444. Remarkable points in a triangle.

by Virgil Nicula, May 21, 2016, 1:15 AM

P1 (Evgeniy Kulanin). $(\forall )\ \triangle ABC\ ,\ N\Gamma\ \parallel\ BC\ \iff\ F\in (AM)$ , where $M$ is midpoint of $[BC]$ , $N$ is Nagel's point, $\Gamma$ is Gergonne's point and $F$ is Feuerbach's point .

Proof. I"ll prove that $\boxed{N\Gamma\ \parallel\ BC\ \iff\ a(b+c)=b^2+c^2\ \iff\ F\in (AM)}\ .$ See aici. Denote the second intersection $L$ of the Euler's circle with the median $AM$ ,

the orthocenter $H$ , the point $D\in AH\cap BC$ and the midpoint $E$ of the segment $[AH]\ .$ Is well known that the points $\{M,D,E\}$ belong to the Euler's circle and from the power

of the point $A$ w.r.t. this circle obtain that $AE\cdot AD=AL\cdot AM\iff$ $R\cos A\cdot h_a=AL\cdot m_a\iff$ $4Rh_a\cdot\cos A=4m_a\cdot AL\iff$ $2bc\cdot\cos A=4m_a\cdot AL\iff$

$\boxed{\ AL=\frac {b^2+c^2-a^2}{4m_a}\ }\ .$ Hence $LM=AM-AL=$ $\frac {4m_a^2-\left(b^2+c^2-a^2\right)}{4m_a}=$ $\frac {2\left(b^2+c^2\right)-a^2-\left(b^2+c^2\right)+a^2}{4m_a}$ , i.e. $\boxed{LM=\frac {b^2+c^2}{4m_a}}\ .$

$\blacktriangleright\ \ \boxed{\ N\Gamma\ \parallel\ BC\ }\ \iff\ \frac {p-a}{p}=$ $\frac {(p-b)(p-c)}{\sum (p-b)(p-c)}\ \iff\ \frac {p}{(p-a)}=$ $1+\frac {(p-a)[(p-b)+(p-c)]}{(p-b)(p-c)}$

$\iff$ $\frac {p}{(p-a)}-1=$ $\frac {a(p-a)}{(p-b)(p-c)}\ \iff\ \boxed{\ (p-a)^2=(p-b)(p-c)\ }\ \iff\ \boxed{\ a(b+c)=b^2+c^2\ }\ .$

$\blacktriangleright\ \ \boxed{\ F\in AM\ }\ \iff\ L\in AM\cap w$ , where $w=C(I,r)\ .$ Denote $\{L,S\}=AM\cap w$ , where $L\in (AS)\ .$ From $AL=\frac {b^2+c^2-a^2}{4m_a}$ and $LM=\frac {b^2+c^2}{4m_a}$ obtain

that $p_w(A)=(p-a)^2=AL\cdot AS\iff$ $AS=\frac {4m_a(p-a)^2}{b^2+c^2-a^2}\ .$ Thus, $MS=MA-AS=$ $m_a-\frac {4m_a(p-a)^2}{b^2+c^2-a^2}$ , i.e. $MS=\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\ .$ Also,

$p_w(M)=\frac {(b-c)^2}{4}=$ $MS\cdot ML=$ $\frac {2m_a\left[a(b+c)-a^2-bc\right]}{b^2+c^2-a^2}\cdot$ $\frac {b^2+c^2}{4m_a}$ , i.e. $\left(b^2+c^2-a^2\right)(b-c)^2=$

$2\left(b^2+c^2\right)\left[a(b+c)-a^2-bc\right]\ \iff\ \left[a(b+c)-\left(b^2+c^2\right)\right]^2=0\ \iff\ \boxed{\ a(b+c)=b^2+c^2\ }\ .$

Observatie. Apropo de subiectul topicului "Puncte importante intr-un triunghi". Pentru cei care nu cunosc semnificatiile punctelor remarcabile mentionate in enuntul problemei propuse le recomand sa consulte Google unde, spre surprinderea lor, vor intalni cel putin 100 de asemenea puncte ceea ce inseamna ca triunghiul a fost si ramane o preocupare de secole a multor matematicieni, multi dintre ei cu rezultate remarcabile la nivel inalt. In demonstratie am folosit urmatoarele relatii metrice dintr-un triunghi care se pot dovedi fara dificultate $:$

$ab+bc+ca=p^2+\sum(p-b)(p-c)=p^2+r(4R+r)\ \ ;\ \ IA^2=\frac {bc(p-a)}{p}\ .$

P2 (Cezar Lupu). Prove that $(\forall )$ $\triangle ABC$ there is the inequality $\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+\frac{(b+c-a)(c+a-b)(a+b-c)}{abc}\ \ge\ 7$ (Cezar Lupu, Mathematical Reflections 2009).

Proof. $\frac{(b+c-a)(c+a-b)(a+b-c)}{abc}+\sum \frac{b+c}{a}\ \ge\ 7\ \iff\ \sum a\cdot\sum \frac 1a+\frac {8pr^2}{4Rpr}\ \ge\ 10\ \iff$ $\frac {p^2+r^2+4Rr}{2Rr}+\frac {2r}{R}\ \ge\ 10\ \iff\ \boxed{p^2+5r^2\ \ge\ 16Rr}\ .$

Lemma. $\{x\ ,\ y\ ,\ z\}\ \subset\ R\ \wedge\ x+y+z=0\ \implies\ yza^2+zxb^2+xyc^2\ \le\ 0\ .$

Proof. $\left\|\ \begin{array}{c} yza^2+zxb^2+xyc^2\ \le\ 0\\\\ z=-(x+y)\end{array}\ \right\|\ \iff\ yza^2\le (y+z)\left(zb^2+yc^2\right)\iff$ $y^2c^2+yz\left(b^2+c^2-a^2\right)+z^2b^2\ge 0\iff$

$y^2c^2+2yzbc\cdot \cos A+z^2b^2\ge 0\iff$ $\left(yc+zb\cdot\cos A\right)^2+z^2b^2\sin^2A\ge 0\ ,$ what is true. Have equality $\Longleftrightarrow\ x=y=z=0 .$

Particular case.. $\left|\begin{array}{c} x=b+c-2a\\\\ y=c+a-2b\\\\ z=a+b-2c\end{array}\ \right|\ .$ Therefore, $\sum (c+a-2b)(a+b-2c)a^2\le0\ \iff s_1^3-5s_1s_2+18s_3\le 0\iff$ $8p^3-10p\left(p^2+r^2+4Rr\right)+18\cdot 4Rpr\le 0$

$\boxed{p^2+5r^2\ \ge\ 16Rr}\ .$ This wel known inequality has an remarkable geometrical interpretation, $\boxed{9\cdot GI^2=p^2+5r^2-16Rr\ \ge\ 0}\ .$ Indeed, from the Leibniz's identity

$\sum XA^2=3\cdot XG^2+\frac 13\cdot\sum a^2$ for $X:=I$ $IA^2=\frac {bc(p-a)}{p}$ a.s.o. obtain $:\ 3\left(p^2+r^2-8Rr\right)=9\cdot IG^2+2\left(p^2-r^2-4Rr\right)\Longleftrightarrow 9\cdot IG^2=p^2+5r^2-16Rr\ .$

P3 (ONM Germany). Let $\triangle\ ABC$ with the orthocenter $H$ , the circumcircle $\Omega=\mathbb C(O,R)$ and the incircle $w=\mathbb C(I,r)\ .$ Denote

$\left|\ \begin{array}{ccccccc} E & \in & BH\cap AC & ; & F & \in & CH\cap AB\\\\ N & \in & in BI\cap AC & ; & P & \in & CI\cap AB\end{array}\ \right|\ .$ Prove that $I\in EF\iff \cos B+\cos C=\cos A\iff O\in NP\ .$

Proof. $I(a,b,c)$ and $\left|\begin{array}{c} H(\tan A,\tan B,\tan C)\\\\ O(\sin 2A,\sin 2B,\sin 2C)\end{array}\right|$ are with barycentrical coordinates w.r.t. $\triangle ABC\ .$ Thus, $N(a,0,c)$ , $P(a,b,0)$ and $\left|\begin{array}{c} E(\tan A\ ,\ 0\ ,\ \tan C)\\\\ F(\tan A,\ \tan B\ ,\ 0)\end{array}\right|\ .$

In conclusion $\odot\begin{array}{cccc} \nearrow & I\in EF & \iff & \left|\begin{array}{ccc} a & b & c \\\\ \tan A & 0 & \tan C\\\\ \tan A & \tan B & 0\end{array}\right|=0\\\\ \searrow & O\in NP & \iff & \left|\begin{array}{ccc} \sin 2A & \sin 2B & \sin 2C \\\ a & 0 & c\\\\ a & b & 0\end{array}\right|=0\end{array}\ \begin{array}{c} \searrow\\\\ \nearrow\end{array}\odot\ \iff\ \left|\begin{array}{ccc} \cos A & \cos B & \cos C \\\\ 1 & 0 & 1\\\\ 1 & 1 & 0\end{array}\right|=0\ \iff\ \cos B+\cos C=\cos A\ .$

P4 Daca paralelele duse prin punctele de contact ale cercului inscris cu laturile unui triunghi la medianele corespunzatoare ale acestuia sunt concurente, atunci triunghiul este isoscel.

Proof. In $\triangle ABC$ denote : the midpoint $M$ of $[BC]$ ; the tangent points $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ of its incircle; $X\in (EF)$ for which $DX\ \parallel\ AM$ and analogous points

$Y\in (DF)$ , $Z\in (DE)\ .$ Show easily that $\boxed{\frac {XF}{XE}=\frac {b(p-b)(2c-a)}{c(p-c)(2b-a)}}$ . Obtain analogously $\frac {YD}{YF}=$ $\frac {c(p-c)(2a-b)}{a(p-a)(2c-b)}$ and $\frac {ZE}{ZD}=\frac {a(p-a)(2b-c)}{b(p-b)(2a-c)}\ .$ Therefore,

$DX\cap EY\cap FZ\ne\emptyset$ $\iff$ (the Ceva's theorem) in $\triangle\ DEF$ $\iff$ $(2a-b)(2b-c)(2c-a)=(2a-c)(2b-a)(2c-b)$ $\iff$ $(a-b)(b-c)(c-a)=0$ (vezi aici).

Remark. Denote $U\in AX\ \cap\ BC$ , $V\in BY\ \cap\ CA$ , $W\in CZ\ \cap\ AB\ .$ Prove easily that $\frac {XF}{XE}=\frac bc\cdot\frac {UB}{UC}$ , i.e. $\frac {UB}{UC}=\frac {(p-b)(2c-a)}{(p-c)(2b-a)}$ and analogous ratios.

Hence $AX\ \cap\ BY\ \cap\ CZ\ne\emptyset\iff$ $DX\ \cap\ EY\ \cap\ FZ\ne\emptyset$ $\iff$ $a=b\ \vee\ b=c\ \vee\ c=a\ .$ See here

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$^{(*)}\ \$ Se arata usor identitatea $a(p-a)\stackrel{(1)}{\ \ =\ \ }(p-b)(2c-a)+c(b-c)\stackrel {(2)}{\ \ =\ \ }(p-c)(2b-a)+b(c-b)\ .$ Notam $T\in DX\cap AB$ si $\ L\in EF\cap AM\ .$ Se arata usor ca

$\ \frac {LF}{b}=\frac {LE}{c}=\frac {EF}{b+c}\ .$ Deoarece $\overline {DXT}\ \parallel\ \overline {MLA}$ obtinem $\ \frac {TA}{TB}=\frac {DM}{DB}=\frac {b-c}{a+c-b}\ \iff$ $\frac {TA}{b-c}=\frac {TB}{a+c-b}=\frac ca\ \iff$ $TA=\frac {c(b-c)}{a}$ $\iff$ $\frac {LX}{LF}=\frac {AT}{AF}=$

$\frac {c(b-c)}{a(p-a)}\ .$ Asadar obtinem sirul de rapoarte (proportia) $\frac {LX}{c(b-c)}=\frac {LF}{a(p-a)}=\frac {XF}{a(p-a)-c(b-c)}\stackrel{(1)}{\ \ =\ \ }\frac {XF}{(p-b)(2c-a)}=\frac {LE}{\frac {ac(p-a)}{b}}=\frac {XE}{c(b-c)+\frac {ac(p-a)}{b}}\ \Longrightarrow$

$\frac {XF}{XE}=\frac {(p-b)(2c-a)}{c(b-c)+\frac {ac(p-a)}{b}}=$ $\frac {b(p-b)(2c-a)}{c\left[b(b-c)+a(p-a)\right]}\stackrel{(2)}{\ \ =\ \ }\frac {b(p-b)(2c-a)}{c(p-c)(2b-a)}\ .$ In concluzie, $\frac {XF}{XE}=\frac {b(p-b)(2c-a)}{c(p-c)(2b-a)}\ .$

P5. the points $G,H,O,I$ are centroid, orthocenter, incenter and circumcenter respectively of $\triangle\ ABC\ \Longrightarrow\ OI\ + \ OG\ \ge\ IH$ .

Proof. $m(\angle HIG)\ge 90^{\circ}\Longleftrightarrow HG^2\ge IH^2 + IG^2\Longleftrightarrow 4\cdot OG^2\ge IH^2 + IG^2$ . Since $I$ belongs to the disk with diameter $[HG]$ obtain $OG\le OI\ .$ Apply Stewart'

relation to cevian $IG$ in $\triangle\ HIO\ : \ \Longrightarrow\ \underline {\underline{3\cdot IG^2}} = IH^2 + 2\cdot IO^2 - 6\cdot OG^2\ (*)$ . Thus, $4\cdot OG^2\ge IH^2 + IG^2\ \Longleftrightarrow\ 12\cdot$ $OG^2\ \ge\ 3\cdot IH^2 + \underline{\underline{3\cdot IG^2}}\ \stackrel {(*)}{\Longleftrightarrow}$

$\boxed {\ 9\cdot OG^2 - OI^2\ge 2\cdot IH^2\ }\ (**)$ . Therefore, $(OG - OI)(7\cdot OG + 3\cdot OI)\le 0\ \Longleftrightarrow$ $2(OG + OI)^2\ge 9\cdot OG^2 - OI^2\stackrel {(**)}{\Longrightarrow}2(OG + OI)^2\ge 2\cdot IH^2$ .

Remark. $\left\{\begin{array}{ccccccc} \overrightarrow{OH}=3\cdot \overrightarrow{OG} & ; & OH^2=9R^2-\left(a^2+b^2+c^2\right) & \implies & a^2+b^2+c^2 & \le & 9R^2\\\\ \overrightarrow{IN}=3\cdot \overrightarrow{IG} & ; & NI^2=s^2+5r^2-16Rr & \implies & 16Rr & \le & s^2+5r^2\\\\ OI^2=R^2-2Rr & ; & HI^2=4R(R+r)+3r^2-s^2 & \implies & s^2 & \le & 4R^2+4Rr+3r^2\end{array}\right\|\ .$

P6. Fie $\triangle ABC$ ascutit cu ortocentrul $H$ , $\triangle DEF$ ortic si paralelogramele $HDYF$ , $HFXE$ , $HEZD\ .$ Aratati ca $AX\cap BY\cap CZ\ne\emptyset$ (Gh. Szollosy, R.M.T. 1/2009).

Comentariu. Cand intalniti o problema, incercati sa o demonstrati folosind cu "zgarcenie" datele din ipoteza. In felul acesta s-ar putea sa nu va bizuiti in demonstratie pe anumite particularitati ale punctelor/liniilor din figura asociata. De fapt in cele din urma ati gasit si totodata ati demonstrat o extindere/generalizare a acesteia. Iata mai jos generalizarea problemei propuse in RMT.

Generalizare. Fie $\triangle ABC$ ascutit, un punct interior $P$ pentru care notam $\left\|\ \begin{array}{c} D\in AP\ \cap BC\\\\ E\in BP\ \cap\ CA\\\\ F\in CP\ \cap\ AB\end{array}\ \right\|$ si paralelogramele $PEXF$ , $PFYD$ , $PDZE\ .$ Aratati ca $AX\cap BY\cap CZ\ne\emptyset\ .$

P7. Sa se arate ca in $\triangle ABC$ avem echivalenta $\boxed{\ A=60^{\circ}\ \Longleftrightarrow\ IH=IO\ }$ si in acest caz $\boxed{\ \sin (\widehat {IOH}) = \frac 12\cdot\sqrt {1 - \frac {2r}{R}}\ }\ .$

Demonstratie. $\boxed{\ A=60^{\circ}\ \Longleftrightarrow\ IH=IO\ }\ .$ Ecuatia $f(x)= p\cdot x^3-(4R+r)\cdot x^2+p\cdot x-r=0$ are solutiile $\left\{\ \tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\ \right\}\ .$ Asadar,

$A=60^{\circ}\ \Longleftrightarrow\ f\left(\frac{\sqrt 3}{3}\right)\ =\ 0\ \Longleftrightarrow\ \boxed{\ p=\sqrt{3}(R+r)\ }\ .$ Dar in orice triunghi sunt adevarate relatiile $\boxed{\ \begin{array}{ccc} OI^2 & = & R^2-2Rr \\\\ IH^2 & = & 4R(R+r)+3r^2-p^2\end{array}\ }\ .$ Deci

$IH=IO\ \Longleftrightarrow\ R^2-2Rr=4R(R+r)+3r^2-p^2$ $\Longleftrightarrow\ p^2=3R^2+6Rr+3r^2\ \Longleftrightarrow\ \boxed{\ p=\sqrt{3}(R+r)\ }\ .$ Prin urmare avem $\boxed{\ A=60^{\circ}\ \Longleftrightarrow\ IH=IO\ }\ .$ Avem

$\widehat{BHC}\equiv \widehat{BIC}\equiv\widehat{BOC}=120^{\circ}$ , adica $B,H,I,O,C$ sunt conciclice $(*)\ .$ Din teorema Sinusurilor aplicata in $\triangle IBO\Longrightarrow\frac{IO}{\sin(\angle IBO)}=\frac{OB}{\sin(\angle BIO)}$ $\Longleftrightarrow$ $\sin(\angle IBO)\ \stackrel{(*)}{=}$

$\sin(\angle IHO)=\sin(\angle IOH)=\frac{\sqrt{R^2-2Rr}\cdot\sin(\angle BIO)}{R}\ \stackrel{(*)}{\iff}\ \sin(\angle IOH)=$ $\frac{\sqrt{R^2-2Rr}\cdot\sin(\angle BCO)}{R}=$ $\frac 12\cdot \frac{\sqrt{R^2-2Rr}}{R}$ $\Longleftrightarrow$ $\boxed{\ \sin (\angle {IOH}) = \frac 12\cdot\sqrt {1 - \frac {2r}{R}}\ }\ .$

P8. Fie $\triangle ABC,\ R$ raza cercului circumscris si $F$ mijlocul lui $[GH]$ (notatii standard). Aratati ca exista relatia $\boxed{AF^2+BF^2+CF^2=3R^2}$ (O.M. 1997 - Ungaria).

Demonstratie. Se poate folosi relatia lui Leibniz pentru orice punct $M$ din plan $:\ \boxed{MA^2+MB^2+MC^2=3MG^2+GA^2+GB^2+GC^2}\ (*)\ .$ Notam puterea

$p_w(X)$ a lui $X$ in raport cu cercul circumscris $w=C(O,R)\ .$ Stim ca $HF=FG=GO$ , $p_w(G)=GO^2-R^2$ si pentru orice punct $M$ exista relatia Leibniz $:$

$\sum MA^2=3\cdot \left[MG^2-p_w(G)\right]\ .$ Pentru $M:=F$ , obtinem $\sum FA^2=3\cdot \left[FG^2-p_w(G)\right]=$ $3\cdot\left[GO^2-p_w(G)\right]$ , adica $\sum FA^2=3R^2\ .$

P9. Fie $\triangle DEF$ ortic al $\triangle ABC\ ,$ unde $D\in BC\ ,$ $E\in CA\ ,$ $F\in AB$ si mijloacele $X\ ,$ $Y\ ,$ $Z$ ale $[EF]\ ,$ $[FD]\ ,$ $[DE]$ respectiv. Sa se arate ca $AX\cap BY\cap CZ\ne\emptyset\ .$

Demonstratie. Notam mijloacele $M\ ,$ $N\ ,$ $P$ ale lui $[BC]\ ,$ $[CA]\ ,$ $[AB]\ .$ Din faptul ca $\triangle AEF\sim\triangle ABC$ rezulta ca $[AX\ ,$ $[AM$ sunt omoloage in asemanare, ceea ce inseamna

ca $\widehat {FAX}\equiv\widehat {CAM}\ ,$ adica $[AX$ este $A$ -simediana in $\triangle ABC\ .$ In concluzie, $AX\ ,$ $BY\ ,$ $CZ$ sunt simediane si se stie ca sunt concurente in centrul simedian (Lemoine) al $\triangle ABC\ .$

Extindere. Fie $\triangle\ ABC$ si $D\in BC\ ,$ $E\in CA\ ,$ $F\in AB$ astfel incat $AD\cap BE\cap CF\ne\emptyset\ .$ Fie $X\in EF\ ,$ $Y\in FD\ ,$ $Z\in DE\ .$ Sa se arate ca

$AX\cap BY\cap CZ\ne \emptyset\iff$ $DX\cap EY\cap FZ\ne\emptyset$ si in acest caz ce se poate spune de cele trei puncte de concurenta, sunt intotdeauna coliniare sau nu ?

P10 (Relatia lui Stewart). Fie $ABC$ un triunghi oarecare si $M\in (BC)\ .$ Sa se demonstreze ca $AM^2\cdot BC=AB^2\cdot MC+AC^2\cdot MB-BC\cdot MB\cdot MC\ .$

Metoda 1. Notam $D\in BC$ pentru care $AD\perp BC\ .$ Presupunem fara a restrange generaltatea ca $m(\widehat{AMB})\le 90^{\circ}\ .$ Aplicam teorema lui Pitagora generalizata in triunghiurile $ABM\ ,$

$ACM\ \ :$ $\left\|\begin{array}{c} AB^2=AM^2+MB^2-2\cdot MB\cdot MD\\\\ AC^2=AM^2+MC^2+2\cdot MC\cdot MD\end{array}\right\|\ \begin{array}{ccc} \odot & MC & \searrow\\\\ \odot & MB & \nearrow\end{array}\bigoplus\Longrightarrow$ $AB^2\cdot MC+AC^2\cdot MB=AM^2\cdot (MB+MC)+MB\cdot MC\cdot (MB+MC)\ \Longrightarrow$

$\boxed{\ AB^2\cdot MC+AC^2\cdot MB=AM^2\cdot (MB+MC)+MB\cdot MC\cdot BC\ }\ .$

Metoda 2 (inedita !). Notam cercul circumscris $w$ al triunghiului $ABC$ si $\{A,N\}=AM\cap w\ ,$ $AB=c\ ,\ AC=b\ ,\ MB=x\ ,\ MC=y\ ,\ NB=u\ ,\ NC=v\ .$

Se stie sau se arata usor ca $\frac {MA}{bc}=\frac {MN}{uv}=\frac {x}{cu}=\frac {y}{bv}\equiv\rho\ (*)\ .$ Aplicam teorema lui Ptolemeu $:\ bu+cv=a\cdot (AM+MN)\ .$ Inmultim aceasta relatie cu $bc\rho$

si folosim relatiile $\ (*)\ :\ b^2\cdot cu\rho +c^2\cdot bv\rho =a\cdot bc\rho\cdot AM+a\cdot bv\rho\cdot cu\rho\ \Longleftrightarrow\ \boxed{\ b^2\cdot x+c^2\cdot y=a\cdot MA^2+a\cdot x\cdot y\ }\ .$

Metoda 3 (inedita !). Vom folosi notatiile din metoda 2 la care adaugam $\left\|\begin{array}{c} U\in AM\ ,\ BU\perp AM\\\\ V\in AM\ ,\ CV\perp AM\end{array}\right\|\ .$ Presupunem f.r.g. $m(\widehat {AMB})\le 90^{\circ}\ ,$ adica $U\in [MA$ si $M\in [AV]\ .$ Astfel,

$\left\|\begin{array}{c} BU\perp AM\ \Longleftrightarrow\ BA^2-BM^2=UA^2-UM^2\ \Longleftrightarrow\ c^2-x^2=MA\cdot (UA-UM)\\\\ CV\perp AM\ \Longleftrightarrow\ CA^2-CM^2=VA^2-VM^2\ \Longleftrightarrow\ b^2-y^2=MA\cdot (VA+VM)\end{array}\right\|\ (**)\ .$ In concluzie, relatia lui Stewart $c^2y+b^2x=a\cdot AM^2+axy$ este

echivalenta cu $c^2y+b^2x=(x+y)\left(AM^2+xy\right)\ \Longleftrightarrow\ x\left(b^2-y^2\right)+y\left(c^2-x^2\right)=(x+y)\cdot AM^2\stackrel{(**)}{\ \Longleftrightarrow\ }$ $x\cdot (VA+VM)+y\cdot (UA-UM)=(x+y)\cdot AM$ $\Longleftrightarrow$

$x\cdot (VA+VM-AM)=y\cdot (UM+AM-UA)\ \Longleftrightarrow$ $2x\cdot VM=2y\cdot UM\ \Longleftrightarrow\ \frac {x}{UM}=\frac {y}{VM}\ ,$ ceea ce este adevarat.

Comentariu. Ultimele doua metode sunt culese din lucrarile elevilor din clasa a IX - a pe care i-am avut de-a lungul timpului din care cele mai frumoase le-am pastrat. Obisnuiam sa le dau ca subiect o chestiune teoretica de rezolvat, cu intentia de a gasi si alta solutie decat cea din manual sau cea oferita de profesor. Astfel elevul realiza ca si problemele remarcabile au aparut tot ca niste probleme propuse, insa avand calitatea de a fi "scurtaturi" in rezolvarea unor probleme mult mai dificile. Voi reveni curand si cu metoda a patra care mie mi se pare cea mai ingenioasa la nivelul unui elev de clasa a IX - a (Mihai Esanu, de multa vreme la Paris !).

P11. Fie triunghiurile isoscele $ABC\ ,$ $MNP$ astfel incat $AB=AC\ ,$ $MN=MP$ si $N\in (AC)\ ,$ $B\in (MP)\ .$ Notam

$R\in AB\cap MN\ ,$ $S\in BC\cap NP\ .$ Aratati ca $BP=CN\ \Longleftrightarrow$ semidreapta $[RS$ este una din bisectoarele unghiului $\widehat {BRN}\ .$

Demonstratie. Notam $X\in BC\cap MN\ ,\ Y\in AB\cap NP\ ,\ Z\in NB\cap RS\ .$ Aplicam teorema lui Menelaus transversalelor $\overline {XBC}\ ,$ $\overline {YNP}$

in $\triangle ANR\ ,\ \triangle BMR\ :$ $\left|\begin{array}{ccc} \frac {XR}{XN}\cdot\frac {CN}{CA}\cdot\frac {BA}{BR}=1 & \Longrightarrow & \frac {XN}{XR}=\frac {CN}{BR}\\\\ \frac {YR}{YB}\cdot\frac {PB}{PM}\cdot\frac {NM}{NR}=1 & \Longrightarrow & \frac {YR}{YB}=\frac {NR}{PB}\end{array}\right|\ .$ Aplicam teorema lui Ceva lui $S$ si la $\triangle BRN\ : \ \ \frac {TB}{TN}\cdot\frac {XN}{XR}\cdot\frac {YR}{YB}=1 \Longrightarrow$

$\frac {TN}{TB}=\frac {XN}{XR}\cdot\frac {YR}{YB}\ \Longrightarrow$ $\frac {TN}{TB}=\frac {NC}{PB}\cdot\frac {RN}{RB}\ .$ In concluzie, $NC=PB\ \Longleftrightarrow\ \frac {TN}{TB}=\frac {RN}{RB} \Longleftrightarrow$ semidreapta $[RS$ este una dintre bisectoarele unghiului $\widehat {BRN}\ .$

P12 (Manuela Prajea: clasa a VII - a lista scurta 2009). Fie $\triangle ABC$ si $D\ ,$ $E\ ,$ $F$ punctele de contact ale cercului inscris $w=\mathbb C(I,r)$ cu $BC\ ,$ $CA\ ,$ $AB$ respectiv.

Bisectoarea unghiului $\widehat {BAC}$ intersecteaza cercul circumscris $\Omega=\mathbb C(O,R)$ in punctul $S$ iar $SD$ taie din nou acelasi cerc in $T\ .$ Demonstrati ca $ATIE$ este inscriptibil.

Demonstratie. Fie diametrul $[AK]$ al cercului $\Omega\ .$ Se observa ca $\triangle IDS\sim\triangle AIK$ deoarece $m\left(\widehat{DIS}\right)=m\left(\widehat{IAK}\right)=\frac {|B-C|}2$ si $\frac {ID}{AI}=\frac {IS}{AK}\iff$ $\frac r{AI}=\frac {IS}{2R}$

$\iff$ $IA\cdot IS=2Rr$ ceea ce este adevarat $($ puterea punctului $I$ fata de $\omega)\ .$ Asadar $\widehat{TSA}\equiv\widehat{DSI}\equiv\widehat{IKA}\equiv \widehat{TKA}$ , adica $\widehat{TSA}\equiv\widehat{TKA}\iff TSKA$ este un

patrulater inscriptibil. Mai exact $T\in KI$ ceea ce inseamna ca $TA\perp TK$ , adica $T$ apartine cercului cu diametrul $[AI]\ .$ In concluzie, patrulaterul $ATIE$ este inscriptibil.

Observatie. Frumoasa problema, insa depaseste cu mult nivelul clasei a VII - a. Probabil acesta a fost motivul pentru care a ramas doar pe lista scurta. Ar fi mers foarte bine in finala la clasa

a IX - a, insa aici geometria se face vectorial in exclusivitate ceea ce inseamna ca avem si aici o "piedica". Imi permit sa adaug si o relatie metrica in plan secund: $\boxed{\ IK^2+bc=4R^2\ }\ .$

Intr-adevar, aplicam teorema medianei $IO$ in $\triangle AIK\ :\ 4\cdot IO^2=2\left(IA^2+IK^2\right)-AK^2\iff$ $2\left(\underline{\underline{R}}^2-2\underline{Rr}\right)=$ $\left[\left(bc-\underline{4Rr}\right)+IK^2\right]-2\underline{\underline{R}}^2\iff$ $IK^2+bc=4R^2\ .$

P13. Aratati ca in orice $\triangle ABC$ exista relatiile $\boxed{OH\ge R\sin|B-C|}\ (1)$ si $\boxed{IH\ge\frac{b+c}{2p}\sqrt{4p(p-a)\left(\frac{bc}{(b+c)^2}-\frac{(p-b)(p-c)}{a^2}\right)}}\ (2)\ .$ Cand are loc egalitatea ?

Proof.

$1\blacktriangleright\ OH\ \ge\ pr_{BC}OH=pr_{BC}OA=R\cdot \left|\cos\left(90^{\circ}-B+C\right)\right|=R\cdot \left|\sin (B-C)\right|\ .$

$2\blacktriangleright\ IH\ \ge\ pr_{BC}IH=\frac {|b-c|(s-a)}a\ .$

P14. Sa se arate ca $:\ \left\{\begin{array}{cccc} A=60^{\circ}\ \Longleftrightarrow\ OI=2R\cdot\sin\frac {|B-C|}{4} & (1)\\\\ OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}{4}} & (2)\\\\ OI\ \ge\ 2R\cdot \sin\frac {|B-C|}{4}\sqrt {2\sin\frac A2} & (3)\end{array}\ \right\|\ .$

Proof.

$1\blacktriangleright$ Identitatea $\boxed{\cos A+\cos B+\cos C=1+\frac rR}$ pentru $A:=60^{\circ}$ devine $\cos B+\cos C=\frac 12+\frac rR$ $\Longleftrightarrow$ $2\cos\frac{B+C}{2}\cdot\cos\frac{B-C}{2}=$ $\frac 12+\frac rR$ $\Longleftrightarrow$ $\cos\frac{B-C}{2}=\frac 12+\frac rR$

$\Longleftrightarrow$ $1-\cos\frac{B-C}{2}=$ $\frac 12-\frac rR$ $\Longleftrightarrow$ $2sin^2\frac{B-C}{4}=$ $\frac{R-2r}{2R}$ $\Longleftrightarrow$ $4R\sin^2\frac{B-C}{4}=R-2r$ $\Longleftrightarrow$ $4R^2\sin^2\frac{B-C}{4}=$ $R^2-2Rr=OI^2$ $\Longleftrightarrow$ $\boxed{OI=2R\cdot\sin\frac{|B-C|}{4}}\ .$

$2\blacktriangleright\ \cos A=1-2\sin^2\frac A2$ $\Longrightarrow$ $\cos A+\cos B+\cos C=1+\frac rR$ $\Longleftrightarrow$ $\cos B+\cos C=\frac rR+2\sin^2\frac A2$ $\Longleftrightarrow$ $2\cos\frac{B+C}2\cos\frac{B-C}2=$ $\frac rR+2\sin^2\frac A2$ $\Longleftrightarrow$

$\left|2\sin\frac A2\cos\frac{B-C}{2}=\frac{r}{R}+2\sin^2\frac A2\right|\ \odot\ 2R^2$ $\Longleftrightarrow$ $4R^2\sin\frac A2\left(1-2\sin^2\frac{B-C}4\right)=$ $2Rr+4R^2\sin^2\frac A2$ $\Longrightarrow$ $OI^2=R^2-2Rr=$ $R^2+4R^2\sin^2\frac A2-$ $4R^2\sin\frac A2+$

$8R^2\sin\frac A2\sin^2\frac{B-C}{4}$ $\Longleftrightarrow$ $OI^2=R^2\left(2\sin^2\frac A2-1\right)^2+$ $8R^2\sin\frac A2\sin^2\frac{B-C}{4}$ $\Longleftrightarrow$ $OI=R\sqrt {\left(2\sin\frac A2-1\right)^2+8\sin\frac A2\sin^2\frac {B-C}{4}}\ .$

$3\blacktriangleright$ Inlocuind pe $OI$ si ridicand la patrat, inegalitatea de demonstrat devine $\left(2\sin\frac A2-1\right)^2\ge 0\ ,$ ceea ce este adevarat. Egalitatea are loc daca si numai daca $A=60^{\circ}\ .$

P15. Consideram $\triangle ABC$ si un punct $P\ .$ Notam proiectiile $D\ ,$ $E\ ,$ $F$ ale lui $P$ pe $BC\ ,$ $CA\ ,$ $AB$ respectiv

si punctul $R\in EF\cap BC\ .$ Cercul circumscris al $\triangle DEF$ taie din nou $BC$ in $M\ .$ Sa se arate ca $RP\perp AM\ .$

Cazuri particulare.

$1.\ P: = H$ (ortocentrul $\triangle ABC$ ). In acest caz punctul $M$ este mijlocul lui $[BC]$ $\Longrightarrow\ RH\perp AM\ .$

$2.\ P: = I$ (centrul cercului inscris in $\triangle ABC$ ). In acest caz punctul $M$ este $D\ \Longrightarrow\ RI\perp AD\ .$

Proof. Patrulaterul $AFPE$ este inscris in cercul de diametru $[AP]\ .$ Notam $Q$ - cel de al doilea punct de intersectie al cercului circumscris patrulaterului $AFPE$ cu cercul circumscris al $\triangle MPD\ .$ Asadar, dreapta $PQ$ este axa radicala a cercurilor $\odot{AFPE}$ si $\odot{MPD}\ .$ Axa radicala a cercului $\odot{AFPE}$ si a cercului $\odot{DEF}$ este dreapta $EF\ ;$ axa radicala a cercurilor $\odot{DEF}$ si $\odot{MPD}$ este dreapta $BC\implies$ punctul $R$ este centrul radical al cercurilor $\odot{AFPE}\ ,$ $\odot{DEF}$ si $\odot{MPD}\implies$ $R\in{PQ}\ .$ Deoarece $\odot{AFPE}$ si $\odot{MPD}$ au diametre respectiv $[AP]$ si $[PM]$ rezulta ca $m\left(\widehat{AQP}\right)=$ $m\left(\widehat{PQM}\right)=$ $90^{\circ}\implies$ punctele $A$ , $Q$ si $M$ sunt coliniare si $AM\perp RP\ .$

P16. Fie $\triangle ABC$ cu bisectoarele $AD$ , $BE$ , $CF$ unde $D\in (BC)$ , $E\in (CA)$ si $F\in (AB)\ .$ Mediatoarele segmentelor

$[AD]\ ,\ [BE]\ ,\ [CF]$ intalnesc dreptele $BC\ ,\ CA\ ,\ AB$ in punctele $X\ ,\ Y\ ,\ Z$ respectiv. Sa se arate ca $Z\in XY\ .$

Demonstratie 1 (sintetic). Notam $A$ -bisectoarea exterioara $AL$ , unde $L\in BC\ .$ Din $AD\perp AL$ obtinem $XL=XA=XD$ si $\frac A2+C=m(\widehat {ADX})=$ $m(\widehat {DAX})=$

$\frac A2+m(\widehat {BAX})$ $\Longrightarrow$ $m(\widehat {BAX})=C\ ,$ adica dreapta $AX$ este tangenta cercului circumscris la $\triangle ABC\ .$ Asadar, $\triangle XAB\sim\triangle XCA$ $\Longrightarrow$ $\frac {XA}{XC}=\frac cb=$ $\frac {BX}{AX}$ $\Longrightarrow$

$\boxed{\begin{array}{c} XB=\frac cb\cdot AX\\\\ XC=\frac bc\cdot AX\end{array}}$ $\Longrightarrow$ $\boxed{\frac {XB}{XC}=\left(\frac cb\right)^2}\ .$ Analog $\frac {YC}{YA}=\left(\frac ac\right)^2$ si $\frac {ZA}{ZB}=\left(\frac ba\right)^2\ .$ Aplicarea teoremei Menelaus la $X\in BC\ ,$ $Y\in CA\ ,$ $Z\in AB$ confirma $Z\in XY\ .$

Demonstratie 2. Fie $U\in AB\ ,$ $V\in AC$ pentru care $X\in UV$ si $\overline {XUV}\perp AD\ .$ Din relatia cunoscuta (lungimea bisectoarei) $AD=\frac {2bc\cdot\cos\frac A2}{b+c}$ obtinem $AU=AV=$

$\frac {AD}{2\cos\frac A2}=\frac {bc}{b+c}\ .$ Aplicam teorema lui Menelaus la $\overline {XUV}$ si $\triangle ABC\ :\ \frac {XB}{XC}\cdot\frac {VC}{VA}\cdot\frac {UA}{UB}=1$ $\Longrightarrow$ $\frac {XB}{XC}=\frac {UB}{VC}=$ $\frac {c-AU}{b-AV}=\frac {c-\frac {bc}{b+c}}{b-\frac {bc}{b+c}}$ $\Longrightarrow$ $\boxed{\frac {XB}{XC}=\left(\frac cb\right)^2}\ .$

Observatie (retineti si dovediti !). Fie cercul inscris $\mathbb C(I,r)$ si $M\in AB\ ,\ IM\perp IA$ , $N\in AI\ ,\ MN\perp MA\ .$ Atunci $AM=\frac {bc}{p}$ si cercul cu centrul $N$ si raza $\rho_a=[NM]$

este tangent arcului de cerc circumscris care este subintins de $\widehat {BAC}$ (cercul $A$ -mixtliniar este cercul tangent la $AB\ ,$ $AC$ si tangent intern cercului circumscris al $\triangle ABC\ .$

Remarca. Proprietatea " $X\in AA\cap BC\ ,$ $Y\in BB\cap CA$ , $Z\in CC\cap AB$ sunt coliniare" este cunoscuta (caz degenerat al teoremei lui Pascal).

Am notat $AA$ - tangenta in $A$ la cercul circumscris al $\triangle ABC\ .$ Va invit sa incercati mai multe solutii si daca se poate cel putin o generalizare.

Demonstratie 3 (metric). Fie $M$ mijlocul lui $[AD]$ , $N\in MX\cap AB$ , $P\in MX\cap AC\ .$ Deci $N$ se afla pe mediatoarea $[AD]$ $\implies AN=ND\implies$ $m(\angle NDA)=\frac A2\implies$

$ND\parallel AC\implies$ $\triangle BND\sim \triangle BAC\implies$ $\frac{ND}b=\frac{BD}a=\frac{c}{b+c}\implies$ $ND=NA=\frac{bc}{b+c}\implies$ $BN=c-NA=c-\frac{bc}{b+c}=\frac{c^2}{b+c}\implies$ $\boxed{\frac{BN}{NA}=\frac{c}{b}}\ (1)\ .$

$\triangle ANP$ este $A$ -isoscel $\implies$ $AP=AN=\frac{bc}{b+c}$ $\implies$ $PC=b-AP=\frac{b^2}{b+c}\implies$ $\boxed{\frac{AP}{PC}=\frac{c}{b}}\ (2)\ .$ In $\triangle ABC$ aplicam teorema lui Menelaus pentru transversala $\overline{PNX}\ :$

$\frac{AP}{PC}\cdot\frac{CX}{XB}\cdot\frac{BN}{NA}=1\ .$ Din $(1)$ si $(2)$ obtinem $\boxed{\frac{BX}{XC}=\frac{c^2}{b^2}}\ (3)\ ,\ \boxed{\frac{AZ}{ZB}=\frac{b^2}{a^2}}\ (4)\ ,\ \boxed{\frac{CY}{YA}=\frac{a^2}{c^2}}\ (5)\ .$ Produsul $(3)\cdot (4)\cdot (5)\implies$ $\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$ $\implies Z\in XY\ .$

P17 (The Mathscope-Vietnam; Nr.5/2010). Let $\triangle ABC$ with the circumcircle $\Omega =\mathbb C(O,R)$ and the incircle $w=\mathbb C(I,r)\ .$ Denote the second

intersections $(D,E,F)$ of $(AI,BI,CI)$ with $\Omega\ .$ Prove that $\boxed{(b+c-a)\cdot ID^2+(a+c-b)\cdot IE^2+(a+b-c)\cdot IF^2=abc}\ .$

Proof. $\sum (b+c-a)\cdot ID^2=$ $2\sum (s-a)\left(2R\sin\frac A2\right)^2=$ $8R^2\cdot\sum \frac {(s-a)(s-b)(s-c)}{bc}=abc\ .$ Can use the identities $IA\cdot ID=2Rr$ and $ID^2=\frac {bc(s-a)}s\ .$

P18 (IRAN, 2011). In $\triangle ABC$ let $X$ and $Y$ be the tangent points of incircle $w=C(I,r)$ with $AB$ and $AC$ respectively. The tangent at $A$ to the circumcircle

$C(O,R)$ of $\triangle ABC$ intersects $BC$ at $D$ . Prove that $D\in XY\iff D\in OI\iff IL\parallel BC\iff\ IO\perp AZ$ $\iff$ $\ \boxed{a(b+c)=b^2+c^2}\ .$

$\iff (s-a)^2=(s-b)(s-c)$ where $2s=a+b+c$ , $Z\in BC\cap w$ and $L$ is the Lemoine's point (symmedian center) .

Proof. Suppose w.l.o.g. $b\ne c$ .

$\blacktriangleright\ \boxed{D\in XY}\iff$ $\frac {DB}{DC}=\frac{ZB}{ZC}\iff$ $\frac {c^2}{b^2}=\frac {s-b}{s-c}\iff$ $\frac {b^2-c^2}{b^2+c^2}=\frac {b-c}a\iff$ $\frac {b+c}{b^2+c^2}=\frac 1a\iff$ ${\boxed{a(b+c)=b^2+c^2}}$ .

$\blacktriangleright\ \boxed{IL\parallel BC}\iff$ $[BLC]=[BIC]\iff$ $\frac {a^2}{a^2+b^2+c^2}=\frac {a}{2s}\iff$ $a^2(a+b+c)=a\left(a^2+b^2+c^2\right)\iff$ $\boxed{a(b+c)=b^2+c^2}$ .

$\blacktriangleright\ \boxed{(s-a)^2=(s-b)(s-c)}\iff$ $a^2-2a(b+c)+(b+c)^2=a^2-(b-c)^2\iff$ $\boxed{a(b+c)=b^2+c^2}$ .

$\blacktriangleright\ \boxed{IO\perp AZ}\iff$ $OA^2-OZ^2=IA^2-IZ^2\iff$ $ZB\cdot ZC=IA^2-IX^2\iff$ $\boxed{(s-b)(s-c)=(s-a)^2}$ .

$\blacktriangleright\ \boxed{D\in XY}\iff$ $Z$ is the conjugate of $D$ w.r.t. $\{B,C\}$ $\iff$ $AZ$ is polar line of $D \iff$ $DO\perp AZ$ $\iff \boxed{D\in OI}$ .

P19. Let $\triangle ABC$ with the centroid $G\ ,$ the incircle $w=\mathbb C(I,r)$ and the Nagel's point $N\ .$ Prove that $\boxed{G\in (IN)\ \wedge\ IN=3\cdot IG}\ .$

Proof. Suppose w.l.o.g. $b>c$ and let the midpoint $M$ of $[BC]\ ,$ $D\in AI\cap BC$ and $P\in AN\cap BC\ ,$ where $\boxed{MP=\frac {b-c}2}\ (1)\ ,$ $MD=MB-DB=\frac a2-\frac {ac}{b+c}=$

$\frac {a(b-c)}{2(b+c)}$ $\implies$ $\boxed{MD=\frac {a(b-c)}{2(b+c)}}\ (2)$ and $DP=DM+MP=$ $\frac {a(b-c)}{2(b+c)}+\frac {b-c}2=\frac {b-c}2\cdot\left(1+\frac a{b+c}\right)\implies$ $\boxed{DP=\frac {s(b-c)}{b+c}}\ (3)\ .$ I"ll use the van Aubel's relation $:$

$\left\{\begin{array}{c} \frac {IA}{b+c}=\frac {ID}a=\frac {AD}{2s}\\\\ \frac {NA}a=\frac {NP}{s-a}=\frac {AP}s\end{array}\right\|\ (4)\ .$ Apply the Cristea's relation $:\ G\in IN\iff$ $\frac {ID}{IA}\cdot MP+\frac {NP}{NA}\cdot MD=\frac {GM}{GA}\cdot DP\iff$ $\frac a{b+c}\cdot \frac {\cancel{b-c}}2+\frac {s-a}{\cancel a}\cdot \frac {\cancel a(\cancel{b-c})}{2(b+c)}=$ $\frac 12\cdot \frac {s(\cancel{b-c})}{b+c}$

what is true, i.e. $G\in (IN)\ .$ Now I"ll apply an well known identity $:\ \frac {GN}{GI}=\frac {MP}{MD}\cdot\frac {AN}{AP}\cdot\frac {AD}{AI}=$ $\frac {\frac {b-c}2}{\frac {a(b-c)}{2(b+c)}}\cdot\frac as\cdot\frac {2s}{b+c}=2\implies GN=2\cdot GI\ ,$ i.e. $\boxed{IN=3\cdot IG}\ .$

P20. Given $\triangle ABC$ and $N$ inside the triangle. Let $(D,E,F)$ be the projections of $N$ onto $BC\ ,$ $AC\ ,$ $AB$ respectively. Find the position of $N$ which satisfies $\frac{ND}a=\frac{NE}b=\frac{NF}c\ .$

Proof. Let $(X,Y,Z)$ be the intersections of $AN\ ,$ $BN\ ,$ $CN$ with the opposite sidelines respectively and $S$ be the area of $\triangle ABC\ .$ Therefore,

$\frac{ND}a=\frac{NE}b=\frac{NF}c\iff$ $\frac{a\cdot ND}{a^2}=\frac{b\cdot NE}{b^2}=\frac{c\cdot NF}{c^2}\iff$ $\frac{[BNC]}{a^2}=\frac{[CNA]}{b^2}=\frac{[ANB]}{c^2}=\frac S{a^2+b^2+c^2}=k\ .$ In conclusion,

$\left\{\begin{array}{ccccc} \frac {b^2}{c^2}=\frac {[CNA]}{[BNA]}=\frac {\delta_{NA}(C)}{\delta_{NA}(B)}=\frac {XC}{XB} & \implies & \frac {XB}{XC}=\frac {c^2}{b^2} & \implies & \mathrm{AX\ is\ the\ A-symmedian}\\\\ \frac {c^2}{a^2}=\frac {[ANB]}{[BNC]}=\frac {\delta_{NB}(A)}{\delta_{NB}(C)}=\frac {YA}{YC} & \implies & \frac {YA}{YC}=\frac {c^2}{a^2} & \implies & \mathrm{BY\ is\ the\ B-symmedian}\end{array}\right\|\implies$ $N\in AX\cap BY$ is the Lemoine's point.

Remark. The Lemoine's point $S$ (symmedian center is isogonal conjugate of the centroid $G$ ) has the barycentrical coordinates $S\left(a^2,b^2,c^2\right)$ w.r.t. $\triangle ABC\ .$

P21. Let $\triangle ABC$ with the orthocenter $H\ ,$ the Lemoine's point $L\ ,$ the midpoint $M$ of $[BC]$ and the intersections $\left\{\begin{array}{ccc} E\in BH\cap CA & ; & F\in CH\cap AB\\\\ Q\in AM\cap EF & ; & D\in AH\cap BC\end{array}\right\|\ .$ Prove that $L\in DQ\ .$

Proof. Denote $S\in AL\cap BC\ .$ Thus, $L\in DQ\iff$ $\overline{DLQ}/\triangle AMS\ :\ \boxed{\frac {DS}{DM}\cdot\frac {QM}{QA}\cdot\frac {LA}{LS}=1}\ (*)\ .$ From the Van Aubel's relation obtain that $\boxed{\frac {LA}{LS}=\frac {b^2+c^2}{a^2}}\ (1)\ .$ Apply

the Cristea's relation $:\ \frac {QM}{QA}\cdot BC=\frac {EC}{EA}\cdot MB+\frac {FB}{FA}\cdot MC\iff$ $\frac {QM}{QA}=\frac {a\cdot\cos C}{2c\cdot\cos A}+\frac {a\cdot \cos B}{2b\cdot\cos A}\iff$ $\frac {QM}{QA}=\frac {a(b\cdot\cos C+c\cdot \cos B)}{2bc\cdot\cos A}=$ $\frac {a^2}{b^2+c^2-a^2}\implies$

$\boxed{\frac {QM}{QA}=\frac {a^2}{b^2+c^2-a^2}}\ (2)\ .$ Prove easily that $\boxed{DM=\frac {\left|b^2-c^2\right|}{2a}}\ (3.1)$ and $\frac {SB}{c^2}=\frac {SC}{b^2}=\frac a{b^2+c^2}\implies$ $\boxed{SB=\frac {ac^2}{b^2+c^2}}\ (3.2)\ .$ Therefore, $DS=|SB-DB|=$

$\left|\frac {ac^2}{b^2+c^2}-\frac {a^2+c^2-b^2}{2a}\right|=$ $\left|\frac {\cancel 2a^2c^2-\left(c^2-b^2\right)\left(c^2+b^2\right)-a^2\left(b^2+\cancel{c^2}\right)}{2a\left(b^2+c^2\right)}\right|=$ $\left|\frac {a^2\left(c^2-b^2\right)-\left(b^2+c^2\right)\left(c^2-b^2\right)}{2a\left(b^2+c^2\right)}\right|=$ $\frac {\left|b^2-c^2\right|\left(b^2+c^2-a^2\right)}{2a\left(b^2+c^2\right)}\implies$ $\boxed{DS=\frac {\left|b^2-c^2\right|\left(b^2+c^2-a^2\right)}{2a\left(b^2+c^2\right)}}\ (3.3)\implies$ $\frac {DS}{DM}=\frac {\frac {\left|b^2-c^2\right|\left(b^2+c^2-a^2\right)}{2a\left(b^2+c^2\right)}}{\frac {\left|b^2-c^2\right|}{2a}}\implies$ $\boxed{\frac {DS}{DM}=\frac {b^2+c^2-a^2}{b^2+c^2}}\ (3)\ .$ In conclusion, $\frac {DS}{DM}\cdot\frac {QM}{QA}\cdot\frac {LA}{LS}\ \stackrel{3\wedge 2\wedge 1}{=}\ 1\ \stackrel{(*)}{\iff}\ L\in DQ\ .$

P22 (Kadir Altintas - Turkey). Let $\triangle ABC$ with orthocenter $H\ ,$ midpoint $M$ of $[BC]$ , $D\in AB\ ,$ $E\in AC$ so that $H\in DE\ .$ Prove that $HE=HD\iff MH\perp DE\ .$

Proof 1. Let orthic $\triangle XYZ$ of $\triangle ABC\ ,$ where $X\in BC\ ,$ $Y\in CA\ ,$ $Z\in AB$ and $U\in HB\ ,$ $V\in HC$ so that $DU\perp DE$ and $EV\perp ED\ .$ Thus, $DUEY\ ,$ $EVDZ$ are cyclic,

i.e. $HU\cdot HY=HD\cdot HE=HV\cdot HZ\implies$ $HU\cdot HY=HV\cdot HZ$ , i.e. $YZUV$ is cyclic $\implies$ $m(\widehat{YUV})=m(\widehat{YZC})$ $=m(\widehat{YBC})\implies$ $m(\widehat{YUV})=m(\widehat{YBC})\implies$

$UV\parallel BC$ and midpoint $W$ of $[UV]$ belongs to $HM\ .$ So $HD=HE\iff HW$ is midline of trapezoid $DEVU\iff HW\parallel DU\iff$ $HW\perp DE\iff HM\perp DE\ .$

Proof 2. I"ll use same notations from previous proof. Denote the projections $P\ ,$ $Q$ of $B\ ,$ $C$ on $DE$ respectively and $\left\{\begin{array}{ccc} m\left(\widehat{YHE}\right) =m\left(\widehat{BHD}\right)=\phi\\\\ m\left(\widehat{ZHD}\right) =m\left(\widehat{CHE}\right)=\psi\end{array}\right\|$ where $\psi +\psi =A\ .$ Thus, $\left\{\begin{array}{ccc} HZ & = & HD\cos \psi\\\\ HY & = & HE\cos \phi\\\\ HZ\cdot HC & = & HY\cdot HB\end{array}\right\|$ and $HD=HE\iff$ $\frac {HZ}{\cos \psi}=\frac {HY}{\cos \phi}\iff$ $\frac {HZ\cdot HC}{HC\cos \psi}=\frac {HY\cdot HB}{HB\cos \phi}\iff$ $HC\cos \psi =HB\cos \phi\iff$ $HQ=HP\iff$ $MH\perp DE\ .$

This post has been edited 321 times. Last edited by Virgil Nicula, Sep 9, 2016, 4:26 PM

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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!
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El_Ectric:
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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!
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Dear Virgil !

Congratulations for this blog !
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444. Remarkable points in a triangle.

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