458. Working page IV.

by Virgil Nicula, Aug 27, 2017, 10:47 AM

$1.\blacktriangleright\ \boxed{f(x)=\frac {1-\cos^3x}{\sin^2x}}=$ $\frac{\cancel{(1-\cos x)}\left(1+\cos x+\cos^2x\right)}{\cancel{(1-\cos x)}(1+\cos x)}=$ $\frac {1+\cos x+\cos^2x}{1+\cos x}\implies$ $\lim_{x\to 0}f(x)=$ $\lim_{x\to 0}\frac {1+\cos x+\cos^2x}{1+\cos x}=\frac 32\implies \boxed{\lim_{x\to 0}f(x)=\frac 32}\ .$

$2.\blacktriangleright\ \boxed{f(x)=\frac {\sin (x-1)}{x^{\alpha}-1}\ ,\ \alpha \ne 0}=$ $\frac {\sin (x-1)}{x-1}\cdot\frac {x-1}{x^{\alpha}-1}$ $\implies$ $\lim_{x\to 1}f(x)=$ $\lim_{x\to 1}\frac {\sin (x-1)}{x-1}\cdot\lim_{x\to 1}\frac {x-1} {x^{\alpha}-1}\ \stackrel{t:=x-1}{=}$ $\lim_{t\to 0}\frac {\sin t}t\cdot \frac 1{\alpha}=$ $1\cdot\frac 1{\alpha}=$ $\frac 1{\alpha}$ $\implies$ $\boxed{\lim_{x\to 1}f(x)=\frac 1{\alpha}}\ .$

You can use the following remarkable limits $:\ \lim_{x\to 0}\frac {\sin x}x=1\ ;\ \lim_{x\to 0}\frac {1-\cos x}{x^2}=\frac 12\ ;\ \lim_{x\to 0}\frac {x^{\alpha}-1}{x-1}=\alpha\ ,\ \alpha\ne 0$ a.s.o. Here is some interesting limits (<= click).

P1. Denote $\left\{\begin{array}{ccc} a+b+c & = & s_1\\\\ ab+bc+ca & = & s_2\\\\ abc & = & s_3\end{array}\right\|$ and $(\forall )\ k\in\mathbb N^*\ ,\ S_k=a^k+b^k+c^k\ ,$ where $S_1=s_1\ .$ Prove that $(\forall )\ k\in\mathbb N^*\ ,\ S_{k+3}=s_1S_{k+2}-s_2S_{k+1}+s_3S_{k}\ .$

Calculate $\left\{\begin{array}{cccc} S_2 & = & s_1^2-2s_2\\\\ S_3 & = & s_1^3-3s_1s_2+3s_3\end{array}\right\|\ \mathrm{and}\ \left\{\begin{array}{cccc} S_4 & = & s_1^4-4s_1^2s_2+4s_1s_3+2s_2^2\\\\ S_5 & = & s_1^5-5s_1^3s_2+5s_1s_2^2+5s_1^2s_3-5s_2s_3\end{array}\right\|\ .$ Prove that $(\forall )\ \{a,b,c\}\subset\mathbb C$ exist the following

$\boxed{\ \underline{\underline{\mathrm{Identities}}}\ :\ \left\{\begin{array}{cccccccccc} \sum\limits_{\mathrm{cyc}}a^2(b+c) & = & \sum\limits_{\mathrm{cyc}}bc(b+c) & = & \sum\limits_{\mathrm{cyc}}a\left(b^2+c^2\right) & \implies & s_1S_2-S_3=s_1s_2-3s_3 & \implies & \boxed{\ s_1\left(S_2-s_2\right)=S_3-3s_3\ } & (1)\\\\ \sum\limits_{\mathrm{cyc}}a^3(b+c) & = & \sum\limits_{\mathrm{cyc}}a\left(b^3+c^3\right) & = & \sum\limits_{\mathrm{cyc}}bc\left(b^2+c^2\right) & \implies & s_1S_3-S_4=s_2S_2-s_1s_3 & \implies & \boxed{\ s_1\left(S_3+s_3\right)=S_4+s_2S_2\ } & (2)\end{array}\right\|\ }$

Proof. The equation $x^3-s_1x^2+s_2x-s_3=0\begin{array}{ccc} \nearrow & x_1=a & \searrow\\\\ \rightarrow & x_2=b &\rightarrow\\\\ \searrow & x_3=c & \nearrow\end{array}\odot\implies$ $(\forall )\ k\in\mathbb N^*$ and $(\forall )\ i\in\{1,2,3\}\ ,\ \left\|x_i^{k+3}=s_1x_i^{k+2}-s_2x_i^{k+1}+s_3x_i^{k}\right\|\ \sum\limits_{i=1}^3$ $\implies$

$(\forall )\ k\in\mathbb N^*\ ,\ \sum\limits_{i=1}^3x_i^{k+3}=s_1\sum\limits_{i=1}^3x_i^{k+2}-s_2\sum\limits_{i=1}^3x_i^{k+1}+s_3\sum\limits_{i=1}^3x_i^{k}$ $\implies$ $(\forall )\ k\in\mathbb N^*\ ,\ S_{k+3}=s_1S_{k+2}-s_2S_{k+1}+s_3S_{k}$ a.s.o.

$\blacktriangleright\ \left\{\begin{array}{ccccc} \sum\limits_{\mathrm{cyc}}a^2(b+c) & = & \sum\limits_{\mathrm{cyc}}a^2[(a+b+c)-a] & = & s_1S_2-S_3\\\\ \sum\limits_{\mathrm{cyc}}bc(b+c) & = & \sum\limits_{\mathrm{cyc}}bc[(a+b+c)-a] & = & s_1s_2-3s_3\\\\ \sum\limits_{\mathrm{cyc}}a\left(b^2+c^2\right) & = & \sum\limits_{\mathrm{cyc}}a\left[\left(b^2+c^2+a^2\right)-a^2\right] & = & s_1S_2-S_3\end{array}\right\|\ \mathrm{\underline{{and}}}\ \left\{\begin{array}{ccccc} \sum\limits_{\mathrm{cyc}}a^3(b+c) & = & \sum\limits_{\mathrm{cyc}}a^3[(a+b+c)-a] & = & s_1S_3-S_4\\\\ \sum\limits_{\mathrm{cyc}}a\left(b^3+c^3\right) & = & \sum\limits_{\mathrm{cyc}}a[(a^3+b^3+c^3)-a^3] & = & s_1S_3-S_4\\\\ \sum\limits_{\mathrm{cyc}}bc\left(b^2+c^2\right) & = & \sum\limits_{\mathrm{cyc}}bc\left[\left(b^2+c^2+a^2\right)-a^2\right] & = & s_2S_2-s_1s_3\end{array}\right\|\ .$

Examples.

$\sum\limits_{\mathrm{cyc}}a^3(b+c)=$ $\sum\limits_{\mathrm{cyc}}a^2[(ab+ac+bc)-bc]=$ $s_2S_2-s_1s_3\ ;\ \sum\limits_{\mathrm{cyc}}a^2(b+c)=$ $\sum\limits_{\mathrm{cyc}}a[(ab+ac+bc)-bc]=$ $s_1s_2-3s_3\ ;\ \boxed{\sum\limits_{\mathrm{cyc}}a^3\left(b^2+c^2\right)=\sum\limits_{\mathrm{cyc}}a^2\left(b^3+c^3\right)=S_2S_3-S_5}\ .$

$\sum\limits_{\mathrm{cyc}}a^4(b+c)=$ $\sum\limits_{\mathrm{cyc}}a\left(b^4+c^4\right)=s_1S_4-S_5\ ;\ \sum\limits_{\mathrm{cyc}}a^4(b+c)=\sum\limits_{\mathrm{cyc}}a^3[(ab+ac+bc)-bc]=$ $s_2S_3-s_3S_2\implies$ $s_1S_4-S_5=s_2S_3-s_3S_2\implies$ $\boxed{s_1S_4+s_3S_2=S_5+s_2S_3\ }\ .$

Problema propusa 7 (<= click).

Proof. Notam $L$ -nr. ladite si $K$ -nr. kg. fructe $\implies L=\frac {K-4}3=\frac {K+9\cdot 4}4=\frac {(K+36)-(K-4)}{4-3}=40\ \odot\begin{array}{ccccc} \nearrow & L & = & 40\ & \searrow\\\\ \searrow & K & = & 124 & \nearrow\end{array}\odot$ Altfel: $3L+4=K=4L-4\cdot 9$ etc.

Cartile lui Ken ROBINSON ar trebui sa fie lecturi obligatorii pentru profesori. Vedeti aici (<= click). Ce inseamna a fi creator?! Sa-l intrebam ("citim") pe Ken ROBINSON, autorul cartii "O lume iesita din minti". Un raspuns imediat se cuvine a veni de la sine: "a reproduce integralul din partial" - prima lege a creatiei. Vom ilustra prima lege a creatiei prin cateva exemple intalnite in matematica $:$

1. Intr-o problema de geometrie: o constructie auxiliara care sa "intregeasca" o figura cunoscuta ale carei proprietati le stim si le putem folosi bine, poate reduce substantial demonstratia problemei. Uneori elevii chiar intreaba "cum v-a venit ideea, dle profesor?!".

2. Sa presupunem ca cerem cu anticipatie elevilor de clasa a VII - a (care nu au invatat inca ecuatia de gradul doi) sa gaseasca doua numere reale x & y pentru care stim suma 10 si produsul 18. Un "act creator" ar fi acela cand elevul ar "intregi" sistemul dat la unul de gradul intai cu doua necunoscute. De exemplu sa afle, daca se poate, diferenta intre x & y. Deoarece stim suma lor si cateva identitati algebrice, atunci incercam sa aflam (x-y)^2=(x+y)^2-4xy, adica (x-y)^2=100-72=28. Asadar |x-y|=2sqrt7. Putem presupune fara a restrange generalitatea ca x este cel putin egal lui y. Asadar sistemul x+y=10 & x-y=2sqrt7 este agreabil si are solutia {x,y}={5±sqrt7}.

3. In analiza matematica de clasa a XI - a: la limite de siruri/functii se foloseste foarte frecvent aceasta "lege". De exemplu vrem sa aflam limita cand x tinde la 0 pentru functia f(x)=sinxsin2x/(1-cos x). Stim ca sinx/x->1 & (1-cosx)/(x^2)->1/2. Vom "intregi" cele doua limite cunoscute in functia f astfel: f(x)=[(sinx)/x]•[(sin2x)/(2x)]•[(2x^2)/(1-cosx)] si prin trecere la limita se obtine 1•1•2•2=4.

4. In analiza matematica de clasa a XII - a: la capitolul "primitive" se foloseste "metoda integrarii prin parti" care nu este altceva decat "intregirea" derivatei unui produs: (fg)'=f'g+fg' etc. Voi continua sa ilustrez "actul de creatie" si in alte domenii de cercetare/proiectare care sa atenueze unele situatii cu dificultate ridicata.

P2 (Leo GIUGIUC). Three parallel lines are drawn through the vertices of an equilateral $\triangle ABC$ . The line perpendicular to the three through $B$

intersects the remaining two at $E$ and $F$ so that $B\in (EF)\ .$ If $BE=m\ ,\ BF=n\ ,$ prove that $m^2+mn+n^2$ is constant for a given triangle.

Proof 1. Denote $AC=a$ and the projection $P$ of $B$ on $AC$ $\implies$ $APBE$ and $CPBF$ are cyclic with the diameters $[AB]$ and $[BC]\ .$ Hence $m\left(\widehat{PEB}\right)= m\left(\widehat{PAB}\right)=60^{\circ}\ ,$ $m\left(\widehat{PCB}\right)=$

$m\left(\widehat{PFB}\right)=60^{\circ}\implies$ $m\left(\widehat{PEB}\right)=$ $m\left(\widehat{PFB}\right)=$ $60^{\circ}\ ,$ i.e. $\triangle PEF$ is equilateral $:\ PE=PF=EF=m+n\ .$ Apply generalized Pythagoras' theorem in $\triangle BPF$ with $m\left(\widehat{BFP}\right)=$

$60^{\circ}\ :\ BP^2=FB^2+FP^2-FB\cdot FP=$ $n^2+(m+n)^2-n(m+n)=$ $n^2+m^2+ n^2+2mn-mn-n^2=$ $m^2+n^2+mn\implies$ $m^2+n^2+mn= BP^2=\frac{ 3a^2}4$ (constant).

Proof 2. Suppose w.l.o.g. $AC=1$ and denote $m\left(\widehat{ABE}\right)=x\ ,$ $m\left(\widehat{CBF}\right)=y\ ,$ where $x+y=120^{\circ}\ .$ Thus, $m=\cos x\ ,$ $n=\cos y$ and

$2\left(m^2+mn+n^2\right)=$ $2\left(\cos^2x+\cos x\cos y+\cos^2y\right)=$ $1+\cos 2x+\cos (x+y)+\cos (x-y)+1+\cos 2y=$ $2+\cos 2x+\cos 2y-$

$-\frac 12+\cos (x-y)=$ $\frac 32+2\cos (x+y)\cos (x-y)+\cos (x-y)=$ $\frac 32-\cos (x-y)+\cos (x-y)=\frac 32\implies$ $\boxed{\ m^2+n^2+mn=\frac 34\ }\ .$

P3 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with the incircle $\mathbb C(I,r)$ what touches given triangle at $D\in BC\ ,$ $E\in CA$ and $F\in AB\ .$ Denote $P\in EF\ ,$ $DP\perp EF\ .$ Prove that $\boxed{\ PD=h_a\sin\frac A2\ }\ .$

Proof. Suppose w.l.o.g. $b>c$ and denote $L\in AI\cap BC$ and $T\in EF\cap BC\ .$ Apply the Menelaus' theorem to the collinear points $\overline{EFT}/\triangle ABC\ :\ \frac {TB}{TC}\cdot \frac {EC}{EA}\cdot\frac {FA}{FB}=1\iff$

$\frac {TD}{TC}\cdot \frac {s-c}{\cancel{s-a}}\cdot\frac {\cancel{s-a}}{s-b}=1\iff$ $\frac {TB}{TC}=\frac {s-b}{s-c}\iff$ $\frac {TB}{s-b}=\frac {TC}{s-c}=\frac a{b-c}$ $\implies$ $TB=\frac {a(s-b)}{b-c}\implies\ TD=TB+BD=$ $\frac {a(s-b)}{b-c}+(s-b)=$ $\frac {2(s-b)(s-c)}{b-c}=$

$\frac {2bc\sin^2\frac A2}{b-c}\implies$ $\boxed{TD=\frac {2bc\sin^2\frac A2}{b-c}}\ (1)\ .$ Observe that $PD\parallel AL\implies$ $\widehat{TDP}\equiv\widehat{TLA}\equiv\widehat{BLA}\implies$ $m\left(\widehat{TDP}\right)=m\left(\widehat{BLA}\right)=\frac A2+C=90^{\circ}-\left(\frac B2-\frac C2\right)\implies$

$\boxed{\ \cos \widehat{TDP}=\sin\frac {B-C}2\ }\ (2)\ .$ Prove easily that $\boxed{\ bc=2Rh_a\ }\ (3)\ .$ Therefore, $DP\perp TE\implies DP\parallel AL\implies$ $PD=TD\cdot\cos\widehat{TDP}\ \stackrel{(1\wedge 2)}{=}\ \frac {2bc\sin^2\frac A2\sin\frac {B-C}2}{b-c}=$

$\frac {2bc\sin \frac A2\cos\frac {B+C}2\sin\frac {B-C}2}{b-c}=\frac {bc\sin\frac A2(\sin B-\sin C)}{b-c}=\frac {bc\sin\frac A2}{2R}\ \stackrel{(3)}{=}\ h_a\sin\frac A2\ .$ In conclusion, $PD=h_a\sin\frac A2\ .$

Remark. Denote $K\in BC\ ,$ $AK\perp BC$ and the diameter $[AS]$ of the circumcircle $\Omega=\mathbb C(O,R)\ .$ Observe that $\triangle ABK\sim\triangle ASC\iff$ $\frac {AB}{AS}=\frac {AK}{AC}\iff$ $\frac c{2R}=\frac {h_a}b\implies bc=2Rh_a\ .$

P4 (Miguel Ochoa Sanchez). Let an $A$ -isosceles $\triangle AFE$ and $\left\|\begin{array}{c} B\in (AF)\\\ C\in (AE)\end{array}\right\|$ so that $FB+EC=BC\ .$ Let $D\in (BC)$ so that $\left\{\begin{array}{c} DB=BF\\\ DC=CE\end{array}\right\|\ .$ Prove that $\frac 1{[FBE]}+\frac 1{[FCE]}=\frac 2{[FDE]}\ .$

Proof. Let the projections $(U,P,V,W)$ of $(B,A,D,C)$ on $EF$ and denote $AF=AE=l\ ,$ $FB=BD=m\ ,$ $EC=CD=n\ ,$ $BU=x\ ,$ $AP=h\ ,$ $DV=z$ and $CW=y\ .$

Thus, $\frac xm=\frac hl=\frac yn\ .$ The relation $\boxed{\ z(m+n)=nx+my\ }\ (*)$ is well known in the trapezoid $BCWU\ .$ Since $(m,n)$ are proportional with $(x,y)$ then the relation $(*)$

becomes $z(x+y)=yx+xy\ ,$ i.e. $\frac 1x+\frac 1y=\frac 2z\ (2)\ .$ But $(x,y,z)$ are proportional with $\left([FBE],[FCE],[FDE]\right)$ $\implies$ $\frac 1{[FBE]}+\frac 1{[FCE]}=\frac 2{[FDE]}\ .$

Remark Suppose w.l.o.g. that $x>z>y$ and let $M\in BU\ ,$ $N\in DV$ so that $\overline{MNC}\perp BU\ .$ Thus, $\frac {BM}{CB}=\frac {DN}{CD}\iff$ $\frac {x-y}{m+n}=\frac {z-y}n=\frac {x-z}m\iff$ $m(z-y)=n(x-z)\iff$ $(*)\ .$

P5 (Miguel Ochoa Sanchez). Let $ABCD$ be a convex quadrilateral where $AB=a\ ,$ $BC=b\ ,$ $BD=x$ and $m\left(\widehat{CAD}\right)=\alpha\ ,$

$m\left(\widehat{ACD}\right)=\beta\ ,$ $m\left(\widehat{ABC}\right)=\phi$ so that $\alpha +\beta \in \left\{\phi\pm90^{\circ}\right\}\ .$ Prove that $\boxed{\ a^2\sin^2\alpha +b^2\sin^2\beta =x^2\cos^2\phi\ }\ .$

Proof. Let $\left\{\begin{array}{c} DA=y\\\ DC=z\end{array}\right\|\implies\boxed{\ \frac y{\sin\beta}=\frac z{\sin\alpha}\ }\ (*)\ .$ Observe that $B+D=\phi +\left[180^{\circ}-\left(\alpha +\beta\right)\right]=$ $\phi +180^{\circ}-\left(\phi \pm 90^{\circ}\right)\in \left\{90^{\circ},270^{\circ}\right\}$ and $AC=y\cos\alpha +z\cos\beta .$ Apply the

generalized Ptolemy's theorem $:\ (AB\cdot CD)^2+(BC\cdot AD)^2=(AC\cdot BD)^2\ ,$ i.e. $\boxed{(az)^2+(by)^2=x^2(y\cos \alpha +z\cos\beta )^2}\ (1)\ .$ Thus, $(y,z)$ are proportional with $(\sin\beta ,\sin\alpha )\implies$

the relation $(1)$ becomes $(a\sin\alpha )^2+(b\sin\beta )^2=x^2(\sin \beta\cos\alpha +\sin\alpha\cos\beta )^2\iff$ $a^2\sin^2\alpha +b^2\sin^2\beta=x^2\sin^2(\alpha +\beta )\iff$ $b^2\sin^2\beta +c^2\sin^2\alpha =x^2\cos^2\phi\ .$ Nice problem!

P6 (Soji NAKAJIMA). Prove that $(\forall )\ \triangle ABC$ there is the inequality $\boxed{\ a^2+b^2+c^2\le 8R^2+\frac {4S}{3\sqrt 3}\ }\ (*)\ .$

Proof 1. The relations $\boxed{\sum a^2=2\left(s^2-r^2-4Rr\right)}$ and $\boxed{s^2\le 4R^2+4Rr+3r^2}$ are well-known. Hence $\sum a^2\le 2\left[\left(4R^2+\cancel{4Rr}+3r^2\right)-r^2-\cancel{4Rr}\right]=$ $4\left(2R^2+r^2\right)\ .$

Observe that $4\left(\cancel{2R^2}+r^2\right)\le \cancel{8R^2}+\frac {4S}{3\sqrt 3}\iff$ $\cancel 4r^2\le \frac {\cancel 4S}{3\sqrt 3}\iff$ $3r\cancel{^2}\sqrt 3\le s\cancel r\iff$ $3r\sqrt 3\le s\ ,$ what is true. In conclusion and the required inequality $(*)$ is true.

Proof 2 (trigonometric). $\underline{\underline{a^2+b^2+c^2\le 8R^2+\frac {4}{3\sqrt 3}\cdot S}}\iff$ $2\cancel{R^2}\cdot\sum 2\sin^2A\le 8\cancel{R^2}+\frac 4{3\sqrt 3}\cdot 2\cancel{R^2}\prod\sin A\iff$ $2\cdot\sum \left(1-\cos 2A\right)\le 8+\frac {8}{3\sqrt 3}\cdot \prod\sin A\iff$

$3-\sum\cos 2A\le 4+\frac 4{3\sqrt 3}\cdot\prod\sin A\iff$ $\left(1+\sum\cos 2A\right)+\frac 4{3\sqrt 3}\cdot \prod\sin A\ge 0\iff$ $-\cancel 4\prod\cos A+\frac {\cancel 4}{3\sqrt 3}\cdot\prod\sin A\ge 0\iff$ $3\sqrt 3\cdot \prod\cos A\le \prod\sin A\ \stackrel{\mathrm{acute}\ \triangle\ ABC}{\iff}$

$\prod\tan A\ge 3\sqrt 3\iff$ $\underline{\underline{\sum\tan A\ge 3\sqrt 3}}\ ,$ what is true only on the first quadrant because the function $\tan$ is convex: $\tan\left(\frac {A+B+C}3\right)\le\frac {\tan A+\tan B+\tan C}3\ .$

P7 (Miguel Ochoa Sanchez). Let a rhombus $ABCD$ with $AB=r\ ,$ $A\le 120^{\circ}$ and the circle $\mathrm w(C,r)\ .$ Prove that $(\forall )\ P\in w$ there is the relation $\boxed{\ PB^2+PD^2=PA^2+PC^2(1-2\cos A)\ }\ (*)\ .$

Proof. Denote $M\in AC\cap BD$ and $PA=x\ ,$ $PB=y$ and $PD=z\ .$ Observe that $AM=r\cos\frac A2$ and $BM=r\sin\frac A2\ .$ Apply the theorem of median in the triangles $:$

$\left\{\begin{array}{cccc} PM/\triangle APC\ : & 4\cdot PM^2=2\left(PA^2+PC^2\right)-AC^2 & \iff & 4\cdot PM^2=2\left(x^2+r^2\right)-4r^2\cos^2\frac A2\\\\ PM/\triangle BPD\ : & 4\cdot PM^2=2\left(PB^2+PD^2\right)-BD^2 & \iff & 4\cdot PM^2=2\left(y^2+z^2\right)-4r^2\sin^2\frac A2\end{array}\right\|$ $\implies$ $x^2+r^2-r^2(\cancel 1+\cos A)=y^2+z^2-r^2(\cancel 1-\cos A)\iff$

$y^2+z^2=x^2+r^2(1-2\cos A)\ ,$ i.e. the required relation $PB^2+PD^2=PA^2+PC^2(1-2\cos A)\ .$ Particular cases $:\ \left\{\begin{array}{ccc} \underline{\underline{A=60}}^{\circ} & \implies & \underline{\underline{PB^2+PD^2=PA^2}}\\\\ \underline{\underline{A=90}}^{\circ} & \implies & \underline{\underline{PB^2+PD^2=PA^2+PC^2}}\\\\ \underline{\underline{A=120}}^{\circ} & \implies & \underline{\underline{PB^2+PD^2=PA^2+2\cdot PC^2}}\end{array}\right\| .$

P8 (Miguel Ochoa Sanchez). (<= click).

Proof. I"ll use only the theorem of Sines. Denote $\left\{\begin{array}{c} m\left(\widehat{ADP}\right)=x\implies m\left(\widehat{PAB}\right)=60^{\circ}-x\\\\ m\left(\widehat{ADQ}\right)=y\implies m\left(\widehat{QAC}\right)=60^{\circ}-y\end{array}\right\|\ .$ Hence $\underline{\underline{\frac {AP}{MD}+\frac {AQ}{ND}}}=$ $\frac {AP}{AM}\cdot\frac {AM}{MD}+\frac {AQ}{AN}\cdot\frac {AN}{ND}=$

$\cos\widehat{PAM}\cdot\frac {\sin\widehat{MDA}}{\sin\widehat{MAD}}+\cos \widehat{QAN}\cdot\frac {\sin\widehat{NDA}}{\sin\widehat{NAD}}=$ $\cos \left(60^{\circ}-x\right)\cdot \frac {\sin x}{\sin 30^{\circ}}+$ $\cos \left(60^{\circ}-y\right)\cdot \frac {\sin y}{\sin 30^{\circ}}=$ $2\cos\left(60^{\circ}-x\right)\sin x+$ $2\cos \left(60^{\circ}-y\right)\sin y=$

$\sin 60^{\circ}-\sin\left(60^{\circ}-2x\right)+\sin 60^{\circ}-\sin \left(60^{\circ}-2y\right)=$ $\sqrt 3-\left[\sin(60^{\circ}-2x)+\sin (60^{\circ}-2y)\right]=$ $\sqrt 3-2\sin\left[60^{\circ}-(x+y)\right]\cos(x-y)\ \stackrel{x+y=60^{\circ}}{=}\ \underline{\underline{\sqrt 3}}\ .$

P9 (Old Greek Mathematical Journal). Prove that $(\forall )\ \triangle\ ABC$ there is the following identity $:\ \left(1+\tan\frac A4\right)\cdot\left(1+\tan\frac B4\right)\cdot \left(1+\tan\frac C4\right)=2\cdot\left(1+\tan\frac A4\tan\frac B4\tan\frac C4\right)\ .$

Proof. Let $f(t)=t^3-s_1t^2+s_2t-s_3=0$ $\begin{array}{ccccc} \nearrow & x & = & \tan \frac A4 & \searrow\\\\ \rightarrow & y & = & \tan \frac B4 & \rightarrow\\\\ \searrow & z & = & \tan \frac C4 & \nearrow\end{array}\odot$ where $\odot\begin{array}{ccccc} \nearrow & s_1 & = & x+y+z & \searrow\\\\ \rightarrow & s_2 & = & xy+yz+zx & \rightarrow\\\\ \searrow & s_3 & = & xyz & \nearrow\end{array}\odot$ Thus, $\sum\frac {\pi+A}4=\frac {\sum\pi +\sum A}4=\frac {3\pi +\pi}4=\pi\implies$

$\boxed{\sum\left(\frac {\pi}4+\frac A4\right)=\pi}\ (*)\ .$ Hence in this case $\sum\tan\left(\frac {\pi}4+\frac A4\right)=\prod\tan\left(\frac {\pi}4+\frac A4\right)\iff$ $\sum\frac {1+x}{1-x}=\prod\frac {1+x}{1-x}\iff$ $\sum [(1+x)(1-y)(1-z)]=\prod (1+x)\iff$

$\sum[1-y-\cancel z+\cancel {yz}+x(\cancel 1-\cancel y-z+yz)]=\prod (1+x)\iff$ $\sum[1-y+x(-z+yz)]=\prod (1+x)\iff$ $3-s_1-s_2+3s_3=1+s_1+s_2+s_3\iff$ $\boxed{\ s_1+s_2=1+s_3\ }\ (1)\ .$

Otherwise. $\frac A4+\frac B4=\frac {\pi}4-\frac C4\implies$ $\tan\left(\frac A4+\frac B4\right)=\tan\left(\frac {\pi}4-\frac C4\right)\implies$ $\frac {x+y}{1-xy}=\frac {1-z}{1+z}\iff$ $(x+y)(1+z)=(1-xy)(1-z)\iff$ $\sum x+\sum yz=1+xyz$ $\iff$

$s_1+s_2=1+s_3\ .$ Hence $\prod\left(1+\tan\frac A4\right)=$ $2\left(1+\prod\tan\frac A4\right)\iff$ $\boxed{\prod (1+x)=2(1+xyz)}$ $\iff$ $1+s_1+s_2+s_3=2+2s_3\iff$ $s_1+s_2=1+s_3\ ,$ what is the true relation $(1)\ .$

P10 (Fleetwood). Sa se determine toate numerele naturale in baza $10$ de forma $N=\overline{xyz}$ stiind ca $\overline{xy}+\overline{yz}+\overline{zx}=66\ .$ Frumoasa problema pentru clasa a IV - a !

Proof. $\overline{xy}+\overline{yz}+\overline{zx}=66\iff$ $(10\underline{x}+\underline{\underline {y}})+(10\underline{\underline{y}}+\underline{\underline{\underline{z}}})+(10 \underline{\underline{\underline{z}}}+\underline x)=66\iff$ $11(x+y+z)=66\iff \boxed{x+y+z=6}\ .$ Cifra $x\ge 1\ ,$ ca prima cifra de la stanga a lui $N\ .$ Apar cazurile $:$

$\blacktriangleright\ x=1\ \implies\ y+z=5\ \implies\ \left\{\begin{array}{cccccccc} x & : & 1 & 1 & 1 & 1 & 1 & 1\\\ y & : & 0 & 1 & 2 & 3 & 4 & 5\\\ z & : & 5 & 4 & 3 & 2 & 1 & 0 \end{array}\right\|\ (6-2=4)\ ;$
$\blacktriangleright\ x=2\ \implies\ y+z=4\ \implies\ \left\{\begin{array}{cccccccc} x & : & 2 & 2 & 2 & 2 & 2\\\ y & : & 0 & 1 & 2 & 3 & 4\\\ z & : & 4 & 3 & 2 & 1 & 0\end{array}\right\|\ (5-2=3)\ ;$
$\blacktriangleright\ x=3\ \implies\ y+z=3\ \implies\ \left\{\begin{array}{cccccccc} x & : & 3 & 3 & 3 & 3\\\ y & : & 0 & 1 & 2 & 3\\\ z & : & 3 & 2 & 1 & 0\end{array}\right\|\ (4-2=2)\ ;$
$\blacktriangleright\ x=4\ \implies\ y+z=2\ \implies\ \left\{\begin{array}{cccccccc} x & : & 4 & 4 & 4\\\ y & : & 0 & 1 & 2\\\ z & : & 2 & 1 & 0\end{array}\right\|\ (3-2=1)\ ;$
$\blacktriangleright\ x=5\ \implies\ y+z=1\ \implies\ \left\{\begin{array}{cccccccc} x & : & 5 & 5\\\ y & : & 0 & 1\\\ z & : & 1 & 0\end{array}\right\|\ (2-2=0)\ ;$
$\blacktriangleright\ x=6\ \implies\ y+z=0\ \implies\ \left\{\begin{array}{cccccccc} x & : & 6\\\ y & : & 0\\\ z & : & 0\end{array}\right\|\ (1-1=0)\ .$
Numerele naturale $xy\ ,$ $yz$ si $zx$ fiind de doua cifre trebuie ca $0\not\in \{x,y,z\}\ .$ Asadar, solutia $S$ are cel mult $\sum_{k=1}^6k=\frac {6\cdot 7}2=21$ de numere naturale $($ scrise de sus

in jos $\downarrow )\ ,$ mai putin cele $5\cdot 2+1=11$ numere care au cel putin o cifra egala cu zero. In concluzie, $S=\{411,321,312,231,222,213,141,132,123,114\}\ .$

P11 (Old Greek Mathematical Journal). Prove that $(\forall )\ w\not\in\left\{\frac{k\pi}6\ ,\ k\in \mathbb Z\right\}\ ,$ $\frac {\cot 3w}{\cot w}\not\in \left(\frac 13,3\right)\ .$

Proof. $\tan 2w=\frac {2\tan w}{1-\tan^2w}\implies$ $\tan 3w=\tan (w+2w)=\frac {\tan w+\tan 2w}{1-\tan w\tan 2w}=\frac {\tan w+\frac {2\tan w}{1-\tan^2w}}{1-\tan w\cdot\frac {2\tan w}{1-\tan^2w}}=$ $\frac {\tan w\left(3-\tan^2w\right)}{1-3\tan^2w}\implies$ $\boxed{\frac {\cot 3w}{\cot w}=\frac {\tan w}{\tan 3w}=\frac {1-3\tan^2w}{3-\tan^2w}}\ (*)\ .$

Let $\boxed{\ \tan w=t\in \mathbb R\ }\ (1)\ .$ Suppose against all reason $\frac {\cot 3w}{\cot w}\in \left(\frac 13,3\right)\ ,$ i.e. $\frac {1-3t^2}{3-t^2}\in \left(\frac 13,3\right)\iff$ $\left(\frac {1-3t^2}{3-t^2}-\frac 13\right)\left(\frac {1-3t^2}{3-t^2}-3\right)<0\iff$ $\left(-8t^2\right)\cdot (-8)<0\ ,$ what is false.

P12. Let $ax^2+bx+c=0\begin{array}{cc} \nearrow & x_1\\\\ \searrow & x_2\end{array}$ be the quadratic equation for what $x_1=2x_2 \ \vee\ x_2=2x_1\ .$ Find the value of $\frac {b^2}{ac}\ .$

Proof. Are well-known the notations $:\ \left\{\begin{array}{ccccc} S & = & x_1+x^2 & = & -\frac ba\\\\ P & = & x_1x_2 & = & \frac ca\end{array}\right\|\ .$ Therefore, $x_1=2x_2\ \stackrel{\mathrm {or}}{\vee}\ x_2=2x_1\ \iff\ \left(x_1-2x_2\right)\left(x_2-2x_1\right)=0\iff$

$2\left(x_1^2+x_2^2\right)=5x_1x_2\iff$ $2\left(S^2-2P\right)=5P\iff$ $2S^2=9P\iff$ $2\left(-\frac ba\right)^2=9\cdot\frac ca\iff$ $\frac {2b^2}{a\cancel{^2}}=\frac {9c}{\cancel a}\iff$ $2b^2=9ac\iff$ $\boxed{\frac {b^2}{ac}=\frac 92}\ .$

P13 (Israel DIAZ). Let $f:D\rightarrow\mathbb R\ ,$ where $D=(4,5)$ and $f(x)=\frac {5x-11}{x-3}\ .$ Find the image $f(D)\ ,$ i.e. $\mathrm{Im}(f)\ .$

Proof. Denote the image $y=\frac {5x-11}{x-3}$ of $x\in (4,5)\ .$ Observe that $xy-3y=5x-11\iff$ $\boxed{x=\frac {3y-11}{y-5}}\ ,\ \underline{y\ne 5}\ .$ Therefore, $x\in (4,5)\iff$ $\frac {3y-11}{y-5}\in (4,5)\iff$

$\left(\frac {3y-11}{y-5}-4\right)\left(\frac {3y-11}{y-5}-5\right)<0\iff$ $\frac {(-y+9)(-2y+14)}{(y-5)^2}<0\iff$ $(y-9)(y-7)<0\iff$ $\underline{y\in (7,9)}\ .$ In conclusion, $\boxed{\ \mathrm{Im}(f)=(7,9)\ }\ .$

Remark. $\lim_{x\searrow 3}f(x)=\infty\ ,\ \lim_{x\nearrow 3}f(x)=-\infty$ and $\lim_{x\rightarrow\pm \infty}f(x)=5\ ,\ \lim_{x\searrow 4}f(x)=9$ and $\lim_{x\nearrow 5}f(x)=7\ .$

P14 (Leo GIUGIUC). Let $A$ -right $\triangle ABC$ with $H\in (BC)$ so that $AH\perp BC\ ,\ F\in (AB)$ so that $CF$ is the $C$ -bisector of $\widehat{ACB}\ ,\ E\in AH\cap CF\ ,\ D\in BE\cap FH\ .$ Prove $[AEDF]=[BDH]$

Proof. Denote $S=[ABC]$ and apply the Menelaus' theorem to the transversal $\overline {AEH}/\triangle BCF\ :\ \frac {AF}{AB}\cdot\frac {HB}{HC}\cdot\frac {EC}{EF}=1\iff$ $\frac {\cancel b}{a+b}\cdot\frac {c^2}{b\cancel{^2}}\cdot\frac {EC}{EF}=1\iff$ $\frac {EC}{b(a+b)}=\frac {EF}{c^2}=\frac {CF}{a(a+b)}\implies$

$\frac {[AEB]}{[ACB]}=\frac {FE}{FC}=\frac {c^2}{a(a+b)}=\frac {a-b}a\implies$ $\boxed{\ [AEB]=\frac {a-b}a\cdot S\ }\ (1)\ .$ Observe that $\frac {[BFH]}{[ABC]}=\frac {BF}{BA}\cdot \frac {BH}{BC}=\frac {\cancel a}{a+b}\cdot \frac {c^2}{a\cancel{^2}}=\frac {c^2}{a(a+b)}=\frac {a-b}a\implies$ $\boxed{\ [BFH]=\frac {a-b}a\cdot S\ }\ (2)\ .$

In conclusion, from the relations $(1\wedge 2)$ obtain that $[AEB]=[BFH]\iff$ $[AEB]-[BDF]=[BFH]-[BDF]\iff$ $[AEDF]=[BDH]\ .$

P15 (Khoa Linh, Vietnam). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)\ ,$ the circumcenter $O$ and the midpoint $M$ of the side $[BC]\ .$ Prove that $OI\perp AM\iff \boxed{\ \frac 1b+\frac 1c=\frac 2a\ }\ .$

Proof. $IA^2=r^2+(s-a)^2\ ,$ $IM^2=r^2+\frac{(b-c)^2}4$ and $OM^2=R^2-\frac {a^2}4\ .$ Hence $IO\perp AM\iff$ $IA^2-IM^2=$ $OA^2-OM^2\iff$ $\left[r^2+(s-a)^2\right]-\left[r^2+\frac {(b-c)^2}4\right]=$

$R^2-\left(R^2-\frac {a^2}4\right)\iff$ $(b+c-a)^2=(b-c)^2+a^2\iff$ $(b+c)^2-2a(b+c)=(b-c)^2\iff$ $4bc=2a(b+c)\iff$ $2bc=a(b+c)\iff$ $\frac 1b+\frac 1c=\frac 2a\ .$ Very nice problem !

P16 (Kadir, ALTINTAS, Turkey). Let $\triangle BC$ with the incircle $I\ ,$ the Nagel's point $N$ and $E\in BI\cap AC\ ,$ $F\in CI\cap AB\ .$

Prove that $\boxed{\ N\in EF\implies\frac {s-b}b+\frac {s-c}c=\frac {s-a}a\iff \frac 1b+\frac 1c=\frac 1a+\frac 1s\iff s^2+r^2=4Rh_a\ }$ (standard notations).

Proof. Denote $D\in AN\cap BC$ and apply the Van Aubel's relation for the point $N\ :\ \frac {NA}{ND}=\frac {FA}{FB}+\frac {EA}{EC}=$ $\frac {s-b}{s-a}+\frac {s-c}{s-a}=\frac {(s-b)+(s-c)}{s-a}=\frac a{s-a}\implies$ $\boxed{\ \frac {NA}{ND}=\frac a{s-a}\ }\ (1)\ .$ Apply the

Cristea's relation $:\ \frac {FB}{FA}\cdot DC+\frac {FC}{FA}\cdot DB=$ $\frac {ND}{NA}\cdot BC\ ,$ i.e. $\frac {\cancel a}b\cdot (s-b)+\frac {\cancel a}c\cdot (s-c)=$ $\frac {s-a}a\cdot \cancel a\iff\boxed{\ \frac {s-b}b+\frac {s-c}c=\frac {s-a}a\ }\ .$ Prove easily that $\frac {s-b}b+\frac {s-c}c=\frac {s-a}a\iff$

$\frac 1b+\frac 1c=$ $\frac 1a+\frac 1s\iff s^2+r^2=4Rh_a\ .$ Indeed $:\ \frac {s-b}b+\frac {s-c}c=$ $\frac {s-a}a\iff$ $\frac sb-1+\frac sc-1=$ $\frac sa-1\iff$ $\frac sb+\frac sc=\frac sa+1\iff$ $\boxed{\frac 1b+\frac 1c=\frac 1a+\frac 1s\ }\ ;$ $\frac {s-b}b+\frac {s-c}c=$ $\frac {s-a}a\iff$

$ac(s-b)+ab(s-c)=bc(s-a)\iff$ $\cancel s(ac+ab-bc)=4R\cancel sr\iff$ $(ab+bc+ca)=2bc+4Rr\iff$ $s^2+r^2+\cancel{4Rr}=4Rh_a+\cancel{4Rr}\iff$ $\boxed{\ s^2+r^2=4Rh_a\ }\ .$

Generalization 1. Let $ABC$ be a triangle with the incircle $I\ ,$ where denote $E\in BI\cap AC\ ,\ F\in CI\cap AB$ and a point $P\left(\alpha ,\beta ,\gamma\right)\in (EF)\ ,$

where $\left(\alpha ,\beta ,\gamma\right)$ are the barycentrical coordinates of $P$ w.r.t. $\triangle ABC\ .$ Prove that there is the following identity $:\ \boxed{\ \frac {\beta}b+\frac {\gamma}c=\frac {\alpha}a\ }\ .$

Particular cases $:\ G(1,1,1)\implies \frac 1b+\frac 1c=\frac 1a\ ;\ N(s-a,s-b,s-c)\implies \frac {s-b}b+\frac {s-c}c=\frac {s-a}a\ ;\ \Gamma\left(r_a,r_b,r_c\right)\implies$ $\frac {r_b}b+\frac {r_c}c=\frac {r_a}a\ ;$

$H\left (\tan A,\tan B,\tan C\right)\implies$ $\frac 1{\cos B}+\frac 1{\cos C}=\frac 1{\cos A}\ ;\ O(\sin 2A,\sin 2B,\sin 2C)\implies \cos B+\cos C=\cos A\ ;\ L\left(a^2,b^2,c^2\right)\implies L\not\in EF\ .$

Proof. Prove easily that $E(a,0,c)\ \wedge F(a,b,0)\ .$ Hence $P\left(\alpha ,\beta ,\gamma\right)\in (EF)\iff$ $\left|\begin{array}{ccc} \alpha & \beta & \gamma\\\\ a & 0 & c\\\\ a & b & 0\end{array}\right|=0\iff$ $ab\gamma+ac\beta=bc\alpha\iff \frac {\beta}b+\frac {\gamma}c=\frac {\alpha}a\ .$

Generalization 2. Let $ABC$ be a triangle and two interior points $M_1\left(x_1,y_1,z_1\right)\ ,$ $M_2\left(x_2,y_2,z_2\right)$ with the barycentrical coordinates w.r.t.

$\triangle ABC$ such that $M_2\in EF\ ,$ where $E\in BM_1\cap AC\ ,$ $F\in CM_1\cap AB\ .$ Prove that there is the following identity $:\ \boxed{\ \frac {y_2}{y_1}+\frac {z_2}{z_1} =\frac {x_2}{x_1}\ }\ .$

Example: $M_1\equiv H\left(\cos A,\cos B,\cos C\right)\ \wedge\ M_2\equiv G(1,1,1)\implies$ $\tan A\tan C+\tan A\tan B=\tan B\tan C\implies$ $\cot B+\cot C=\cot A\implies$

$\frac {\cos b}b+\frac {\cos C}c=\frac {\cos A}a\iff$ $a(c\cdot \cos B+b\cdot \cos C)=bc\cdot\cos A\iff$ $a^2=bc\cdot \cos A\iff$ $2a^2=b^2+c^2-a^2\implies$ $\boxed{\ b^2+c^2=3a^2\ }\ .$

P17 (Ercole SUPPA). Let $ABC$ be a triangle with the midpoint $M$ of $[BC]\ ,$ the incircle $w=\mathbb C(I,r)\ ,$ the circumcenter $O$ and $D\in BC\cap w\ .$ Prove that $\boxed{\ IO\perp AD\implies\frac 1b+\frac 1c\ge\frac 2a\ }\ .$

Proof. Prove easily that $OD^2=OM^2+MD^2=$ $\left(OC^2-MC^2\right)+MD^2=$ $R^2-\left(MC^2-MD^2\right)=$ $R^2-\left[\frac {a^2}4-\frac {(b-c)^2}4\right]=$ $R^2-(s-b)(s-c)\implies$ $\boxed{\ OD^2=R^2-(s-b)(s-c)\ }\ .$

Therefore, $IO\perp AD\iff$ $IA^2-ID^2=OA^2-OD^2\iff$ $(bc-4Rr)-r^2=\cancel{R^2}-\left[\cancel{R^2}-(s-b)(s-c)\right]\iff$ $bc=(s-b)(s-c)+r^2+4Rr\iff$ $s(s-a)+\cancel{(s-b)(s-c)}=$

$\cancel{(s-b)(s-c)}+r^2+4Rr\iff$ $\boxed{\ s(s-a)=r^2+4Rr\ }\iff$ $s^2+s(s-a)=s^2+r^2+4Rr\iff$ $s(b+c)=ab+bc+ca\iff$ $(a+b+c)(b+c)=2a(b+c)+2bc\iff$

$(b+c)^2=a(b+c)+2bc\iff$ $\boxed{\ b^2+c^2=a(b+c)\ }\ .$ In conclusion, $a(b+c)=b^2+c^2\ge 2bc\iff$ $a(b+c)\ge 2bc\iff$ $\boxed{ \frac 1b+\frac 1c\ge \frac 2a\ }\ .$

P18 (Kadir Altintas, Turkey). Let $\triangle ABC$ with $b\ne c$ and its interior point $P$ with the barycentrical coordinates $(\alpha ,\beta ,\gamma )$ w.r.t. $\triangle ABC\ ,$ where $D\in AP\cap BC\ .$

Denote the midpoints $E,F$ of the sides $[AC],[AB]$ respectively. Prove that $:\ \boxed{\ DE=DF\iff 2a^2(\beta -\gamma )=(\beta +\gamma )\left(b^2-c^2\right)\ }$ (standard notations).

Proof.

P19 (Mehmet Sahin, Turkey). $\boxed{\ \mathrm{Prove\ that}\ \forall\ \triangle ABC\ \mathrm{there\ is\ the\ identity}\ :\ ar_a+br_b+cr_c=2\left(\left[I_aI_bI_c\right]-\left[ABC\right]\right)\ \mathrm{(standard\ notations)}\ .}$

Proof. $[ABC]=sr=(s-a)r_a\implies$ $ar_a=s\left(r_a-r\right)\implies$ $\sum ar_a=s\sum\left(r_a-r\right)=s(4R+r)-3sr=s(4R-2r)=2s(2R-r)\implies$

$\boxed{\sum ar_a=2s(2R-r)}\ (*)\ .$ Hence $\left[I_aI_bI_c\right]=\sum\left([BIC]+\left[BI_aC\right]\right)=[ABC]+\frac 12\cdot \sum ar_a\implies$ $2\left(\left[I_aI_bI_c\right]-[ABC]\right)\ \stackrel{(*)}{=}\ \sum ar_a\ .$

P20 (Mehmet Sahin, Turkey). $\boxed{\ \mathrm{Prove\ that}\ \forall\ \triangle ABC\ \mathrm{there\ is\ the\ identity}\ :\ \frac a{r_b+r_c}+\frac b{r_c+r_a}+\frac c{r_a+r_b}=\frac {2\left(r_a+r_b+r_c\right)}{a+b+c}\ \mathrm{(standard\ notations)}\ .}$

Proof. $\frac a{r_b+r_c}=\frac {a(s-b)(s-c)}{r_b(s-b)\cdot (s-c)+r_c(s-c)\cdot(s-b)}=$ $\frac {a(s-b)(s-c)}{S[(s-c)+(s-b)]}=$ $\frac {\cancel a(s-b)(s-c)}{\cancel aS}=$ $\frac S{s(s-a)}=$ $\frac {r_a}s$ $\implies$ $\sum\frac a{r_b+r_c}=\sum\frac {r_a}s\implies$ $\sum\frac a{r_b+r_c}=\frac {2\left(r_a+r_b+r_c\right)}{a+b+c}\ .$

P21 (USAMO 2001). Let $\triangle ABC$ and let $w=\mathbb C(I,r)$ be its incircle. Denote by $D_{1}$ and $E_{1}$ where $w$ is tangent to $BC$ , $AC$ respectively. Denote by $D_{2}$ , $E_{2}$ the points on $BC$ and $AC$ respectively,

such that $CD_{2}=BD_{1}$ , $CE_{2}=AE_{1}$ and denote by $P\in AD_{2}\cap BE_{2}\ .$ Circle $w$ intersects $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$ . Prove that $AQ=PD_{2}\ .$

Proof. Is well-known property $Q\in ID_1\ \ (*)$ . Therefore, $\frac {AQ}{AD_2}=\frac {h_a-2r}{h_a}=\frac {ah_a-2ar}{ah_a}=\frac {2pr-2ar}{2pr}=\frac {p-a}{a}\ \ (1)$ . Apply Menelaus' theorem to $\overline {BPE_2}/AD_2C\ \ :$

$\frac {BD_2}{BC}\cdot\frac {E_2C}{E_2A}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {p-b}{a}\cdot\frac {p-a}{p-c}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {PA}{p}=\frac {PD_2}{p-a}=\frac {AD_2}{a}$ $\implies$ $\frac {PD_2}{AD_2}=\frac {p-a}{a}\ \ (2)$ . From the relations $(1)$ , $(2)$ obtain $AQ=PD_2$ .

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

$(*)\ \blacktriangleright$ Let $\{D_1,Q_1\}=w\cap D_1I$ .Prove easily that $\frac {AD}{Q_1D_1}=\frac {h_a}{2r}=\frac {ah_a}{2ar}=\frac {2pr}{2ar}=\frac pa$ and $\frac {DD_2}{D_1D_2}=\frac {p|b-c|}{a}\cdot \frac {1}{|b-c|}=\frac pa$ . Hence $\frac {AD}{Q_1D_1}=\frac {DD_2}{D_1D_2}$ , i.e. $Q_1\in w\cap AD_2$ $\implies$ $Q_1\equiv Q$ .

This post has been edited 509 times. Last edited by Virgil Nicula, Jun 24, 2018, 9:27 AM

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October 2011

327. Some geometrical/algebraic extremum problems.

326. Some problems from trigonometry.

325. Without the l'Hospital's rules.

324. Some determinants.

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

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

321. Harmonical quadrilateral.

320. Some problems with metrical relations.

September 2011

319. Some problems with induction.

318. IRAN - 2011 (geometrie).

317. Some interesting geometry problems for middle school.

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

August 2011

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

314. USAMTS 2009 Round 2 #5.

313. For middle school - some nice problems.

312. Very nice (famous trigonometrical problems) !

311. Some problems from USAMO and other contests.

310. Some properties of the tangential quadrilateral.

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

308. Some nice and easy properties in a triangle.

July 2011

307. Very nice proofs !

306. Some easy and nice "slicing" problems.

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

304. Integrals ("brothers") III.

303. Integrals II.

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

301. Some trigonometrico-geometrical inequalities.

300. Lagrange's multiplier. Examples and problems.

299. A class of the recurrent sequencies.

298. Some nice problems from Crux Mathematicorum.

297. Nice inequalities with concrete numbers.

296. Some nice Vittasko's problems.

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

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

June 2011

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

292. Length of the angled bisector and some means.

291. Some integrals.

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

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

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

287. Some metrical problems with or without trigonometry.

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

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

284. Ellipse (definition).

283. Some nice and easy problems.

282. Factorize of polynomials.

May 2011

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

280. Common roots of two polynomial equations.

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

278. In memoria profesorului Alexandru Lupas.

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

276. Chinese TST 2009 and Second Round 2006.

275. Some interesting algebraic equations and systems.

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

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

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

271. Relations concerning Fermat points.

April 2011

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

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

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

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

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

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

260. Some nice, simple identities and their applications.

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

258. Some difficult and nice problems with complex numbers.

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

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

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

253. An old algebraic inequality.

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

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

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

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

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

February 2011

232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

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

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

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

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

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

January 2011

219. A very nice geometrical inequality.

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

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

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

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

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

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

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

December 2010

202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

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

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

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

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

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

181. Two special inequalities from CRUX (own).

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

November 2010

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

178. Some nice and simple geometry problems.

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

176. Another nice problems for middle school.

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

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

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

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

October 2010

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

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

168. Some properties of the Lemoine's point.

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

166. Some constant geometrical expressions.

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

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

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

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

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

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

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

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

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

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

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

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

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

138. Generalized "slicing" problems (own).

137. Stewart's relation.

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

135. Three strong geometrical inequalities.

September 2010

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

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

132. Asupra unei inegalitati intr-un triunghi.

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

130. A triangle and a fixed point.

129. BX=CY and three properties.

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

127. Pagina educativa.

126. Routh's theorem.

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

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

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

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

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

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

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

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

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

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

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

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

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

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

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

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

93. The Appolonius' construct problems.

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

91. JBMO 2010, Problem 3.

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

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

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

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

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

81. Sawayama and Thebault’s theorem.

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

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

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

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

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

July 2010

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

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

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

69*1. Slicing problem 3.

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

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

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

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

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

62. Nice and easy problem with a right triangle.

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

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

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

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

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

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

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

June 2010

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

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

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

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

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

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

38. Minimum area of triangle touching with semicircle.

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

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

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

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

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

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

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

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

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

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

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

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

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

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

10. Walker's inequality and its extension.

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

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

Submit

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391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
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the visits tho
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long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

458. Working page IV.

by Virgil Nicula, Aug 27, 2017, 10:47 AM

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