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440. Piece of Poetry.

by Virgil Nicula, Feb 25, 2016, 6:58 AM

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http://i944.photobucket.com/albums/ad288/GemenLeu/9ab9a8bd10e9d2d3be232a59bf0d7f131c26629b_zpsj1csk4bh%202_zpsuwumd4gm.png

O scurtă notă matematică cu aplicaţii bine alese şi îngrijit ordonate (în funcţie de gradul de dificultate) ale unei cunoscute leme "Incenter Excenter Lemma" (<= click).

Evaluare Nationala 2011 ==> "Problema cu albina" - Marcel Chirita ... http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=367104

$\begin{array}{c} \underline{\mathrm{PIECE\ OF\ POETRY}}\\\\ \boxed{\ \tan\frac {\pi}{24}=\left(\sqrt 3-\sqrt 2\right)\left(\sqrt 2-1\right)=\sqrt 6-\sqrt 3+\sqrt 2-2\ }\\\\ \tan\frac {\pi}{24}=4\left(\sin\frac {\pi}{3}-\sin\frac {\pi}{4}\right)\left(\sin\frac {\pi}{4}-\sin\frac {\pi}{6}\right)\\\\ \tan\frac {\pi}{24}=16\sin\frac {\pi}{24}\cos\frac {7\pi}{24}\sin\frac {\pi}{24}\cos\frac {5\pi}{24}\\\\ \tan\frac {\pi}{24}=8\sin^2\frac {\pi}{24}\cdot 2\sin\frac {5\pi}{24}\cos\frac {5\pi}{24}\\\\ \tan\frac {\pi}{24}=8\sin^2\frac {\pi}{24}\sin\frac {5\pi}{12}\\\\ 8\sin\frac {\pi}{24}\sin\frac {5\pi}{12}\cos\frac {\pi}{24}=1\\\\ 4\sin\frac {\pi}{12}\sin\frac {5\pi}{12}=1\\\\ 4\sin\frac {\pi}{12}\cos\frac {\pi}{12}=1\\\\ 2\sin\frac {\pi}{6}=1\\\\ \mathrm{O.K.}\end{array}$
Lemma 1. Let $ABC$ be a triangle for which $b\ne c$ . Denote: the middlepoint $M$ of the side $[BC]$ ; the points $D,E,F$ where the incircle $C(I,r)$ touches the side-lines $BC,CA,AB$

respectively; the intersections $K,L$ between the incircle and the median $AM$ ; the intersection $S\in EF\cap AM$ . Then $I\in SD$ and the division $(A,K,S,L)$ is harmonically.

Remark. \sum\cos A=\cos A+(\cos B+\cos C)=\cos A+2\cos\frac {B+C}2\cos\frac {B-C}2=

Proof. Apply an well known relation $\frac{FB}{FA}\cdot MC+\frac{EC}{EA}\cdot MB=\frac{SM}{SA}\cdot BC$ $\Longrightarrow$ $\frac {s-b}{s-a}\cdot \frac a2+\frac {s-c}{s-a}\cdot\frac a2=\frac {SM}{SA}\cdot a\iff$ $\frac{SM}{SA}=\frac{a}{b+c-a}\ \ (1)$ . If $T\in BC,\ AT\perp BC$

then $MD=\frac 12\left| b-c\right|\ ,$ $DT=\frac{(p-a)|b-c|}{a},\ \frac{DM}{DT}=\frac{a}{b+c-a}\ \ (2)$ . From the relations $(1)$ and $(2)$ results that $\frac{SM}{SA}=\frac{DM}{DT}$ , i.e. $SD\parallel AT$ what means $I\in SD$ .

Have immediately and the second part of the conclusion (the line $EF$ is the polar of the point $A$ w.r.t. the incircle).

PP1. Let $\triangle ABC$ and the midpoint $M$ of its side $[BC]$ . Let the incircle $\gamma$ of $\triangle ABC$ . The median $AM$ of $\triangle ABC$ intersects $\gamma$ at $K$ and $L$ . Suppose w.l.o.g. $c<b$ . Hence $K\in (AL)$ .

Let the lines passing through $K$ and $L$ , parallel to $BC$ , intersect the incircle $\gamma$ again in two points $X$ and $Y$ . Let $P\in AX\cap BC$ and $Q\in AY\cap BC$ . Prove that $BP = CQ$ .

Proof. Let $S\in EF\cap AM$ . Thus, $KX\parallel LY\iff$ $KXLY$ is an isosceles trapezoid $\iff$ $S\in XY\cap KL\implies$ $\frac {AK}{AL}=\frac {SK}{SL}=\frac {KX}{YL}\implies$ $\boxed{\frac {AK}{AL} =\frac {KX}{YL}}\ (1)$ .

Thus, $\left\{\begin{array}{cccc} \frac{AM}{AK} & = & \frac{MP}{KX} & (2)\\\\ \frac{AL}{AM} & = & \frac{LY}{MQ} & (3)\end{array}\right\|$ . From the product of $(1)$ , $(2)$ and $(3)$ we obtain $\frac {AK}{AL}\cdot \frac{AM}{AK}\cdot \frac{AL}{AM}=\frac {KX}{YL}\cdot \frac{MP}{KX} \cdot \frac{LY}{MQ}\implies$ $MP=MQ$ , i.e. $BP=CQ\ .$

Lemma 1. Let $w=C(O)$ be a circumcircle of the acute triangle $ABC$ , where $AB<AC$ . Denote the point $A'\in w\cap (AO$ . Define the points $X\in AB$ , $Y\in AC$

so that $B\in (AX)\ ,\ AX=AC$ and $Y\in (AC)\ ,\ AY=AB$ . Then $m(\widehat {A'BY})=m(\widehat {A'CX})=\frac A2\ .$ Standard notation. $\boxed {\ (XY\ }$ - the ray without the point $X$ .

Lemma 2. 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, $\frac{AN}{AP}=\frac{AH}{AH+MP}=$ $\frac{2\cos A}{1+\cos A}$ $\Longrightarrow$ $\boxed {AN=\frac{2\cos A}{1+\cos A}\cdot AP}\ .$

Denote $N'\in AP\cap w$ , 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 the circumcircle of the triangle $ADE\ .$

PP2. Let an acute $\triangle ABC$ with $AB \not= AC$ . Let $H$ be the orthocenter of $\triangle ABC$ , and let $M$ be the midpoint of $[BC]$ . Let $D\in [AB]$ and $E\in [AC]$

such that $AE=AD$ and $H\in DE$ . Prove that $HM$ is perpendicular to the common chord of the circumcircles of $\triangle ABC$ and $\triangle ADE$ .

Proof. Let: $A'\in c\cap (AO$ ; the middlepoint $A_1$ of $[AH]$ ; $N\in w\cap (AO_a$ . From lemma 2 results $N\in MH$ . But $AA_1=HA_1=OM$ ( $M$ is the middlepoint of $[HA']$ ), $OA=OA'$

and $O_aA=O_aN$ . Thus, $A_1$ , $O$ , $O_a$ are the middlepoints of $[AH]$ , $[AA']$ , $[AN]$ respectively and $N\in HM\equiv HA'$ $\Longrightarrow$ $O_a\in OA_1$ and $OA_1\parallel MH$ . Thus, $MH\parallel OO_a$ .

PP3. $(\forall )$ $\triangle ABC$ there is the relations $\frac {l_al_bl_c}{h_ah_bh_c}=\frac {8R^2}{s^2+2Rr+r^2}\ge 1$ , where $(\forall )\ x\in\{a,b,c\}\ ,\ l_x$ is the length of $X$ -bisector and $h_x$ is the length of $X$ - altitude (standard notations).

Proof. Observe that $(a+b)(b+c)(c+a)=\prod(2s-a)=$ $(2s)^3-(2s)^2(a+b+c)+(2s)(ab+bc+ca)-abc=$ $(2s)\left(s^2+r^2+4Rr\right)-4Rrs=$

$2s\left[\left(s^2+r^2+4Rr\right)-2Rr\right]\implies$ $\boxed{(a+b)(b+c)(c+a)=2s\left(s^2+r^2+2Rr\right)}\ (*)$ . I"ll use the well known relation $\boxed{l_a=\frac {2bc\cos\frac A2}{b+c}}$ . Therefore,

$\prod l_a=\prod \frac {2bc\cos\frac A2}{b+c}=$ $\frac {8a^2b^2c^2}{(a+b)(b+c)(c+a)}\cdot\prod\sqrt{\frac {s(s-a)}{bc}}\ \stackrel{(*)}{=}\ \frac {8abc}{2s\left(s^2+r^2+2Rr\right)}\cdot$ $\sqrt{s^3(s-a)(s-b)(s-c)}=$ $\frac {8(4Rrs)}{2s\left(s^2+r^2+2Rr\right)}\cdot s^2r\implies$

$\boxed{l_al_bl_c=\frac {16Rr\cdot s^2r}{s^2+r^2+2Rr}}\ (1)$ . I"ll use the well known identity $\boxed{2Rh_a=bc}$ a.s.o. Thus, $(2R)^3\cdot \prod h_a=\prod (bc)\iff$ $\prod h_a=\frac {(abc)^2}{8R^3}=$ $\frac {(4Rrs)^2}{8R^3}\iff$ $\boxed{\prod h_a=\frac {2s^2r^2}{R}}\ (2)$ .

$\frac {l_al_bl_c}{h_ah_bh_c}\ \stackrel{1\wedge 2}{=}\ \frac {16Rr\cdot s^2r}{s^2+r^2+2Rr}\cdot \frac R{2s^2r^2}\implies$ $\frac {l_al_bl_c}{h_ah_bh_c}=\frac {8R^2}{s^2+2Rr+r^2}$ . I"ll use inequalities $\boxed{R\ge 2r\ \wedge\ s^2\le R^2+4Rr+3r^2}\ (3)$ . Hence $s^2\le 4R^2+4Rr+3r^2=$

$\left(8R^2-2Rr-r^2\right)-\left(4R^2-6Rr-4r^2\right)=$ $\left(8R^2-2Rr-r^2\right)-2(R-2r)(2R+r)\le 8R^2-2Rr-r^2\implies$ $s^2\le 8R^2-2Rr-r^2\implies$ $\frac {8R^2}{s^2+2Rr+r^2}\ge 1\ .$

PP4 (sqing). Let $\{a, b, c\}\subset R^+$ so that $abc=2$ and $a^2+b^2+c^2=6$ . Show that $a+b+c\ge 4\ .$

Proof 1. Suppose w.l.o.g. $a=\min\{a,b,c\}$ . Hence $0<a\le\sqrt[3]{2}<2$ and $\boxed{a+b+c\ge 4}\iff$ $b+c\ge 4-a\iff$ $(b+c)^2\ge (4-a)^2\iff$ $\left(b^2+c^2\right)+2bc\ge (4-a)^2$

$\iff$ $\left(6-a^2\right)+\frac 4a\ge 16-8a+a^2\iff$ $\frac 4a\ge 10-8a+2a^2\iff$ $a\left(a^2-4a+5\right)-2\le 0\iff$ $a^3-4a^2+5a-2\le 0\iff$ $\boxed{(a-1)^2(a-2)\le 0}$ what is truly.

PP5 (M. Ochoa Sanchez). Let a line $d$ , the points $(A,C,P,D,B)\subset d$ in the mentioned order and a circle $w$ so that $\{C,D\}\subset w$ . Denote

$\{Q,T\}\subset w$ so that $A\in QQ$ and $B\in TT$ . Prove that $\frac 1{PA}-\frac 1{PB}=\frac 1{PC}-\frac 1{PD}\ ($ denote $XX$ - the tangent to $w$ at $X\in w\ )$ .

Proof. I"ll use an well known relation $\boxed{\frac {PT}{PQ}=\frac {TC\cdot TD}{QC\cdot QD}}\ (*)$ . Therefore, $\frac 1{PA}-\frac 1{PB}=\frac 1{PC}-\frac 1{PD}\iff$ $\frac 1{PC}-\frac 1{PA}=\frac 1{PD}-\frac 1{PB}\iff$ $\frac {PA-PC}{PA\cdot PC}=\frac {PB-PD}{PB\cdot PD}\iff$

$\frac {AC}{PA\cdot PC}=\frac {BD}{PB\cdot PD}$ $\iff$ $\frac {AC}{AP}=\frac {BD}{BP}\cdot\frac {PC}{PD}\iff$ $\frac {QC}{QP}\cdot \frac {QC}{QT}=\left(\frac {TD}{TP}\cdot\frac {TD}{QT}\right)\cdot\left(\frac {CQ\cdot CT}{DQ\cdot DT}\right)\iff$ $\frac {QC}{QP}=\frac {TD}{TP}\cdot\frac {TC}{QD}$ , i.e. the relation $(*)$ what is truly.

PP6. Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$ , $AB<CD$ and the circumcircle $w=\mathbb C(O,R)$ . Denote the midpoint $M$

of $[CD]$ and the point $P\in CD$ so that $PA$ is tangent to $w$ . Prove that $\boxed{PC^2+PD^2+2\cdot AB\cdot PM=PA^2+PB^2+CD^2}\ (*)$ .

Proof. $\left\{\begin{array}{cc} \left|\begin{array}{ccc} PC\cdot PD & = & PA^2\\\\ PD-PC & = & CD\end{array}\right|\implies PC^2+PD^2=(PD-PC)^2+2\cdot PC\cdot PD=CD^2+2\cdot PA^2 \implies \boxed{PC^2+PD^2=CD^2+2\cdot PA^2} & (1)\\\\ \mathrm{Euler's\ relation\ to\ ABMP}\implies PA^2+BM^2+2\cdot AB\cdot PM=PB^2+AM^2\ \mathrm{and}\ MA=MB \implies \boxed{PA^2+2\cdot AB\cdot PM=PB^2} & (2)\end{array}\right|$ $\bigoplus\implies (*)$

Particular case. If $O\in CD$ , then prove easily that $CD^2=2\cdot AB\cdot PM$ , i.e. $PA^2+PB^2=PC^2+PD^2$ .

PP7 (Miguel O. Sanchez). Let $w$ be the circumcircle of $\triangle ABC$ and $\{D,E\}\subset (BC)$ so that $\widehat{DAB}\equiv\widehat{EAC}$ .

Prove that $\boxed{AD\cdot AE= AB\cdot AC-\sqrt{p_w(D)\cdot p_w(E)}}\ (*)$ , where $p_w(X)$ is power of $X$ w.r.t. $w\ .$

Proof 1. Let $\left\{\begin{array}{ccc} AD\cap w=\{A,X\} & ; & BY=CX=m\\\\ AE\cap w=\{A,Y\} & ; & BX=CY=n\end{array}\right\|$ and Ptolemy's theorem to $\left\{\begin{array}{cccc} ABXC\ : & BC\cdot AX=AB\cdot XC+AC\cdot XB & \implies & a\cdot AX=cm+bn\\\\ ABYC\ : & BC\cdot AY=AB\cdot YC+AC\cdot YB & \implies & a\cdot AY=cn+bm\end{array}\right\|$ .

$\blacktriangleright\ \odot\begin{array}{ccc} \nearrow AXB\sim ACE & \implies & \frac {AX}b=\frac n{CE}=\frac c{AE}\searrow\\\\ \searrow AXC\sim ABE & \implies & \frac {AX}c=\frac m{BE}=\frac b{AE}\nearrow\end{array}$ $\odot\implies AX\cdot AE=bc\ \odot\begin{array}{c} \nearrow AX\cdot CE=bn\searrow\\\\ \searrow AX\cdot BE=cm\nearrow\end{array}\odot\ AX\cdot (BE+CE)=bn+cm\implies a\cdot AX=bn+cm$

$\blacktriangleright\ \odot\begin{array}{ccc} \nearrow AYC\sim ABD & \implies & \frac {AY}c=\frac n{BD}=\frac b{AD}\searrow\\\\ \searrow AYB\sim ACD & \implies & \frac {AY}b=\frac m{CD}=\frac c{AD}\nearrow\end{array}$ $\odot\implies AY\cdot AD=bc\ \odot\begin{array}{c} \nearrow AY\cdot BD=cn\searrow\\\\ \searrow AY\cdot CD=bm\nearrow\end{array}\odot\ AY\cdot (BD+CD)=bm+cn\implies a\cdot AY=bm+cn$

Observe that $BC\parallel XY\iff \frac {AD}{DX}=\frac {AE}{EY}$ , i.e. $\boxed{AD\cdot EY=DX\cdot AE}\ (1)$ . Thus, $(DB\cdot DC)(EB\cdot EC)=$ $p_w(D)p_w(E)=$ $(AD\cdot DX)(AE\cdot EY)=$

$(AD\cdot EY)(DX\cdot AE)\ \stackrel{(1)}{=}\ (AD\cdot EY)^2$ $\implies$ $\boxed{AD\cdot EY=\sqrt {p_w(D)p_w(E)}}\ (2)$ . In conclusion, $bc=AD\cdot AY=AD\cdot (AE+EY)=$ $AD\cdot AE+AD\cdot EY\ \stackrel{(2)}{=}$

$AD\cdot AE+$ $\sqrt {p_w(D)p_w(E)}\implies$ $AD\cdot AE=bc-\sqrt {p_w(D)p_w(E)}$ (It is a nice Stan Fulger's proof).

Proof 2. Let $\left\{\begin{array}{ccc} BY=CX=m\\\\ BX=CY=n\end{array}\right\|$ . Apply the Ptolemy's theorem to $:\ \left\{\begin{array}{cc} ABXC\ : & a\cdot AX=bn+cm\\\\ ABYC\ : & a\cdot AY=bm+cn\end{array}\right\|$ $\implies$ $\boxed{AX\cdot AY=\frac {(bn+cm)(bm+cn)}{a^2}}\ (1)$ . Are well known

relations $\left\{\begin{array}{ccc} ABXC\ :\ \frac {DA}{bc}=\frac {DX}{mn}=\frac {AX}{bc+mn}=\frac {DB}{cn}=\frac {DC}{bm}=\frac a{bm+cn}=\sqrt{\frac {-p_w(D)}{bcmn}}\\\\ ABYC\ :\ \frac {EA}{bc}=\frac {EY}{mn}=\frac {AY}{bc+mn}=\frac {EB}{cm}=\frac {EC}{bn}=\frac a{bn+cm}=\sqrt{\frac {-p_w(E)}{bcmn}}\end{array}\right\|$ $\implies$ $AX\cdot AY=bc+mn\ \stackrel{(1)}{\implies}\ \boxed{(bm+cn)(bn+cm)=a^2(mn+bc)}\ (2)$ .

Otherwise. Apply the Pythagoras' theorem $:\ \left\{\begin{array}{c} \triangle BAC\ : a^2=b^2+c^2-2bc\cdot\cos A\\\\ \triangle BXC\ : a^2=m^2+n^2+2mn\cdot\cos A\end{array}\right\|$ $\implies$ $mn\left(b^2+c^2-a^2\right)+bc\left(m^2+n^2-a^2\right)=0\implies$ the relation $(2)$ .

Thus, $AD\cdot AE=\frac {abc}{bm+cn}\cdot\frac {abc}{bn+cm}\ \stackrel{(2)}{\implies}\ \boxed{AD\cdot AE= \frac {b^2c^2}{bc+mn}}\ (3)\ ;\ \frac {\sqrt{p_w(D)p_w(E)}}{bcmn}=$ $\frac {a^2}{(bm+cn)(bn+cm)}\ \stackrel{(2)}{\implies}\ \boxed{\sqrt{p_w(D)p_w(E)}=\frac {bcmn}{bc+mn}}\ (4)$ .

In conclusion, $bc-AD\cdot AE\ \stackrel{(3)}{=}\ bc-\frac {b^2c^2}{bc+mn}=bc\cdot\left(1-\frac {bc}{bc+mn}\right)=\frac {bcmn}{bc+mn}\ \stackrel{(4)}{=}\ \sqrt{p_w(D)p_w(E)}\implies$ $bc-AD\cdot AE=\sqrt{p_w(D)p_w(E)}$ .

Particular case. If $D\equiv E$ , then $AD$ is $A$ -bisector of $\triangle ABC$ and the relation $(*)$ becomes well-known relation $AD^2=bc-DB\cdot DC$ , what get length of $A$ -bisector in $\triangle ABC$ .

If $E$ is the midpoint of $[BC]$ , then $\frac {DB}{c^2}=\frac {DC}{b^2}=\frac a{b^2+c^2}$ and $s_am_a=bc-\frac {a^2bc}{2\left(b^2+c^2\right)}=bc\cdot\left[1-\frac {a^2}{2\left(b^2+c^2\right)}\right]=\frac {bc}{2\left(b^2+c^2\right)}\cdot 4m_a^2\implies$ $\boxed{s_a=\frac {2bc}{b^2+c^2}\cdot m_a}$ (well-known).

PP8 (Miguel O. Sanchez). Let an $A$ -right $\triangle ABC$ and $D\in (BC)$ , $E\in (CA)$ and $F\in (AB)$ so that $P\in AD\cap BE\cap CF$ and $\widehat{AEB}\equiv\widehat{CED}$ . Prove that $FB=2\cdot FA$ .

Proof. Let $K\in DE\cap AB$ . Observe that $\widehat{AEB}\equiv\widehat{CED}\equiv\widehat{AEK}$ and $EA\perp BK\iff$ $\triangle BEK$ is $E$ -isosceles $\iff$ $\boxed{AB=AK}\ (*)$ .

Is well known that the points $\{K,F\}$ are conjugate w.r.t. the points $\{A,B\}$ , i.e. $\frac {FB}{FA}=\frac {KB}{KA}=2\ \stackrel{(*)}{\implies}\ \frac {FB}{FA}=2\iff FB=2\cdot FA$ .

Remark. $\left\{\begin{array}{cccccc} \mathrm{Menelaus'\ theorem\ to} & \overline {DEK}/\triangle ABC & : & \frac {KA}{KB}\cdot\frac{DB}{DC}\cdot\frac {EC}{EA} & = & 1\\\\ \mathrm{The\ Ceva's\ theorem\ to} & \mathrm{the\ point}\ P & : & \frac {EA}{EC}\cdot\frac {DC}{DB}\cdot\frac {FB}{FA} & = & 1\end{array}\right\|\bigodot\implies \frac {FA}{FB}=\frac {KA}{KB}\ ,$ i.e. $\{K,F\}$ are conjugate w.r.t. $\{A,B\}$

PP9. Let $P$ be an interior interior point of $\triangle ABC$ with the circumcircle $w$ and $a\ne b$ . The lines $AP$ , $BP$ , $CP$ meet again $w$ at $K$ , $L$ ,

$M$ respectively. Denote $S\in CC\cap AB$ , where $XX$ is the tangent line to $w$ at $X\in w$ .Prove that $SC = SP\iff MK = ML$ .

Proof 1. Consider the circle which is tangent to $w$ in $C$ and to $SP$ in $P$ . Denote $\{X,Y\}=SP\cap w$ , where $X\in \overarc{AL}$ . From the well-known property

obtain that $\widehat {MCX}\equiv\widehat {MCY}$ , i.e. $\overarc{MX}=\overarc{MY}$ . Since $SP^2=SC^2=SA\cdot SB$ obtain that $SP^2=SA\cdot SB$ $\iff$ $\widehat {SPA}\equiv\widehat{SBP}\implies$ the line $SP$

is tangent to the circumcircle of $\triangle APB$ ) , i.e. $\overarc{XL}=\overarc{YK}$ . In conclusion, $\overarc{ML}=\overarc{MX}+\overarc{XL}=\overarc{MY}+\overarc{YK}=\overarc{MK}$ $\implies$ $\overarc{ML}=\overarc{MK}$ , i.e. $ML=MK$ .

Proof 2. Using the power point theorem we get that $SP^2=SA\cdot SB$ which means that the triangles $SPB$ , $PAS$ are similarly. A bit angle chasing shows that $\widehat{SPL}\equiv$

$\widehat{BAP}\equiv\widehat{BLK}$ so $SP\parallel LK\ (1)\ .$ But $OC\bot CS$ and $\widehat{SPC}\equiv\widehat{SCP}$ , so $OM\bot SP\ (2)\ .$ From the relations $(1)$ and $(2)$ we have that $OM\bot LK$ and we are done.

PP10 (Chinese MO 1996). Let $\triangle{ABC}$ be a triangle with orthocenter $H$ . The tangent lines from $A$ to the circle $w$ with diameter $[BC]$ touch it on $P$ and $Q$ . Prove that $H\in PQ$ .

Proof 1 (pole/polar). Denote the circumcircle $(O)$ of $\triangle ABC$ , $\{A,D_1\}=AH\cap (O)$ , $D\in BC$ for which $AD\perp BC$ and $\{U,V\}=AH\cap w$ . Observe that

$DU^2=DB\cdot DC=DA\cdot DD_1=DA\cdot DH$ , i.e. $\boxed {\ DU^2=DA\cdot DH\ }$ what is a characterization of $(A,H,U,V)$ - harmonical division, i.e. $H$ is harmonical

conjugate of $A$ w.r.t. $\{U,V\}$ . In conclusion, $H$ belongs to polar of $A$ w.r.t. $(O)$ , i.e. $H\in PQ$ .

Proof 2 (metric). Suppose that $AB$ separates $P$ , $C$ . Denote $E\in AC$ so that $BE\perp AC$ and $F\in AB$ so that $CF\perp AB$ . Observe that $\frac {PF}{PB}=\sqrt {\frac {AF}{AB}}$ and $\frac {QE}{QC}=\sqrt {\frac {AE}{AC}}$ .

Thus $\frac {PF}{PB}\cdot\frac {QE}{QC}=|\cos A|=\frac {EF}{BC}$ , i.e. $\frac {PF}{PB}\cdot\frac {QE}{QC}=\frac {EF}{BC}$ $\Longleftrightarrow$ $BP\cdot FE\cdot QC=PF\cdot EQ\cdot CB$ $\stackrel{(*)}{\Longleftrightarrow}$ $BE$ , $PQ$ , $FC$ are concurrently $\Longleftrightarrow$ $H\in PQ$ .

Lemma. Let $ABCDEF$ be a cyclic hexagon. Prove that $\boxed {\ AD\cap BE\cap CF\ne\emptyset \Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA\ }\ (*)$ .

An easy extension. Let $\alpha=\mathbb C(O)$ be circumcircle of $\triangle ABC$ and $w=\mathbb C(K)$ be a circle so that $\{B,C\}\subset w$ and exists $D\in (BC)$ such that $DA\perp DK\ .$

Let $\{A,D_1\}=\{A,D\}\cap \alpha$ and the symmetric $L$ of $D_1$ w.r.t. $D\ .$ The tangent lines from $A$ to the circle $w$ touch it on $P$ and $Q\ .$ Prove that $L\in PQ\ .$

Proof. Denote $\{U,V\}=AD\cap w$ . Prove easily that $D$ is the common midpoint of $[UV]$ and $[D_1L]\ .$ Since $D\in BC$ what is the radical axis of $w$ and

$\alpha$ obtain that $DU^2=DB\cdot DC=DD_1\cdot DA=DL\cdot DA\implies$ $DU^2=DL\cdot DA$ . The last relation is a characterization A2 of " $(A,L;U,V)$ is

an harmonic division". In conclusion, the point $L$ is the harmonical conjugate of $A$ w.r.t. $\{U,V\}\subset w$ . Hence $L\in PQ$ what is the polar of $A$ w.r.t. $\alpha$ .

PP11 (P. Paulino). Let a square $ABCD$ , $\{M,N\}\subset (BC)$ so that $M\in (BM)$ , $P\in AM\cap DN$ and $x=m\left(\widehat{APD}\right)$ . Prove that $(A,M,N,C)$ is harmonically $\iff x=45^{\circ}$ .

Proof 1. Suppose w.l.o.g. $AB=1$ and let $\left\{\begin{array}{ccc} MB=m\ ;\ m\left(\widehat{MAB}\right)=y & \implies & \tan y=m\\\\ NC=n\ ;\ m\left(\widehat{NDC}\right)=z & \implies & \tan z=n\end{array}\right\|$ $\implies$ $\tan x=\tan (y+z)=\frac{\tan y+\tan z}{1-\tan y\tan z} \implies$ $\boxed{\tan x=\frac {m+n}{1-mn}}\ (*)$ .

Thus, $(A,M,N,C)$ is harmonically $\iff$ $\frac {MB}{MN}=\frac {CB}{CN}\iff$ $\frac m{1-m-n}=\frac 1n\iff$ $1-m-n=mn\iff$ $m+n=1-mn\ \stackrel{*}{\iff}\ \tan x=1\iff x=45^{\circ}$ .

Proof 2. Let the circumcircle $w$ of $ABCD$ , $\{A,R\}=\{A,M\}\cap w$ and $S\in DR\cap BC$ . Observe that $RM\perp RC$ and the ray $[RM$ is the bisector of $\widehat{BRS}$ , what means

that the point $S$ is harmonical conjugate of $B$ w.r.t. $\{M,C\}$ , i.e. $(B,M,S,C)$ is harmonically $\iff S\equiv N$ $\iff$ $R\equiv P$ . In conclusion, $P\in w$ and $m\left(\widehat{APD}\right)=45^{\circ}$ .

Proof 3. Denote $\left\{\begin{array}{ccc} X\in (PB) & ; & XM\perp BC\\\\ Y\in (PC) & ; & YN\perp BC\end{array}\right\|$ . Thus, $\boxed{\frac {XM}{BA}}=\frac {PM}{PA}=\boxed{\frac {MN}{AD}}=\frac {PN}{PD}=\boxed{\frac {YN}{CD}}\implies$ $\frac {XM}{BA}=\frac {MN}{AD}=\frac {YN}{CD}\ \stackrel{BA=AD=CD}{\implies}$

$XM=MN=YN\implies$ $XMNY$ is a square. Since $(B,M,N,C)$ in this order is an harmonical division then $\frac {BM}{BC}=\frac {NM}{NC}\iff$ $\frac {BM}{BA}=\frac {NY}{NC}\iff$

$\triangle BMA\sim\triangle NYC\iff$ $\widehat{BAM}\equiv\widehat{NCY}\iff$ $\widehat{BAP}\equiv\widehat{BCP}\iff$ $ABPC$ is cyclically $\iff$ $PA\perp PC\iff m\left(\overarc{AD}\right)=90^{\circ}\iff$ $m\left(\widehat{APD}\right)=45^{\circ}\ .$

PP12 (Edson Curahua Ortega). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ for what denote the intersections $D\in (BC)\cap AI$ , $E\in (CA)\cap BI$ ,

$F\in (AB)\cap CI$ . Denote $M\in EF\cap AD$ and $N\in (AC)\ ,P\in (AB)$ so that $NP\parallel BC$ and $M\in NP$ . Prove that $BP+CN=2\cdot NP$ .

Proof 1. $\frac {AP}{AB}=\frac {AN}{AC}=\frac {PN}{BC}=\frac {AM}{AD}\iff$ $\frac {BP}{AB}=\frac {CN}{AC}=\frac {BC-PN}{BC}=\frac {MD}{AD}=k\iff$ $\frac {BP}c=\frac {CN}b=$ $1-\frac {PN}a=\frac {MD}{AD}=k\implies$

$\left\{\begin{array}{ccc} BP+CN & = & k(b+c)\\\\ PN & = & a(1-k)\end{array}\right\|\implies$ $\boxed{\frac {BP+CN}{PN}=\frac k{1-k}\cdot \frac {b+c}a}\ (*)$ . Apply an well known relation (the Cristea's identity) $:$

$\frac {MD}{MA}=\frac {DC}{BC}\cdot\frac {FB}{FA}+\frac{DB}{BC}\cdot\frac {EC}{EA}\iff$ $\frac k{1-k}=\frac b{b+c}\cdot \frac ab+\frac c{b+c}\cdot\frac ac\iff$ $\frac k{1-k}=\frac {2a}{b+c}\ \stackrel{*}{\implies}\ \boxed{\frac {BP+CN}{PN}=2}\ .$

Proof 2. I"ll use upper notations. Is well known that the division $(\ A\ ,\ I\ ;\ M\ ,\ D\ )$ is harmonically and apply its $\mathrm{A}8^{th}$ -characterization (<= click) $:\ \frac {IA}{ID}=2\cdot\frac {MA}{MD}$ , i.e. $\frac {MA}{MD}=\frac {b+c}{2a}\ .$

Therefore, $\frac {BP}c=\frac {CN}b=$ $\frac {DM}{DA}=1-\frac {AM}{AD}=\frac {2a}{b+c+2a}=$ $\boxed{\frac {BP+CN}{b+c}}=1-\frac {NP}{a}=$ $\frac {2a-2\cdot NP}{2a}=\frac {2a-(2a-2\cdot NP)}{(b+c+2a)-2a}=$ $\boxed{\frac {2\cdot NP}{b+c}}\implies$ $BP+CN=2\cdot NP\ .$

PP13. Folosind in exclusivitate formulele de calcul prescurtat in doua variabile (predate in lectiile anterioare) $\left\{\begin{array}{cccc} (a+b)^2 & = & a^2+b^2+2ab & (1)\\\\ a^3+b^3 & = & (a+b)^3-3ab(a+b) & (2)\\\\ (a+b)(a+c) & = & a^2+a(b+c)+bc & (3)\end{array}\right\|$

sa se demonstreze urmatoarele identitati algebrice in trei variabile $:\ \left\{\begin{array}{cccc} a^3+b^3+c^3 & = & 3abc+(a+b+c)\left[\left(a^2+b^2+c^2\right)-(ab+bc+ca)\right] & (\alpha )\\\\ a^3+b^3+c^3 & = & (a+b+c)^3-3(a+b)(b+c)(c+a) & \beta )\\\\ a^3+b^3+c^3 & = & (a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc & (\gamma )\end{array}\right\|\ .$
Proof. $\begin{array}{cc} \boxed{\alpha} & \boxed{\beta}\\\\ a^3+\underline{b^3+c}^3-3abc & a^3+\underline{b^3+c}^3\\\\ a^3+\underline{(b+c)^3-3bc(b+c)}-3abc & a^3+\underline{(b+c)^3-3bc(b+c)}\\\\ \underline{(a+b+c)^3-3a(b+c)(a+b+c)}-3bc(a+b+c) & \underline{(a+b+c)^3-3a(b+c)(a+b+c)}-3bc(b+c)\\\\ (a+b+c)\left[\underline{a^2+(b+c)^2+2a(b+c)}-3a(b+c)-3bc\right] & (a+b+c)^3-3(b+c)\left[\underline{a(a+b+c)+bc}\right]\\\\ \boxed{a^3+b^3+c^3-3abc=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(ab+bc+ca)\right]} & \boxed{a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)}\end{array}$ $\boxed{(a+b)(b+c)(c+a)+abc=(a+b+c)(ab+bc+ca)}$ $\boxed{\begin{array}{ccccc} \sum a^2(b+c) & = & \sum a\left(b^2+c^2\right) & = & \sum bc(b+c)\\\\ \sum a^3(b+c) & = & \sum a\left(b^3+c^3\right) & = & \sum bc\left(b^2+c^2\right)\\\\ \sum a^4(b+c) & = & \sum a\left(b^4+c^4\right) & = & \sum bc(b^3+c^3)\end{array}}$
PP14 (IMO shortlist 2003). Let $\triangle ABC$ with the circumcircle $w$ and a point $D\in\overarc{AC}\subset w$ so that $AC$ separates $D$ , $B$ . Denote the projection $P,Q,R$ of $D$ on $BC$ , $CA$ ,

$AB$ respectively. Prove that $QP=QR\iff$ the intersection between the $B$ -bisector of $\triangle ABC$ with the $D$ -bisector of $\triangle ADC$ belongs to $AC$ , i.e. $ABCD$ is harmonic.

Proof. Is well-known that ${PQR}$ is the Simson's line of $D$ w.r.t. $w$ . Let $\{X,Y\}\subset (AC)$ so that $\left\{\begin{array}{ccc} \widehat{XBA}\equiv\widehat{XBC} & \iff & \frac {XA}{XC}=\frac {BA}{BC}\\\\ \widehat {YDA}\equiv\widehat {YDC} & \iff & \frac {YA}{YC}=\frac {DA}{DC}\end{array}\right\|\ (*)$ . Therefore, $CDQP$

and $AQDR$ are cyclically and their circumcircles have the diameters $[DC]$ and $[DA]$ respectively. Hence $\left\{\begin{array}{ccc} QR=AD\sin\widehat {QAR} & \implies & QR=AD\sin A\\\\ QP=CD\sin\widehat{PCQ} & \implies & QP=CD\sin C\end{array}\right\|\ .$

In conclusion, $QP=QR\iff$ $AD\sin A=CD\sin C\iff$ $DA\cdot BC=DC\cdot BA\ \stackrel{(*)}{\iff}\ \frac {YA}{YC}=\frac {DA}{DC}=\frac {BA}{BC}=\frac {XA}{XC}\iff$ $\frac {YA}{YC}=\frac {XA}{XC}\iff$ $X\equiv Y$ .

PP15. $\{a,b,c\}\subset\mathbb R^*_+$ and $a^2+b^2+c^2+2abc=1\implies a+b+c\le \frac 32\ .$

Proof 1. Denote $k=a+b+c$ and observe that $k^2-2k+1=2(1-a)(1-b)(1-c)\le 2\left(\frac{3-k}{3}\right)^3\implies$ $27(k-1)^2\le 2(3-k)^3\implies$

$27\left(k^2-2k+1\right)\le 2\left(27-27k+9k^2-k^3\right)\implies$ $2k^3+9k^2-27\le 0\implies (2k-3)(k+3)^2\le 0\implies k\le\frac 32\implies a+b+c\le \frac 32\ .$

Proof 2. Is well known the identity $\cos ^2A+\cos^2 B+\cos^2C+2\cos A\cos B\cos C=1$ in any $\triangle ABC$ . Since $\{a,b,c\}\subset (-1,1)$

exists the triangle $ABC$ so that $a=\cos A$ , $b=\cos B$ , $c=\cos C$ . Prove easily that $\cos A+\cos B+\cos C\le\frac 32$ , i.e. $a+b+c\le \frac 32\ .$

Remark. $\left\{\begin{array}{ccc} \cos A=1-2\sin^2\frac A2\\\\ \cos\frac {B+C}2=\sin\frac A2\end{array}\right\|$ $\implies$ $\sum\cos A=\cos A+(\cos B+\cos C)=$ $1-2\sin^2\frac A2+2\cos\frac {B+C}2\cos\frac {B-C}2=$ $1+2\sin\frac A2\left(\cos\frac {B-C}2-\cos\frac {B+C}2\right)=$

$1+4\sin\frac A2\sin\frac B2\sin\frac C2=$ $1+4\sqrt{\frac {(s-b)(s-c)}{bc}\cdot\frac {(s-a)(s-c)}{ac}\cdot\frac {(s-a)(s-b)}{ab}}=$ $1+\frac {4(s-a)(s-b)(s-c)}{abc}=$ $1+\frac {4sr^2}{4Rrs}=1+\frac rR\ .$

In conclusion, $\boxed{\sum\cos A=1+\frac rR\le\frac 32}$ and $\boxed{\prod\sin\frac A2=\frac r{4R}\le\frac 18}$ because $\boxed{R\ge 2r}\ .$ Indeed, $a^2\ge a^2-(b-c)^2\implies$ $a^2\ge 4(s-b)(s-c)$ a.s.o.

In conclusion, $\left\{\begin{array}{ccc} a^2 & \ge & 4(s-b)(s-c)\\\\ b^2 & \ge & 4(s-c)(s-a)\\\\ c^2 & \ge & 4(s-a)(s-b)\end{array}\right\|\ \bigodot\ \implies$ $abc\ge 8(s-a)(s-b)(s-c)\implies$ $4Rrs\ge 8sr^2\implies R\ge 2r\ .$

PP16. Solve the following system over $\mathbb R^*_+\ :\ \left\{\begin{array}{ccc} x+y^2+z^3 & = & 3\\\\ x^2+y^3+z & = & 3\\\\ x^3+y+z^2 & = & 3\end{array}\right\|\ .$

Proof. Apply $\mathrm{C.B.S.}$ over $\mathbb R^*_+\ :\ \boxed{\left(\sqrt{ax}+\sqrt{by}+\sqrt{cz}\right)^2\le (a+b+c)(x+y+z)}\ (*)\ .$ Thus, $(x^2+\sqrt{y^3}+\sqrt{z^5})^2\le (x^3+y+z^2)*(x+y^2+z^3)=9\implies$

$x^2+\sqrt{y^3}^2 + \sqrt{z^5}\le 3=x^2+y^3+z\implies$ $\sqrt{y^3}+\sqrt{z^5}\le y^3+z\ .$ Prove easily that $y\le 1\implies z\le 1\implies x\le 1$ and $y\ge 1\implies z\ge 1\implies x\ge 1$ ,

i.e. $\{x,y,z\}\subset (0,1]$ or $\{x,y,z\}\subset [1,\infty )$ . In conclusion, $x+y^2+z^3=3\implies x=y=z=1\ .$

PP17 (Miguel O. Sanchez). Let $\triangle ABC$ with $A\ge 120^{\circ}\ .$ Prove that $\sqrt 3(b+c)\left(b^2+bc+c^2\right)\le 2a^2$ (standard notations).

Proof. Suppose w.l.o.g. $\boxed{a=1}\ (*)$ and $B\le 30^{\circ}\le C<60^{\circ}\ .$ Therefore, $A\ge 120^{\circ}\iff$ $\cos A\le -\frac 12\iff$ $2\cos A+1\le 0\iff$ $bc(2\cos A+1)\le 0\iff$

$b^2+c^2-a^2+bc\le 0\ \stackrel{(*)}{\iff}\ \boxed{b^2+bc+c^2\le 1}\ (1)\ \implies$ $\left\{\begin{array}{cccccccc} 3bc & \le & b^2+bc+c^2 & \stackrel{(1)}{\le} & 1 & \implies & \boxed{bc\le \frac 13} & (2)\\\\ (b+c)^2 & \stackrel{(1)}{\le} & bc+1 & \stackrel{(2)}{\le} & \frac 43 & \implies & \boxed{b+c\le \frac 2{\sqrt 3}} & (3)\end{array}\right\|\ \stackrel{(1\wedge 3)}{\implies}\ (b+c)$ $\left(b^2+bc+c^2\right)\le \frac 2{\sqrt 3}\ .$

Lemma. $(\forall )\ \triangle ABC$ and $\phi\in\mathbb R$ , there is the identity $\sum^{cyc}\sin A\sin (B+\phi )\sin (C-\phi )=(1+2\cos 2\phi )\prod\sin A\ .$

Proof. Apply $\boxed{S=2R^2\sin A\sin B\sin C}\ (*)$ for the area of $\triangle ABC$ and the identity $\boxed{\alpha +\beta +\gamma =\pi\implies 4\sin\alpha\sin\beta\sin\gamma=\sin 2\alpha +\sin 2\beta +\sin 2\gamma}\ (1)\ .$

Therefore, $A+(B+2\phi )+(C-2\phi )=\pi\ \stackrel{(1)}{=}\ 4\sin A\sin (B+\phi )\sin (C-\phi )=\sin 2A+\sin (2B+2\phi )+\sin (2C-2\phi )\implies$

$\underline{\underline{4\sum^{cyc}\sin A\sin (B+\phi )\sin (C-\phi )}}=\sum\sin 2A+\sum^{cyc}[\sin (2B+2\phi )+\sin (2C-2\phi )]=\sum\sin 2A+\sum [\sin (2A+2\phi )+\sin (2A-2\phi )]=$

$\sum\sin 2A+2\cos 2\phi\sum\sin 2A=$ $(1+2\cos 2\phi )\sum\sin 2A=\underline{\underline{4(1+2\cos 2\phi )\prod\sin A}}\implies$ $\boxed{\sum^{cyc}\sin A\sin (B+\phi )\sin (C-\phi )=(1+2\cos 2\phi )\prod\sin A}\ (1)\ .$

PP18 (Miguel O. Sanchez). Let $\triangle ABC$ and the midpoints $(M,N,P)$ of its sides $[BC]$ , $[CA]$ , $[AB]$ . Prove that for any $\triangle DEF$ so that

$M\in (DE)$ , $N\in (EF)$ , $P\in (DF)$ and $\widehat {APF}\equiv\widehat{BMD}\equiv\widehat{CNE}$ , there is the inequality $[DEF]\le S$ , where $S$ is the area of $\triangle ABC$ .

Proof. $AFPN$ , $BDMP$ , $CENM$ are cyclically with same radius $\frac R2\ .$ Thus, $\sum [PFN]=[PFN]+[MDP]+[NEM]\ \stackrel{(*)}{=}\ \sum\frac {R^2}2\sin A\sin (B+\phi )\sin (C-\phi )\ \stackrel {(1)}{=}$

$\frac {R^2}2\cdot (1+2\cos 2\phi )\prod\sin A=$ $\frac {1+2\cos 2\phi}4\cdot 2R^2\prod\sin A\ \stackrel{(*)}{=}\ \frac {(1+2\cos 2\phi )S}4\le \frac {3S}4\implies$ $\sum [PFN]\le\frac {3S}4\implies$ $[DEF]=[MNP]+\sum [PFN]\le S\ .$

P19. Find the maximum value of the sum $x^2+y^2$ , where $(x,y)\in \mathbb Z^2$ and $2xy+2x-5y=40\ .$

Proof. I"ll find all ordered pairs $(x,y)\in \mathbb Z^2$ , where $2xy+2x-5y=40$ . Thus, $y\ne -1$ and $2xy+2x-5y=40\iff$ $2x(y+1)=5(y+8)\iff$ $2x=\frac {5(y+8)}{y+1}\ (*)\ .$

Thus, $x\in \mathbb Z\iff$ $2x=5+\frac {35}{y+1}\in\mathbb Z\iff$ $(y+1)|35\iff$ $(y+1)\in(\pm 1,\pm 5,\pm 7\pm 35)$ , i.e. $y\in (\underline{0,-2,4,-6,6,-8,34,-36})$ . Let $k(y+1)=35$ , i.e. $x=\frac {k+5}2$ .

Hence $k\in (35,-35,7,-7,5,-5,1,-1)$ and $x\in (\underline{20,-15,6,-1,5,0,3,2})$ , i.e. $\boxed{(x,y)\in \{(20,0);(-15,-2);(6,4);(-1,-6);(5,6) ; (0,-8); (3,34);(2,-36)\}\subset\mathbb Z^2}$

In conclusion, $\odot\begin{array}{ccccccccccc} \nearrow & x\ : & 20 & -15 & 6 & -1 & 5 & 0 & 3 & 2 & \searrow\\\\ \searrow & y\ : & 0 & -2 & 4 & -6 & 6 & -8 & 34 & -36 & \nearrow\end{array}\odot$ Prove easily that the maximum value of the sum $x^2+y^2$ is $36^2+2^2$ , i.e. $1300.$

P20. Sa se arate ca intr-un triunghi ascutitunghic are loc inegalitatea: $\frac{1}{b^{2}+c^{2}-a^{2}}\ +\frac{1}{c^{2}+a^{2}-b^{2}}\ +\frac{1}{a^{2}+b^{2}-c^{2}}\geq\frac{1}{4r^{2}}\ .$

Metoda 1. Reamintim relatia remarcabila in $\triangle ABC\ :\ \sum a\cdot\cos A=\frac {2pr}{R}\ .$ Deci $\sum\frac {1}{b^2+c^2-a^2}=$ $\frac {1}{2abc}\cdot\sum \frac {a^2}{a\cdot\cos A}\ \stackrel{(C.B.S.)}{\ \ge\ }\ \frac {1}{2abc}\cdot\frac {\left(\sum a\right)^2}{\sum a\cos A}=\frac {1}{8Rpr}\cdot \frac {4p^2}{\frac {2pr}{R}}=\frac {1}{4r^2}\ .$

Metoda 2. Folosind $\boxed{4\cdot S=\left(b^2+c^2-a^2\right)\cdot\tan A}$ inegalitatea devine $\boxed{\tan A+\tan B+\tan C\ge\frac pr}\ .$ Cateva relatii remarcabile in $\triangle ABC\ :\ \sum\sin A=$ $\frac pR\ ;\ \sum\sin 2A=$

$4\prod\sin A=\frac {2pr}{R^2}\ .$ Asadar, $\sum\tan A=2\cdot\sum \frac {\sin^2A}{\sin 2A}\ \stackrel{(C.B.S.)}{\ge}\ 2\cdot\frac {\left(\sum\sin A\right)^2}{\sum\sin 2A}=\frac {2\left(\frac pR\right)^2}{\frac {2pr}{R^2}}=\frac pr\ .$ Inegalitatea $\sum \frac{1}{b^{2}+c^{2}-a^{2}}\ \stackrel{(C.B.S.)}{\ \ge\ }\ \frac {9}{a^2+b^2+c^2}$ este slaba

deoarece $\frac{1}{4r^{2}}\ge \frac {9}{a^2+b^2+c^2}\ (*)\ .$ Intr-adevar, $a^2+b^2+c^2=2\left(p^2-r^2-4Rr\right)\ge 2\left[\left(16Rr-5r^2\right)-r^2-4Rr\right]=12r(2R-r)\ge 36r^2\ .$

Problema propusa. Sa se arate ca intr-un triunghi ascutitunghic exista urmatorul lant de relatii $:$

$\boxed{\frac {a}{b^2+c^2-a^2}+\frac {b}{c^2+a^2-c^2}+\frac {c}{a^2+b^2-c^2}\ge\frac {(a+b+c)^2}{3abc}\ge\frac {3(a+b+c)}{ab+bc+ca}\ge\frac {9}{a+b+c}\ge\frac {3(a+b+c)}{a^2+^2+c^2}}\ .$

Proof. $\sum\frac a{b^2+c^2-a^2}=\frac 1{4S}\cdot \sum a\cdot\tan A=$ $\frac 1{4sr}\cdot \sum \frac {a^2}{2R\cos A}=\frac 1{8Rsr}\cdot \sum\frac {a^2}{\cos A}\ \stackrel{C.B.S}{\ge}\ \frac s{2r(R+r)}\implies$ $\boxed{\sum\frac a{b^2+c^2-a^2}\ge \frac s{2r(R+r)}\ge \frac s{3Rr}=\frac{4s^2}{12Rsr}=\frac {(a+b+c)^2}{3abc}}\ .$

PP21. Let $\triangle ABC$ with the orthocenter $H\ ,$ the circumcircle $\mathbb C(O,R)\ ,$ $A=60^0$ and $b\ne c\ .$ Prove that $OH$ (Euler's line) is the exterior bisector of $\widehat{BHC}$ and $HO=|b-c|\ .$

Proof. Suppose w.l.o.g. $c<b$ and let $E\in BH\cap AC\ .$ Hence $\left\{\begin{array}{ccccc} m\left(\widehat{BHC}\right) & = & 180^{\circ}-A & = & 120^{\circ}\\\\ m\left(\widehat{BOC}\right) & = & 2A & = & 120^{\circ}\end{array}\right\|$ $\implies BHOC$ is cyclic $\implies$ $\left\{\begin{array}{c} m\left(\widehat{OHE}\right)=m\left(\widehat{BCO}\right)=30^{\circ}\\\\ m\left(\widehat{OHC}\right)=m\left(\widehat{OBC}\right)=30^{\circ}\end{array}\right\|$

Thus, the ray $[HO$ is the exterior bisector of $\widehat{BHC}\ .$ Observe that $A=60^{\circ}\implies$ $3R^2=a^2=b^2+c^2-bc\implies$ $\boxed{HO^2=9R^2-\left(a^2+b^2+c^2\right)}=$ $3a^2-a^2-(a^2+bc)=$

$a^2-bc=\left(b^2+c^2-bc\right)-bc=$ $(b-c)^2\implies \boxed{HO=b-c}\ .$ Otherwise. $\left\{\begin{array}{c} AH=AO=R\\\\ m\left(\widehat{HAO}\right)=B-C\end{array}\right\|\implies$ $HO=2R\sin\frac {B-C}2\implies$ $2R(\sin B-\sin C)=b-c\ .$

PP22 (O.M. Serbia 1998). Let $\triangle ABC$ with incircle $w=\mathbb C(I,r)\ ,$ its tangent points $(D,E,F)$ with $[BC]\ ,\ [AC]\ ,\ [AB]$ and $\left\{\begin{array}{c} M\in BI\cap DE\\\\ N\in CI\cap DF\end{array}\right\|\ .$ Prove that $MN\parallel BC\ .$

Proof. Is well known or prove easily that $\left\{\begin{array}{c} MA\perp MB\\\\ NA\perp NC\end{array}\right\|\ .$ Let the midpoints $X\ ,$ $Y$ of $[DF]\ ,$ $[DE]$ respectively. Thus, $MNXY$ is inscribed in the circle with the

diameter $[MN]$ and $IXDY$ is inscribed in the circle with the diameter $[ID]\ .$ In conclusion, $\widehat{MNC}\equiv\widehat{MXY}\equiv$ $\widehat{IXY}\equiv\widehat{IDY}\equiv$ $\widehat{NCB}$ $\implies MN\parallel BC\ .$

PP23. Let $ABCD$ be a parallelogram and $\left\{\begin{array}{ccc} M\in [AB] & ; & AB=3\cdot AM\\\\ N\in [BC] & ; & BC=2\cdot BN\\\\ P\in [AD] & ; & AD=4\cdot AP\end{array}\right\|\ .$ Prove that $BD\cap CM\cap NP\ne\emptyset\ \ (*)\ .$

Proof. $\left\{\begin{array}{c} AB=CD=a\ ;\ X\in PN\cap BD\\\\ AD=BC=b\ ;\ Y\in MC\cap BD\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c} XNB\sim XPD\ \implies\ \frac {XB}{XD}=\frac {NB}{PD}=\frac {\frac b2}{\frac {3b}4}\implies\frac {XB}{XD}=\frac 23\\\\ YMB\sim YCD\ \implies\ \frac {YB}{YD}=\frac {MB}{CD}=\frac {\frac {2a}3}a\ \implies\ \frac {YB}{YD}=\frac 23\end{array}\right\|$ $\implies\frac {XB}{XD}=$ $\frac {YB}{YD}\implies X\equiv Y\implies (*)\ .$

Extindere. Let $ABCD$ be a parallelogram and $\left\{\begin{array}{ccc} M\in[AB] & ; & AB=m\cdot AM\\\\ N\in[BC] & ; & BC=p\cdot BN\\\\ P\in[AD] & ; & AD=n\cdot AP\end{array}\right\|\ .$ Prove that $BD\cap CM\cap NP\ne\emptyset\ \iff\ \left(1-\frac 1m\right)\left(1-\frac 1n\right)=\frac 1p\ .$

PP24. Let $\{a,b,c,d\}\subset\mathbb R^*$ so that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ and $\left(1+\frac{a}{bc}\right)\left(1+\frac{b}{ac}\right)\left(1+\frac{c}{ab}\right)=$ $\dfrac{1}{d^2}\ .$ Ascertain $a+b+c\ .$

Proof. $\left(1+\frac{a}{bc}\right)\left(1+\frac{b}{ac}\right)\left(1+\frac{c}{ab}\right)=\frac{1}{d^2}\iff$ $\underline{\underline 1}+\sum \frac {a}{bc}+\underline{\sum \frac {1}{a^2}}+\frac {1}{abc}=\underline{\underline 1}-2\sum \frac 1a+\underline {\sum \frac {1}{a^2}}+2\sum\frac {1}{bc}$ $\iff$ $\sum \frac {a}{bc}+\frac {1}{abc}=$

$-2\sum \frac 1a+2\sum\frac {1}{bc}\iff$ $\sum a^2+1=-2\sum bc+2\sum a\iff$ $1+\left(\sum a\right)^2-2\sum a=0\iff$ $\left(\sum a-1\right)^2=0\iff a+b+c=1\ .$

$\mathrm{END}$

This post has been edited 602 times. Last edited by Virgil Nicula, Sep 9, 2017, 9:20 PM

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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