrt 3}{2r}\ }}}\ .$ Prove easily that $\left\{\begin{array}{ccc} r_a+r_b+r_c & = & 4R+r\\\\ r_ar_b+r_br_c+r_cr_a & = & s^2\\\\ (s-a)(s-b)(s-c) & = & sr^2\end{array}\right\| \ .$ I"ll prove the inequality $s\sqrt 3\le 4R+r$ and the chain of the inequalities $\boxed{\ 3r\sqrt 3\le s\le \frac {3R\sqrt 3}2\ }\ :$

$\left\{\begin{array}{ccccc} 3s^2=3\sum r_br_c\le \left(r_a+r_b+r_c\right)^2=(4R+r)^2 & \iff & \boxed{s\sqrt 3\le 4R+r} & \iff & 9s\sqrt 3\le 9(4R+r)\\\\ 3\cdot\sqrt[3]{\prod (s-a)}\le\sum (s-a)=s\iff 27sr^2\le s^3 & \iff & \boxed{3r\sqrt 3\le s} & \implies & 9r\le s\sqrt 3\end{array}\right\|$ $\bigoplus\implies$ $8s\sqrt 3\le 36R\iff 2s\sqrt 3\le 9R\iff \boxed{\ s\le \frac {3R\sqrt 3}2\ }\ .$

This post has been edited 577 times. Last edited by Virgil Nicula, Feb 27, 2019, 11:09 AM

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Dear sir Virgil Nicula, thanks. Furthermore, I have in doubt that can $\boxed{OH^2= R^2(1-8\cos A \cos B \cos C)}$ ?

This post has been edited 1 time. Last edited by Virgil Nicula, Aug 12, 2017, 8:13 AM

by Mathskidd, Aug 11, 2017, 1:08 PM

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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.

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232. Some geometrical inequalities (own).

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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.

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219. A very nice geometrical inequality.

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200. Some nice applications of the inversion.

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

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

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176. Another nice problems for middle school.

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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).

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

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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).

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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).

Submit

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by mathMagicOPS, Jan 9, 2025, 3:40 AM
this css is sus
by ihatemath123, Aug 14, 2024, 1:53 AM
391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
by tigerzhang, Aug 1, 2021, 12:02 AM
Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
by GoogleNebula, Apr 14, 2021, 5:25 AM
Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

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

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

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

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

72 shouts

My (math)blog

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

by Virgil Nicula, Jul 9, 2017, 2:12 PM

1 Comment