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

by Virgil Nicula, May 11, 2016, 6:14 PM

Cezar . 296.060 . GIL

$\bf\color{red}Problems\ of\ the\ type\ "instant"$ $\bf\color{black}====================$
P0.0. Find $\{a,b,c\}\subset\mathbb N$ so that $a \cdot b \cdot c = 70$ and $a + b + c = 14\ .$

Proof. Suppose w.l.o.g. that $1\le a\le b\le c\ .$ Since $a\le \frac {14}{3}\ ($ or a "stronger" $a\le \sqrt[3]{70})$ . But $a\in\mathbb N$ and $a$ divides $70$ , i.e. $a\ |\ 70$ , obtain $\boxed{\ a\in\{1,2\}\ }\ .$ Appear

two cases $:$ IF $a=1$ , then $\left\|\begin{array}{ccc} bc & = & 70\\\\ b+c & = & 13\end{array}\right\|\ \stackrel{\frac {13}2<7<\sqrt{70}}{\implies}\ \mathrm{absurd}\ ;$ IF $a=2$ , then $\left\|\begin{array}{ccc} bc & = & 35\\\\ b+c & = & 12\end{array}\right\|$ $\iff$ $\{b,c\}=\{5,7\}\ .$ In conclusion, $\boxed{\ \{a,b,c\}=\{2,5,7\}\ }\ .$

P0.1 (variations on the same theme). Find $\{a,b,c\}\subset\mathbb N$ so that $a \cdot b \cdot c = 70$ and $ab+bc+ca=59\ .$

Proof. Suppose w.l.o.g. that $1\le a\le b\le c\ .$ Since $a\le \frac 3{\frac 1a+\frac 1b+\frac 1c}=\frac {3abc}{ab+bc+ca}=\frac {210}{59}<\frac{18}5=3,6\ .$ But $a\in\mathbb N$ and $a\ |\ 70\implies\boxed{\ a\in\{1,2\}\ }\ .$ Appear two cases $:$

$a=1\implies \left\|\begin{array}{ccc} bc & = & 70\\\\ bc+b+c & = & 13\end{array}\right\|\implies\mathrm{absurd}\ ;\ a=2\implies\left\|\begin{array}{ccc} bc & = & 35\\\\ bc+2(b+c) & = & 59\end{array}\right\|$ $\implies$ $\left\|\begin{array}{ccc} bc & = & 35\\\\ b+c & = & 12\end{array}\right\|$ $\iff$ $\{b,c\}=\{5,7\}$ $\implies$ $\boxed{\ \{a,b,c\}=\{2,5,7\}\ }\ .$

P0.2. $\{a,b,c\}\subset\mathbb R^*_+\ \implies\ \boxed{\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\ge abc(a+b)(b+c)(c+a)}\ (*)\ .$

Proof 1. $\left\|\ \begin{array}{ccc} a^4+b^2c^2 & \ge & 2a^2bc\\\\ a^2\left(b^2+c^2\right) & = & a^2\left(b^2+c^2\right) \end{array}\right\|\ \bigoplus$ $\implies$ $\left(a^2+b^2\right)\left(a^2+c^2\right)\ge a^2(b+c)^2\implies$ $\left\{\begin{array}{ccc} \left(a^2+b^2\right)\left(a^2+c^2\right) & \ge & a^2(b+c)^2\\\\ \left(b^2+c^2\right)\left(b^2+a^2\right) & \ge & b^2(c+a)^2\\\\ \left(c^2+a^2\right)\left(c^2+b^2\right) & \ge & c^2(a+b)^2\end{array}\ \right\|\ \bigodot$ $\implies\ (*)\ .$

Proof 2. From the product of $\left|\begin{array}{ccc} 2(a^2+b^2) & \geq & (a+b)^2\\\\ 2(b^2+c^2) & \geq & (b+c)^2\\\\ 2(c^2+a^2) & \geq & (c+a)^2\end{array}\right|$ and $\left|\begin{array}{ccc} a^2+b^2 & \geq & 2ab\\\\ b^2+c^2 & \geq & 2bc\\\\ c^2+a^2 & \geq & 2ca\end{array}\right|$ obtain $(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2 \ge$ $[abc(a+b)(b+c)(c+a)]^2\implies\ (*)\ .$

Remark. See and here, where I"ll use the remarkable C.B.S - inequality.

P0.3. Prove $\forall\ \{a,b,c \}\subset \mathbb{R}$ there is the inequality $\left(a^2+c^2\right)\left(b^2+c^2\right)\ge c(a+b)\left(ab+c^2\right)$ and $\left(\frac {ab}c+\frac {bc}a+\frac {ca}b\right)^2\ge 3\left(a^2+b^2+c^2\right)\ .$

Proof. Apply the C.B.S - inequality $:\ \left\|\begin{array}{ccc} (a\cdot c+c\cdot b)^2 & \le & \left(a^2+c^2\right)\left(c^2+b^2\right)\\\\ (a\cdot b+c\cdot c)^2 & \le & \left(a^2+c^2\right)\left(b^2+c^2\right) \end{array}\ \right\|\ \bigodot\ \implies\ \left|c(a+b)\left(ab+c^2\right)\right|\le\left(a^2+c^2\right)\left(b^2+c^2\right)\ .$ From $x\le |x|$

obtain $c(a+b)\left(ab+c^2\right)\ \le\ \left(a^2+c^2\right)\left(b^2+c^2\right)\ .$ We have equality iff $a=b=c\ .$ For $c:=1$ obtain the particular inequality $\left(a^2+1\right)\left(b^2+1\right)\geq (a+b)(ab+1)\ .$

$\blacktriangleright\ \left(\frac {ab}c+\frac {bc}a+\frac {ca}b\right)^2\ge 3\left(\frac {ab}c\cdot\frac {bc}a+\frac {bc}a\cdot\frac {ca}b+\frac {ca}b\cdot\frac {ab}c\right)=3\left(a^2+b^2+c^2\right)\implies$ $\boxed{\left(\frac {ab}c+\frac {bc}a+\frac {ca}b\right)^2\ge 3\left(a^2+b^2+c^2\right)}\ .$

P0.4. Prove that $\left\|\begin{array}{c}0<a\le b\le c\\\\ 0<m\le n\le p\end{array}\right\|\ \implies\ ma^2+nb^2+pc^2\ge mbc+nca+pab\ .$

Proof 1 (F.C). Aplicam convenabil teorema majorizarii (Karamata, Polya sau cum o numeste fiecare) pentru functia convexa $f(t)=e^t\ .$ Spunem ca $\mathrm a = (a_1, ..., a_n)$ majoreaza

$\mathrm b= (b_1, ..., b_n)$ in $\mathbb{R}^n$ daca $a_1 \ge a_2 \ge ... \ge a_n$ , $b_1 \ge b_2 \ge ... \ge b_n$ , $a_1 + \cdots + a_k \ge b_1 + \cdots + b_k$ , $(\forall )\ k\in\overline{1, ..., n-1}$ si $\sum a_j = \sum b_j\ .$ Atunci, pentru functia

convexa $f : [a,b] \rightarrow \mathbb{R}$ convexa si $\mathbf{a}$ majoreaza pe $\mathbf{b}$ , avem si $f(a_1) + \cdots + f(a_n) \ge f(b_1) + \cdots + f(b_n)\ .$ Demonstratia se face prin inductie, aplicand Jensen.

Proof 2 (M.M). Inegalitatea rearanjamentelor $:\ \left\{\begin{array}{ccc} ma^2+nb^2+pc^2 & \ge & mb^2+nc^2+pa^2\\\\ ma^2+nb^2+pc^2 & \ge & mc^2+na^2+pb^2\end{array}\right\|\ \bigoplus\ \implies$ $2\left(ma^2+nb^2+pc^2\right)\ge$

$m\left(b^2+c^2\right)+n\left(c^2+a^2\right)+p\left(a^2+b^2\right)\ge$ $m\cdot 2bc+n\cdot 2ca+p\cdot 2ab\implies$ $ma^2+nb^2+pc^2\ge mbc+nca+pab\ .$

Proof 3 (N.V). Apply the Chebyshev's theorem $:\ \left\{\begin{array}{cc} a_1\le a_2\le\ \cdots\ \le a_{n-1}\le a_n & \uparrow\\\\ b_1\le b_2\le\ \cdots\ \le b_{n-1}\le b_n & \uparrow\end{array}\right\|\implies$ $\sum_{k=1}^na_k\cdot\sum_{k=1}^nb_k\le n\cdot \sum_{k=1}^n\left(a_kb_k\right)\implies$

$\left\{\begin{array}{ccc} bc+ca+ab & \le & a^2+b^2+c^2\\\\ \left(\underrightarrow{m+n+p}\right)\left(\underrightarrow{a^2+b^2+c^2}\right) & \stackrel{\mathrm{Cheb}\ \uparrow\uparrow}{\le} & 3\left(ma^2+nb^2+pc^2\right)\\\\ 3\left(mbc+nca+pab\right) & \stackrel{\mathrm{Cheb}\ \uparrow\downarrow}{\le} & \left(\underrightarrow{m+n+p}\right)\left(\underleftarrow{bc+ca+ab}\right)\end{array}\right\|\ \bigodot\ \implies$ $mbc+nca+pab\le ma^2+nb^2+pc^2\ .$

F.C., permite-mi te rog un mic comentariu ... Tehnica asta "tare" o poti enunta sa o invete, cel putin pe dinafara, si ceilalti ?! Si cand te gandesti ca de fapt inegalitatea propusa de mine este foarte usoara, se face intr-un rand. Iata ce inseamna ca un elev de clasa a X - a sa fie "dopat" cu tot felul de "bombe atomice" (carora deseori nu le stie nici macar o demonstratie) pentru "a omori o musca sau un fluture". Sper ca nu te vei supara pe mine, P.C ! Am dreptul si eu la o parere personala.

P0.5. Prove that for any $\{a,b,c\}\subset \mathbb R$ the equation $f(x)=0$ , where $f(x)\equiv (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$ has only real roots.

Proof 1. Denote the roots $\odot\begin{array}{ccc} \nearrow & x_1 & \searrow\\\\ \searrow & x_2 & \nearrow\end{array}\odot$ of the equation $f(x)=0$ and suppose w.l.o.g. that $a\le b\le c$ . Observe that the dominant coefficient $3$ is positive and

$f(a)=(a-b)(a-c) \ge 0$ , $f(b)=(b-c)(b-a) \le 0$ and $f(c)=(c-a)(c-b) \ge 0$ . In conclusion, the roots $x_1$ and $x_2$ are real and si $a \le x_1 \le b\le x_2\le c$ .

Proof 2. Observe that the equation becomes $f(x)=3x^2-2(a+b+c)x+(ab+bc+ca)=0$ and its discriminant

$\Delta^{\prime} (a,b,c)=(a+b+c)^2-3(ab+bc+ca)=\frac 12\cdot [(a-b)^2+(b-c)^2+(c-a)^2]\ge 0$ , i.e. $\left\{x_1,x_2\right\}\subset\mathbb R\ .$

P0.6. Prove that $\{a,b,c\}\subset\mathbb R^*_+$ and $a+b+c=1\implies$ $\frac{a^2+a}{a+bc}+\frac{b^2+b}{b+ca}+\frac{c^2+c}{c+ab}=3\ .$

Proof. $\sum a^2(b+c)=\sum bc(b+c)$ , i.e. $\sum \frac {a^2-bc}{(a+b)(a+c)}=0\ .$ Hence $\sum\left(\frac {a^2+a}{a+bc}-1\right)=$ $\sum \frac {a^2-bc}{(a+b)(a+c)}=0\iff$ $\sum\frac {a^2+a}{a+bc}=3\ .$

P0.7 (M.O. Sanchez). Let an $A$ -right $\triangle ABC$ and $E\in (AC)$ , $D\in (BE)$ so that $m\left(\widehat{EDC}\right)=m\left(\widehat{ECD}\right)=B$ . Prove that $BD=2\cdot AE$ .

Let the symmetric $L$ of $E$ w.r.t. $A$ . Prove easily that $\left\{\begin{array}{ccc} m\left(\widehat{AFC}\right)=m\left(\widehat{LBC}\right)=C & ; & m\left(\widehat{BDF}\right)=m\left(\widehat{EDC}\right)=m\left(\widehat{ECD}\right)=B\\\ m\left(\widehat{AEB}\right)=2B\ ;\ m\left(\widehat{CBE}\right)=2B-C & ; & m\left(\widehat{LBA}\right)=m\left(\widehat{FBE}\right)=m\left(\widehat{FCB}\right)=90^{\circ}-2B\end{array}\right\|$ .

Proof 1.Observe that $\triangle DEC$ is $E$ -isosceles, $\triangle LBE$ is $B$ -isosceles and $\triangle BLC$ is $L$ -isosceles. Therefore, $ED=EC$ , $BL=BE$ , $LB=LC$ and

$LE=2\cdot AE$ . In conclusion, $BE=LC\iff$ $BD+DE=LE+EC\iff$ $BD=LE\iff$ $BD=2\cdot AE$ (Jose Luis F. Ballena). See here.

Proof 2. $\triangle ACF\sim\triangle ABC\iff$ $\frac bc=\frac {CF}a=\frac {AF}b$ $\iff$ $\begin{array}{cccc} \nearrow & CF=\frac {ab}c & (1) & \searrow\\\\ \searrow & AF=\frac {b^2}c & (2) & \nearrow\end{array}\odot$ Hence $FB=AB-AF\stackrel{(2)}{=}\ c-$ $\frac {b^2}c\implies$ $\boxed{FB=\frac {c^2-b^2}c}\ (3)$ .

See analogously $\triangle DFB\sim \triangle BFC\iff$ $\frac {DB}{BC}=\frac{FB}{FC}\ \stackrel{(1\wedge 3)}{=}\ \boxed{DB=\frac {c^2-b^2}b}\ (4)\ .$ Thus, $\tan\widehat{AEB}=\frac {AB}{AE}\iff$ $AE=c\cdot\cot 2B\iff$ $2\cdot AE=$

$c\cdot 2\cot 2B=c\cdot\left(\cot B-\tan B\right)=$ $c\cdot\left(\frac cb-\frac bc\right)\iff$ $\boxed{AE=\frac {c^2-b^2}{2b}}\ (5)\ .$ In conclusion, from the relations $(4)$ and $(5)$ obtain that $BD=2\cdot AE\ .$

Remark. $\frac {BD}{\sin\widehat{BCD}}=\frac {BC}{\sin \widehat{BDC}}\iff$ $\frac {BD}{\cos 2B}=\frac a{\sin B}\iff$ $BD=\frac {a\left(1-2\sin^2B\right)}{\sin B}=\frac a{\frac ba}\cdot\left[1-2\cdot\left(\frac ba\right)^2\right]=$ $\frac {a^2}b\cdot\frac {a^2-2b^2}{a^2}=$ $\frac {c^2-b^2}b\implies$ $\boxed{BD=\frac {c^2-b^2}b}\ .$

P0.8 (IMO TST 2015) Let an acute $\triangle ABC$ wuth $AB>AC\ .$ Denote $T\in BB\cap CC$ and $S\in (AB)$ so that $TS\parallel AC\ .$ Prove that $SC=SA\ .$

Proof. Observe that $TS\parallel AC\implies$ $\widehat{SCA}\equiv \widehat{CST}$ and $\widehat{BST}\equiv\widehat{BAC}\equiv\widehat{BCT}\implies$ $\widehat{BST}\equiv\widehat{BCT}\implies$

$BSCT$ is cyclic $\implies$ $\widehat{SCA}\equiv\widehat{CST}\equiv \widehat{CBT}\equiv\widehat{SAC}\implies$ $\widehat{SCA}\equiv\widehat{SAC}\implies$ $SA=SC\ .$

P0.9. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and $E\in AC\cap w$ , $F\in AB\cap w$ . Let the midpoints $M$ , $N$ of the sides $[BC]$ , $[AB]$ respectively. Prove that $MN\cap EF\cap CI\ne\emptyset .$

Proof. Denote $S\in CI\cap EF$ . Observe yhat $\widehat{AFS}\equiv \widehat{BIS}$ , i.e. $BFSI$ is cyclic $\implies$ $SB\perp SC$ , i.e. $\triangle BSC$ is right and for its

median $SM$ have $\widehat{ECS}\equiv\widehat{MCS}\equiv$ $\widehat{MSC}\implies$ $\widehat{ECS}\equiv\widehat{MSC}\implies$ $MS\parallel AC\implies N\in MS\implies$ $MN\cap EF\cap CI\ne\emptyset .$

P0.10. Prove that $\triangle ABC$ is acute $\implies\tan A\tan B+\tan B\tan C+\tan C\tan A\ge 3+\frac 1{\cos A}+\frac 1{\cos B}+\frac 1{\cos C}$ and $\sin^2x-\sin^2y=\sin(x+y)\sin (x-y)\ .$

Proof. $\sum\cos^2A\ge$ $\sum\cos B\cos C\iff$ $-\sum \cos (B+C)\cos A\ge$ $\sum \cos B\cos C\iff$ $\sum (\sin B\sin C-\cos B\cos C)\cos A\ge$

$\sum \cos B\cos C\iff$ $\sum \sin B\sin C\cos A\ge$ $3\cos A\cos B\cos C+\sum\cos B\cos C\iff$ $\sum\tan B\tan C\ge$ $3+\sum\frac 1{\cos A}\ .$

$\blacktriangleright\ 2\left(\sin^2x-\sin^2y\right)=(1-\cos 2x)-(1-\cos 2y)=$ $\cos 2y-\cos 2x=$ $2\sin(x+y)\sin (x-y)\implies$ $\boxed{\sin^2x-\sin^2y=\sin(x+y)\sin (x-y)}\ .$

P0.11 (O.L. Iasi 2008). Prove that in any $A$ -right $\triangle ABC$ with $b\ne c$ the line $IG$ cann't be parallel with the hypotenuse $BC$ , but can be parallel with a cathetus (standard notations).

Proof 1. Prove easily that in any $\triangle ABC$ there is the equivalencies $\boxed{\begin{array}{ccc} IG\perp BC & \iff & b+c=3a\\\\ IG\parallel BC & \iff & b+c=2a\end{array}}\ (*)\ .$ Indeed, denote the projections $K$ , $D$ , $S$ of $A$ , $I$ , $G$ on $BC\ :$

$\blacktriangleright\ IG\perp BC\iff$ $\mathrm{pr}_{BC}(I)=\mathrm{pr}_{BC}(G)\iff$ $D\equiv S\iff$ $DM=SM=\frac 13\cdot KM\iff$ $\frac {|b-c|}2=\frac 13\cdot \frac {|b^2-c^2|}{2a}\iff$ $b+c=3a$ , where $M$ is the midpoint of $[BC]\ .$

$\blacktriangleright\ IG\parallel BC\iff$ $\delta_{BC}(I)=\delta_{BC}(G)\iff$ $r=\frac {h_a}3\iff$ $3r=h_a\iff$ $3ar=2S=2sr\iff$ $3a=2s=a+b+c\iff b+c=2a\ .$

Suppose that $AB\perp AC\ .$ In this case $(b+c)^2\le 2\left(b^2+c^2\right)=$ $2a^2$ , i.e. $b+c\le a\sqrt 2\ .$ Appear three cases $:$

$\blacktriangleright$ If $IG\parallel BC$ , then $b+c=2a\ \iff\ 2a\le a\sqrt 2$ , what is absurd. If $IG\perp BC$ , then $b+c=3a\ \iff\ 3a\le a\sqrt 2$ , what is absurd.

$\blacktriangleright\ IG\ \parallel\ AC\ (\ IG\perp AB\ )\ \iff\ \odot\begin{array}{ccccc} \nearrow & a+c & = & 2b & \searrow\\\\ \searrow & a+b & = & 3c & \nearrow \end{array}\odot \iff\ a=2b-c=3c-b\ \iff\ 3b=4c\iff\ \frac a5=\frac b4=\frac c3\ .$

$\blacktriangleright\ IG\ \parallel\ AB\ (\ IG\perp AC\ )\ \iff\ \odot\begin{array}{ccccc} \nearrow & a+b & = & 2c & \searrow\\\\ \searrow & a+c & = & 3b & \nearrow \end{array}\odot \iff\ a=2c-b=3b-c\ \iff\ 4b=3c\iff\ \frac a5=\frac b3=\frac c4\ .$

Proof 2. Let $D\in IG\cap BC$ . Thus, $IG\perp BC\iff$ https://scontent.fomr1-1.fna.fbcdn.net/v/t1.0-9/17362649_1888040991464614_817726152431194852_n.jpg?oh=f66f0d7493dcd9887079d79d2271b141&oe=59519C90

P0.12. Let $\triangle ABC$ and mobile $M\in [AC]$ , $N\in [AB]$ so that $\frac {MA}{MC}+\frac {NA}{NB}=2\ .$ Find the locus of $L\in BM\cap CN$ and its position for what $LB+LC$ is minimum.

Proof. Denote the centroid $G$ of $\triangle ABC$ and the intersection $K\in BC\cap AL$ . Apply the Aubel's relation $\frac {LA}{LK}=\frac{MA}{MC}+\frac {NA}{NB}=2$ (constant), i.e. $LG\parallel BC$

(the centroid $G$ is fixed point). Hence the locus of $L$ is the segment $[UV]$ for what $G\in UV\parallel BC\ .$ Let $E\in AC\cap LG$ and $F\in AB\cap LG\ .$ The required point

$L_{\mathrm{min}}$ from $[EF]$ has the property $\widehat{BLE}\equiv\widehat{CLF}$ , i.e. the symmetric point of $C$ w.r.t. $EF$ belongs to $BL_{\mathrm{min}}\ .$

P0.13. Prove that in any acute $\triangle ABC$ there are the inequalities $:\ \left\{\begin{array}{cccc} \frac{b+c}{\cos A}+\frac{c+a}{\cos B}+\frac{a+b}{\cos C} & \ge & 4(a+b+c) & (1)\\\\ \frac a{\cos B+\cos C}+\frac b{\cos C+\cos A}+\frac c{\cos A+\cos B} & \ge & a+b+c & (2)\end{array}\right\|\ .$

Proof 1 $\sum\frac {b+c}{\cos A}=\sum\frac {(b+c)^2}{(b+c)\cos A}\ \stackrel{\mathrm{C.B.S}}{\ge}\ \frac {\left[\sum(b+c)\right]^2}{\sum (b+c)\cos A}=$ $\frac {(4s)^2}{\sum(b\cos C+c\cos B)}=$ $\frac {16s^2}{\sum a}=$ $\frac {16s^2}{2s}=8s=4(a+b+c)\implies$ $\sum\frac {b+c}{\cos A}\ge 4(a+b+c)\ .$

$\sum\frac a{\cos B+\cos C}=\sum\frac {a^2}{a(\cos B+\cos C)}\ \stackrel{C.B.S}{\ge}\ \frac {\left(\sum a\right)^2}{\sum [a(\cos B+\cos C)]}=$ $\frac {4s^2}{\sum (b\cos C+c\cos B)}=\frac {4s^2}{\sum a}=\frac {4s^2}{2s}=a+b+c\implies \sum\frac a{\cos B+\cos C}\ge a+b+c\ .$

P0.14 (clasa a VII - a). Let $\triangle ABC$ and $D\in (AB)$ so that $CDFE$ is a parallelogram where $C\in (BE)$ . For $Q\in (CD)$ denote $N\in DE\cap (QF)\ .$ Prove that $\frac {DE}{DN}=2+\frac {QC}{QD}\ .$

Proof. Denote $S\in FQ\cap BC$ and observe that $\frac {DE}{DN}=\frac {DN+NE}{DN}=1+\frac {NE}{ND}=$ $1+\frac {SE}{DF}=1+\frac {SC+CE}{CE}=2+\frac {CS}{CE}=2+\frac {SC}{DF}=2+\frac {QC}{QD}$ , i.e. $\frac {DE}{DN}=2+\frac {QC}{QD}\ .$

P0.15. Solve over $\mathbb R$ the following system: $\left\{\begin{array}{c} 2-x=\frac {2}{y^2+1}\\\\ 2-y=\frac {2}{z^2+1}\\\\ 2-z=\frac {2}{x^2+1}\end{array}\right\|$ (at least two methods).

Proof . Observe that $\{x,y,z\}\subset [0,2)$ and $x=0\iff y=0\iff z=0\ .$ Suppose w.l.o.g. $xyz\ne 0\ .$ Our system is equivalently with

$\left\{\begin{array}{c} x\left(y^2+1\right)=2y^2\\\\ y\left(z^2+1\right)=2z^2\\\\ z\left(x^2+1\right)=2x^2\end{array}\right\|\ \bigodot\implies$ $xyz\left(x^2+1\right) \left(y^2+1\right)\left(z^2+1\right)=8x^2y^2z^2\implies$ $\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)=8xyz\implies$

$\boxed{x=y=z=1}$ because over $\mathbb R^*_+$ there is the inequality $\left(x^2+1\right)\cdot \left(y^2+1\right)\cdot\left(z^2+1\right)\ge 2x\cdot 2y\cdot 2z$ with equality iff $x=y=z\ .$

P0.16. Discutia pozitiei radacinilor reale ale unei ecuatii de gradul doi fata de un numar real dat.

Demonstratie. Studiul pozitiei radacinilor $\{\ x_1\ ,\ x_2\ \}\subset\mathbb R^*_+$ ale ecuatiei $f(x)\equiv ax^2+bx+c=0\ ,\ a\ne 0$ fata de $\alpha\in\mathbb R$

impune studiul semnului pentru expresiile $\Delta$ (discriminant) , $af(\alpha )$ si $S-2\alpha$ , unde $S=x_1+x_2=-\frac ba\ .$

$\boxed{\begin{array}{cccc} \Delta & af(\alpha ) & S-2\alpha & \mathrm{Pozitia\ radacinilor\ reale\ fata\ de}\ \alpha\in\mathbb R\\ == & === & ==== & ======================\\ + & + & + & \alpha\ <\ x_1\ <\ x_2\\\\ + & + & - & x_1\ <\ x_2\ <\ \alpha\\\\ + & 0 & + & \alpha\ =\ x_1\ <\ x_2\\\\ + & 0 & - & \ x_1\ <\ x_2\ =\ \alpha\\\\ + & - & \Box & x_1\ <\ \alpha\ <\ x_2\\\\ 0 & + & + & \alpha \ <\ x_1\ =\ x_2\\\\ 0 & + & - & x_1\ =\ x_2\ <\ \alpha\\\\ 0 & 0 & 0 & x_1\ =\ x_2\ =\ \alpha\\\\ - & \Box & \Box & x_1\ ,\ x_2\ \not\in\ \mathbb R\end{array}}\ .$

P0.17. Let the equation $t^3-28t+48=0\implies\odot \begin{array}{ccc} \nearrow & a & \searrow\\\\ \rightarrow & b & \rightarrow\\\\ \searrow & c & \nearrow\end{array}\odot$ Solve the equation $\frac {x-(b+c)}a+\frac {x-(c+a)}b+\frac {x-(a+b)}c=10\ .$

P0.18. Sa se arate ca in orice $\triangle\ ABC$ avem $\begin{array}{ccccc} \nearrow & 1. & A=30^{\circ} & \Longleftrightarrow & b^2+c^2=a^2+4S\sqrt 3.\\\\ \searrow & 2. & A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right] & \Longrightarrow & b^2+c^2\ \ge\ a^2+4S\sqrt 3.\end{array}\ .$ Se stie ca in orice $\triangle ABC$ avem $\boxed{a^2+b^2+c^2\ge 4S\sqrt 3}\ .$

Demonstratie. Se stie ca $2bc\cdot\cos A=b^2+c^2-a^2=4S\cot A\ .$ Asadar, $b^2+c^2=a^2+4S\sqrt 3\iff$ $2bc\cos A=2bc\sin A\sqrt{3}$ $\Longleftrightarrow$

$\tan A=\frac{1}{\sqrt{3}}$ $\Longleftrightarrow$ $A=30^{\circ}\ .$ Se arata usor ca $A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right]$ $\Longrightarrow$ $\tan A\le\frac{1}{\sqrt{3}}$ $\Longrightarrow$ $\cot A\ge \sqrt 3$ $\implies$ $b^2+c^2-a^2\ge 4S\sqrt 3\implies$ $b^2+c^2\ \ge\ a^2+4S\sqrt 3\ .$

P0.19. Sa se arate ca in orice $\triangle ABC$ exista $9r\le l_a+l_b+l_c\le s\sqrt 3\ ,$ unde $l_a$ este lungimea bisectoarei interioare din $A\ .$

Demonstratie. $\sum l_a=$ $\sum \frac{2\sqrt{bcs(s-a)}}{b+c}\le \sum\sqrt{s(s-a)}=\sqrt s\cdot\sum\sqrt{s-a}\le$ $\sqrt s\cdot\sqrt{3\sum(s-a)}=$ $\sqrt s\cdot\sqrt {3s}=s\sqrt 3$ $\implies$ $l_a+l_b+l_c\le s\sqrt 3\ .$

$h_a=\frac {2rs}{a}\implies$ $\sum l_a\ge \sum h_a=$ $r\sum a\sum \frac 1a\ge 9r\implies$ $\sum l_a\ge 9r\ .$ In concluzie, $\boxed{9r\le\sum l_a\le s\sqrt 3}\ .$ Se obtine astfel inegalitatea cunoscuta $\boxed{3r\sqrt 3\le s}\ .$

P0.20. Prove that for any $x\in\mathbb R$ there is the inequality $\boxed{\ f(x)\equiv 1+(x+1)(x+2)(x+3)(x+4)\ \ge 0\ }\ .$

Proof. $f(x)=1+(x+1)(x+2)(x+3)(x+4)=1+(x+1)(x+4)\cdot (x+2)(x+3)=1+(x^2+5x+4)(x^2+5x+6)=1+(t-1)(t+1)=t^2\ge 0\ ,$ where

$x^2+5x+5=t\ge -\frac 54$ for any $x\in\mathbb R$ because the equation $g(x)=x^2+5x+(5-t)=0$ has real roots, i.e. $\Delta (t)\equiv 5^2-4(5-t)\ge 0\iff 4t+5\ge 0\iff t\ge -\frac 54\ .$

P0.21. Let $ABCD$ be a trapezoid with $AB\parallel CD,$ $O\in AC\cap BD$ and the area $S.$ Denote $\left\{\begin{array}{ccc} a=[AOB] & ; & b=[COD]\\\\ c=[AOD] & ; & d=[BOC]\end{array}\right\|.$ Prove that $c=d=2\sqrt{ab}$ and $\sqrt S=\sqrt a+\sqrt b.$

=================================================================================================================================================

P1 (Mathtime). Let an $A$ -right $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ and the Nagel's point $\mathcal N$ . Let $\left\{\begin{array}{ccc} D & \in & BC\cap A\mathcal N\\\ E & \in & CA\cap B\mathcal N\\\ F & \in & AB\cap C\mathcal N\end{array}\right\|$ .

Prove that the circumcenter of $\triangle DEF$ is $S\in w$ for which $SO\perp BC$ and $BC$ doesn't separate $S$ and $A$ .

Proof 1. $\left\{\begin{array}{ccc} FB=EC=s-a\\\\ DB+c=DC+b=s\end{array}\right\|\ ,\ DO=\frac {|b-c|}2$ . Let $\{X,Y\}\subset BC$ and $\{U,V\}\subset SO$ so that $EX$ , $FY$ are perpendicular on $BC$ and $EU$ , $FV$ are perpendicular

on $SO$ . Thus, $\triangle FBY\equiv\triangle CEX\implies$ $\left\{\begin{array}{ccccc} BY & = & EX & = & \frac ca\cdot (s-a)\\\\ FY & = & CX & = & \frac ba\cdot (s-a)\end{array}\right\|$ and $\left\{\begin{array}{ccccc} EX+CX & = & BY+FY & = & \frac 1a\cdot (s-a)(b+c)\\\\ EX^2+CX^2 & = & BY^2+FY^2 & = & (s-a)^2\end{array}\right\|\ (*)\ .$ Observe that

$SD^2=OS^2+OD^2=$ $\left(\frac a2\right)^2+\left(\frac {b-c}2\right)^2=$ $\frac {2a^2-2bc}4=$ $\frac {a^2-bc}2\implies$ $\boxed{SD^2=\frac {a^2-bc}2}\ (1)\ .$ Now we"ll prove easily that and $SE^2=SF^2=SD^2=\frac {a^2-bc}2\ .$ Indeed $:$

$\begin{array}{cc} SE^2=US^2+UE^2=\left(\frac a2-EX\right)^2+\left(\frac a2-CX\right)^2=\frac {a^2}2-a(EX+CX)+\left(EX^2+CX^2\right) & \searrow\\\\ SF^2=VS^2+VF^2=\left(\frac a2-FY\right)^2+\left(\frac a2-BY\right)^2=\frac {a^2}2-a(FY+BY)+\left(FY^2+BY^2\right) & \nearrow\end{array}\odot\stackrel{(*)}{=}\ \frac {a^2}2-(b+c)(s-a)+(s-a)^2=$ $\frac {a^2}2-s(s-a)=SD^2\ .$

P2 (Canadian Mathematical Olympiad, 2015) Let $ABC$ be an acute-angled triangle with the orthocenter $H$ . Prove that $\frac{ab+bc+ca}{h_a\cdot AH+h_b\cdot BH+h_c\cdot CH}\le 2$ (standard notations)

Proof. I"ll use two well known and usual identities $\boxed{AH=2R\cos A}\ (1)\ ,\ \boxed{2Rh_a=bc}\ (2)$ and one simple inequality $\sum a^2\ge\sum (bc)\ (3)\ .$ Therefore, $\frac{\sum (bc)}{\sum \left(h_a\cdot AH\right)}\ \stackrel{(1)}{=}$

$\frac{\sum (bc)}{\sum \left(h_a\cdot 2R\cos A\right)}=$ $\frac {\sum (bc)}{\sum (2Rh_a\cdot\cos A)}\ \stackrel{(2)}{=}$ $\frac {\sum (bc)}{\sum (bc\cdot\cos A)}=$ $\frac {2\cdot\sum (bc)}{\sum (2bc\cdot\cos A)}=$ $\frac {2\cdot\sum (bc)}{\sum \left(b^2+c^2-a^2\right)}=$ $2\cdot\frac {\sum (bc)}{\sum a^2}\ \stackrel{(3)}\le\ 2\implies$ $\frac{ab+bc+ca}{h_a\cdot AH+h_b\cdot BH+h_c\cdot CH}\le 2\ .$

P3. Let an acute $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ , the midpoints $M$ , $N$ of $[AC]$ , $[AB]$ and $\left\{\begin{array}{ccc} P\in AA\cap BC\\\\ E\in AC\cap PO\\\\ F\in AB\cap PO\end{array}\right\|$ . Prove that $AO\cap EN\cap FM\ne\emptyset\ .$

Proof. Suppose $b>c$ and denote $S\in BC\cap AO\ .$ Prove easily that $\frac {SB}{SC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{SAB}}{\sin\widehat{SAC}}$ , i.e. $\frac {SB}{c\cdot \cos C}=\frac {SC}{b\cdot\cos B}=\frac ak\ ,$ where $k=b\cdot\cos B+c\cdot \cos C\ .$ Denote $\left\{\begin{array}{ccc} \frac {EC}x=\frac {EA}1=\frac b{x+1} & \implies & ME=\frac {b(1-x)}{2(x+1)}\\\\ \frac {FB}y=\frac {FA}1=\frac c{y+1} & \implies & NF=\frac {c(1-y)}{2(y+1)}\end{array}\right\|\ .$ Apply the Menelaus' theorem to the transversal $\overline{PEF}/\triangle ABC\ :\ \frac {PB}{PC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1\iff$ $\boxed{\frac {c^2}{b^2}\cdot \frac xy=1}$ , i.e.

$\boxed{\frac {cx}b=\frac {by}c}\ (1)\ .$ Prove easily that $\frac {OS}{OA}=\frac {BS}{BA}\cdot\frac {\sin\widehat{OBS}}{\sin\widehat{OBA}}=$ $\frac {ac\cdot\cos C}{kc}\cdot\frac {\cos A}{\cos C}=\frac {a\cdot\cos A}k$ , i.e. $\boxed{\frac {OS}{OA}=\frac {a\cdot\cos A}k}\ (2)\ .$ Apply the Cristea's theorem to the cevian $[AO$

in $\triangle ABC\ :\ \frac {FB}{FA}\cdot SC+\frac {EC}{EA}\cdot SB=\frac {OS}{OA}\cdot BC$ $\iff$ $\boxed{yb\cdot\cos B+xc\cdot\cos C=a\cdot\cos A}\ (3)\ .$ From $(1)$ and $(3)$ obtain that $\boxed{\frac {cx}b=\frac {by}c=\cos A}\ (4)\ .$ Indeed, the

relation $c\cdot \cos B+b\cdot\cos C=a$ is true. Apply the remarkable relation $\frac {OF}{OE}=\frac {SB}{SC}\cdot \frac {AF}{AB}\cdot\frac {AC}{AE}\implies$ $\boxed{\frac {OF}{OE}=\frac {c\cdot\cos C}{b\cdot\cos B}\cdot \frac {x+1}{y+1}}\ (5)\ .$ In conclusion, $\frac {OF}{OE}\cdot\frac {ME}{MA}\cdot\frac {NA}{NF}=$

$\frac {c\cdot\cos C}{b\cdot\cos B}\cdot \frac {x+1}{y+1}\cdot \frac {1-x}{1+x}\cdot \frac {1+y}{1-y}=$ $\frac {\cos C}{\cos B}\cdot\frac {c-cx}{b-by}=$ $\frac {\cos C}{\cos B}\cdot\frac {c-b\cdot\cos A}{b-c\cdot\cos A}=$ $\frac {\cos C}{\cos B}\cdot\frac {a\cdot\cos B}{a\cdot\cos C}=1\implies$ $\frac {OF}{OE}\cdot\frac {ME}{MA}\cdot\frac {NA}{NF}=1\implies$ $AO\cap EN\cap FM\ne\emptyset\ .$

P4. Let the bisectors $[BE$ , $[CF$ in $\triangle ABC$ , where $E\in (AC)$ and $F\in (AB)\ .$ For $M\in [EF]$ denote the distancies $x$ , $y$ , $z$ from $M$ to $BC$ , $CA$ , $AB$ . Prove that $x = y + z\ .$

Extension. Let $\triangle ABC$ and two positive numbers $\beta$ , $\gamma$ so that exist $\left\|\ \begin{array}{c} F\in (AB)\ ,\ \frac {FA}{FB} = \beta\\\\ E\in (AC)\ ,\ \frac {EA}{EC} = \gamma\end{array}\ \right\|\ .$ For $M\in [EF]$ denote the

projections $X$ , $Y$ , $Z$ of $M$ on $BC$ , $CA$ , $AB$ respectively and $MX=x$ , $MY=y$ , $MZ=z\ .$ Prove that $\boxed{ax =\frac {by}{\beta} + \frac {cz}{\gamma}}\ .$

Proof. Denote $:$ the projections $U$ , $P$ of $E$ on $BC$ , $AB$ and $EU=u$ , $EP=p\ ;$ the projections $V$ , $N$ of $F$ on $BC$ , $AC$ and $FV=v$ , $FN=n\ .$ I"ll use an well

known relation $\frac {EM}{EF}\cdot FV+\frac {FM}{FE}\cdot EU=MX\ ,$ i.e. $\boxed{v\cdot\frac {EM}{EF}+u\cdot \frac {FM}{FE}=x}\ (*)\ .$ Therefore, $\left\{\begin{array}{ccccc} MY\parallel FN & \implies & \frac {EM}{EF}=\frac {MY}{FN} & \implies & \frac {EM}{EF}=\frac yn\\\\ MZ\parallel EP & \implies & \frac {FM}{FE}=\frac {MZ}{EP} & \implies & \frac {FM}{FE}=\frac zp\end{array}\right\|$ $\stackrel{(*)}{\implies}$

$v\cdot \frac yn+u\cdot \frac zp=x$ $\implies$ $\boxed{y\cdot \frac vn+z\cdot \frac up=x}\ (1)\ .$ On other hand $\left\{\begin{array}{ccc} \beta =\frac {FA}{FB}=\frac {[FAC]}{[FBC]}=\frac {bn}{av} & \implies & \frac vn=\frac b{a\beta}\\\\ \gamma =\frac {EA}{EC}=\frac {[EAB]}{[ACB]}=\frac {cp}{au} & \implies & \frac up=\frac c{a\gamma}\end{array}\right\|$ $\stackrel{(1)}{\implies}$ $y\cdot \frac {b}{a\beta} + z\cdot \frac {c}{a\gamma}=x\implies$ $ax=\frac {by}{\beta} + \frac {cz}{\gamma}\ .$

P5 Let $\triangle ABC$ and the points $\left\|\ \begin{array}{ccc} E\in (AC) & , & \widehat {EBA}\equiv\widehat {EBC}\\\\ F\in (AB) & , & \widehat {FCA}\equiv\widehat {FCB}\end{array}\ \right\|\ .$ Prove that there is the inequality $\frac {2\cdot EF}{a\sqrt {bc}}\ \le\ \frac {1}{a+b}+\frac {1}{a+c}$ with equality when $b=c\ .$

Proof. Observe that $\left\{\begin{array}{ccc} \frac {EA}c=\frac {EC}a=\frac b{a+c} & \implies & AE=\frac {bc}{a+c}\\\\ \frac {FA}b=\frac {FB}a=\frac c{a+b} & \implies & AF=\frac {bc}{a+b}\end{array}\right\|\ .$ Apply the theorem of Cosines $EF^2=AE^2+AF^2-2\cdot AE\cdot AF\cdot \cos A$ , where $\cos A=\frac{b^2+c^2-a^2}{2bc}\ .$

The required inequality becomes $2\sqrt{bc[(a+b)^2+(a+c)^2]-(a+b)(a+c)(b^2+c^2-a^2)}\le a(2a+b+c)\implies$ $0\le a(b-c)^2(5a+4b+4c)\ ,$ what is true.

P6. Let acute $\triangle ABC$ with bisectors $BE$ , $CF$ where $E\in AC$ , $F\in AB$ and the projections $X$ , $Y$ of $E$ , $F$ on $BC.$ Prove that $\boxed{\begin{array}{cccc} EF^2 & \le & BF\cdot CE & (1)\\\\ XY & \le & \frac a2\left(\frac b{a+b}+\frac c{a+c}\right) & (2)\end{array}}\ \bf\color{red}Wonderful !$

Proof. Apply the Stewart's relation to the cevian $EF$ in $\triangle AEF\ :\ AB\cdot EF^2=EB^2\cdot FA+EA^2\cdot FB-FA\cdot FB\cdot AB\iff$ $c\cdot EF^2=\frac {ac\left[(a+c)^2-b^2\right]}{(a+c)^2}\cdot \frac {bc}{a+b}+$

$\left(\frac {bc}{a+c}\right)^2\cdot \frac {ac}{a+b}-c\cdot \frac {bc}{a+b}\cdot\frac {ac}{a+b}\iff$ $\frac {(a+c)^2(a+b)^2}{abc}\cdot EF^2=(a+b)\left[(a+c)^2-b^2\right]+bc(a+b)-c(a+c)^2\equiv \Delta\ .$ Thus, $\boxed{EF^2\le BF\cdot CE}\iff$

$\frac {abc}{(a+b)^2(a+c)^2}\cdot\Delta\le$ $\frac {ac}{a+b}\cdot\frac {ab}{a+c}\iff$ $\Delta\le a(a+b)(a+c)\iff$ $(a+b)(a+c)^2+bc(a+b)\le b^2(a+b)+c(a+c)^2+a(a+b)(a+c)\iff$

$(a+b)(a+c)^2-a(a+b)(a+c)\le b^2(a+b)-bc(a+b)+c(a+c)^2\iff$ $c(a+b)(a+c)\le b(a+b)(b-c)+c(a+c)^2\iff$ $c(a+b)(a+c)-c(a+c)^2\le$

$b(a+b)(b-c)\iff$ $c(a+c)(b-c)\le b(a+b)(b-c)\iff$ $(b-c)\left[b(a+b)-c(a+c)\right]\ge 0\iff$ $(b-c)\left[\left(b^2-c^2\right)+a(b-c)\right]\ge 0$

$\iff$ $\boxed{(b-c)^2(b+c+a)\ge 0}$ what is true. We have equality iff $b=c\ .$ The segment $[XY]$ is the projection of $[EF]$ on $BC\implies XY\le EF\ .$

$\blacktriangleright\ \left\{\begin{array}{ccc} BX & = & \frac {ac}{a+b}\cdot\cos B\\\\ CY & = & \frac {ab}{a+c}\cdot\cos C\end{array}\right\|\implies$ $(a+b)(a+c)\left(BX+CY\right)=$ $a\left[c(a+c)\cos B+b(a+b)\cos C\right]=$ $a\left[a(c\cdot\cos B+b\cdot\cos C)+c^2\cos B+b^2\cos C\right]=$

$a\left(a^2+c^2\cos B+b^2\cos C\right)=$ $a^3+c\cdot ac\cos B+b\cdot ab\cos C=$ $a^3+\frac 12\cdot c\left(a^2+c^2-b^2\right)+\frac 12\cdot b\left(a^2+b^2-c^2\right)=$ $a^3+\frac 12\cdot a^2(b+c)+\frac 12\cdot\left(b^2-c^2\right)(b-c)=$

$\frac 12\cdot a^2(2a+b+c) +\frac 12\cdot (b+c)(b-c)^2\ge \frac 12\cdot a^2(2a+b+c)\ .$ Therefore, $\boxed{BX+CY\ge \frac {a^2(2a+b+c)}{2(a+b)(a+c)}}\implies$ $XY=a-(BX+CY)\le a-\frac {a^2(2a+b+c)}{2(a+b)(a+c)}=$

$\frac {a[2bc+a(b+c)]}{2(a+b)(a+c)} =$ $\frac a2\left(\frac b{a+b}+\frac c{a+c}\right)\ .$ In conclusion, $\boxed{XY\ \le\frac a2\left(\frac b{a+b}+\frac c{a+c}\right)}\ .$

Remark. This inequality $(1)$ is stronger than $\frac {2\cdot EF}{a\sqrt {bc}}\ \le\ \frac {1}{a+b}+\frac {1}{a+c}$ from P5. Indeed, $EF^2\le BF\cdot CE$ $\iff$ $EF^2\le \frac {ac}{a+b}\cdot\frac {ab}{a+c}\iff$ $EF\le \frac {a\sqrt{bc}}{\sqrt{(a+b)(a+c)}}$ $\iff$

$\frac {2\cdot EF}{a\sqrt{bc}}\le\frac 2{\sqrt{(a+b)(a+c)}}=$ $2\cdot\sqrt{\frac 1{a+b}}\cdot\sqrt{\frac 1{a+c}}\le$ $\frac 1{a+b}+\frac 1{a+c}\ .$ With other words, the inequality $EF^2\le BF\cdot CE$ $\implies$ the inequality $\frac {2\cdot EF}{a\sqrt {bc}}\ \le\ \frac {1}{a+b}+\frac {1}{a+c}\ .$

P7 (Stefan Smarandache). Let semiperimeter $p$ of trapezoid $ABCD$ , $AB\parallel CD$ what is inscribed in circle $w=C(O,1).$ Is known $AD=\sqrt 3.$ Prove that $p\le 1+\sqrt{3}$ (clasa a VII - a).

Proof. Let $AB=2x\ ,\ CD=2y$ so that $x<y$ and $AD=BC=\sqrt 3\ .$ Observe that $p=(x+y)+\sqrt 3$ and must prove the maximum of $p$ is equal with $1+\sqrt 3$ , i.e. the maximum

of the sum $x+y$ is equal to $1\ .$ Prove easily that in the extremum situation the center $O$ is interior for the trapezoid with the maximum perimeter. I"ll find the relation between $x$ and $y$ so that

the trapezoid is inscribed in a circle with the length of its radius is equal to $1\ .$ Prove easily that it is $\left(\sqrt 3\right)^2=(x-y)^2+\left(\sqrt {1-x^2}+\sqrt {1-y^2}\right)^2$ , i.e. $x^2+xy+y^2=\frac 34$ , what

means $(x+y)^2=xy+\frac 34\ .$ Thus the sum $(x+y)$ is maximum $\iff$ $(x+y)^2$ is maximum $\iff$ the product $xy$ is maximum $\iff$ $(xy)^3=x^2\cdot {xy}\cdot y^2$ is maximum. But the sum

$x^2+xy+y^2=\frac 34$ is constant. Hence $xy$ is maximum $\iff$ $x^2=xy=y^2=\frac 14$ , i.e. $x=y=\frac 12$ and in this case $p\ \le\ 1+\sqrt 3\ .$

Comentariu. Am evitat trigonometria, desi era mai simplu si redactarea mai scurta. Intr-adevar, notand mijloacele $M$ , $N$ ale bazelor $[AB]$ si $[CD]$ respectiv

si $m(\angle AOM)=\alpha$ , $m(\angle DON) =\beta$ se obtine $\alpha +\beta =60^{\circ}$ si $x=\sin\alpha$ , $y=\sin\beta\ .$ Deci suma $x+y$ este maxima $\iff$ $\sin\alpha +\sin\beta=$

$\frac 12\cdot\sin\frac {\alpha +\beta}{2}\cdot\cos\frac {\alpha -\beta}{2}=$ $\cos\frac {\alpha -\beta}{2}$ este maxim $\Longleftrightarrow\ \alpha =\beta =30^{\circ}\ \Longleftrightarrow\ x=y=\frac 12\ .$ Daca $AB=l$ , unde $l\ <\ 2$ , atunci notam cercul circumscris $w=C(O,1)$ ,

$I\in AC\ \cap\ BD$ si $\phi =m(\angle AC .$ Se observa c $\frac {BC}{BI}=\frac {AD}{ID}=k=$ $\frac {BC+AD}{BI+ID}=$ $\frac {BC+AD}{BD}$ , unde $k=2\cos\phi$ (constant). Deci $BC+AD$ este maxim daca si numai daca $BD$

este maxim, adica $I\equiv O\ .$ In concluzie, $\boxed{\ p\ \le\ l\ +\ 2\cdot\cos\phi\ }\ .$

P8. Fie un paralelogram $ABCD$ si un punct $E\in BD\ .$ Construim simetricul $S$ al lui $C$ fata de $E\ .$

si paralelogramul $SGAF$ , unde $F \in AB$ si $G\in AD\ .$ Aratati ca $EF\ ||\ AC$ si $GE\in FG\ .$

Demonstratie (vectorial). Fixam originea sistemului de vectori in $O\in AC\cap BD$ , adica $\vec{OX}=X$ si $\vec{XY}=Y-X\ .$ Deci

$C=-A$ , $D=-B$ , exista $m\in\mathbb R$ astfel incat $\boxed{E=mB}$ si $S=2E-C$ , adica $\boxed{S=A+2mB}\ .$

$\blacktriangleright\ \left|\begin{array}{ccccc} SF\ \parallel\ AD & \iff & F=S+u(A-D) & \iff & F=(1+u)A+(2m+u)B\\\\ F\in AB & \iff & F=A+v(A-B) & \iff & F=(1+v)A-vB\end{array}\ \right|\ \iff\ \left|\begin{array}{c} 1+u=1+v\\\\ 2m+u=-v\end{array}\right|\ .$

Deci $u=v=-m$ , adica $\boxed{F=(1-m)A+mB}\ .$ Am folosit ca $F\in AB\ \Longleftrightarrow\ AF\ \parallel\ AB\ .$

$\blacktriangleright\ \ S+A=F+G\ \Longleftrightarrow\ G=S+A-F=A+2mB+A-(1-m)A-mB$ , adica $\boxed{G=(1+m)A+mB}\ .$

Se observa ca $\left|\begin{array}{c} \overrightarrow{EF}=F-E=(1-m)A\\\\ \overrightarrow{EG}=G-E=(1+m)A\end{array}\right|$ , adica $EF\ \parallel\ AC$ si $(1+m)\cdot \overrightarrow{EF}=(1-m)\cdot\overrightarrow{EG} \iff\ E\in FG\ .$

P9. Let an acute $\triangle ABC$ and an its interior $D$ so that $\frac {DA}{DB}=\frac {CA}{CB}$ and $m\left(\angle ADB\right)=C+\phi\ <\ 180^{\circ}\ .$ Prove that $\frac {AB}{AC}\cdot\frac{DC}{DB}=2\cdot \sin\frac {\phi}{2}\ .$

Proof. Let the interior $E$ of $\widehat{ADB}$ so that $m(\angle ADE)=C$ and $DE=DB\ .$ Thus, $m(\angle BDE)=\phi$ , $\frac {EB}{DB}=2\cdot \sin\frac {\phi}{2}$ and $\triangle ACB\sim\triangle ADE$

because $\widehat{ACB}\equiv\widehat{ADE}$ and $\frac {CA}{CB}=\frac {DA}{DB}\ .$ Obtain $\widehat{BAC}\equiv\widehat{EAD}$ , $\widehat{BAE}\equiv\widehat{CAD}$ and $\frac {AB}{AE}=\frac {AC}{AD}.$ So $\triangle BAE\sim\triangle CAD\implies$

$\frac {AB}{AC}=\frac {EB}{DC}$ , $\frac {AB}{AC}\cdot\frac{DC}{DB}=$ $\frac {EB}{DC}\cdot\frac {DC}{DB}=$ $\frac {EB}{DB}=2\cdot \sin\frac {\phi}{2}$ , i.e. $\frac {AB}{AC}\cdot\frac{DC}{DB}=2\cdot \sin\frac {\phi}{2}\ .$

P10. Fie $\triangle\ ABC$ cu $AB=AC$ si $D\in [AC]$ astfel incat $CD=2DA\ .$ Fie $M\in [BD]\ .$ Sa se arate ca $\widehat{MCB}\equiv\widehat{MBA}\ \Longleftrightarrow\ AM\perp MC$ (Marcel Chirita, GM 9/2008).

Dem. Pentru $E\in (BD$ pentru care $AE\ \parallel\ BC$ avem $2\cdot AE=BC\ ,$ adica $EA\ \perp\ EC$ si $\widehat {CAE}\ \equiv\ \widehat {ACB}\ \equiv\ \widehat {CME}\ .$ Asadar $AMCE$ este inscriptibil, i.e.

$AM\ \perp\ MC\ .$ Astfel $\frac {MB}{MD}=\frac {3a^2}{4b^2}\ \ \wedge\ \ \tan\ \left(\widehat{MCB}\right)=\frac {4S}{a^2+4b^2}\ ,$ unde $S=[ABC]$ si daca $F$ este simetricul lui $C$ in raport cu $A\ ,$ atunci $\widehat{AMF}\equiv\widehat{AMD}\ .$

P11. Let acute $\triangle ABC$ , $D\in (AC)$ so that $\frac {DC}{DA}=m$ and $M\in (BD)$ so that $\widehat{MCA}\equiv\widehat{DBC}\ .$

Prove that $\tan\widehat{AMC}=\frac {m\tan B\tan C}{(m-1)\tan B-\tan C}\ .$ For example, if $m=1+\frac {\tan C}{\tan B}$ then $MA\perp MC\ .$

Proof. Let $m\left(\widehat{AMC}\right)=\phi$ and $E\in BD$ so that $AE\parallel BC\ .$ Observe that $m\left(\widehat {CME}\right)=m\left(\widehat {ACB}\right)=m\left(\widehat {CAE}\right)\implies$ $\widehat {CME}\equiv\widehat {CAE}$ , i.e. $AMCE$ is a cyclic

quadrilateral. Thus, $m=\frac {DC}{DA}=\frac a{AE}\implies$ $\boxed{AE=\frac am}\ (1)\ .$ Apply the theorem of Sines in the triangle $ACE\ :\ \frac {AE}{\sin(\phi -C)}=\frac b{\sin\phi}\ \stackrel{(1)}{\iff}\ a\sin\phi =$ $mb\sin (\phi -C)\iff$

$\sin (B+C)\sin\phi =m\sin B\sin (\phi -C)\iff$ $\left(\tan B+\tan C\right)\tan\phi =m\tan B\left(\tan \phi -\tan C\right)\iff$ $\boxed{\tan\phi =\tan\widehat{AMC}=\frac {m\tan B\tan C}{(m-1)\tan B-\tan C}}\ (*)\ .$

P12. Let an acute $\triangle ABC$ with crcumcircle $w=\mathbb C(O,R)$ and $D\in (BC),$ $\{A,H\}=\{A,D\}\cap w,$ $\left\{\begin{array}{ccc} E\in AC & ; & DE\perp AC\\\\ F\in AB & ; & DF\perp AB\end{array}\right\|$ and $G\in EF\cap AD.$ Prove that $\boxed{AG\cdot AH=AD^2}\ (*).$
$\mathrm{END}$

This post has been edited 458 times. Last edited by Virgil Nicula, Aug 6, 2017, 8:34 AM

Comment

0 Comments

Submit

orzzzzzzzzz
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

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

by Virgil Nicula, May 11, 2016, 6:14 PM

0 Comments