381. Probleme date la diverse concursuri.

by Virgil Nicula, Jul 24, 2013, 12:57 AM

PP0. Examen National de Definitivat - 2013.

1. Fie o prog. aritm. $a_n\in\mathbb R$ , $n\in\mathbb N^*$ cu ratia $r>0$ si o prog. geom. $b_n\in\mathbb R$ , $n\in\mathbb N^*$ cu ratia $q$ , $a_1=b_1>0$ si $a_2=b_2$ . Sa se arate ca $q>1$ si $(\forall )\ k\ge 3\ ,\ b_k\ge a_k$ .

2. In $\triangle ABC$ fie $[AD$ bisectoarea lui $\widehat{BAC}$ , $D\in (BC)$ . Sa se calculeze $AD$ stiind ca $b=c=3$ si $m\left(\widehat{BAC}\right)=120^{\circ}\ ;$ sa se arate ca $c+DC=b+DB\ \iff\ b=c$ .

3. Se considera functia $f_n(x)=\frac {e^x}{1+x^n}$ , unde $x\in [0,1]$ ai $n\in\mathbb N^*$ . Sa se calculeze $\int_0^1(1+x)f_1(x)\ \mathrm{dx}$ si sa se arate ca $\lim_{n\to\infty}\int_0^1f_n(x)\ \mathrm{dx}=e-1$ .

An easy extension. Se considera functia $f_n(x)=\frac {g(x)}{1+x^n}$ , unde $g$ este o functie continua pe $[0,1]$ si $n\in\mathbb N^*$ . Sa se arate ca $\lim_{n\to\infty}\int_0^1f_n(x)\ \mathrm{dx}=\int_0^1g(x)\ \mathrm{dx}$ .

Demonstratie.

1. Notam $a_1=b_1=a>0$ . Astfel, $a_2=b_2\iff a+r=aq\implies q=1+\frac ra>1$ . Se observa ca $\boxed{r=a(q-1)}\ (*)$ . Fie $k\ge 3$ . Asadar, $b_k\ge a_k\iff$

$aq^{k-1}\ge a+(k-1)r\stackrel{(*)}{\iff}$ $q^{k-1}-1\ge(k-1)(q-1)\ \stackrel{(q-1>0)}{\iff}\ \sum_{s=0}^{k-2}$ $q^s\ge k-1$ , ceea ce este evident deoarece $(\forall )\ s\in\overline{0,k-2}\ ,\ q^s>1$ si suma a $(k-1)$

numere mai mari decat $1$ este un numar real mai mare decat $(k-1)$ .

Remarca. Dupa parerea mea, s-ar fi potrivit si o a treia cerinta: sa se determine una din sumele $\sum_{k=1}^n\frac {a_k}{b_k}$ sau $\sum_{k=1}^na_kb_k$ in functie de $\{a,q,n\}$ .

2. Daca $b=c=3$ si $m\left(\widehat{BAC}\right)=120^{\circ}$ , atunci $\triangle ABD$ este $D$ -dreptunghic cu $m\left(\widehat{BAD}\right)=60^{\circ}$ , adica $AD=\frac 12\cdot AB\iff$ $AD=\frac 32$ . Revenim la cazul

general. Asadar, $\frac{DB}{c}=\frac {DC}{b}=\frac {a}{b+c}$ si $c+DC=b+DB\ \iff\ c+\frac {ab}{b+c}=b+\frac {ac}{b+c}\iff$ $b-c=\frac {a(b-c)}{b+c}\iff$ $b=c$ deoarece $b+c\ne a$ .

Remarca. $D\in (BC)\ ,\ \frac {DB}{DC}=m\in \mathbb R^*_+$ $\implies$ $c+DC=b+DB\iff$ $c+\frac {a}{m+1}=b+\frac{ma}{m+1}\iff$ $m=\frac {s-b}{s-c}\iff$ $D\in BC\cap w$ , unde $w$ este cercul inscris al

$\triangle ABC$ , adica $ABDC$ este tangential (la limita!) $\iff$ $AB+DC=AC+DB\iff$ $D\in BC\cap w$ , adica $D$ este punctul de tangenta intre $BC$ si cercul inscris in $\triangle ABC$ .

3. $\int_0^1(1+x)f_1(x)\ \mathrm{dx}=\int_0^1(1+x)\cdot\frac {e^x}{1+x}\ \mathrm{dx}=$ $\int_0^1e^x\ \mathrm{dx}=$ $\left|e^x\right|_0^1=e-1$ si $\int_0^1e^x\ \mathrm{dx}-\int_0^1f_n(x)\ \mathrm{dx}=$ $\int\left[e^x-f_n(x)\right]\ \mathrm{dx}=$ $\int_0^1\left(e^x-\frac {e^x}{1+x^n}\right)\ \mathrm{dx}=$

$\int_0^1\frac {x^ne^x}{1+x^n}\ \mathrm{dx}$ . Asadar, $0\le \int_0^1\frac {x^ne^x}{1+x^n}\ \mathrm{dx}\le$ $\int_0^1ex^n\ \mathrm{dx}=$ $\left|\frac {ex^{n+1}}{n+1}\right|_0^1=\frac e{n+1}\rightarrow 0$ . In concluzie, $\lim_{n\to\infty}\int_0^1\frac {x^ne^x}{1+x^n}\ \mathrm{dx}=0\implies$ $\lim_{n\to\infty}\int_0^1f_n(x)\ \mathrm{dx}=\int_0^1e^x\ \mathrm{dx}=e-1$ .

PP1. Prove that the Mandelbrot's relation $IA\cdot IB\cdot IC=4Rr^2$ .

Proof 1. Can use the well-known relation $\boxed{IA^2=\frac {bc(s-a)}{s}}\ (*)$ . Therefore, $IA^2\cdot IB^2\cdot IC^2=$ $\frac {(abc)^2\cdot (s-a)(s-b)(s-c)}{s^3}=$ $\frac {(4Rsr)^2\cdot sr^2}{s^3}\implies$ $\boxed{IA\cdot IB\cdot IC=4Rr^2}$ .

I"ll prove $(*)$ . Let circumcircle $w$ of $\triangle ABC\ ,\ \{A,I\}\cap w=\{A,S\}\implies$ $SB=SC=SI=SI_a$ and $\triangle ABI_a\sim \triangle AIC\iff$ $\frac {AB}{AI}=\frac {AI_a}{AC}\iff$ $AI\cdot AI_a=bc\iff$

$AI\cdot (AI+2\cdot IS)=bc\iff$ $AI^2=bc-2\cdot IA\cdot IS=bc-2\cdot 2Rr=$ $bc-4Rr=$ $\frac {bcs-abc}{s}\implies$ $IA^2=\frac {bc(s-a)}{s}=bc-4Rr$ .

Proof 2. $\prod\sin\frac A2=\prod\sqrt{\frac {(s-b)(s-c)}{bc}}=$ $\frac {(s-a)(s-b)(s-c)}{abc}=\frac {sr^2}{4Rrs}\implies$ $\boxed{\prod\sin\frac A2=\frac r{4R}}\ (*)$ . Thus, $\sin\frac A2=\frac r{IA}$ $\iff$

$IA\cdot\sin\frac A2=r$ and analogously $\implies$ $\prod \left(IA\cdot\sin\frac A2\right)=$ $r^3\ \stackrel{(*)}{\iff}\ \prod IA\cdot \frac r{4R}=$ $r^3\iff IA\cdot IB\cdot IC=4Rr^2$ .

Remark. Since $\left\{\begin{array}{ccccc} \frac {I_aA}{IA} & = & \frac {s}{s-a} & = & \frac {r_a}{r}\\\\ \frac {AI}{s-a} & = & \frac {AI_a}{s} & = & \frac 1{\cos\frac A2}\end{array}\right\|$ we can prove easily that $I_aA\cdot I_bB\cdot I_cC=4Rs^2$ .

PP2. Equilateral triangles are constructed on the sides of the same rectangle in two different ways as shown below. The area of the bold triangle on the left

is two times the area of the bold triangle on the right. Find the ratio of the length of the shorter side of the rectangle to the length of the longer side (sursa).

Proof Let a rectangle $ABCD$ with $1=AB>AD=a$ . Denote the points $\{M,N,P,R\}$ so that: $AMB\ ,\ AND\ ,\ ARB\ ,\ APD$ are equilateral triangles; $AB$ separates $M\ ,\ C$

and doesn't separate $R\ ,\ C\ ;\ AD$ separates $N\ ,\ C$ and doesn't separate $P\ ,\ C$ (see the lower figure). Suppose that the area $[MNC]=2\cdot [RPC]$ . I"ll prove that $MNC$ , $RPC$

are equilateral triangles and find the ratio $\frac {AD}{AB}=a$ . Observe that $\left\{\begin{array}{c} ND\parallel AP\perp AM\\\\ PD\parallel AN\perp AR\end{array}\right\|$ and by $\mathrm{s.a.s.,}\ \left\{\begin{array}{ccc} \triangle AMN\equiv\triangle BMC\equiv\triangle DCN & \implies & MN=MC=CN\\\\ \triangle BRC\equiv\triangle ARP\equiv\triangle DCP & \implies & RC=RP=CP\end{array}\right\|$ .

Therefore, the triangles $MNC$ si $PRC$ are equilateral. Apply the generalized Pythagoras' theorem $:$

$\left\{\begin{array}{ccccccccccc} MN^2 & = & AM^2+AN^2-2\cdot AM\cdot AN\cdot\cos\widehat{MAN} & \implies & MN^2 & = & 1+a^2-2a\cos 150^{\circ} & \implies & MN^2 & = & a^2+a\sqrt 3+1\\\\ PR^2 & = & AP^2+AR^2-2\cdot AP\cdot AR\cdot\cos\widehat{PAR} & \implies & PR^2 & = & a^2+1-2a\cos 30^{\circ} & \implies & PR^2 & = & a^2-a\sqrt 3+1\end{array}\right\|$ . Therefore,

$[MNC]=2\cdot [RPC]\iff$ $MN^2=2\cdot PR^2\iff$ $a^2+a\sqrt 3+1=2\left(a^2-a\sqrt 3+1\right)\iff$ $a^2-3a\sqrt 3+1=0\ ,\ 0<a<1\iff$ $\boxed{\ a=\frac {3\sqrt 3-\sqrt {23}}2\ }$ .

PP3 Let a rectangle $ABCD$ with the center $O$ and $AB\ne BC$ . A line through $O$ cut $AB\ ,\ BC$ in $E\ ,\ F$ respectively.

Denote the midpoints $M\ ,\ N$ of $[CD]\ ,\ [AD]$ respectively. Prove that $FM\perp EN\ \iff\ EF\perp BD$ .

Proof 1. I"ll a metrical characterization of the perpendicularity between two lines. Suppose w.l.o.g. that $F\in (BC)$ si $BF>FC$ . Denote $G\in EF\cap AD$ , $AB=b$ ,

$AD=a$ , $FC=x$ , $AE=y$ . Thus, $FM\perp EN\iff$ $FE^2-FN^2=ME^2-MN^2\iff$ $\left[(a-x)^2+(b+y)^2\right]-$ $\left[b^2+\left(\frac a2-x\right)^2\right]=$

$\left[a^2+\left(y+\frac b2\right)^2\right]-\frac {a^2+b^2}{4}\ \iff\ ax=by\iff$ $\frac {AB}{AG}=\frac {AD}{AE}\iff$ $ABD\sim AGE\iff$ $\widehat {AGE}\equiv\widehat {ABD}\iff ABOG$ este ciclic $\iff EF\perp BD$ .

Proof 2. Suppose w.l.o.g.that $F\in (BC)$ and $BF>FC$ . Let the midpoint $P$ of $[BC]$ and $\left\{\begin{array}{c} R\in EN\cap BD\\\\ S\in EF\cap CD\\\\ X\in FM\cap BD\end{array}\right\|$ $\implies\frac {RN}{RE}=$ $\frac {ON}{BE}=$ $\frac {OP}{BE}=\frac {FP}{FB}$ $\iff \left\{\begin{array}{c} \frac {RN}{RE}=\frac {FP}{FB}\\\\ (\ NP\parallel AB\ )\end{array}\right\|$

$\iff$ $FR\parallel AB$ $\iff FR\perp AD$ . Apply Menelaus' theorem of $\overline {FMX}/\triangle BCD\ :\ \frac {XD}{XB}\cdot\frac {FB}{FC}\cdot\frac {MC}{MD}=1$ $\iff\frac {XB}{XD}=\frac {FB}{FC}=$ $\frac {EB}{SC}=\frac {EB}{EA}\iff$ $\frac {XB}{XD}=\frac {EB}{EA}\iff$

$XE\parallel AD$ $\iff XE\perp AB$ $\iff XE\perp RF$ $\implies RF$ is altitude $\triangle EFX\implies$ $\boxed{FM\perp EN}$ $\iff FX\perp ER\iff\ R$ is orthocenter of $\triangle EFX\iff$ $\boxed{EF\perp BD}$

$\blacktriangleright$ Prove easily that if $ABCD$ is a parallelogram and $O\in EF$ , then $\left\{\begin{array}{c} FR\parallel AB\ ;\ EX\parallel AD\\\\ \frac {RB}{RD}\ =\ \frac {XB}{XD}\end{array}\right\|$ , i.e. $(B,D,R,X)$ is harmonically. If $ABCD$ is a rectangle and $O\in EF\perp BD$ ,

then $AD$ , $CD$ are the bisectors of $\widehat {RAX}$ , $\widehat {RCX}$ respectively, $\left\|\begin{array}{ccc} RO\cdot RX& = & RB\cdot RD\\\ OA^2 & = & OR\cdot OX\end{array}\right\|$ and a circle which pass through $R$ , $X$ is ortogonal to the circumcircle of $ABCD$ .

PP4. Let $\triangle ABC$ and for $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ denote $\left\{\begin{array}{ccc} P\in AD\cap CF & ; & R\in AD\cap BE\\\\ \ \frac {EA}{EC}=m & ; & \frac {FA}{FB}=n\end{array}\right\|$ .

Prove that $\frac {m}{AR} + \frac {n}{AP}=\frac {1+m+n}{AD}$ . For $m=n=1$ get that $\frac {1}{\overline {AR}} + \frac {1}{\overline {AP}}=\frac {3}{\overline {AD}}$ .

Proof. Apply the Menelaus' theorem to the transversals $:\ \left\{\begin{array}{cccc} \overline{CPF}/\triangle ABD\ : & \frac{CB}{CD}\cdot\frac{PD}{PA}\cdot \frac {FA}{FB}=1 & \implies & \frac{DC}{BC}=n\cdot\frac{PD}{PA}\\\\ \overline{BRE}/\triangle ACD\ : & \frac{BC}{BD}\cdot\frac{RD}{RA}\cdot \frac {EA}{EC}=1 & \implies & \frac{DB}{BC}=m\cdot\frac{RD}{RA}\end{array}\right\|$ . Thus, $\frac{DB}{BC}+\frac{DC}{BC}=1\implies$

$m\cdot \frac{RD}{RA}+n\cdot \frac{PD}{PA}=1\iff$ $m\cdot \frac{AD-RA}{RA}+n\cdot \frac{AD-PA}{PA}=1\iff$ $m\cdot \frac{AD}{RA}+n\cdot \frac{AD}{PA}=m+n+1\iff$ $\frac {m}{AR} + \frac {n}{AP}=\frac {1+m+n}{AD}$ .

PP5. Let a trapezoid $ABCD$ , where $AB\parallel CD$ and $O\in AC\cap BD$ . Denote the second intersection $K$ between the circumcircles

of $\triangle AOD$ , $\triangle BOC$ and $S\in KO\cap AB$ . Prove $\frac {SA}{SB}=\left(\frac {OA}{OB}\right)^2$ , i.e. the ray $[OS$ is the $O$ -symmedian of $\triangle AOB$ .

Proof. $AB$ cut again the circumcircles $w_1$ , $w_2$ of $\triangle AOD$ , $\triangle BOC$ in $E$ , $F$ respectively. Thus, $\frac {AO}{AC}=\frac {BO}{BD}\iff$ $\boxed{AO\cdot BD=BO\cdot AC}\ (1)$ . Since $S$ belongs to the

radical axis of $w_1$ , $w_2$ obtain that $SA\cdot SE=SB\cdot SF\iff$ $SA\cdot (BE-SB)=SB\cdot (AF-SA)\iff$ $\boxed{SB\cdot AF=SA\cdot BE}\ (2)$ . I"ll use the powers of $A$ , $B$ w.r.t.

$w_2$ , $w_1$ respectively $:\ \left\{\begin{array}{c} AO\cdot AC=AF\cdot AB\\\\ BE\cdot BA=BO\cdot BD\end{array}\right\|\ (3)$ . In conclusion, $\left\{\begin{array}{cc} AO\cdot BD=BO\cdot AC & (1)\\\\ SB\cdot AF=SA\cdot BE & (2)\\\\ \left|\begin{array}{c} AO\cdot AC=AF\cdot AB\\\\ BE\cdot BA=BO\cdot BD\end{array}\right| & (3)\end{array}\right\|\bigodot\implies$ $AO^2\cdot SB=BO^2\cdot SA\implies$ $\frac {SA}{SB}=\left(\frac {OA}{OB}\right)^2$ .

PP6. Let an acute and $A$ -isosceles $\triangle ABC$ . Denote the midpoints $M$ , $N$ of the sides $[AB]$ , $[AC]$ respectively. Prove that $BN\perp CM\iff \tan A=\frac 34\ .$

Proof. Prove easily that in any triangle $ABC$ there is the equivalence $BN\perp CM\iff b^2+c^2=5a^2$ . If $\triangle ABC$ is isosceles, i.e. $b=c$ , then $2b^2=5a^2\iff \frac {a}{\sqrt 2}=\frac {b}{\sqrt 5}=\frac {c}{\sqrt 5}$

Suppose w.l.o.g. $b=c=\sqrt 5$ and $a=\sqrt 2$ . Apply the generalized Pythagoras' theorem $a^2=b^2+c^2-2bc\cdot \cos A\iff \cos A=\frac 45$ . Thus, $\sin A=\frac 35$ and $\boxed{\tan A=\frac 34}$ .

Observe that $\triangle ABB^{\prime}$ , where $B^{\prime}$ is the projection of $B$ on $AC$ has the lengths of the sides proportionally with $3$ , $4$ , $5$ , more exactly $\frac {AB}{5}=\frac {B^{\prime}A}{4}=\frac {BB^{\prime}}{3}$ . Remark. The relation

$\tan A=\frac 34$ means there is a synthetical proof. Can use the chain of the equivalencies $:\ BN\perp CM\iff$ $G$ belongs to the circle with the diameter $[BC]\iff G$ is the incenter of

$\triangle ABS$ , where $S$ is the projection of $B$ on $AC$ and $H\in AG\cap BS\iff$ $GA=BC\iff$ $3\cdot AH=4\cdot BC\iff$ $6\cdot HL=BC\iff$ $2\cdot LT=BC$ , where $L\in BC$ ,

$AL\perp BC$ and $T$ is the symmetrical point of $G$ w.r.t. $BC$ . With other words, on the segment $\overline {AGHLT}$ have the chain of the equal ratios $\frac {AG}{6}=\frac {GH}{2}=\frac {HL}{1}=$ $\frac {LT}{3}=\frac {BC}{6}$ and the

division $(A,G,H,T)$ is harmonically, i.e. $\frac {GA}{GH}=\frac {TA}{TH}$ . For example, from the relation $\frac {BC}{HA}=\frac 34$ obtain that $\frac {2R\cdot \sin A}{2R\cdot \cos A}=\frac 34$ , i.e. $\boxed{\tan A=\frac 34}$ .

An easy extension. Let an acute $\triangle ABC$ . Let the midpoints $M$ , $N$ of $[AB]$ , $[AC]$ respectively. Prove that $BN\perp CM\implies\tan A\le \frac 34$ with equality if $b=c$ .

PP7. Let an $A$ -rightangled $\triangle ABC$ . Prove that $\frac {a^2}{bc}+\frac {b+c}{a}\ \ge\ 2+\sqrt 2$ (Crux Mathematicorum).

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

On other hand $\frac {a^4}{b^2c^2}+\frac {(b+c)^2}{a^2}=$ $\frac {a^4}{2b^2c^2}+\frac {a^4}{2b^2c^2}+\frac {(b+c)^2}{a^2}\ge$ $3\sqrt[3]{\frac{a^6(b+c)^2}{4b^4c^4}}=$ $6\cdot \frac {a^2}{2bc}\cdot\sqrt [3]{\frac {(b+c)^2}{4bc}}\ge 6\cdot 1\cdot\sqrt [3] 1=6\implies$ $\boxed{\frac {a^4}{b^2c^2}+\frac {(b+c)^2}{a^2}\ge 6}\ (2)$ because

$\left\{\begin{array}{ccc} b^2+c^2\ge 2bc & \implies & \frac {a^2}{2bc}\ge 1\\\\ b+c\ge 2\sqrt{bc} & \implies & \frac {(b+c)^2}{4bc}\ge 1\end{array}\right\|$ . In conclusion, from $(1)$ , $(2)$ and $(*)$ obtain that $\left(\frac {a^2}{bc}+\frac {b+c}{a}\right)^2\ge 6+4\sqrt 2=$ $\left(2+\sqrt 2\right)^2\implies$ $\frac {a^2}{bc}+\frac {b+c}{a}\ \ge\ 2+\sqrt 2$ .

Generalization (own). Prove that in any triangle $ABC$ there is the implication $a=\max\ \{a,b,c\}\ \Longrightarrow\ \frac {b^2+c^2}{bc}+$ $\frac {b+c}{a}\ \ge\ 2+\frac {1}{\sin\frac A2}\cdot\sqrt {1+\left(\frac {b-c}{2a}\right)^2}\ .$

PP8 (Stefan Smarandache). Let $p$ be the semiperimeter of $ABCD$ with $AB\parallel CD$ , $AD=BC=\sqrt 3$ and which is inscribed in the circle $w=C(O,1)$ . Prove that $p\le 1+\sqrt{3}.$

Proof. Denote $\left\{\begin{array}{c} AB=2x\ ;\ CD=2y\ ;\ x\ge y\\\\ AD=BC=\sqrt 3\end{array}\right\|\ .$ Thus, $p=(x+y)+\sqrt 3$ and we must prove that the maximum of $p$ is $1+\sqrt 3$ , i.e. the maximum of $(x+y)$ is equally to $1\ .$

Prove easily that in the extremum case the center $O$ is an interior point and $\sqrt {\left(\sqrt 3\right)^2-(x-y)^2}=$ $\sqrt {1-x^2}+\sqrt {1-y^2}$ , i.e. $x^2+xy+y^2=\frac 34\implies$ $(x+y)^2=xy+\frac 34\ .$

Thus, the sum $(x+y)$ is $\max \Longleftrightarrow\ (x+y)^2$ is $\max \iff$ $xy$ is $\max\Longleftrightarrow$ $(xy)^3=x^2\cdot {xy}\cdot y^2$ is $\max\ .$ Since $x^2+xy+y^2=\frac 34$ (constant) obtain that $x$ is $\max\iff$

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

Otherwise. Denote the midpoints $M$ , $N$ of $[AB]$ , $[CD]$ respectively and $m(\angle AOM)=\alpha$ , $m(\angle DON) =\beta$ . Thus, $\alpha +\beta =60^{\circ}$ si $x=\sin\alpha$ , $y=\sin\beta$

and the sum $x+y$ is $\max\Longleftrightarrow$ $\sin\alpha +\sin\beta=$ $2\sin\frac {\alpha +\beta}{2}\cdot\cos\frac {\alpha -\beta}{2}=$ $\cos\frac {\alpha -\beta}{2}$ is $\max\Longleftrightarrow\ \alpha =\beta =$ $30^{\circ}\ \Longleftrightarrow\ x=y=\frac 12\ .$

Concursul Naţional de Matematică "N.N. Mihăileanu" - Constanţa, 2015.

PP9 (Florian Gache). Let $ABCD$ be a square and $E$ be an interior point so that $\triangle EDC$ is

equilateral. Let $F\in AC\cap DE$ and the symmetrical $G$ of $B$ w.r.t. $A$ . Prove that $AE\parallel FG$ .

Proof 1. $\left\{\begin{array}{ccc} m\left(\widehat{EAF}\right) & = & 30^{\circ}\\\\ m\left(\widehat{AEF}\right) & = & 75^{\circ}\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} AG & = & EC\\\\ AF & = & EA\\\\ \widehat{GAF} & \equiv & \widehat{CEA}\end{array}\right\|$ $\implies$ $\triangle AGF\ \stackrel{(s.a.s)}{\sim}\ \triangle ECA\implies$ $\widehat{AFG}\equiv\widehat{EAC}\implies AE\parallel GF$ .

Proof 2. Denote $\left\{\begin{array}{ccc} K & \in & DE\cap AB\\\\ L & \in & GF\cap AD\end{array}\right\|$ and suppose w.l.o.g. $AB=1$ . Thus, $m\left(\widehat{ADK}\right)=\left(\widehat{ABF}\right)=30^{\circ}\implies$ $\boxed{AK=\frac 1{\sqrt 3}}\ ,$ $\boxed{KB=1-\frac 1{\sqrt 3}}$ . Observe

that $KB=KF\implies$ $FB=2KB\cos 30^{\circ}\implies$ $FB=KB\sqrt 3\implies \frac {BK}{BF}=\frac 1{\sqrt 3}$ . Thus, $m\left(\widehat{EBK}\right)=m\left(\widehat{EBF}\right)=m\left(\widehat{FBD}\right)=15^{\circ}$ . In conclusion,

$\frac {EK}{EF}=\frac {BK}{BF}=\frac 1{\sqrt 3}=\frac {AK}{AG}\implies$ $\frac {EK}{EF}=\frac {AK}{AG}\implies AE\parallel GF$ . Otherwise. $BF\perp CE$ , $GF=GD$ and $DF=DL$ . Thus, $DA=DE\implies GF\parallel AE$ .

PP10 (O.M. Ukraina, 2013). Let an obtuse $\triangle ABC$ with circumcircle $w$ for which denote $:$ circle $w_c$ so that $C\in w_c$ and $\{A\}=AB\cap w_c\ ;$

circle $w_b$ so that $B\in w_b$ and $\{A\}=AC\cap w_b\ ;\ \{A,D\}=w_b\cap w_c\ ;\ \{A,E\}=AD\cap w$ . Prove that $D$ is the midpoint of $[AE]$ .

Proof. $\left\{\begin{array}{cccccccc} \odot\begin{array}{ccc} \nearrow & \widehat{BAD}\equiv\widehat{ACD} & \searrow\\\\ \searrow & \widehat{ABD}\equiv\widehat{CAD} & \nearrow\end{array}\odot & \implies & \triangle ABD \sim\triangle CAD & \implies & \frac{AD}{CD}=\frac{BD}{AD} & \implies & AD^2=BD\cdot CD & (1)\\\\ \odot\begin{array}{ccc} \nearrow & \widehat{DBE}\equiv\widehat{DEC} & \searrow\\\\ \searrow & \widehat{BED}\equiv\widehat{ECD} & \nearrow\end{array}\odot & \implies & \triangle BED \sim\triangle ECD & \implies & \frac{ED}{CD}=\frac{BD}{ED} & \implies & ED^2=BD\cdot CD & (2)\end{array}\right\|$ $\implies \triangle DBE\sim\triangle ABC \sim\triangle DEC$ .

PP11 (O.M. Romania). Let the parallelogram $ABCD$ for which denote $:$ the circle $w_1=\mathbb C\left(O_1,r_1\right)$ so that $A\in w_1$ and $\{B\}=BD\cap w_1 ;$ the circle

$w_2=\mathbb C\left(O_2,r_2\right)$ so that $A\in w_2$ and $\{D\}=BD\cap w_2\ ;\ \{A,I\}=w_1\cap w_2$ . Prove that $:\ I\in AC\ ;\ BIDC$ is cyclically $;\ \frac {r_1}{r_2}=\left(\frac {AB}{BC}\right)^2$ .

Proof. Denote $AB=CD=a$ and $AD=BC=b$ . Observe that $BD$ is the common exterior tangent of the circles $w_1$ , $w_2$ and $AI$ is their radical axis. Hence the midpoint

$O$ of $BD$ , i.e. the midpoint of $AC$ belongs to $AI$ , i.e. $I\in AC$ . Since $\widehat {BCI}\equiv$ $\widehat{BCA}\ \stackrel{(AD\parallel BC)}{\equiv}\ \widehat{CAD}$ $\equiv\widehat{IAD}\equiv$ $\widehat{BDI}$ obtain that $\widehat{BCI}\equiv\widehat{BDI}$ , i.e. the quadrilateral

$BIDC$ is cyclically. Thus, $\left\{\begin{array}{ccc} a & = & 2r_1\sin\widehat{BIC}\\\\ b & = & 2r_2\sin\widehat{CID}\end{array}\right\|$ $\implies$ $\frac ab=\frac {r_1}{r_2}\cdot\frac {\sin\widehat{BIC}}{\sin\widehat{CID}}$ . From $\left\{\begin{array}{ccc} \widehat{BIC} & \equiv & \widehat{BDC}\\\\ \widehat{CID} & \equiv & \widehat{CBD}\end{array}\right\|$ and $\frac {\sin\widehat{BDC}}{\sin\widehat{CBD}}=\frac ba$ obtain that $\frac ab=\frac {r_1}{r_2}\cdot\frac ba$ , i.e. $\frac {r_1}{r_2}=\left(\frac ab\right)^2$ .

Lemma. Let $ABC$ be an acute triangle. Define: the circumcircle $c=C(O)$ and the orthocentre $H$ of the triangle $ABC$ ; the middlepoint $M$ of the side $[BC]$ ; the intersection $N$ between

$MH$ and the bisector of the angle $\widehat {BAC}$ ; $D\in AB$ and $E\in AC$ so that $H\in DE$ and $AD=AE$ . Then the point $N$ belongs to the circumcircle $w=C(O_a)$ of the triangle $ADE$ .

Proof. $AH=2R\cos A$ and $P\in c\cap (AO_a\Longrightarrow$ $AP=2R\cos \frac{B-C}{2}\ ,$ $MP=R(1-\cos A)$ . Thus, $\frac{AN}{AP}=\frac{AH}{AH+MP}=\frac{2\cos A}{1+\cos A}$ $\Longrightarrow$

$\boxed {AN=\frac{2\cos A}{1+\cos A}\cdot AP}\ .$ Let $N'\in AP\cap w$ , i.e. $DN'\perp AB$ . Thus, $\frac{AD}{\sin \left(B+\frac A2\right)}{=\frac{AH}{\cos \frac A2}}$ $\Longrightarrow$ $AD=\frac{\cos\frac{B-C}{2}}{\cos \frac A2}\cdot AH$ and $AN'=\frac{AD}{\cos \frac A2}$ $\Longrightarrow$

$AN'=\frac{\cos \frac{B-C}{2}}{\cos^2\frac A2}\cdot AH=$ $\frac{2\cos \frac{B-C}{2}}{1+\cos A}\cdot AH=$ $\frac{AP}{R}\cdot\frac{2R\cos A}{1+\cos A}=$ $\frac{2\cos A}{1+\cos A}\cdot AP$ $\Longrightarrow AN'=AN$ $\Longrightarrow$ $N\equiv N'$ , i.e. $N$ belongs to circumcircle of $\triangle ADE\ .$

PP12 (Swiss IMO Selection 2006). Let an acute $\triangle ABC\ ,\ b\ne c$ with the orthocenter $H$ , the midpoint $M$ of $[BC]$ and two points $D\in [AB]\ ,$

$E\in [AC]$ such that $AE=AD\ ,\ H\in DE$ . Prove that $HM$ is perpendicular to common chord of circumscribed circles of $\triangle ABC\ ,\ \triangle ADE$ .

Proof Denote: $A'\in c\cap (AO$ ; the midpoint $A_1$ of $[AH]$ ; $N\in w\cap (AO_a$ . From the above lemma results $N\in MH$ . But $AA_1=HA_1=OM$

$(M$ is midpoint of $[HA'])$ , $OA=OA'$ and $O_aA=O_aN$ . Thus $A_1$ , $O$ , $O_a$ are midpoints of $AH$ , $AA'$ , $AN$ respectively and $N\in HM\equiv HA'$

$\Longrightarrow$ $O_a\in OA_1$ and $OA_1\parallel MH$ . In conclusion, $MH\parallel OO_a$ . Remark. I denote ray $(XY$ without the point $X$ .

PP13 (NMO England 2005, level VII). Let $ABC$ be a triangle with $A=120^{\circ}$ and the incenter $I$ for what denote $\left\{\begin{array}{ccc} D\in AI\cap BC\\\ E\in BI\cap CA\\\ F\in CI\cap AB\end{array}\right\|$ . Prove that $DE\perp DF$ .

Proof. Nice property and v.v. easy proof, without words, maybe for Pisa's test. Prove easily that $E$ is $B$ -excenter of $\triangle ABD$ and $F$ is $C$ -excenter of $\triangle ACD$ . In conclusion, $DE\perp DF\ .$

PP13 (NMO England 2008, level VII). Solve over $\mathbb R$ the system $\left\{\begin{array}{ccc} xy(z+1) & = & 4\\\\ yz(x+1) & = & 4\\\\ zx(y+1) & = & 12\end{array}\right\|\ .$

Proof 1. Denote $\boxed{xyz=p\ne 0}\ (*)\ .$ Thus, our system becomes $\left\{\begin{array}{ccc} xy & = & 4-p\\\\ yz & = & 4-p\\\\ zx & = & 12-p\end{array}\right\|\ \bigodot\ \stackrel{(*)}{\implies}\ p^2=(4-p)^2(12-p)\iff$ $p^2+(p-4)^2(p-12)=0\iff$

$(p-3)(p-8)^2=0\implies\odot$ $\begin{array}{ccccccccccc} \nearrow & p=3 & \implies & \odot & \begin{array}{ccc} \nearrow & xy=1 & \searrow\\\\ \rightarrow & yz=1 & \rightarrow\\\\ \searrow & zx=9 & \nearrow\end{array} & \odot & \begin{array}{ccc} \nearrow & x=3 & \searrow\\\\ \rightarrow & y=\frac 13 & \rightarrow \\\\ \searrow & z=3 & \nearrow \end{array} & \odot & \implies & \left(\ 3\ ,\ \frac 13\ ,\ 3\ \right) & \searrow\\\\ \searrow & p=8 & \implies & \odot & \begin{array}{ccc} \nearrow & xy=-4 & \searrow\\\\ \rightarrow & yz=-4 & \rightarrow\\\\ \searrow & zx=4 & \nearrow\end{array} & \odot & \begin{array}{ccc} \nearrow & x=-2 & \searrow\\\\ \rightarrow & y=2 & \rightarrow\\\\ \searrow & z=-2 & \nearrow\end{array} & \odot & \implies & (\ -2\ ,\ 2\ ,\ -2\ ) & \nearrow\end{array}\odot\ \implies\ \left[\begin{array}{ccc} 3 & \frac 13 & 3\\\\ -2 & 2 & -2\end{array}\right]\ .$

Proof 2. Denote $\boxed{xyz=p\ne 0}\ (*)\ .$ Thus, our system becomes $\left\{\begin{array}{ccc} p(z+1) & = & 4z\\\\ p(x+1) & = & 4x\\\\ p(y+1) & = & 12y\end{array}\right\|$ $\implies$ $\frac x{x+1}=\frac {3y}{y+1}=\frac z{z+1}=\frac p4\implies$ $x(z+1)=z(x+1)\implies$ $\boxed{z=x}\ (1)$

Therefore, $\frac x{x+1}=\frac {3y}{y+1}=\frac {x^2y}4\implies$ $\left\{\begin{array}{ccc} y(2x+3) & = & x\\\\ x^2(y+1) &= & 12\end{array}\right\|\implies$ $\frac {12}{x^2}=y+1=$ $\frac x{2x+3}+1=\frac {3(x+1)}{2x+3}$ $\implies$ $\frac {12}{x^2}=\frac {3(x+1)}{2x+3}$ $\implies$ $3x^2(x+1)=12(2x+3)$ $\implies$

$\boxed{x^3+x^2-8x-12=0}\ (2)\ \implies\ (x-3)(x+2)^2=0\ .$ In conclusion, $x\in \{\ 3\ ,\ -2\ \}\ \stackrel{y(2x+3)=x}{\implies}\ \left\{\begin{array}{ccccc} z=x=3 & \implies & y=\frac 13 & \implies & \left(\ 3\ ,\ \frac 13\ ,\ 3\ \right)\\\\ z=x=-2 & \implies & y=2 & \implies & (\ -2\ ,\ 2\ ,\ -2\ )\end{array}\right\|$

PP14 (Contest 2015, Hungary). Let $\{x,y,z\}\subset\mathbb R$ so that $\left\{\begin{array}{ccc} x+y+z & = & 5\\\\ yz+zx+xy & = & 8\end{array}\right\|$ . Prove that $\{x,y,z\}\subset \left[1,\frac 73\right]\ .$

Proof 1. Observe that $\boxed{x+y=5-z}\ (1)$ and $xy=8-z(x+y)=8-z(5-z)$ , i.e. $\boxed{xy=z^2-5z+8}\ (2)\ .$ Apply the simple inequality $:$

$(x+y)^2\ge 4xy\ \stackrel{1\wedge 2}{\iff}\ (5-z)^2\ge 4\left(z^2-5z+8\right)\iff$ $3z^2-10z+7\le 0\iff$ $(z-1)(3z-7)\le 0\iff$ $\{x,y,z\}\subset \left[1,\frac 73\right]\ .$

An easy extension. Let $\{x,y,z\}\subset\mathbb R$ so that $\left\{\begin{array}{ccc} x+y+z & = & a\\\\ yz+zx+xy & = & b\end{array}\right\|$ where $\Delta\equiv a^2-3b\ge 0$ (is obviously!). Prove that $\{x,y,z\}\subset \left[\frac {a-2\sqrt{\Delta}}3,\frac {a+\sqrt{\Delta}}3\right]\ .$

PP15. Let an $A$ -right-angled $\triangle ABC$ so that $\frac a{2r}\le 1+\sqrt 2$ . Find the value of the expression $E\equiv E(a,b,c)=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ .$

Proof. Is well-known that $r=s-a$ and the inequality $\boxed{(b+c)^2\le 2\left(b^2+c^2\right)}\ (*)\ .$ Thus, $\frac a{2r}=\frac a{b+c-a}$ and $\frac a{2r}\le 1+\sqrt 2\iff$

$\frac {b+c-a}a\ge \frac 1{1+\sqrt 2}\iff$ $\frac {b+c}a-1\ge \sqrt 2-1\implies$ $\frac {b+c}a\ge \sqrt 2\iff$ $(b+c)^2\ge 2a^2=$ $2\left(b^2+c^2\right)\ \stackrel{(*)}{\ge}\ (b+c)^2$ $\implies$

$(b+c)^2=2\left(b^2+c^2\right)$ , i.e. $b=c$ . Therefore, $\frac b1=\frac c1=\frac a{\sqrt 2}\ .$ Hence, $E=\frac {\sqrt 2}2+\frac 2{1+\sqrt 2}=$ $\frac {\sqrt 2}2+2\left(\sqrt 2-1\right)\implies$ $\boxed{E=\frac {5\sqrt 2-4}2}\ .$

PP16. Prove that if $\alpha,\beta,\gamma$ are the angles of an acute-angled triangle $ABC$ , then $\mathrm{tg}^3\alpha+\mathrm{tg}^3\beta+\mathrm{tg}^3\gamma\ge 9\sqrt{3}\ .$ Please, use only $\mathrm{A.M.\ \ge\ G.M.}$

Proof. Denote $\tan\alpha=x\ ,\ \tan\beta=y\ ,\ \tan\gamma=z$ , where $x+y+z=xyz$ and $\{\alpha ,\beta ,\gamma\}\subset \left(0,\frac {\pi}2\right)\implies\{x,y,z\}\subset \mathbb R^*_{+}\equiv (0,\infty )$ . Thus,

$xyz=x+y+z\ge 3\sqrt[3]{xyz}\implies$ $(xyz)^3\ge 27(xyz)\implies$ $\boxed{xyz\ge 3\sqrt 3}\ (*)$ . Hence $x^3+y^3+z^3\ge 3xyz\ \stackrel{(*)}{\implies}\ x^3+y^3+z^3\ge 9\sqrt 3$ .

Remark. $3\sqrt 3\le x+y+z\implies$ $81\sqrt 3\le (x+y+z)^3\le 9\left(x^3+y^3+z^3\right)\implies$ $x^3+y^3+z^3\ge 9\sqrt 3$ .

PP17 (selection test, Brazil). The real numbers $a,b,c,d$ verify the relations $:\ \left\{\begin{array}{ccc} a=\sqrt{4-\sqrt{5-a}} & ; & b=\sqrt{4+\sqrt{5-b}}\\\\ c=\sqrt{4-\sqrt{5+c}} & ; & d=\sqrt{4+\sqrt{5+d}}\end{array}\right\|$ . Calculate the product $abcd\ .$

Proof 1. $\left\{\begin{array}{cccccccccc} a=\sqrt{4-\sqrt{5-a}} & \iff & a^2=4-\sqrt{5-a} & \iff & \sqrt{5-a}=4-a^2 & \iff & 5-a=16-8a^2+a^4 & \iff & a^4-8a^2+a+11=0 & (1)\\\\ b=\sqrt{4+\sqrt{5-b}} & \iff & b^2=4+\sqrt{5-b} & \iff & b^2-4=\sqrt{5-b} & \iff & b^4-8b^2+16=5-b & \iff & b^4-8b^2+b+11=0 & (2)\\\\ c=\sqrt{4-\sqrt{5+c}} & \iff & c^2=4-\sqrt{5+c} & \iff & \sqrt{5+c}=4-c^2 & \iff & 5+c=16-8c^2+c^4 & \iff & c^4-8c^2-c+11=0 & (3)\\\\ d=\sqrt{4+\sqrt{5+d}} & \iff & d^2=4+\sqrt{5+d} & \iff & d^2-4=\sqrt{5+d} & \iff & d^4-8d^2+16=5+d & \iff & d^4-8d^2-d+11=0 & (4)\end{array}\right\|$ .

Observe that the real numbers $\{a,b,-c,-d\}$ are the roots of the equation $x^4-8x^2+x+11=0$ and $\boxed{abcd=ab\cdot (-c)\cdot (-d)=11}\ .$

Proof 2. Observe that $\left\{\begin{array}{c} x^2-4+\sqrt{5-x}=0\begin{array}{cc} \nearrow & a\\\\ \searrow & -c\end{array}\\\\ x^2-4-\sqrt{5-x}=0\begin{array}{cc} \nearrow & b\\\\ \searrow & -d\end{array}\end{array}\right\|\implies$ $\left(x^2-4+\sqrt{5-x}\right)\cdot\left(x^2-4-\sqrt{5-x}\right)=0\ .$ In conclusion,

$\{a,b,-c,-d\}$ are the roots of the equation $\left(x^2-4\right)^2-(5-x)=0$ , i.e. $x^4-8x^2+x+11=0\implies abcd=11$ .

PP18 (Hungary, 2013). Solve over $\mathbb {R}$ the equation $:\ x+\dfrac{x}{x-1}+\dfrac{x^2}{x^2-x+1}=\dfrac{49}{6}\ .$

Proof. I"ll use the substitution $y=x+\frac{x}{x-1}=\frac{x(x-1)+x}{x-1}=\frac{x^2}{x-1}$ , i.e. $\boxed{y=\frac {x^2}{x-1}}\ (*)\ ,$ where $y\not\in (0,4)$ because the equation $x^2-yx+y=0$ has the discriminant

$\Delta (y)=y^2-4y\ge 0\implies y\not\in (0,4)$ . Therefore, $\frac{x^2}{x^2-x+1}=\dfrac{\dfrac{x^2}{x-1}}{\dfrac{x^2}{x-1}-1}=\dfrac{y}{y-1}$ and given equation become $y+\dfrac{y}{y-1}=\dfrac{49}{6}\iff$ $6y^2-49y+49=0\begin{array}{cccccc} \nearrow & 7\\\\ \searrow & \frac 76\in (0,4)\end{array}$ .

In conclusion, $y=7$ and $\frac {x^2}{x-1}=7\iff$ $x^2-7x+7=0\iff$ $x\in\left\{\frac {7\pm\sqrt{21}}2\right\}$ .

PP19 (ONM England, 2005). Prove that $\{a,b,c\}\subset\mathbb R^*_+\implies \left(\frac ab+\frac bc+\frac ca\right)^2\ge (a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)\ .$

Proof (bitrak). I"ll apply C.B.S. - inequality $:\ \left\{\begin{array}{ccccc} \frac ab+\frac bc+\frac ca & = & \frac {a^2}{ab}+\frac {b^2}{bc}+\frac {c^2}{ac} & \ge & \frac {(a+b+c)^2} {ab+bc+ca}\\\\ \frac ab+\frac bc+\frac ca & = & \frac {\left(\frac 1b\right)^2}{\frac 1{ab}}+\frac {\left(\frac 1c\right)^2}{\frac 1{bc}}+\frac {\left(\frac 1a\right)^2}{\frac 1{ca}} & \ge & \frac {\left(\frac 1a+\frac 1b+\frac 1c\right)^2}{\frac 1{ab}+\frac 1{bc}+\frac 1{ca}}\end{array}\right\|\ \bigodot\ \implies$ $\left(\frac ab+\frac bc+\frac ca\right)^2\ge$ $\frac {(a+b+c)^2}{ab+bc+ca}\cdot\frac {\left(\frac 1a+\frac 1b+\frac 1c\right)^2}{\frac 1{ab}+\frac 1{bc}+\frac 1{ca}}=$

$\frac {(a+b+c)^2}{ab+bc+ca}\cdot\frac {\left(ab+bc+ca\right)^2}{abc(a+b+c)}=$ $(a+b+c)\cdot\frac {ab+bc+ca}{abc}=(a+b+c)\cdot\left(\frac 1a+\frac 1b+\frac 1c\right)\implies$ $\left(\frac ab+\frac bc+\frac ca\right)^2\ge (a+b+c)\left(\frac 1a+\frac 1b+\frac 1c\right)\ .$ Very nice proof !

An easy extension. Prove that $\{a,b,c,x,y,z\}\subset\mathbb R^*_+$ so that $abc=xyz$ $\implies$ $\left(\frac {\sum a\cdot\sum \frac 1x}{\sum\frac ax}\right)^2\le \sum ax\cdot\sum\frac 1{ax}$ .

Particular case. $x=b\ ,\ y=c\ ,\ z=a\implies \left(\frac ab+\frac bc+\frac ca\right)^2\ge (a+b+c)\cdot\left(\frac 1a+\frac 1b+\frac 1c\right)$ .

Proof.

PP20 (selection test, Pakistan). Prove that $\boxed{\{a,b,c\}\subset\mathbb R^*_+}\ (*)$ and $\frac {a(b-c)}{b+c}+\frac {b(c-a)}{c+a}+\frac {c(a-b)}{a+b}=0\implies$ $(a-b)(b-c)(c-a)=0\ .$

Proof.I"ll apply the substitution $\left\{\begin{array}{ccc} b+c & = & x\\\\ c+a & = & y\\\\ a+b &= & z\end{array}\right\|$ $\implies a+b+c=\frac {x+y+z}2\implies\odot \begin{array}{cccccc} \nearrow & a=\frac {x+y+z}2-x & \implies & a=\frac {-x+y+z}2 & b-c=z-y & \searrow\\\\ \rightarrow & b=\frac {x+y+z}2-y & \implies & b=\frac {x-y+z}2 & c-a= x-z & \rightarrow\\\\ \searrow & c=\frac {x+y+z}2-z & \implies & c=\frac {x+y-z}2 & a-b=y-x & \nearrow\end{array}\odot$ .

Therefore, $\sum \frac {a(b-c)}{b+c}=0\iff$ $\sum \frac {(z-y)(-x+y+z)}x=0\iff$ $\sum (y-z)+\sum \frac {z^2-y^2}x=0\iff$ $yz\left(z^2-y^2\right)+zx\left(x^2-z^2\right)+xy\left(y^2-x^2\right)=0\iff$

$z^3(y-x)+z\left(x^3-y^3\right)+xy\left(y^2-x^2\right)=0\iff$ $(y-x)\left[z^3-z\left(x^2+xy+y^2\right)+xy(x+y)\right]=0\iff$ $(y-x)\left[x^2(y-z)+x\left(y^2-yz\right)+z^3-y^2z\right]=0\iff$

$(y-x)(y-z)\left[x^2+xy-z(y+z)\right]=0\iff$ $\left[\left(x^2-z^2\right)+y(x-z)\right]=0\iff$ $(y-x)(y-z)(x-z)(x+y+z)=0\ \stackrel{(*)}{\iff}\ (x-y)(y-z)(z-x)=0\ .$

PP21 (concurs Canada). Solve over $\mathbb R$ the following system $:\ \left\{\begin{array}{ccccccc} x & + & y & + & z & = & 2\\\\ x^2 & - & y^2 & - & z^2 & = & 2\\\\ x & + & z & - & 3y^2 & = & 0\end{array}\right\|$

Proof. I"ll isolate the variable $y\ :\ \left\{\begin{array}{cccc} x+z & = & 2-y & (1)\\\\ x^2-z^2 & = & 2+y^2 & (2)\\\\ x+z & = & 3y^2 & (3)\end{array}\right\|$ . From the equations $(1)$ and $(3)$ obtain that $3y^2=2-y\iff$ $3y^2+y-2=0$ $\implies$

$\odot\begin{array}{cccc} \nearrow & y=-1 & \implies & \odot\begin{array}{ccc} \nearrow & x+z=3 & \searrow\\\\ \searrow & x-z=1 & \nearrow\end{array}\odot\begin{array}{ccc} \nearrow & x=2 & \searrow\\\\ \searrow & z=1 & \nearrow\end{array}\odot\\\\ \searrow & y=\frac 23 & \implies & \odot\begin{array}{ccc} \nearrow & x+z=\frac 43 & \searrow\\\\ \searrow & x-z=\frac {11}{6} & \nearrow \end{array}\odot \begin{array}{ccc} \nearrow & x=\frac {19}{12} & \searrow\\\\ \searrow & z=-\frac 14 & \nearrow\end{array}\odot\end{array}$ $\begin{array}{c} \searrow \\\\ \nearrow\end{array}\odot\implies$ $\begin{array}{ccccc} \nearrow & x=2 & y=-1 & z=1 & \searrow\\\\ \searrow & x=\frac {19}{12} & y=\frac 23 & z=-\frac 14 & \nearrow\end{array}\odot$

PP22. Sa se determine $\{x,y\}\subset\mathbb R$ stiind ca $\tan^4x + \tan^4y +2\cot^2x \cot^2y = 3+\sin^2(x+y)\ .$

Proof. Se observa ca $\tan^4x+\tan^4y+\frac {2}{\tan^2x\tan^2y}=\left(\tan^2x-\tan^2y\right)^2+2\left(\tan^2x\tan^2y+\frac {1}{\tan^2x\tan^2y}\right)\ge 4\ge 3+\sin^2(x+y)\ .$ Deci solutiile

ecuatiei date sunt aceleasi cu solutiile sistemului de ecuatii $\left\|\begin{array}{c} \tan^2x=\tan^2y=1\\\\ \sin^2(x+y)=1\end{array}\right\|\ .$ In concluzie, $\{x,y\}\subset \pi \mathbb Z+\frac {\pi}4$ astfel incat $x+y\in \pi\mathbb Z+\frac {\pi}2\ .$

PP23 (O.M. England round 1 2011). Let $D\ ,$ $E\ ,$ and $F$ be the feet of altitudes from $A\ ,$ $B$ and $C$ respectively in an acute $\triangle ABC\ .$ Prove that $DE+DF\le BC\ .$ When is equality?

Proof. $DE+DF=HC\cdot \sin C+HB\cdot\sin B=2R\cos C\sin C+2R\cos B\sin B=c\cdot\cos C+b\cdot\cos B\ ,$ i.e. $\boxed{DE+DF=c\cdot\cos C+b\cdot\cos B}\ (*)\ .$ Since

$BC=$ $b\cdot\cos C+c\cdot\cos B$ our required inequality becomes $c\cdot\cos C+b\cdot\cos B\le b\cdot\cos C+c\cdot\cos B\iff$ $(b-c)(\cos B-\cos C)\le 0\ ,$ what is true. Have equality iff $b=c$

PP24. Aratati ca daca $a+c\neq 0\ ,$ atunci $:\ \boxed{\div\ a\ ;\ b\ ;\ c\Longleftrightarrow \div\ a^2\ ;\, ab+bc-ac\ ;\ c^2}\ .$

Demonstratie. $\div\ a^2\ ;\ ab+bc-ac\ ;\ c^2$ $\iff$ $2\left(ab+bc-ac\right)=a^2+c^2\iff$ $(a+c)(a+c-2b)=0\ \stackrel{(a+c\ne 0)}{\iff}\ a+c=$ $2b\iff$ $\div\ a\ ;\ b\ ;\ c\ .$

PP24 (Pham Van Thuan). Let $\{x,y\}\subset\mathbb R$ so that $x^2+y^2=1\ .$ Find the maximum value of $f(x,y)=\mid x-y\mid +\mid x^3-y^3\mid\ .$

Proof. Can use the substitution $|x-y|=t\in \left[0,\sqrt 2\right]$ because $x^2+y^2=1\ .$ Therefore, $t^2=|x-y|^2=x^2+y^2-2xy=1-2xy\implies$ $\boxed{xy=\frac {1-t^2}2}$ and

$f(x,y)=t+t\cdot\left(1+\frac {1-t^2}2\right)=\frac {t\left(5-t^2\right)}2\ .$ In conclusion, our problem becomes to find the maximum of $f(t)=\frac 12\cdot\left(5t-t^3\right)\ .$ Thus, $f(t)\le f\left(\sqrt{\frac 53}\right)=\left(\frac 53\right)^{\frac 32}\ .$

PP25. Let $ABC$ be a triangle. Its $A$ -bisector meet again its circumcircle $w$ in $L\ .$ Suppose that exists $E\in (AB)$ so that $AE=\frac {b+c}{2}\ .$ Prove that $LE\perp AB\ .$

Proof. Get $BE=\frac {c-b}{2}$ and $\boxed{EA^2-EB^2=bc}\ .$ Denote $D\in AL\cap BC\ .$ Thus, $\left\{\begin{array}{cccc} \triangle LBD\sim LAB & \iff & LB^2=LA\cdot LD & (1)\\\\ \triangle LBA\sim\triangle CDA & \iff & AD\cdot AL=bc & (2)\end{array}\right|\implies$

$LA^2-LB^2\stackrel{(1)}{=}LA^2-LA\cdot LD=$ $LA\cdot (LA-LD)=AL\cdot AD\stackrel{(2)}{\implies}$ $\boxed{LA^2-LB^2=bc}\implies$ $LA^2-LB^2=EA^2-EB^2\implies$ $LE\perp AB\ .$

Remark 1. $LA^2-LB^2=$ $4R^2\left[\sin^2\left(B+\frac A2\right)-\sin^2\frac A2\right]=$ $4R^2\sin (A+B)\sin B=$ $4R^2\sin B\sin C\implies$ $LA^2-LB^2=bc\ .$

Remark 2. If $M$ is the midpoint of $[BC]\ ,$ then can prove easily that $EM\perp \overline {ADL}\ .$ Here are some usual metrical relations $:\ \left\{\begin{array}{c} LA^2-LB^2=AD\cdot AL=AI\cdot AI_a=bc\\\\ LB^2-LD^2=bc-AD^2=DB\cdot DC\\\\ \frac {LA}{b+c}=\frac {LB}{a}=\frac {LA+LB}{2s}=\frac {LA-LB}{2(s-a)}=\frac {bc}{(b+c)\cdot AD}\end{array}\right|\ .$

This post has been edited 324 times. Last edited by Virgil Nicula, Jun 19, 2016, 2:37 AM

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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

381. Probleme date la diverse concursuri.

by Virgil Nicula, Jul 24, 2013, 12:57 AM

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