449. Working page I.

by Virgil Nicula, Oct 1, 2016, 4:24 PM

Lemma. Sa se arate ca $\forall\ x\in \mathbb R$ exista inegalitatea $|\sin x|\le |x|\ .$

Dem. Fie primul cadran $\overarc{AB}$ al cercului trigonometric $w=\mathbb C(O,1)$ unde $A(0)$ si $B\left(\frac{\pi}2\right)\ .$ Se stie ca $\sin x$ este lungimea distantei lui $X(x)\in\overarc{AB}$ la $OA$ si $x$ este masura lui $\widehat{AOX}$ exprimata in radiani, adica lungimea arcului $\overarc{AX}\ .$ Deci $\sin x=\mathrm{pr}_{OA}(X)\le\mathrm{l}\left(\overarc{AX}\right)=x\ ,$ adica $\sin x\le x$ pentru orice $\underline{0\le x\le \frac {\pi}2}\ .$ Daca $\underline{x>\frac{\pi}2}$ atunci $|\sin x|<1<\frac {\pi}2<|x|\ ,$ adica $|\sin x|\le |x|\ .$ Daca $\underline{x<0}$ atunci notam $y=-x>0\ .$ Deci $|\sin x|=|\sin (-y)|=$ $|-\sin y|=|\sin y|\le |y|=$ $|-x|=|x|\ ,$ adica si in acest caz $|\sin x|<|x|\ .$ Deci $|\sin x|\le |x|\ ,\ \forall\ x\in (-\infty ,0)\cup\left[0,\frac {\pi}2\right]\cup \left(\frac {\pi}2,\infty\right)=\mathbb R\ .$

P0 .Sa se arate fara a folosi derivatele ca pentru orice $\forall\ x\in \mathbb R$ exista inegalitatea $\cos x\ge 1-\frac {x^2}2\ .$ Atentie: maine se va sterge!

Dem. $\cos x\ge 1-\frac {x^2}2\iff$ $\frac {x^2}2\ge 1-\cos x\iff$ $\frac {x^2}2\ge 2\sin^2\frac x2\iff$ $\left(\frac x2\right)^2\ge \sin ^2\frac x2\iff$ $\left|\frac x2\right|\ge \left|\sin\frac x2\right|$ ceea ce este adevarat (vezi lema precedenta).

TEMA. Sa se demonstreze fara derivate urmatoarele doua inegalitati: $\boxed{\sin x<x<\tan x}\ (1)\ ,\ \forall\ x\in\ \left(0,\frac {\pi}2\right)\ ;\ \boxed{|\sin nx|\le n\cdot |\sin x|}\ (2)\ ,\ \forall\ n\in\mathbb N\ ,\ n\ge 2$ si $\forall\ x\in\mathbb R\ .$

Demonstratie.

$1\blacktriangleright$ Fie primul cadran $\overarc{AB}$ al cercului trigonometric $w=\mathbb C(O,R)$ unde $R=1\ ,$ $A(0)$ si $B\left(\frac{\pi}2\right)$ si un punct $X(x)\in \overarc{AB}\ .$ Notam proiectia $P$ a lui $X$ pe $OA$ si punctul $T\in OX\cap t$ unde $t$ este tangenta la $w$ in $A\ .$ Prima parte a inegalitatii $(1)$ este demonstrata prin lema de mai sus. Se stie ca $:$ lungimea $l$ a arcului $\overarc{AB}$ este $l=R\cdot x=x$ unde $x$ este masura unghiului $\widehat{AOX}$ exprimata in radiani $;$ aria sectorului $\overarc{AOB}$ este $\frac {R\cdot l}2=\frac x2\ ;$ $AT=\tan x$ si aria $[OAT]=\tan x\ .$ Se observa ca $[OAX]<[\overline O\overarc{AX}\overline O]<[OAT]$ $\iff$ $\frac {OA\cdot XP}2<\frac {Rl}2<\frac {AO\cdot AT}2\ ,$ adica $\sin x<x<\tan x\ ,\ \forall\ x\in\ \left(0,\frac {\pi}2\right)\ .$

$2\blacktriangleright$ Vom folosi inductia completa relativ la $n\ge 2\ .$ Pentru $n:=2$ obtinem $|\sin 2x|=|2\sin x\cos x|=2\cdot |\sin x|\cdot|\cos x|\le 2|\sin x|\implies$ $|\sin 2x|\le 2\cdot |\sin x|\ .$ Presupunem ca exista $n\ge 2$ astfel incat $|\sin nx|\le n\cdot |\sin x|\ (*)\ .$ Deci $|\sin (n+1)|=$ $|\sin nx\cdot \cos x+\sin x\cdot \cos nx|\le$ $|\sin nx|\cdot |\cos x|+|\sin x|\cdot |\cos nx|\ \stackrel{(*)}{\le}\ n\cdot |\sin x|\cdot 1+|\sin x|\cdot 1=$ $(n+1)\cdot |\sin x|\ ,$ adica $|\sin (n+1)|\le(n+1)\cdot |\sin x|\ .$ In concluzie, $|\sin nx|\le n\cdot |\sin x|\ ,\ \forall\ n\in\mathbb N\ ,\ n\ge 2$ si $\forall\ x\in\mathbb R\ .$

$\bf\color{black}Rezolvarea\ unei\ inecuatii\ bilaterale\ peste\ R.$ Lemma 1. $(\forall )\ \{a,b\}\subset\mathbb R$ there is the chain of equivalencies $\boxed{\ x\in [a,b]\cup [b,a]\iff (x-a)(x-b)\le 0\iff \left|x-\frac {a+b}2\right|\le \frac{|a-b|}2\iff |x-a|+|x-b|=|a-b|\ }$

Lemma 2. Let $f(x)=ax^2+bx+c$ with $\{a,b,c\}\subset \mathbb R$ , where $a\ne 0$ . Then $\left\{\begin{array}{cccccc} (\exists ) & \alpha\in\mathbb R & \mathrm{so\ that} & a\cdot f(\alpha )\le 0 & \implies & \Delta\ge 0\\\\ (\exists ) & \{\alpha ,\beta\}\subset\mathbb R & \mathrm{so\ that} & f(\alpha )\cdot f(\beta )\le 0 & \implies & \Delta\ge 0\end{array}\right\|$

Example 0. Prove that $(\forall )\ x\in\mathbb R\ ,\ x>-1$ exists the equivalence $|x-3|<x+1\iff x>1\ .$

Proof. $0\le |x-3|<x+1\iff$ $x+1>0$ and $|x-3|^2<(x+1)^2\iff$ $\boxed{x>-1}$ and $(x-3)^2<(x+1)^2\iff$ $(x-3)^2-(x+1)^2<0\iff$

$[(x-3)+(x+1)]\cdot [(x-3)-(x+1)]<0\iff$ $(2x-2)\cdot (-4)<0\iff$ $8(x-1)>0\iff$ $x>1$ and $x>-1\iff$ $x>1\ .$

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

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

Example 2. Prove that $(\forall )\ x\in\mathbb R\ ,\ |\sin x+\cos x|\le \sqrt 2\ \mathrm{and}\ (\forall )\ x\in\left[0,\frac {\pi}2\right]\ ,\ 1\le \sin x+\cos x\le \sqrt 2\ .$

Proof. $\left\{\begin{array}{ccccccccc} x\in\mathbb R & : & |\sin x+\cos x|\le \sqrt 2 & \iff & (\sin x+\cos x)^2\le 2 & \iff & 1+\sin 2x\le 2 & \iff & \sin 2x\le 1\\\\ 2x\in [0,\pi ] & : & 1\le \sin x+\cos x\le \sqrt 2 & \iff & 1\le (\sin x+\cos x)^2\le 2 & \iff & 1\le 1+\sin 2x\le 2 & \iff & 0\le \sin 2x\le 1\end{array}\right\|$ what are truly.

Example 3. Denote $f(x)=\frac {x^2-x+1}{x^2+x+1}$ , where $x\in\mathbb R$ . Prove that $\frac 13\le f(x) \le 3\ ,\ (\forall )\ x\in\mathbb R$ and $\mathbb Im(f)=\mathbb Pr_{Oy}\mathbb{G}_f=\left[\frac 13,1\right)\cup (1,3]$ . See here.

Example 4. $\{a,b,c,d\}\subset \mathbb R$ so that $a<b<c<d$ and $\{A,B,C\}\subset\mathbb R^*_+$ $\implies$ $\left\{\begin{array}{c} h(x)=A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b)=0\\\\ g(x)=(x-a)(x-c)+2(x-b)(x-d)=0\end{array}\right\|$ have real roots.

Proof. First equation has coefficient $A+B+C>0$ for $x^2.$ Thus, $\left\{\begin{array}{ccccc} h(a)=A(a-b)(a-c)>0\\\\ h(b)=B(b-c)(b-a)<0\\\\ h(c)=C(c-a)(c-b)>0\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} h(a)h(b)<0 & \implies & (\exists )\ x_1\in (a,b)\ \mathrm{so\ that}\ h\left(x_1\right)=0\\\\ h(b)h(c)<0 & \implies & (\exists )\ x_2\in (b,c)\ \mathrm{so\ that}\ h\left(x_2\right)=0\end{array}\right\|$

In conclusion, exists $\left\{x_1,x_2\right\}\subset\mathbb R$ so that $a<x_1<b<x_2<c$ and $h\left(x_1\right)=h\left(x_2\right)=0\ ,$ i.e. the equation $h(x)=0$ has two real roots.

Second equation has coefficient $3>0$ of $x^2.$ Thus, $\left\{\begin{array}{ccccc} g(a)=2(a-b)(a-d)>0\\\\ g(b)=(b-a)(b-c)<0\\\\ g(c)=2(c-b)(c-d)<0\\\\ g(d)=(d-a)(d-c)>0\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} g(a)g(b)<0 & \implies & (\exists )\ x_1\in (a,b)\ \mathrm{so\ that}\ g\left(x_1\right)=0\\\\ g(c)g(d)<0 & \implies & (\exists )\ x_2\in (c,d)\ \mathrm{so\ that}\ g\left(x_2\right)=0\end{array}\right\|$

In conclusion, exists $\left\{x_1,x_2\right\}\subset\mathbb R$ so that $a<x_1<b<c<x_2<d$ and $g\left(x_1\right)=g\left(x_2\right)=0\ ,$ i.e. the equation $g(x)=0$ has two real roots.

Remark. $h(x)=\sum A(x-b)(x-c)=$ $\sum A\left[x^2-x(b+c)+bc\right]=$ $x^2\cdot \sum A-x\cdot\sum A(b+c)+\sum Abc$ . Thus,

$\Delta =\left[\sum A(b+c)\right]^2-4\cdot\sum A\cdot\sum Abc=$ $\sum A^2(b+c)^2+2\sum BC(a+b)(a+c)-4\cdot\sum A^2bc-4\cdot\sum BCa(b+c)=$

$\sum A^2\left[(b+c)^2-4bc\right]+2\cdot\sum BC\left[(a+b)(a+c)-2a(b+c)\right]=$ $\sum A^2(b-c)^2+$ $2\cdot \sum BC(a-b)(a-c)=$

$\left(B|a-c|+C|a-b|+A|b-c|\right)^2-4AC|(b-a)(b-c)|\ge$ $(A|b-c|+C|b-a|)^2-4AC|(b-a)(b-c)|=(A|b-c|-C|b-a|)^2\ge 0$

P1 (clasa a VI - a). Fie paralelogramul $ABCD$ si $\left\{\begin{array}{c} G\in (AB)\ ;\ F\in (CD)\ :\ GF\parallel AD\\\\ E\in (BC)\ ;\ I\in (DA)\ :\ EI\parallel AB\end{array}\right\|\ ,\ P\in GF\cap EI\ .$ Sa se arate ca $[BCFG]+[PFDI]=2\cdot [AEF]\ .$

Demonstratie. Vom folosi o proprietate cunoscuta care conserva aria unui triunghi $[ABC]\ :\ \boxed{AX\parallel BC \iff [ABC]=[XBC]}\ (*)\ .$ Prin urmare,

$\left\{\begin{array}{ccccccc} AG\parallel PE & \iff & [APE]=[GPE] & \iff & 2\cdot [APE]=2\cdot [GPE]=[GBEP] & \iff & 2\cdot [APE]=[GBEP]\\\\ EC\parallel PF & \iff & [EPF]=[CPF] & \iff & 2\cdot [EPF]=2\cdot [CPF]=[ECFP] & \iff & 2\cdot [EPF]=[ECFP]\\\\ AD\parallel PF & \iff & [APF]=[DPF] & \iff & 2\cdot [APF]=2\cdot [DPF]=[PFDI] & \iff & 2\cdot [APF]=[PFDI]\end{array}\right\|\ \bigoplus\implies$

$2\cdot [AEF]=2\cdot\left([APE]+[EPF]+[APF]\right)=$ $\left([GBEP]+[ECFP]\right)+[PFDI]=$ $[BCFG]+[PFDI]$ $\implies$ $2\cdot [AEF]=[BCFG]+[PFDI]\ .$

P2. Prove that $(\forall )\ ABC$ with the circumcircle $w=C(O,R)$ and the Nagel point $N$ there is the relation $\boxed{\ ON=R-2r\ }$ .

Proof (metric). Let $\left\{\begin{array}{cc} D\in AN\cap BC\\\\ E\in BN\cap CA\\\\ F\in CN\cap AB\end{array}\right\|$ . Is well-known that $\left\{\begin{array}{c} BF=CE=s-a\\\\ CD=AF=s-b\\\\ AE=BD=s-c\end{array}\right\|$ , $\boxed{\frac {NA}a=\frac {ND}{s-a}=\frac {AD}{s}}$ and identity $\boxed{\sum a^2(s-b)(s-c)=4s^2r(R-r)}\ (*)$ .

I denoted $\{A,L\}=\{A,N\}\cap w$ . Apply the Stewart's relation to the cevian $AD$ in $\triangle ABC\ :\ \boxed{a\cdot AD^2+a(s-b)(s-c)=c^2(s-b)+b^2(s-c)}\ (1)$ .

Since $\boxed{R^2-ON^2=NA\cdot NL}\ (2)$ and $DB\cdot DC=DA\cdot DL$ obtain that $NA\cdot NL=\frac as\cdot AD\cdot(ND+DL)=$ $\frac as\cdot AD\cdot \frac {s-a}{s}\cdot AD+\frac as\cdot DB\cdot DC=$

$\frac {s-a}{s^2}\cdot a\cdot AD^2+\frac {a(s-b)(s-c)}{s}$ . Therefore, $NA\cdot NL\stackrel{(1)}{=}$ $\frac {s-a}{s^2}\cdot \left[c^2(s-b)+b^2(s-c)-a(s-b)(s-c)\right]+$ $\frac {a(s-b)(s-c)}{s}=$ $\frac {1}{s^2}\cdot\sum a^2(s-b)(s-c)\stackrel{(*)}{=}$

$4r(R-r)$ . In conclusion, $NA\cdot NL=4r(R-r)$ and from the relation $(2)$ obtain that $ON^2=R^2-NA\cdot NL=$ $R^2-4r(R-r)=(R-2r)^2\implies$ $ON=R-2r$ .

$(*)$ Remark. $\sum a^2(s-b)(s-c)=\sum a^2\left[bc-s(s-a)\right]=$ $abc\sum a-s\cdot\sum a^2(s-a)=$ $8Rs^2r-s^2\cdot \sum a^2+$ $s\cdot\sum a^3=$

$8Rs^2r-s^2\cdot 2\left(s^2-r^2-4Rr\right)+s\left[8s^3-6s\left(s^2+r^2+4Rr\right)+12Rsr\right]=$ $4Rs^2r-4s^2r^2\implies$ $\boxed{\sum a^2(s-b)(s-c)=4s^2r(R-r)}$ .

P3. Find $m\in\mathbb R$ so that the equations $\boxed{\ \begin{array}{c} x^3-mx^2-4=0\\\\ x^3+mx+2=0\end{array}\ }$ have at least a common root.

Proof. Observe that $m\ne 0$ . Apply the method of the elimination of the variable x between given equations until obtain two at most second degree equations and

put the condition so these equations have at least a common root $:\ \left|\begin{array}{ccc} x^3-mx^2-4 & \odot & -1\\\\ x^3+mx+2 & \odot & 1\end{array}\right|\ \oplus\implies$ $\left|\begin{array}{ccc} \underline{mx^2+mx+6} & \odot & x\\\\ x^3+mx+2 & \odot & -m\end{array}\right|\ \oplus\implies$

$\left|\begin{array}{ccc} \underline{mx^2+\left(6-m^2\right)x-2m}\\\\ mx^2+mx+6\end{array}\right|\ .$ The last two equations have at least a common root $\iff$ $\left|\begin{array}{cc} m & -2m\\\\ m & 6\end{array}\right|^2=$ $\left|\begin{array}{cc} m & 6-m^2\\\\ m & m\end{array}\right|\cdot$ $\left|\begin{array}{cc} 6-m^2 & -2m\\\\ m & 6\end{array}\right|\iff$

$m(m+3)^2\left(m^2-4m+6\right)=0$ $\implies$ $m=-3$ and in this case $\left\{\begin{array}{ccc} x^3+3x^2-4=0 & \implies & x\in\{+1,-2,-2\}\\\\ x^3-3x+2=0 & \implies & x\in\{+1,+1,-2\}\end{array}\right\|\ .$

P4. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and the centroid $G$ . Prove that $IG\perp BC\iff b+c=3a$ or $b=c.$

Proof 1 (metric). Let $D\in BC\cap w.$ Thus, $IG\perp BC\iff$ $DG\perp BC\iff$ $DB^2-DC^2=GB^2-GC^2$ $\iff$ $(s-b)^2-(s-c)^2=$

$\frac 49\cdot\left(m_b^2-m_c^2\right)$ $\iff$ $9a(c-b)=$ $\left[2\left(a^2+c^2\right)-b^2\right]-$ $\left[2\left(a^2+b^2\right)-c^2\right]$ $\iff$ $9a(c-b)=3\left(c^2-b^2\right)\iff$ $b=c\ \vee\ b+c=3a.$

Proof 2 (synthetic). Suppose w.l.o.g. $b\ne c.$ Let the midpoint $M$ of $[BC]$ and the projections $K$ , $D$ , $S$ of the points $A$ , $I$ , $G$ on $BC.$ Thus, $IG\perp BC\iff$

$\mathrm{pr}_{BC}(I)=\mathrm{pr}_{BC}(G)\iff$ $D\equiv S\iff$ $DM=SM=\frac 13\cdot KM\ \stackrel{(*)}{\iff}\ \frac {|b-c|}2=\frac 13\cdot \frac {|b^2-c^2|}{2a}\iff$ $b+c=3a,$ i.e. $\boxed{IG\perp BC\iff b+c=3a}\ .$

Remark. Suppose w.l.o.g. $K\in (BM).$ Therefore, $AK\perp BC\iff$ $AB^2-AC^2=KB^2-KC^2\iff$ $c^2-b^2=(KB+KC)(KB-KC)\iff$

$c^2-b^2=a\cdot \left[(MB-MK)-(MC+MK)\right]\iff$ $b^2-c^2=2a\cdot MK\iff$ $\boxed{MK=\frac {b^2-c^2}{2a}}\ (*).$

Proof 3 (synthetic). Keep the notations of the upper proof. Let $E\in CA\cap w,$ $F\in AB\cap w,$ $R\in MI\cap AK$ and $L\in AM\cap EF.$ Suppose w.l.o.g. $b>c.$ Thus:

$\blacktriangleright\ ID\parallel AK\iff$ $\frac {MD}{MK}=\frac {ID}{KR}\iff$ $\frac {\frac {b-c}2}{\frac {b^2-c^2}{2a}}=$ $\frac {r}{KR}\iff$ $\frac a{b+c}=\frac {r}{KR}\iff$ $KR=\frac {r(b+c)}a\iff$ $AR=AK-KR=$ $h_a-\frac {r(b+c)}a=$

$\frac {ah_a-r(b+c)}a=$ $\frac {2sr-r(b+c)}a=$ $\frac ra\cdot [2s-(b+c)]=r\implies$ $\boxed{\ AR\ =\ r\ }\ (1).$

$\blacktriangleright$ Apply the well-known Cristea's relation $:\ \frac {LM}{LA}\cdot BC=\frac{FB}{FA}\cdot MC+\frac {EC}{EA}\cdot MB\iff$ $\frac {LM}{LA}\cdot a=\frac {s-b}{s-a}\cdot \frac a2+\frac {s-c}{s-a}\cdot\frac a2\iff$

$\frac {LM}{LA}=\frac {(s-b)+(s-c)}{2(s-a)}=$ $\frac{2s-(b+c)}{2(s-a)}=\frac a{b+c-a}\implies$ $\boxed{\frac {ML}{MA}=\frac a{b+c}}\ .$ Since $\frac {MD}{MK}=\frac a{b+c}$ obtain that $\frac {ML}{MA}=\frac {MD}{MK}\ ,$

i.e. $LD\parallel AK\iff \boxed{\ L\in ID\ }\ .$ In conclusion, $IG\perp BC\iff L\equiv G\iff$ $\frac 13=\frac {MG}{MA}=\frac a{b+c}\iff$ $b+c=3a.$

P5. Prove that $(\forall )$ an acute $\triangle ABC$ there are the inequalities $\left\{\begin{array}{cccc} \sum\frac {b+c}{\cos A} & \ge & 4(a+b+c) & (1)\\\\ \sum\frac {a}{\cos B+\cos C} & \ge & a+b+c & (2)\end{array}\right\|\ .$

Proof. $\begin{array}{ccccccc} \nearrow & \sum\frac {b+c}{\cos A}=\sum\frac {(b+c)^2}{(b+c)\cos A} & \stackrel{\mathrm{C.B.S}}{\ge} & \frac {\left[\sum(b+c)\right]^2}{\sum (b+c)\cos A}=\frac {(4s)^2}{\sum(b\cos C+c\cos B)}=\frac {16s^2}{\sum a}=8s=4(a+b+c) & \implies & \sum\frac {b+c}{\cos A}\ge 4(a+b+c) & \searrow\\\\ \searrow & \sum\frac a{\cos B+\cos C}=\sum\frac {a^2}{a(\cos B+\cos C)} & \stackrel{C.B.S}{\ge} & \frac {\left(\sum a\right)^2}{\sum [a(\cos B+\cos C)]}=\frac {4s^2}{\sum (b\cos C+c\cos B)}=\frac {4s^2}{\sum a}=2s=a+b+c & \implies & \sum\frac a{\cos B+\cos C}\ge a+b+c & \nearrow\end{array}\odot$

Remark. $(\forall )\ \triangle ABC$ there are the identities $\left\{\begin{array}{ccccccccc} HO^2=9\cdot OG^2 & = & 9R^2-\left(a^2+b^2+c^2\right) & ; & OI^2 & = & R^2-2Rr\\\\ IN^2=9\cdot IG^2 & = & s^2+5r^2-16Rr & ; & OI_a^2 & = & R^2+2Rr_a\\\\ IH^2 & = & 4R^2+4Rr+3r^2-s^2 & ; & ON & = & R-2r\end{array}\right\|$ (standard notations).

P6 (Adil Abdullayev). Let the circumcircle $w=\mathbb C(O,R)$ of $\triangle ABC$ and the circle $w_a=\mathbb C\left(O_a,\rho\right)$ so that $O_a\in (OA)$ and $w_a$ is tangent to $BC$ at $D.$ Prove that $\frac R{R-\rho}\ge \frac {4bc}{a^2}.$

Proof. Let $I$ be the incircle of $\triangle ABC.$ Prove easily that $D\in AI,$ $O_aD=O_aA=\rho$ and $O_aD\parallel OS,$ where $AI\cap w=\{A,S\}.$ Observe that $\triangle ABS\sim\triangle ADC,$ i.e. $\frac {AB}{AD}=\frac {AS}{AC}\iff$

$\boxed{AD\cdot AS=AB\cdot AC}\ (1).$ In conclusion, $\frac R{R-\rho}=$ $\frac {OA}{OA-O_aA}=$ $\frac {OA}{O_aO}=$ $\frac {SA}{SD}=$ $\frac {AS\cdot AD}{SD\cdot AD}\ \stackrel{(1)}{=}\ \frac {AB\cdot AC}{DB\cdot DC}=$ $\frac {\cancel b\cancel c}{\frac {\cancel ca}{b+c}\cdot\frac {\cancel ba}{b+c}}=$ $\frac {(b+c)^2}{a^2}\ge\frac {4bc}{a^2}$ $\implies$ $\frac R{R-\rho}\ge \frac {4bc}{a^2}.$ Nice and easy problem!

P7. Let $\triangle ABC$ with midpoint $M$ of $[BC],$ altitude $AD$ where $D\in (BC)$ and incircle $\mathbb C(I,r)$ where $E\in BI\cap AC,$ $F\in CI\cap AB.$ Prove that $MI\cap EF\cap AD\ne \emptyset\iff A=90^{\circ}$ (Zaslavsky).

Proof. Let $R\in MI\cap AD,$ $S\in EF\cap AD,$ $J\in AI\cap BC.$ Suppose w.l.o.g. $b>c.$ Observe that $\frac {JB}c=\frac {JC}b=\frac a{b+c},$ $\frac {IA}{IJ}=\frac {BA}{BJ}=\frac c{\frac {ac}{b+c}}=\frac {b+c}a,$ i.e. $\frac {IA}{IJ}=\frac {b+c}a,$ $MB=MC=\frac a2$

and $MJ=MB-JB=$ $\frac a2-\frac {ac}{b+c}=$ $\frac {a(b-c)}{2(b+c)} \implies \boxed{MJ=\frac {a(b-c)}{2(b+c)}}\ (1).$ From $AD\perp BC\iff AC^2-AB^2=DC^2-DB^2$ obtain that $b^2-c^2=(DC+DB)(DC-DB)=$

$a\left[(MD+MC)-(MB-MD)\right]=$ $2a\cdot MD,$ i.e. $\boxed{MD=\frac {b^2-c^2}{2a}}\ (2).$ Thus, $\boxed{\frac {MJ}{MD}\ \stackrel{1\wedge 2}{=}\ \left(\frac a{b+c}\right)^2}\ (*)\ .$ Apply the Menelaus's theorem to $\overline{MIR}/\triangle ADJ\ :\ \frac {MJ}{MD}\cdot\frac {RD}{RA}\cdot\frac {IA}{IJ}=1\ \stackrel{*}{\iff}$

$\frac {a^2}{(b+c)^2}\cdot \frac {RD}{RA}\cdot \frac {b+c}a=1$ $\iff$ $\frac {RA}{RD}=\frac a{b+c}$ $\iff$ $\frac {RA}a=\frac {RD}{b+c}=$ $\frac {h_a}{2s}=\frac {ah_a}{2as}=$ $\frac {2rs}{2as}=\frac ra,$ i.e. $\boxed{RA=r}\ (3).$ Apply Cristea's identity $\boxed{\frac {SD}{SA}\cdot BC=\frac {FB}{FA}\cdot DC+\frac {EC}{EA}\cdot DB}$

i.e. $\frac {SD}{SA}\cdot a=\frac ab\cdot DC+\frac ac\cdot DB\iff$ $\frac {SD}{SA}=\frac {DC}b+\frac {DB}c=\cos C+\cos B\implies$ $\frac {h_a}{SA}=1+\cos B+\cos C\iff$ $\boxed{SA=\frac {h_a}{1+\cos B+\cos C}}\ (4).$ Hence $MI\cap EF\cap AD\ne \emptyset\iff$

$S\equiv R\iff$ $SA=RA\iff$ $h_a=r(1+\cos B+\cos C)\iff$ $2sr=ar(1+\cos B+\cos C)\iff$ $b+c=a(\cos B+\cos C)\ \stackrel{5}{\iff}\ (\underline{\underline{a\cos C}}+c\cos A)+$ $(\underline{a\cos B}+b\cos A)=$

$a(\underline{\cos B}+\underline{\underline{\cos C}})$ $\iff$ $(b+c)\cos A=0$ $\iff$ $A=90^{\circ}\ .$ In conclusion, $MI\cap EF\cap AD\ne \emptyset\iff A=90^{\circ}\ .$ I used and the identities $\boxed{a=b\cdot\cos C+c\cdot\cos B}\ (5)$ a.s.o.

An easy extension (own). For $\triangle ABC$ denote $:$ the midpoint $M$ of the side $[BC]\ ;$ the point $D\in (BC)$ so that $AD\perp BC\ ;$ the incenter $I$ for which denote

$J\in AI\cap BC\ ;$ a point $X\in (AJ)$ for which denote $E\in BX\cap AC\ ,$ $F\in CX\cap AB$ and $S\in EF\cap AD\ .$ Prove that $X\in MS\iff A=90^{\circ}\ .$

Proof. Denote the ratios $\frac {XA}{XJ}=k,$ $\frac {EA}{EC}=u$ and $\frac {FA}{FB}=v.$ From the Aubel's relation obtain $u+v=k$ and from the Menelaus' relation obtain that $\frac {JB}{JC}\cdot\frac {EC}{EA}\cdot \frac {FA}{FB}=1$ $\iff$ $\frac cb\cdot \frac vu=1\iff$

$\boxed{\frac uc=\frac vb=\frac k{b+c}}\ (1)\ .$ Apply the Cristea's relation for $S\in EF\cap AD\ :\ \frac {SD}{SA}\cdot BC=\frac {FB}{FA}\cdot DC+\frac {EC}{EA}\cdot DB,$ i.e. $\frac {SD}{SA}\cdot a=\frac {b\cdot\cos C}v\cdot DC+\frac {c\cdot\cos C}u=\frac {b+c}k\cdot\left(\cos B+\cos C\right)\iff$

$\boxed{\frac {SD}{SA}=\frac {(b+c)\left(\cos B+\cos C\right)}{ka}}\ (2)\ .$ Menelaus' theorem $:\ X\in MS\iff$ $\frac {MJ}{MD}\cdot \frac {SD}{SA}\cdot\frac {XA}{XJ}=1$ $\iff$ $\frac {a^{\cancel 2}}{(b+c)^{\cancel 2}}\cdot \frac {(\cancel{b+c})\left(\cos B+\cos C\right)}{\cancel k\cancel a}\cdot \cancel k=1\iff$ $\boxed{a(\cos B+\cos C)=b+c}\ (3).$

Apply the identities $\boxed{a=b\cdot\cos C+c\cdot\cos B}$ to the relation $(3)\ :\ a\cdot\cos B+a\cdot \cos C=b+c\iff$ $(\cancel c-b\cdot \cos A)+(\cancel b-c\cdot \cos A)=\cancel{b+c}\iff\cos A=0\iff A=90^{\circ}\ .$

An difficult extension (own). For an interior point $X$ of $\triangle ABC$ denote $J\in AX\cap BC,$ $E\in BX\cap AC$ and $F\in CX\cap AB.$ Prove that for any points $D\in (JB)$ and $M\in (JC)$

there is the following equivalence $\boxed{X\in MS\iff \frac {DB}{DC}\cdot\frac {MB}{MC}=\left(\frac {JB}{JC}\right)^2}\ ,$ where $S\in EF\cap AD.$ See the extension of the proposed problem P1 from here.

Remark. The previous extension is a particular case of the upper estension. Indeed, $\frac {DB}{DC}=\frac {c\cdot\cos B}{b\cdot\cos C}\ ,$ $MB=MC$ and $\frac {JB}{JC}=\frac cb.$ In conclusion, $X\in MS\iff$

$\frac {c\cdot\cos B}{b\cdot\cos C}=\left(\frac cb\right)^2\iff$ $b\cdot\cos B=c\cdot\cos C\iff$ $\sin 2B=\sin 2C\iff$ $B=C\ \vee\ 2B+2C=180^{\circ}\iff$ $B=C\ \vee\ B+C=90^{\circ}\iff b=c\ \vee\ A=90^{\circ}\ .$

P8 (M.O. Sanchez). Let $:\ \triangle ABC$ with $AB\perp AC\ ;$ $D\in BC\ ,$ $AD\perp BC\ ;$ the midpoint $M$ of $[AB]\ ;$ $E\in AD\cap CM\ ;$ $m\left(\widehat{ACM}\right)=2\cdot m\left(\widehat{BCM}\right).$ Find the ratio $\frac {DB}{DC}\ .$

Proof 1 (trigonometric). Denote $m\left(\widehat{BCM}\right)=x\implies$ $m\left(\widehat{ACM}\right)=2x$ and $C=3x\ .$ Thus, $MA=MB\iff$ $b\cdot\sin 2x=a\sin x\iff$ $\frac ba\cdot 2\sin x\cos x=\sin x\iff$

$\boxed{2\cos 3x\cos x=1}\ (1)\iff$ $\cos 4x+\cos 2x=1$ $\iff$ $2\cos ^22x+\cos 2x-2=0\iff$ $\boxed{\cos 2x=\frac {-1+\sqrt {17}}4}\ (2)\ .$ Hence $\frac {DB}{DC}=\frac {a\cdot DB}{a\cdot DC}=$ $\left(\frac cb\right)^2=$ $\tan^2C=$

$\tan^23x=$ $\frac 1{\cos^23x}-1\ \stackrel{(1)}{=}\ 4\cos^2x-1=$ $2(1+\cos 2x)-1=$ $1+2\cos 2x\ \stackrel{(2)}{=}\ 1+2\cdot \frac {-1+\sqrt {17}}4=$ $1+\frac {-1+\sqrt {17}}2=$ $\frac {1+\sqrt {17}}2\implies$ $\boxed{\frac {DB}{DC}=\frac {1+\sqrt {17}}2}\ ;\$

Proof 2 (synthetic).

P9. Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$ . Prove that the area of one of the triangles $AOH$ , $BOH$ and $COH$ is equal to the sum of the areas of the other two.

Proof. Suppose w.l.o.g. $a\le b\le c\ ,$ i.e. $A\le B\le C\ .$ Thus, $m\left(\widehat{OAB}\right)=m\left(\widehat{HAC}\right)=90^{\circ}-C$ and $m\left(\widehat{OAH}\right)=A-m\left(\widehat{OAB}\right)-m\left(\widehat{HAC}\right)=A-2\cdot \left(90^{\circ}-C\right)=C-B\implies$

$\boxed{m\left(\widehat{OAH}\right)=C-B}\ (*)\ .$ Therefore, $[AOH]=\frac 12\cdot AO\cdot AH\cdot \sin \widehat {OAH}=$ $\frac 12\cdot R\cdot 2R\cos A\cdot \sin(C-B)=$ $R^2 \cdot\cos(B+C)\cdot\sin (B-C)=$ $\frac 12\cdot R^2\cdot (\sin 2B-\sin 2C)$ a.s.o.

In conclusion, $a\le b\le c\implies$ $A\le B\le C<90^{\circ}\implies$ $180^{\circ}-2A\ge 180^{\circ}-2B\ge 180^{\circ}-2C>0\implies$ $\frac {R^2}2=\frac {[AOH]}{\sin 2B-\sin 2C}=$ $\frac {[BOH]}{\sin 2A-\sin 2C}=$ $\frac {[COH]}{\sin 2A-\sin 2B}=$

$\frac {[AOH]+[COH]}{\sin 2A-\sin 2C}$ $\implies$ $[BOH]=[AOH]+[COH]\ .$

Remark 1. $16S[AOH]=\left(b^2+c^2-a^2\right)\cdot \left|b^2-c^2\right|$ a.s.o. So $a<b<c\implies$ $\left\{\begin{array}{ccc} \left(b^2+c^2-a^2\right)\left(c^2-b^2\right)=16S[AOH]\\\\ \left(c^2+a^2-b^2\right)\left(c^2-a^2\right)=16S[BOH]\\\\ \left(a^2+b^2-c^2\right)\left(b^2-a^2\right)=16S[COH]\end{array}\right\|\implies$ $\frac {[AOH]+[COH]-[BOH]}{16S}=$ $\left(b^2+c^2-a^2\right)\left(c^2-b^2\right)+$

$\left(a^2+b^2-c^2\right)\left(b^2-a^2\right)- \left(c^2+a^2-b^2\right)\left(c^2-a^2\right)=$ $\left(c^4-b^4\right)+a^2\left(b^2-c^2\right)+\left(b^4-a^4\right)+c^2\left(a^2-b^2\right)-\left(c^4-a^4\right)+b^2\left(c^2-a^2\right)=0\implies$ $[BOH]=[AOH]+[COH]\ .$

Remark 2. If $a<b<c\ ,$ then $\left\{\begin{array}{cc} \bullet & [AOH]=[AOC]-[AOB]\\\\ \bullet & [BOH]=[BOC]-[AOB]\\\\ \bullet & [COH]=[BOC]-[AOC]\end{array}\right\|\implies [AOH]+[COH]=$ $(\cancel{[AOC]}-[AOB])+([BOC]-\cancel{[AOC]})=$ $[BOC]-[AOB]=[BOH]\ .$

P10 (Cosmin Pohoata). Let $a$ be positive real number such that $a^{3}=6(a+1)$ . Prove that the equation $x^{2}+ax+a^{2}-6=0$ has no real solutions.

Proof I (generally). Show easily that $\left \{\begin{array}{c}f\ : \ (0,\infty )\rightarrow\mathcal R\\\\ f(x)=x^{2}(x+1)\end{array}\right\|$ is strict $(\uparrow )$ . Thus $f\left(\frac{1}{a}\right)=\frac{1}{6}$ , i.e. $a^{3}=6(a+1)$ has at most one positive root. Observe that for $\left\{\begin{array}{c}g\ : \ (0,\infty )\rightarrow \mathcal R\\\\ g(a)=a^{3}-6(a+1)\end{array}\right\|$ there is the

relation $g(2\sqrt 2)=2(2\sqrt 2-3)<0<3=g(3)$ , i.e. $g(2\sqrt 2)<0<g(3)$ . In conclusion, the single root $a\in (2\sqrt 2, 3)$ . The equation $x^{2}+ax+a^{2}-6=0$ has no real roots iff $a^{2}> 8$ , what is truly.

Proof II. $f(a)=a^{3}-6(a+1)$ $\implies$ $f(0)\cdot f(3)<0$ $\implies$ $(\exists)\ a>0$ so that $a^{3}=6(a+1)$ . But $0<a\le 2\sqrt 2$ $\implies$ $a^{2}\le 8\implies a^{3}\le 8a$ $\implies 6(a+1)\le 8a\implies 3\le a$ , what is falsely.

Thus $a>2\sqrt 2$ and $a\ge 3\implies$ $a^{3}\ge 3a^{2}\implies$ $6(a+1)\ge 3a^{2}\implies$ $3a\le a^{2}\le 2(a+1)\implies$ $a\le 2$ , what is falsely. Hence $a<3$ . Thus, $a\in \left(2\sqrt 2,3\right)$ and $x^{2}+ax+a^{2}-6=0$ has no real

roots iff $a^{2}> 8$ , what is truly. Hencr $a^{3}-8=6a+6-8=2(3a-1)\implies \left(a-2\sqrt 2\right)(3a-1)>0$ $\implies a\in \left(0,\frac{1}{3}\right)\cup \left(2\sqrt 2,\infty \right)$ . If $a\in \left(0,\frac{1}{3}\right)$ then $6<6(a+1)=a^{3}<\frac{1}{27}$ $\implies$

$6<\frac{1}{27}$ , what is falsely. In conclusion, $a>2\sqrt 2$ a.s.o.

P11. Let $\triangle ABC$ with $b=c$ . The bisectors of $\widehat{CAB},$ $\widehat{ABC}$ meet $BC,$ $CA$ at $D,$ $E$ respectively. Let $K$ be the incentre of $\triangle ADC$ and $m\left(\widehat{BEK}\right)=45^{\circ}.$ Find all possible values of $m\left(\widehat{CAB}\right).$

Proof. Let $\boxed {\ A = 4x\ }.$ Thus, $\left\|\begin{array}{cc} \triangle IDC\ : & \frac {KI}{KC} = \frac {DI}{DC} = \tan \widehat {DCI} = \tan \left(45^{\circ} - x\right) \\ \\ \triangle IEC\ : & \frac {KI}{KC} = \frac {\sin\widehat {ECI}}{\sin\widehat {EIC}}\cdot \frac {\sin\widehat {KEI}}{\sin\widehat {KEC}} = \frac {\sin\left(45^{\circ} - x\right)}{\sin \left(90^{\circ} - 2x\right)}\cdot\frac {\sin 45^{\circ}}{\sin 3x}\end{array}\right\|\ \implies$ $\cos\left(45^{\circ} - x\right)\sin 45^{\circ} =$ $\sin 3x\cos 2x\ \Longleftrightarrow\ \cos x + \sin x =$

$2\sin 3x\cos 2x = \sin 5x + \sin x\ \Longleftrightarrow$ $\cos x = \cos\left(90^{\circ} - 5x\right)\ \Longleftrightarrow\ \left(\ x = 90^{\circ} - 5x\ \right)\ \ \vee\ \ \left(\ x + 90^{\circ} - 5x = 0\ \right)$ $\stackrel{(A=4x)}{\ \Longleftrightarrow\ }\boxed {\ A\ \in\ \{\ 60^{\circ}\ ,\ 90^{\circ}\ \}\ }$ .

An easy extension. Let $\triangle ABC$ with the incentre $I$ . Denote $D\in AI\cap BC$ , $E\in BI\cap AC$ and the incentre $K$ of $\triangle ADC$ . Prove that $\widehat {B E K}\equiv\widehat {ADK}\ \implies\ A\ \in\ \{B\ ,\ 2C\ \}$ .

Proof. $\left\|\begin{array}{cc} \triangle IDC\ : & \frac {KI}{KC} = \frac {DI}{DC} = \frac {\sin \widehat {DCI}}{\sin\widehat {DIC}}=\frac {\sin \frac C2}{\sin\left(90^{\circ}-\frac B2\right)}\\ \\ \triangle IEC\ : &{ \frac {KI}{KC} = \frac {\sin\widehat {ECI}}{\sin\widehat {EIC}}\cdot \frac {\sin\widehat {KEI}}{\sin\widehat {KEC}} = \frac {\sin\frac C2}{\sin\frac {B+C}{2}}\cdot\frac {\sin \frac {A+2B}{4}}{\sin \frac {3A}{4}}}\end{array}\right\|\ \implies$ $\cos\frac B2\sin\left(\frac A4+\frac B2\right)=\cos \frac A2\sin\frac {3A}{4}\ \Longleftrightarrow\ \cos\left(\frac A4+B\right)+\sin\frac A4=$

$\sin\frac {5A}{4}+\sin\frac A4 \Longleftrightarrow$ $\cos\left(\frac A4+B\right)=$ $\sin\frac {5A}{4}$ $\Longleftrightarrow$ $\left(\frac A4+B=\frac {5A}{4}\right)\ \ \vee\ \ \left(\frac A4+B+\frac {5A}{4}=180^{\circ}\right)$ $\Longleftrightarrow$ $\boxed {\ A\ \in\ \left\{\ B\ ,\ 2C\ \right\}\ }$ .

$\boxed{\begin{array}{c} \underline{\mathrm{Proposed\ problem}}.\ \mathrm{Prove\ that}\ \forall\ \{a,b,c\}\subset\mathbb R^*_+\ \mathrm{and}\ a+b+c=1\ \implies\ \frac{a^2+a}{a+bc}+\frac{b^2+b}{b+ca}+\frac{c^2+c}{c+ab}=3\\\\ \underline{\mathrm{Proof}}.\ \mathrm{I"ll\ use\ the\ following\ identities\ :}\ a+bc=a(a+b+c)+bc=(a+b)(a+c)\ (1)\ \mathrm{and}\ \sum a^2(b+c) = \sum bc(b+c)\ (2).\ \mathrm{Hence:}\\\\ \sum \left(\frac {a^2+a}{a+bc}-1\right)\ \stackrel {(1)}{=}\ \sum \frac {a^2-bc}{(a+b)(a+c)}=\frac 1{\prod (b+c)}\cdot \sum \left(a^2-bc\right)(b+c)=\frac 1{\prod (b+c)}\cdot \left[\sum a^2(b+c)-\sum bc(b+c)\right]\ \stackrel{(2)}{=}\ 0\ \end{array}}$ .

$\boxed{\begin{array}{ccc} \mathrm{E.N.} & \mathrm{A-right}\ \triangle ABC\ ,\ \left|\begin{array}{ccc} D\in BC & \mathrm{so\ that} & AD\perp BC\\\\ E\in AB & \mathrm{so\ that} & DE\perp AB\end{array}\right|\ \mathrm{and}\ AD=h\ .\ \mathrm{Then\ there\ is\ the\ relation}\ h^2=b\cdot DE\ .\\\\ \mathrm{Proof.} & \mathrm{Indeed,}\ \triangle ADE\sim \triangle CAD\ \iff\ \frac {AD}{CA}=\frac {DE}{AD}\ \iff\ \frac hb=\frac {DE}h\ \iff\ h^2=b\cdot DE\ \ \mathrm{(sesiunea\ 2017)\ .}\end{array}}$

P12. Let $\triangle$ $ABC$ with the $A$ -excircle $w_a=\mathbb C\left(I_a,r_a\right)$ what touches $AB$ at $T$ and the midpoints $M\ ,\ S$ of $[AB]\ ,\ [CT]$ respectively. Prove that $I_a\in MS\ .$

Proof. Denote $P\in MS\cap BC,$ $D\in AI\cap BC$ and $F\in CI\cap AB,$ where is well-known that $\boxed{\frac {IC}{IF}=\frac {a+b}c}\ (*).$ Apply the Menelaus' theorem to the transversal $\overline{MPS}/\triangle BCT :$

$\frac {MB}{MT}\cdot \frac {\cancel{ST}}{\cancel{SC}}\cdot \frac {PC}{PB}=1\iff$ $\frac {PC}{PB}=\frac {MT}{MB}=\frac{s-\frac c2}{\frac c2}=\frac {a+b}c\ \stackrel{(*)}{=}\ \frac {IC}{IF}\implies PI\parallel AB$ and $\frac {PB}c=\frac {PC}{a+b}=\frac a{a+b+c}\ .$ Observe that $PI\parallel AB\implies\frac {PB}{PD}\ \stackrel {(*)}{=}\ \frac {IA}{ID}=\frac {b+c}a\ .$

Let $\triangle ABD$ and $M\in AB\ ,\ P\in BD$ and $I_a\in DA.$ Observe that $\frac {I_aD}{I_aA}\cdot\frac {\cancel{MA}}{\cancel{MB}}\cdot\frac {PB}{PD}=$ $\frac {r_a}{h_a+r_a}\cdot\frac {b+c}a=$ $\frac {r_a(b+c)}{ah_a+ar_a}=\frac {\cancel{r_a}(b+c)}{2\cancel{r_a}(s-a)+a\cancel{r_a}}=\frac {b+c}{2(s-a)+a}=1\implies I_a\in MP\equiv MS\ .$

Remark. $(A,D;I,I_a)$ is an harmonical division $\iff$ $\boxed{\frac {AI}{AI_a}=\frac {DI}{DI_a}\iff \frac {s-a}{s}=\frac r{r_a}}\iff$ $\boxed{\frac {IA}{ID}=\frac {I_aA}{I_aD}\iff\frac {b+c}a=\frac {r_a+h_a}{r_a}}\iff$ $\boxed{\widehat{IBA} \equiv \widehat{IBD}\ \mathrm{and}\ BI_a \perp BI}\ .$

P13. What is greater between $a=\sqrt 3^{{\sqrt 5}^{\sqrt 5}}$ and $b=\sqrt 5^{{\sqrt 3}^{\sqrt 3}}\ ?$ Answer : $a\ >\ b\ .$

Proof. In conclusion, $a=\sqrt 3^{{\sqrt 5}^{\sqrt 5}}\ \ \wedge\ \ b=\sqrt 5^{{\sqrt 3}^{\sqrt 3}}\ \implies\ a\ >\ b\ \mathrm{(with\ derivatives).}$

P14 (Kunihiko Chikaya). Let $\{a,b,c,d,k\}\in\mathbb R_+^*$ be the five positive constants. Find the range of the function $f(x,y)=ax+by,$ where $(x,y)\in \mathbb R^2$ and $cx^2+dy^2=k\ .$

Proof. $(ax+by)^2=\left(\frac a{\sqrt c}\cdot x\sqrt c +\frac b{\sqrt d}\cdot y\sqrt d\right)\ \stackrel{C.B.S}{\le}\ \left[\left(\frac a{\sqrt c}\right)^2+\left(\frac b{\sqrt d}\right)^2\right]\cdot\left[\left(x\sqrt c\right)^2+\left(y\sqrt d\right)^2\right]=$ $\left(\frac {a^2}c+\frac {b^2}d\right)\cdot\left(cx^2+dy^2\right)\iff$ $|ax+by|\le \sqrt {k\cdot\left(\frac {a^2}c+\frac {b^2}d\right)}\ .$ Particular case. For $k=cd$ obtain that $\frac {x^2}d+\frac {y^2}c=1\implies$ $|ax+by|\le \sqrt {da^2+cb^2}\ .$

P15 (Generalization). Let $\triangle ABC$ with $AB=1$ and ${M,N}$ so that $\left\{\begin{array}{ccc} B\in (MC) & ; & MB=b\\\\ C\in (BN) & ; & CN=c\end{array}\right\|$ and $m\left(\widehat{MAN}\right)=\phi\ge 60^{\circ}\ .$ Prove that $\boxed{(2bc+b+c-1)\tan\phi +(b+c+1)\sqrt 3=0}\ (*)\ .$

Proof. $S=[ABC]\ \stackrel{AB=1}{\implies}\ S=\frac {\sqrt 3}4$ and $\left\{\begin{array}{ccc} AM=u & ; & m\left(\widehat{BAM}\right)=x\\\\ AN=v & ; & m\left(\widehat{CAN}\right)=y\end{array}\right\|\ ,$ where $x+y+\left(60^{\circ}-\phi\right)=0$ $\implies$ $\boxed{\tan x+\tan y+\tan\left(60^{\circ}-\phi\right)=\tan x\cdot \tan y\cdot\tan\left(60^{\circ}-\phi\right)}\ (1)\ .$

Apply the generalized Pythagoras' theorem to $:\ \left\{\begin{array}{ccc} \triangle ABM\ : & u^2=b^2+b+1\\\\ \triangle ACN\ : & v^2=c^2+c+1\end{array}\right\|\ (2)\ .$ Observe that $\frac {S_1}b=\frac {S_2}c=S=\frac {\sqrt 3}4\ (3)\ .$ I"ll use identity $\boxed{4S=\left(b^2+c^2-a^2\right)\tan A}$ for any $\triangle ABC\ :$

$\left\{\begin{array}{cccccc} \triangle ABM\ : & 4S_1=\left(u^2+1-b^2\right)\tan x & \stackrel{2\wedge 3}{\implies} & \cancel 4\cdot\frac {b\sqrt 3}{\cancel 4}=(b+2)\tan x & \implies & \tan x=\frac {b\sqrt 3}{b+2}\\\\ \triangle ACN \ : & 4S_2=\left(v^2+1-c^2\right)\tan y & \stackrel{2\wedge 3}{\implies} & \cancel 4\cdot\frac {c\sqrt 3}{\cancel 4}=(c+2)\tan y & \implies & \tan y=\frac {c\sqrt 3}{c+2}\end{array}\right\|\ \stackrel{(1)}{\implies}\ \frac {b\sqrt 3}{b+2}+\frac {c\sqrt 3}{c+2}+\frac {\sqrt 3-\tan\phi}{1+\sqrt 3\tan\phi}=$ $\frac {b\sqrt 3}{b+2}\cdot \frac {c\sqrt 3}{c+2}\cdot\frac {\sqrt 3-\tan\phi}{1+\sqrt 3\tan\phi}$ $\implies$

$b\sqrt 3(c+2)\left(1+\sqrt 3\tan\phi\right)+c\sqrt 3(b+2)\left(1+\sqrt 3\tan\phi\right)+(b+2)(c+2)\left(\sqrt 3-\tan\phi\right)=3bc\left(\sqrt 3-\tan\phi\right)\implies$ $\left[3b(c+2)+3c(b+2)-(b+2)(c+2)+3bc\right]\tan\phi +$

$\sqrt 3\left[b(c+2)+c(b+2)+(b+2)(c+2)-3bc\right]=0\implies$ $\left[8bc+4(b+c)-4\right]\tan\phi +4\sqrt 3(b+c+1)=0\implies$ $(2bc+b+c-1)\tan\phi +\sqrt 3(b+c+1)=0\ .$

Particular cases. $\left\|\begin{array}{ccccc} 1\blacktriangleright\ \phi =90^{\circ} & \implies & x+y=30^{\circ} & \implies & 2bc+b+c=1\\\\ 2\blacktriangleright\ \phi =120^{\circ} & \implies & x+y=60^{\circ} & \implies & bc=1\end{array}\right\|\ \ \wedge\ \ \left\|\begin{array}{ccccc} 3\blacktriangleright\ \phi =135^{\circ} & \implies & x+y=75^{\circ} & \implies & \frac {b+c+1}{bc-1}=\sqrt 3+1\\\\ 4\blacktriangleright\ \phi =150^{\circ} & \implies & x+y=90^{\circ} & \implies & b+c+2=bc\end{array}\right\|\ .$

Remark. If in the last particular case $c=2b\ ,$ then $2b^2-3b-2=0\implies (b-2)(2b+1)=0\implies b=2\ ,$ i.e. $MB=2\cdot AB\ .$ See the proposed problem P14 from (here).

P16 (Nelson Deza Velasquez). Let $\triangle ABC$ with the incentre $I$ and the $B$ -excentre $I_b\ .$ Prove that $II_b=2c\iff B=2C$ (standard notations).

Proof. Can prove that $\boxed{B=2C\iff b^2=c(c+a)}\ (*)\ .$ Cyclic $AICI_b$ has diameter $[II_b]\implies$ $b=II_b\sin\widehat{AI_bC}\implies$ $\boxed{b=II_b\cdot\cos\frac B2}\ (1)\ .$ Thus, $II_b=2c\ \stackrel{(*)}{\iff}\ b=2c\cdot\cos\frac B2\iff$

$b^2=4c^2\cdot\frac {s(s-b)}{ac}\iff$ $ab^2=4cs(s-b)\iff$ $ab^2=c\left[(a+c)^2-b^2\right]\iff$ $ab^2=c(a+c)^2-b^2c\iff$ $b^2(a+c)=c(a+c)^2\iff$ $b^2=c(a+c)\ \stackrel{(*)}{\iff}\ B=2C\ .$ In conclusion,

$II_b=2c\iff B=2C\ .$ Otherwise. If $X\ ,$ $Y$ are the projections on $AB$ of $I\ ,$ $I_b\ ,$ then $XY=BY-BX=s-(s-b)=b\ ,\ XY=II_b\cos\widehat{ABI_b}\implies$ $\boxed{b=II_b\cos\frac B2}$ a.s.o.

Remark. I"ll prove the equivalence $(*)\ .$ If denote $E\in BI\cap AC\ ,$ then $\frac {EA}c=\frac {EC}a=\frac b{c+a}\implies$ $EC=\frac {ab}{a+c}\ (1)\ .$ Hence $B=2C\iff$

$\triangle BEC\sim\triangle CBA\iff$ $\frac {BC}{CA}=\frac {EC}{BA}\iff$ $\frac ab=\frac {EC}c\ \stackrel{(1)}{\iff}\ \frac {\cancel a}b=\frac {\cancel ab}{c(a+c)}\iff\ b^2=c(a+c)\ .$ Hence $B=2C\iff b^2=c(a+c)\ .$

P17 (Adil Abdullayev). Prove $(\forall )\ \triangle ABC$ there is the following inequality $:\ \boxed{\frac {bc}{r_br_c}+\frac {ca}{r_cr_a}+\frac {ab}{r_ar_b}\ge 5-\frac {2r}R}\ (*)\ .$ (standard notations).

Proof. $\frac {bc}{r_br_c}=$ $abc\cdot \frac 1{ar_br_c}=$ $abc\cdot\frac {(s-b)(s-c)}{aS^2}=$ $4Rr\cancel s\cdot\frac 1{a\cancel s(s-a)}=$ $\frac {4Rr}s\cdot \left(\frac 1a+\frac 1{s-a}\right)\implies$ $\boxed{S_0=\frac {4Rr}s\cdot\left(S_1+S_2\right)}\ (1)\ ,$ where $S_0=\sum\frac {bc}{r_br_c}\ ,$ $S_1\equiv\sum \frac 1a$ and $S_2\equiv \sum \frac 1{s-a}\ .$

Thus, $S_1=\sum \frac 1a=\frac {ab+bc+ca}{abc}=\frac {s^2+r^2+4Rr}{4Rrs}\implies$ $\boxed{S_1=\sum \frac 1a=\frac {s^2+r^2+4Rr}{4Rrs}}\ (2)$ and $S_2=\sum\frac 1{s-a}=$ $\frac {\sum (s-b)(s-c)}{\prod (s-a)}=$ $\frac {r(4R+r)}{sr^2}=$ $\frac {4R+r}{sr}\implies$ $\boxed{S_2=\frac {4R+r}{sr}}\ (3)$

From $(2)$ and $(3)$ obtain that $S_0=S_1+S_2=\frac {s^2+r(4R+r)}{4Rrs}+\frac {4R+r}{sr}=$ $\frac {s^2+r(4R+r)+4R(4R+r)}{4Rrs}=\frac {s^2+(4R+r)^2}{4Rrs}\implies$ $S_0=\frac {\cancel{4Rr}}s\cdot\frac{s^2+(4R+r)^2}{\cancel{4Rr}s}=$

$\frac {s^2+(4R+r)^2}{s^2}=1+\left(\frac {4R+r}s\right)^2\implies$ $\boxed{S_0=1+\left(\frac {4R+r}s\right)^2}\ .$ Must to prove only $\boxed{\left(\frac {4R+r}s\right)^2+\frac {2r}R\ge 4}\ (4)\ .$ Hence $\left(\frac {4R+r}s\right)^2+\frac {2r}R\ge 4\iff$

$\frac {(4R+r)^2}{s^2}\ge\frac {2(2R-r)}R\iff$ $s^2\le \frac {R(4R+r)^2}{2(2R-r)}\ ,$ what is the remarkable Blundon & Gerretsen's inequality (<= click).

P18 (Ruben AUQUI, Peru). Let an O-rightangled $\triangle COD$ and construct the squares $OCBA\ ,$ $ODEF$ so that $O\in (AD)\cap (CF)\ .$ Let $I\in BD\cap CE\ .$ Prove that $OI\perp CD\ .$

Proof. Denote $OA=a\ ,$ $OD=b\ ;\ X\in BD\cap OC\ ,$ $Y\in CE\cap OD\ ;\ P\in OI\cap CD\ ,$ $R\in CD$ so that $OR\perp CD\ .$ Prove easily that $\boxed{\ \frac {XC}{XO}=\frac ab=\frac {YO}{YD}\ }\ (*)$

and $\boxed{\ \frac {RC}{RD}=\frac {a^2}{b^2}\ }\ (1)\ .$ Apply the Ceva's relation to the interior point $I$ of $\triangle COD\ :\ \frac {XC}{XO}\cdot\frac {YO}{YD}\cdot\frac {PD}{PC}=1\stackrel{*}{\iff} \frac {PC}{PD}=\frac {a^2}{b^2}\stackrel{(1)}{=}\frac {RC}{RD}\iff P\equiv R\iff$ $OI\perp CD.$

$\mathrm{END}$

This post has been edited 504 times. Last edited by Virgil Nicula, Mar 14, 2019, 3:01 PM

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August 2018

462. Diverse probleme de geometrie.

January 2018

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

October 2017

460. Working page VI.

September 2017

459.Working page V.

August 2017

458. Working page IV.

457. Relatii metrice si inegalitati geometrice.

456. Probleme de Algebra.

July 2017

456* Integrals.

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

May 2017

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

March 2017

453. Working page III.

December 2016

452. Working page II.

November 2016

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

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

October 2016

449. Working page I.

September 2016

448. Extensions or generalizations.

July 2016

447. Locuri geometrice remarcabile.

June 2016

446. Mathematics "instant" for the middle school.

445. Mathematics "instant" for the high school.

May 2016

444. Remarkable points in a triangle.

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

April 2016

442. Problems from and for baccalaureate II.

441. LaTeX (common simbols).

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440. Piece of Poetry.

January 2016

439. Properties of medians in triangle ABC and applications.

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438. Tangential quadrilateral. Zaslavsky's problem.

437. An equivalence relation over R.

436. Two characterizations of A-right triangle ABC.

435. Geometry problems in space.

November 2015

434. Problems with areas II.

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

432. Geometry problems with perpendicularities II.

431. Trigonometry in geometry II.

430. Trigonometrical identities.

October 2015

429. Bretschneider's_formula

September 2015

428. Some metrical problems from the contests II.

July 2015

427. Selected nice or well-known geometry problems.

June 2015

426. Problems from and for baccalaureate.

May 2015

425. Geometrical extremity II.

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

March 2015

423. Geometry problems.

422. Geometrical problems.

420. Some geometry problems for high school II.

419. Some geometry problems for hight school I

418. Some nice problems with polynomials/equations III

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

417. The incircle of ABC and other two incircles.

416. Cateva solutii ale unei probleme de tip slicing.

415. Geometry 8.

414. Geometry 7.

413. Geometry 6.

412. Geometry 5.

411. Geometry 4.

410. Geometry 3.

409. Geometry 2.

408. Geometry 1.

February 2015

407. Cyclical quadrilaterals.

November 2014

406. Trigonometry in geometry (middle or high school) II.

405. Art of problem solving (algebra).

403. Some nice geometry problems for the high school. II

402. Some interesting inequalities I.

401. Some nice problems with polynomials/equations II.

October 2014

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

September 2014

399. Algebra (I).

398. Geometry problems (II).

August 2014

July 2014

396. Own inegalities stronger than the Panaitopol's.

April 2014

395. Metrical relations in a quadrilateral.

394. A cevian in a triangle and two incircles.

393. Some problems with circles.

March 2014

392. Art of Problem Solving (geometry).

December 2013

391. CRUX Mathematicorum.

November 2013

390. Probleme de analiza matematica.

389. Some geometric properties in a triangle I.

October 2013

388. Some interesting "slicing" problems.

September 2013

387. Algebra (II).

August 2013

386. Barycentrical coordinates.

385. Some nice geometry problems.

384. Geometry problems (I).

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

July 2013

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

381. Probleme date la diverse concursuri.

380. Functions. Range. Bijection and inverse.

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379. Geometrical extremity I.

378. Some metrical relations.

377. Some remarkable properties of the symmedian.

May 2013

376. Trigonometrical identities.

374. Some proofs of the Law of Cosines.

373. Extension of Chebyshev's inequality.

April 2013

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

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

February 2013

370. Some metrical relations in a triangle.

January 2013

369. Mixtilinear circles of ABC.

364. Some interesting inequalities I

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363. Geometry problems with perpendicularities.

November 2012

362. Fermat's problem and other metrical problems.

October 2012

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

360. Some problems for the middle school.

359. Ptolemy's theorem and generalized Ptolemy's relation.

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

September 2012

357. Morley theorem.

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

August 2012

355. Some problems with polynomials or equations I

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

353. Some problems of the analytical geometry.

July 2012

352. Some definite integrals.

351. Problems with areas.

350. Another (easy) integrals.

349. Another interesting limits.

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

June 2012

347. Two isogonal lines in a triangle.

346. Some easy integrals.

345. Characteristic function of a set.

344. Some trigonometry problems.

343. Problems with collinear points and concurrent lines.

May 2012

342. Exponential or logarithmic equations/inequations.

April 2012

341. Some nice and interesting "slicing" problems.

340. Geometry problems for middle school.

339. Derivatives. Rolle's function.

338. ROU National Math Olympiad 2012.

February 2012

337. Nice metrical problems.

January 2012

336. Some problems of the analytical geometry.

335. Some difficult geometric/trigonometric inequalities.

334. 2011 Japan Mathematical Olympiad Finals Problem 1.

333. Problems with the common tangent for two circles.

332. Some metrical problems from the contests I.

December 2011

331. Some nice problems with polynomials/equations I.

November 2011

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

329. Some problems with complex numbers.

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

October 2011

327. Some geometrical/algebraic extremum problems.

326. Some problems from trigonometry.

325. Without the l'Hospital's rules.

324. Some determinants.

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

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

321. Harmonical quadrilateral.

320. Some problems with metrical relations.

September 2011

319. Some problems with induction.

318. IRAN - 2011 (geometrie).

317. Some interesting geometry problems for middle school.

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

August 2011

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

314. USAMTS 2009 Round 2 #5.

313. For middle school - some nice problems.

312. Very nice (famous trigonometrical problems) !

311. Some problems from USAMO and other contests.

310. Some properties of the tangential quadrilateral.

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

308. Some nice and easy properties in a triangle.

July 2011

307. Very nice proofs !

306. Some easy and nice "slicing" problems.

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

304. Integrals ("brothers") III.

303. Integrals II.

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

301. Some trigonometrico-geometrical inequalities.

300. Lagrange's multiplier. Examples and problems.

299. A class of the recurrent sequencies.

298. Some nice problems from Crux Mathematicorum.

297. Nice inequalities with concrete numbers.

296. Some nice Vittasko's problems.

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

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

June 2011

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

292. Length of the angled bisector and some means.

291. Some integrals.

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

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

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

287. Some metrical problems with or without trigonometry.

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

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

284. Ellipse (definition).

283. Some nice and easy problems.

282. Factorize of polynomials.

May 2011

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

280. Common roots of two polynomial equations.

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

278. In memoria profesorului Alexandru Lupas.

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

276. Chinese TST 2009 and Second Round 2006.

275. Some interesting algebraic equations and systems.

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

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

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

271. Relations concerning Fermat points.

April 2011

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

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

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

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

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

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

260. Some nice, simple identities and their applications.

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

258. Some difficult and nice problems with complex numbers.

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

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

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

253. An old algebraic inequality.

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

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

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

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

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

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

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

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

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

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

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

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

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

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

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

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

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

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

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

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

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

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

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

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

181. Two special inequalities from CRUX (own).

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

November 2010

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

178. Some nice and simple geometry problems.

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

176. Another nice problems for middle school.

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

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

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

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

October 2010

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

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

168. Some properties of the Lemoine's point.

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

166. Some constant geometrical expressions.

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

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

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

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

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

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

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

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

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

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

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

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

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

138. Generalized "slicing" problems (own).

137. Stewart's relation.

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

135. Three strong geometrical inequalities.

September 2010

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

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

132. Asupra unei inegalitati intr-un triunghi.

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

130. A triangle and a fixed point.

129. BX=CY and three properties.

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

127. Pagina educativa.

126. Routh's theorem.

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

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

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

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

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

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

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

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

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

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

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

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

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

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

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

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

93. The Appolonius' construct problems.

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

91. JBMO 2010, Problem 3.

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

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

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

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

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

81. Sawayama and Thebault’s theorem.

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

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

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

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

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

July 2010

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

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

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

69*1. Slicing problem 3.

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

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

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

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

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

62. Nice and easy problem with a right triangle.

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

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

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

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

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

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

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

June 2010

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

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

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

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

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

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

38. Minimum area of triangle touching with semicircle.

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

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

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

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

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

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

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

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

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

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

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

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

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

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

10. Walker's inequality and its extension.

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

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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El_Ectric:
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You're welcome.
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Dear Virgil !

Congratulations for this blog !
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