453. Working page III.

by Virgil Nicula, Mar 4, 2017, 5:51 PM

P0. Sa se determine pozitia punctului mobil $M$ situat pe axa ordonatelor $Oy$ pentru care suma $MA +MB$ este minima, unde $A(6,7)$ si $B(4,2)$ .

Metoda 1. Se obseva ca axa ordonatelor nu separa punctele fixe $A$ si $B$ . Alegem un punct din perechea $\{A,B\},$ de exemplu $A,$ construim simetricul $C$ al acestuia fata de axa

ordonatelor si $D\in BC\cap Oy.$ Se arata usor ca $(\forall ) M\in OY$ exista lantul de relatii $DA+DB=DC+DB=BC\le MC+MB=MA+MB,$ adica $(\forall ) M\in OY,$

$MA+MB\ge DA+DB,$ ceea ce inseamna ca suma $MA+MB$ este minima $\iff M:=D,$ unde (analitic) $D(0,4)\ ,\ MA+MB\ge DA+DB=5\sqrt 5.$

Metoda 2. Notam proiectiile $C=\mathrm{pr}_{Oy}B$ si $D=\mathrm{pr}_{Oy}A,$ unde $CD=5.$ Se arata usor ca punctul de minim cautat $E\in (CD).$ Asadar putem presupune ca $M\in (CD).$

Denote $\left\{\begin{array}{ccc} u & = & MD\\\ v & = & MC\end{array}\right\|$ unde $u+v=5.$ Aplicam Inegalitatea Minkowski (<= click): $\boxed{MA+MB=\sqrt{u^2+6^2}+\sqrt{v^2+4^2}\ge \sqrt{(u+v)^2+(6+4)^2}=5\sqrt 5}\ ,$

adica $MA+MB\ge 5\sqrt 5.$ Avem egalitate $\iff \frac u6=\frac v4=\frac {u+v}{6+4}=\frac 5{10}=\frac 12\implies$ $u=3$ si $v=2$ $\implies$ $OE=OC+CE=2+2=4,$ adica $OE=4.$

Metoda 3. Notam $M(0,m)\ ,\ m\in\mathbb R$ si $f(m)=MA+MB=\sqrt{(m-7)^2+6^2}+\sqrt {(m-2)^2+4^2}\implies$ $\boxed{f(m)=\sqrt{m^2-14m+85}+\sqrt {m^2-4m+20}}\ (*).$

Se arata usor ca $f'(m)=\frac {m-7}{\sqrt {m^2-14m+85}}+\frac {m-2}{\sqrt {m^2-4m+20}}$ si $f'(m)=0\iff$ $(m-7)\cdot\sqrt {m^2-4m+20}+(m-2)\cdot \sqrt {m^2-14m+85}=0\iff$

$(m-7)(m-2)\le 0$ si $(m-7)^2\cdot \left(m^2-4m+20\right)=(m-2)^2\cdot\left(m^2-14m+85\right)\iff$ $m\in [2,7]$ si $20m^2+80m-640=0\iff$ $m\in (2,7)$ si

$m^2+4m-32=0\begin{array}{ccc} \nearrow & m=-8 & \searrow\\\\ \searrow & m=4 & \nearrow\end{array}\odot\iff$ $m\in (2,7)\cap \{-8,4\}\iff \boxed{\ m\ =\ 4\ }$ si $(\forall )\ m\in \mathbb R\ ,\ f(m)\ge f(4)=5\sqrt 5\implies$ $\boxed{\ (\forall )\ m\in \mathbb R\ ,\ f(m)\ge 5\sqrt 5\ }\ .$

Extindere. Sa se determine pozitiile $\{M,N\}\subset Oy$ astfel incat $MN=k>0$ (constant) si suma $MA +NB$ este minima, unde $A,B$ sunt fixe si $[AB]\cap Oy=\emptyset.$ .

Ptolemy's inequality. Let $\triangle ABC$ and a point $M$ which belongs to the plane of $ABC$ . Prove that there is the relation $\boxed{MA\cdot BC\le MB\cdot CA+MC\cdot AB}\ (*).$

Proof 1. I"ll use the complex numbers. Thus, denote $X(x)$ - the point $X$ with the affix $x\in\mathbb C$ . Hence $A(a)$ , $B(b)$ , $C(c)$ and $M(m)$ . Denote $\left\{\begin{array}{ccc} z_1=(m-a)(b-c)\\\ z_2=(m-b)(c-a)\\\ z_3=(m-c)(a-b)\end{array}\right\|$

Observe that $z_1+z_2+z_3=0$ $\iff$ $z_1=-\left(z_2+z_3\right)$ $\implies \left|z_1\right|=\left|z_2+z_3\right|\le \left|z_2\right|+\left|z_3\right|$ $\implies \left|z_1\right|\le \left|z_2\right|+\left|z_3\right|$ $\iff$ $MA\cdot BC\le MB\cdot CA+MC\cdot AB$ .

P1. Let $I$ be the incenter of the non-isosceles $\triangle ABC$ and let $A',B',C'$ be the tangency points of the incircle with the sides $BC$ ,

$CA$ and $AB$ respectively. Let $P\in AA'\cap BB'$ , $M\in AC\cap A'C'$ and $N\in B'C'\cap BC$ . Prove that the lines $IP\perp MN$ .

Alternative formulation. The incircle of a non-isosceles triangle $ABC$ has center $I$ and touches the sides $BC$ , $CA$ and $AB$ in $A^{\prime}$ , $B^{\prime}$ and $C^{\prime}$ , respectively. The lines

$AA^{\prime}$ and $BB^{\prime}$ intersect in $P$ , the lines $AC$ and $A^{\prime}C^{\prime}$ intersect in $M$ , and the lines $BC$ and $B^{\prime}C^{\prime}$ intersect in $N$ . Prove that the lines $IP$ and $MN$ are perpendicular.[/size]

Proof. I note the points $X\in AA'\cap IN,\ Y\in BB'\cap IM$ . From the wellknown relations $IN\perp AA'$ , $IM\perp BB'$ results $IX\cdot IN=IY\cdot IM=r^2$ ,

i.e. the line $MN$ is the image of the circumcircle with the diameter $[IP]$ through the inversion with the pole I and the constant $r^2$ . Therefore, $IP\perp MN.$

P2 (Alexandru Lupas). Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ and an its interior point $P$ for

which denote $\left\{\begin{array}{ccc} D\in AP\cap BC & ; & X\in AP\cap w\\\ E\in BP\cap CA & ; & Y\in BP\cap w\\\ F\in CP\cap AB & ; & Z\in CP\cap w\end{array}\right\|\ .$ Prove that $\boxed{\frac {AD}{DX}+\frac {BE}{EY}+\frac {CF}{FZ}\ge 9}\ .$

Proof. $K\equiv \sum\frac {AD}{DX}=\sum\frac {AD^2}{AD\cdot DX}=\sum\frac {AD^2}{DB\cdot DC}.$ Stewart's relation to $AD/\triangle ABC: a\cdot AD^2+a\cdot DB\cdot DC=c^2\cdot DC+b^2\cdot DB\stackrel{:(a\cdot DB\cdot DC)}{\iff}$

$1+\frac {AD^2}{DB\cdot DC}=$ $\frac {c^2}{a\cdot DB}+\frac {b^2}{a\cdot DC}\ \stackrel{\sum}{\implies}$ $3+K=$ $\sum\frac 1a\cdot \left(\frac {c^2}{DB}+\frac {b^2}{DC}\right)\stackrel{C.B.S.}{\ge}\sum\frac 1a\cdot \frac {(b+c)^2}{DB+DC}=$ $\sum\left(\frac {b+c}a\right)^2\stackrel{AM-GM}{\ge}4\cdot \sum \frac {bc}{a^2}\stackrel{AM-GM}{\ge}$

$4\cdot 3\cdot \sqrt[3]{\prod\frac {bc}{a^2}}=12\implies$ $3+K\ge 12\implies K\ge 9.$ Observe that $K=9\iff P\equiv O$ and $a=b=c.$

P3. Find the minimum of the expression $E\equiv E(x,y)$ (without utilization of derivatives!), where $E\equiv\frac {1-xy}{(1-x)(1-y)},\ \{x,y\}\subset (0,1)$ and $x+y+xy=1\ .$ Success!

Metoda 1. Using $\boxed{y=\frac {1-x}{1+x}}$ the expression $E$ becomes $E=\frac {1-x\cdot\frac {1-x}{1+x} }{(1-x)\left(1-\frac {1-x}{1+x}\right)}=\frac {1+x^2}{2x(1-x)} ,$ i.e. $E=\frac {1+x^2}{2x(1-x)} , x\in (0,1) .$ I"ll use the substitution $x=\tan\phi ,$ where $\phi \in \left(0,\frac {\pi}4\right) .$

Thus, $E=\frac {1+\tan^2\phi}{2\tan\phi(1-\tan\phi)}=$ $\frac 1{\sin 2\phi -2\sin^2\phi}=$ $\frac 1{\left(\sin 2\phi +\cos 2\phi\right) -1}\ .$ Since $:\ (\forall )\ \phi\in \left(0,\frac {\pi}4\right)\ ,\ 1<\sin 2\phi +\cos 2\phi \le\sqrt 2$ obtain that $E\ge\frac 1{\sqrt 2-1}=1+\sqrt 2 .$ Therefore,

$\boxed{\min_{\begin{array}{c} \{x,y\}\subset (0,1)\\\ x+y+xy=1\end{array}}\frac {1-xy}{(1-x)(1-y)}=1+\sqrt 2}\ .$ I"ll prove the bilateral inequality $1\le \sin x+\cos x\le \sqrt 2$ for any $x\in \left[0,\frac {\pi}2\right] .$ Indeed, it is equivalent with $1\le (\sin x+\cos x)^2\le 2\iff$

$1\le \sin^2x+\cos^2x+2\sin x\cos x\le 2\iff$ $1\le 1+\sin 2x\le 2\iff$ $0\le \sin 2x\le 1 ,$ what is truly for any $x\in \left[0,\frac {\pi}2\right] ,$ i.e. $2x\in [0,\pi ] .$

Metoda 2. Obtain easily $E=\frac {1+x^2}{2x(1-x)} .$ Therefore, $E$ is $\min\iff$ $2E=\frac {1+x^2}{x(1-x)}$ is $\min\iff$ $\frac 1{2E}=\frac {x(1-x)}{1+x^2}$ is $\max\iff$ $1+\frac 1{2E}=1+ \frac {x(1-x)}{1+x^2}=\frac {1+x}{1+x^2}$ is $\max\iff$ $\frac {1}{1+\frac {1}{2E}}=$ $\frac {1+x^2}{1+x}=x+1+\frac 2{x+1}-2$ is $\min .$ From $(x+1)+\frac 2{x+1}\ge 2\sqrt 2$ obtain that $\frac 1{1+\frac 1{2E}}\ge 2(\sqrt 2-1)\iff$ $1+\frac {1}{2E}\le$ $\frac 1{2\left(\sqrt 2-1\right)}=$ $\frac {1+\sqrt 2}2\iff$ $\frac {1}{2E}\le \frac {\sqrt2-1}2\iff E\ge 1+\sqrt 2\ .$

P4. Prove the proposed problem from here.

Proof. Prove easily that $\frac r{S_x}=\frac {r_a}{S_1}=\frac {r_b}{S_2}=\frac {r_c}{S_3}=\frac {2R}{S}\ ,$ i.e. $\left(r,r_a,r_b,r_c\right)$ are direct proportionally with $\left(S_x,S_1,S_2,S_3\right)\ .$ Hence $\frac 1{S_x}=$ $\frac 1{S_1}+\frac 1{S_2}+\frac 1{S_3}$ $\iff$ $\frac 1r=\frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c}\ ,$ what is true.

P5. Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R),$ the diameter $[A'A]$ and the midpoint $M$ of the side $[BC].$ Prove that there are the following inequalities :

$\boxed{4\cdot MA\cdot MA'\ge a^2}\ (*)$ with equality iff $b=c\ ;$ the Tereshin's inequality $\boxed{\ 4Rm_a\ge b^2+c^2\ }\ (**)$ with equality iff $b=c\ \vee A=90^{\circ}$ (standard notations).

Proof 1. Let diameter $[B'B]$ of $w,$ i.e. $2\cdot OM=B'C=2\sqrt {4R^2-a^2}.$ Thus, $4Rm_a\ge b^2+c^2\iff$ $4\left(m_a-R\right)^2\le 4R^2-a^2\iff$ $2\left|m_a-R\right|\le B'C\iff$ $\left|AM-AO\right|\le OM$ what is true.

Proof 2. Denote $\{A,S\}=AM\cap w.$ Observe that $SA'\perp SM$ $\implies MS\le$ $MA'$ $\implies$ $\frac {a^2}4=$ $MB\cdot MC=MA\cdot MS$ $\le MA\cdot MA'$ $\implies$ $4\cdot MA\cdot MA'\ge a^2\ .$ Apply the theorem

of median to $A'M$ in $\triangle BA'C\ :\ 4\cdot A'M^2=2\left(A'B^2+A'C^2\right)-BC^2\iff$ $4\cdot A'M^2=$ $2\left[\left(4R^2-c^2\right)+\left(4R^2-b^2\right)\right]-a^2$ $\implies$ $\boxed{4\cdot A'M^2=2\cdot \left[8R^2-\left(b^2+c^2\right)-a^2\right]}\ (1)\ .$

Therefore, $4MA^2\cdot 4A'M^2\ge a^4\iff$ $\left[2 \left(b^2+c^2\right)-a^2\right]\cdot\left[2\left(8R^2-b^2-c^2\right)-a^2\right]\ge a^4\iff$ $4\left(b^2+c^2\right)\left[8R^2-\left(b^2+c^2\right)\right]\ge 2a^2\left[\left(\cancel{b^2+c^2}\right)+8R^2-\left(\cancel{b^2+c^2}\right)\right]\iff$

$4R^2\left[2\left(b^2+c^2\right)-a^2\right]\ge \left(b^2+c^2\right)^2\iff$ $16R^2m_a^2\ge \left(b^2+c^2\right)^2\iff$ $4Rm_a\ge b^2+c^2\ .$

Remark (Adil Abdullayev). $h_a\le s_a=\frac {2bc}{b^2+c^2}\cdot m_a=$ $\frac {4Rh_a}{b^2+c^2}\cdot m_a\implies$ $h_a\le\frac {4Rh_a}{b^2+c^2}\cdot m_a\implies$ $b^2+c^2\le 4Rm_a\ .$

Application. Prove that in any acute triangle $\triangle ABC$ there is the inequality $\boxed{\ \sum \frac {m_a\cos A}{h_a}\ge \frac 32\ }\ .$

Proof. Let $\left\{\begin{array}{ccc} x: & = & a^2\\\ y: & = & b^2\\\ z: & = & c^2\end{array}\right\|$ . Therefore, $\sum\frac {m_a\cos A}{h_a}\ \stackrel{(**)}{\ge}\ \sum\frac {\left(b^2+c^2\right)\cos A}{4Rh_a}=$ $\sum\frac {\left(b^2+c^2\right)\cos A}{2bc}=$ $\sum\frac {\left(b^2+c^2\right)\left(b^2+c^2-a^2\right)}{4b^2c^2}=$ $\sum\frac {a^2\left(b^2+c^2\right)\left(b^2+c^2-a^2\right)}{4a^2b^2c^2}=\frac 32$

because $\sum\frac {x(y+z)(y+z-x)}{4xyz}=\frac 32\iff$ $\sum [x(y+z)(y+z-x)]=6xyz\iff$ $\sum x\left(y^2+z^2\right)+\cancel{\sum (2xyz)}-\sum x^2(y+z)=\cancel{6xyz}\iff$ $\sum x\left(y^2+z^2\right)=$

$\sum x^2(y+z)\iff$ $\sum x\left[\left(x^2+y^2+z^2\right)-\cancel{x^2}\right]=\sum x^2[(x+y+z)-x]\iff$ $\sum x^2\cdot\sum x=\sum x\cdot \sum x^2$ what is true. In conclusion, $\sum\frac {m_a\cos A}{h_a}\ge\frac 32\ .$

P6. Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R),$ the diameter $[A'A]$ and the midpoint $M$ of the side $[BC].$ Prove that there are the following inequalities $:$ $\boxed{m_a^2\le R^2+h_a^2}\ (*)$

with equality iff $|B-C|=90^{\circ}\ ;\ \boxed{4S+\left|b^2-c^2\right|\le 2am_a\sqrt 2}\ (**)$ with equality iff $m\left(\widehat{BMA}\right)=\frac {\pi}4\ ;\ \boxed{\frac {a^2+b^2+c^2}{m_a+m_b+m_c}\le 2R}\ (***)$ with equality iff $a=b=c\ .$

Lemma. Let an acute $\triangle ABC$ with the incircle $\omega = C(I,r)$ and the circumcircle $\Omega = C(O,R).$ The circles $c_{1}= \mathbb C\left(P,r_1\right)$ and $c_{2}= \mathbb C\left(Q,r_{2}\right)$ are tangent internally to $\Omega$ in the same point $A;$ the circle $w$ is tangent externally to the circle $c_1$ and is tangent internally to the circle $c_2.$ Prove that $PQ =\frac{a^{2}(p-a)}{4S}$ , where $2p=a+b+c$ and $S=[ABC]$ is the area of $\triangle ABC.$

Remark. Prove easily that $\left \{\begin{array}{ccc} IP = r+r_{1} & ; & IQ = r_{2}-r\\\\ PO = R-r_{1} & ; & PA = r_{1}\\\\ QO = R-r_{2} & ; & QA = r_{2}\\\\ OA = R & ; & PQ = r_{2}-r_{1}\end{array}\right\|$ . The relations $\left \{\begin{array}{ccc} IO^{2}= R(R-2r) & ; & IA^{2}=\frac{bc(p-a)}{p}=\frac{4Rr(p-a)}{a}\\\\ IA^{2}-r^{2}= (p-a)^{2} & ; & IA^{2}+4Rr = bc\\\\ p(p-a)+(p-b)(p-c) = bc.& ; & p(p-a)(p-b)(p-c) = S^2\end{array}\right\|$ are well-known.

Proof I. Denote $\left\{\begin{array}{ccc} IO & = & m\\\ IA & = & n\end{array}\right\|$ . Apply the Stewart's theorem in $\triangle OIA$ for the cevian-rays $[IP$ and $[IQ\ :\ \left\{\begin{array}{ccc} m^{2}r_{1}^{2}+n^{2}(R-r_{1}) & = & R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\ m^{2}r_{2}+n^{2}(R-r_{2}) & = & R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\right\|$ $\implies$

$r_{1}(R^{2}+2Rr+n^{2}-m^{2}) = R(n^{2}-r^{2}) = r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $\implies$ $r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr) =$ $R(p-a)^{2}=$ $r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $\implies$

$r_{1}bc = R(p-a)^{2}= r_{2}(bc-4Rr)$ $\implies$ $PQ = r_{2}-r_{1}=$ $\frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $\frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $\frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $\implies$ $\boxed{\ PQ =\frac{a^{2}(p-a)}{4S}\ }$ .

Proof II. Apply the Pythagoras' theorem in the triangles :
$\triangle\ IOP\blacktriangleright$ $(r+r_{1})^{2}= (R-r_{1})^{2}+(R^{2}-2Rr)-2(R-r_{1})\cdot$ $OI\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $\boxed{r_{1}=\frac{R(p-a)^{2}}{bc}}\ .$

$\triangle\ IOQ\blacktriangleright$ $(r_{2}-r)^{2}= (R-r_{2})^{2}+(R^{2}-2Rr)-2(R-r_{2})\cdot$ $IO\cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot IO}\implies$ $\boxed{r_{2}=\frac{Rp(p-a)}{bc}}\ .$ Therefore,

$\frac{r_{1}}{p-a}=\frac{r_{2}}{p}=\frac{R(p-a)}{bc}=\frac{PQ}{a}$ , i.e. $\boxed{PQ =\frac{aR(p-a)}{bc}}=\frac{a^{2}(p-a)}{4S}$ . Observe that $\frac{r_{1}}{r}=\frac{r_{2}}{r_{a}}$ , where $r_{a}$ is the $A$ - exinradius of $\triangle ABC$ .

Remark. Prove easily that two more interesting relations : $\boxed{\sum PQ = R+r\ }$ and $r_{1}= R-\frac{a(4R+r)}{4p}$ , $\boxed{r_{2}=\frac{ r_{b}+r_{c}}{4}}$ .

P7 (N.T.Tuan). Let an acute $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and the circumcircle $\Omega =\mathbb C(O,R).$ Denote $:\ \omega_a$ - the circle which is tangent internally to $\Omega$

at $A$ and tangent externally to $\omega\ ;$ circle $\Omega_a$ is tangent internally to $\Omega$ at $A$ and tangent internally to $\omega .$ Denote the centers $P_{a}$ and $Q_{a}$ of $\omega_{a}$ and $\Omega_{a}$ respectively.

Define points $P_{b}, Q_{b}, P_{c}, Q_{c}$ analogously. Prove that $8P_{a}Q_{a}\cdot P_{b}Q_{b}\cdot P_{c}Q_{c}\le R^{3}$ with equality if and only if triangle $ABC$ is equilateral (figure).

Proof. $8\cdot\prod P_aQ_a\ \stackrel{lemma}{=}\ 8\cdot\prod\frac{aR(p-a)}{bc}=$ $\frac {8\cancel{abc}R^3(p-a)(p-b)(p-c)}{a\cancel{^2}b\cancel{^2}c\cancel{^2}}=$ $\frac {8R^3sr^2}{4Rsr}=$ $2R^2r\le R^3$ because $R\ge 2r.$

P8. Prove that in any $\triangle ABC$ there is the chain of inequalities $27r^2\le 3r(4R+r)\le s^2\le \frac 13\cdot (4R+r)^2\le \frac {27}4\cdot R^2$ (standard notations).

Proof (with derivatives).

$\blacktriangleright\ \left\{\begin{array}{cccc} a+b+c & = & 2s\\\\ ab+bc+ca & = & s^2+4Rr+r^2\\\\ abc & = & 4RS=4Rsr\end{array}\right\|\implies f(x)\equiv x^3-2sx^2+\left(s^2+r^2+4Rr\right)x-4Rsr=0\ \odot\begin{array}{ccc} \nearrow & a & \searrow\\\\ \rightarrow & b & \rightarrow\\\\ \searrow & c & \nearrow\end{array}\odot$ $\implies f'(x)=3x^2-4sx+\left(s^2+r^2+4Rr\right)\ .$ Since

$\{a,b,c\}\subset\mathbb R\ ,$ from the Rolle's theorem obtain that the equation $f'(x)=0$ has real roots, i.e. $\Delta^{\prime}=4s^2-3\left(s^2+r^2+4Rr\right)\ge 0\ ,$ i.e. $s^2-3r(4R+r)\ge 0\implies \boxed{s^2\ge 3r(4R+r)}\ (1)\ .$

Otherwise (without derivatives). Apply the well-known inequality $(a+b+c)^2\ge 3(ab+bc+ca)\iff 4s^2\ge 3\left(s^2+r^2+4Rr\right)\iff s^2\ge 3r(4R+r)\ .$

$\blacktriangleright\ \left\{\begin{array}{cccc} r_a+r_b+r_c & = & 4R+r\\\\ r_ar_b+r_br_c+r_cr_a & = & s^2\\\\ r_ar_br_c & = & \frac{S^2}r=rs^2\end{array}\right\|\implies g(x)\equiv x^3-(4R+r)x^2+s^2x-rs^2=0\ \odot\begin{array}{ccc} \nearrow & r_a & \searrow\\\\ \rightarrow & r_b & \rightarrow\\\\ \searrow & r_c & \nearrow\end{array}\odot$ $\implies g'(x)=3x^2-2(4R+r)x+s^2\ .$ Since

$\{a,b,c\}\subset\mathbb R\ ,$ from the Rolle's theorem obtain that the equation $g'(x)=0$ has real roots, i.e. $\Delta^{\prime}=(4R+r)^2-3s^2\ge 0\ ,$ i.e. $(4R+r)\ge s\sqrt 3\implies \boxed{s\sqrt 3\le 4R+r}\ (2)\ .$

Otherwise (without derivatives). Apply the well-known inequality $\left(r_a+r_b+r_c\right)^2\ge 3\left(r_ar_b+r_br_c+r_cr_a\right)\iff (4R+r)^2\ge 3s^2\iff s\sqrt 3\le (4R+r)\ .$

Remark. $\odot$ $\begin{array}{ccccccc} \nearrow & 27r^2\le 3r(4R+r) & \iff & 9r\le 4R+r & \iff & 8r\le 4R & \searrow\\\\ \searrow & \frac 13\cdot (4R+r)^2\le \frac {27}4\cdot R^2 & \iff & 4(4R+r)^2\le 81R^2 & \iff & 2(4R+r)\le 9R & \nearrow\end{array}$ $\odot\implies 2r\le R\ ,$ what is true.

P9 (Ruben Dario). Let an $A$ -right $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ which touches it at $D\in BC,$ $E\in CA$ and $F\in AB.$ Let $\left\{\begin{array}{ccc} m\left(\widehat{BED}\right) & = & x\\\ m\left(\widehat{CFD}\right) & = & y\end{array}\right\|\ .$ Prove that $\cot x+\cot y=\frac {3a}{a+r}\ .$

Proof. Is wellknown that $:\ a+r=s\ ;\ (s-a)(s-b)(s-c)=sr^2\ ;\ \left\{\begin{array}{ccccc} AE & = & AF & = & s-a\\\\ BD & = & BF & = & s-b\\\\ CD & = & CE & = & s-c\end{array}\right\|\ ;\ \left\{\begin{array}{ccccc} \cot \frac B2 & = & \frac {s-b}r & = & \frac {s-b}{s-a}\\\\ \cot\frac C2 & = & \frac {s-c}r & = & \frac {s-c}{s-a}\end{array}\right\|\ .$

$\blacktriangleright\ m\left(\widehat{CED}\right)=90^{\circ}-\frac C2$ and $\frac {BC}{BD}=\frac {EC}{ED}\cdot\frac {\sin\widehat{BEC}}{\sin\widehat{BED}}\iff$ $\frac {2a}{s-b}=\frac {\cos\left(\frac C2-x\right)}{\sin\frac C2\sin x}=\cot \frac C2\cot x+1\iff$ $\frac {2a}{s-b}=\frac {s-c}{s-a}\cdot\cot x+1\iff$ $\boxed{\cot x=\frac {(s-a)(2a+b-s)}{(s-b)(s-c)}}\ (1)\ .$

$\blacktriangleright\ m\left(\widehat{CFD}\right)=90^{\circ}-\frac B2$ and $\frac {CB}{CD}=\frac {FB}{FD}\cdot\frac {\sin\widehat{CFB}}{\sin\widehat{CFD}}\iff$ $\frac {2a}{s-c}=\frac {\cos \left(\frac B2-x\right)}{\sin\frac B2\sin y}=\cot\frac B2\cot y+1\iff$ $\frac {2a}{s-c}=\frac {s-b}{s-a}\cdot\cot y+1\iff$ $\boxed{\cot y=\frac {(s-a)(2a+c-s)}{(s-c)(s-b)}}\ (2)\ .$

From the sum of $(1)$ and $(2)$ get $\cot x+\cot y=$ $\frac {(s-a)(4a+b+c-2s)}{(s-b)(s-c)}=$ $\frac {3a(s-a)}{(s-b)(s-c)}=\frac {3a}s=\frac {3a}{a+r}\implies$ $\boxed{\cot x+\cot y=\frac {3a}{a+r}}\ .$ See here an easy extension P12.

P10 (Kadir Altintas). Let $ABC$ be a triangle so that $a<b<c.$ Denote : its incircle $w=\mathbb C(I,r)\ ;\ D\in BC,$ $E\in CA,$ $F\in AB$ so that $\{D,E,F\}\subset w\ ;$

the midpoints $M,$ $N$ of the sides $[AB],$ $[BC]$ respectively $;$ the circumcircle $W=\mathbb C\left(O_1,R_1\right)$ of $\triangle MBN.$ Prove that $I\in \mathbb W\iff$ $O_1\in IE\iff$ $a+c=2b.$

Proof. $\left\{\begin{array}{ccccc} F\in (MB) & \implies & MF=AF-AM=(s-a)-\frac c2=\frac {(b+c-a)-c}2=\frac {b-a}2 & \implies & MF=\frac {b-a}2\\\\ D\in (CN) & \implies & ND=CN-CD=\frac a2-(s-c)=\frac {a-(a+b-c)}2=\frac {c-b}2 & \implies & ND=\frac {c-b}2\end{array}\right\|\ (1)\ .$ Therefore, $I\in \mathbb W\iff$ $MINB$ is cyclic $\iff$ $\widehat{IMF}\equiv\widehat{IND}\iff$

$\triangle IMF\equiv\triangle IND\iff$ $MF=ND\ \stackrel{(1)}{\iff}\ \frac {b-a}2=\frac {c-b}2\iff$ $a+c=2b\ .$ Hence $\boxed{I\in \mathbb W\iff\ a+c=2b}\ .$ From the power of the points $\{A,C\}$ w.r.t. the circle $W$ obtain that

$\left\{\begin{array}{ccccc} AM\cdot AB & = & AO_1^2-R_1^2 & \implies & AO_1^2=\frac {c^2}2 +R_1^2\\\\ CN\cdot CB & = & CO_1^2-R_1^2 & \implies & CO_1^2=\frac {a^2}2+R_1^2\end{array}\right\|\implies$ $\boxed{O_1A^2-O_1C^2=\frac {c^2-a^2}2}\ (2)\ .$ Therefore, $O_1\in IE\iff$ $O_1I\perp AC\iff$ $O_1A^2-O_1C^2=IA^2-IC^2\ \stackrel{(2)}{\iff}$

$\frac {c^2-a^2}2=(s-a)^2-(s-c)^2\iff$ $c^2-a^2=2b(c-a)\iff$ $a+c=2b.$ Hence $\boxed{O_1\in IE\iff\ a+c=2b}\ .$ In conclusion, $\boxed{I\in \mathbb W\ \iff\ O_1\in IE\ \iff\ a+c=2b}\ .$

P11 (Cristian Tello). Let $\triangle ABC$ with $K\in (AC)$ and $L\in (BC)$ so that $ABLK$ is cyclic. Denote $G\in (KL),$ $M\in AG\cap BC$ and $N\in CG\cap AB\ .$ Prove that $\frac {GK}{GL}=\frac {NA}{NB}\cdot \left(\frac {CB}{CA}\right)^2\ .$

Proof. Apply the Menelaus' theorem to the transversals $\overline{1,3}\ :\ \left\{\begin{array}{cccc} & \mathrm{ABLK\ is\ cyclic} & \iff & \cancel{\frac {CL}{CK}}\cdot \frac {CB}{CA}=1\\\\ (1) & \overline{LGK}/\triangle AMC & \iff & \frac {LM}{\cancel{LC}}\cdot\frac {\cancel{KC}}{KA}\cdot\cancel{\frac {GA}{GM}}=1\\\\ (2) & \overline{AGM}/\triangle CLK & \iff & \frac {AK}{AC}\cdot\frac {\cancel{MC}}{ML}\cdot\frac {GL}{GK}=1\\\\ (3) & \overline{CGN}/\triangle ABM & \iff & \frac {CB}{\cancel{CM}}\cdot\cancel{\frac {GM}{GA}}\cdot\frac {NA}{NB}=1\end{array}\right\|\bigodot\implies \left(\frac {CB}{CA}\right)^2\cdot \frac {GL}{GK}\cdot\frac {NA}{NB}=1\implies\frac {GK}{GL}=\frac {NA}{NB}\cdot \left(\frac {BC}{AC}\right)^2\ .$

Remark. If $G$ is the centroid of the $A$ -right $\triangle ABC$ and $KL\perp BC,$ then $N$ is the midpoint of $[AB],$ $ABLK$ is cyclic, i.e. $\frac {CK}{CL}=\frac {CB}{CA}\ .$ Hence $\frac {GK}{GL}=\frac {CK}{CL}\cdot\frac {CB}{CA}=\left(\frac {CB}{CA}\right)^2\ .$

An easy extension. Let $\triangle ABC$ with $K\in (AC),\ L\in (BC),\ G\in (KL),\ M\in AG\cap BC$ and $N\in CG\cap AB\ .$ Prove that $\frac {GK}{GL}=\frac {NA}{NB}\cdot \frac {CK}{CL}\cdot\frac {CB}{CA}\ .$

Proof. Apply the Menelaus' theorem to the transversals $\overline{1,3}\ :\ \left\{\begin{array}{cc} (1) & \overline{LGK}/\triangle AMC\iff\frac {\cancel{LM}}{LC}\cdot\frac {KC}{\cancel{KA}}\cdot\cancel{\frac {GA}{GM}}=1\\\\ (2) & \overline{AGM}/\triangle CLK\iff \frac {\cancel{AK}}{AC}\cdot\frac {\cancel{MC}}{\cancel{ML}}\cdot\frac {GL}{GK}=1\\\\ (3) & \overline{CGN}/\triangle ABM\iff\frac {CB}{\cancel{CM}}\cdot\cancel{\frac {GM}{GA}}\cdot\frac {NA}{NB}=1\end{array}\right\|\bigodot\implies\frac {CK}{CL}\cdot \frac {CB}{CA}\cdot \frac {GL}{GK}\cdot \frac {NA}{NB}=1\implies\frac {GK}{GL}=\frac {NA}{NB}\cdot \frac {CK}{CL}\cdot\frac {CB}{CA}\ .$

Remark. If $ABLK$ is cyclic, then there is the relation $CK\cdot CA=CL\cdot CB,$ i.e. $\boxed{\frac {CK}{CL}=\frac {CB}{CA}}\ (*)$ and $\frac {GK}{GL}=\frac {NA}{NB}\cdot \frac {CK}{CL}\cdot\frac {CB}{CA}\ \stackrel{(*)}{\implies}\ \frac {GK}{GL}=\frac{NA}{NB}\cdot \left(\frac {CB}{CA}\right)^2\ .$

P12 (Kadir Altintas, Turkey). Let $\triangle ABC$ with the interior point $D$ so that $\left\{\begin{array}{ccccccc} AC=b & ; & BC=d & ; & m\left(\widehat{DAB}\right)=40^{\circ} & ; & m\left(\widehat{DAC}\right)=20^{\circ}\\\\ DA=c & ; & DB=a & ; & m\left(\widehat{DBA}\right)=10^{\circ} & ; & m\left(\widehat{DBC}\right)=10^{\circ}\end{array}\right\|\ .$ Profe that $a^2+b^2=c^2+d^2\ .$ .

Proof. Denote $m\left(\widehat{ACD}\right)=x.$ Hence $m\left(\widehat{BCD}\right)=100-x.$ Apply the trigonometrical form of the Ceva's theorem $:\ \sin\widehat{DAB}\cdot\sin\widehat{DBC}\cdot\sin\widehat{DCA}=\sin\widehat{DAC}\cdot\sin\widehat{DCB}\cdot\sin\widehat{DBA}\iff$

$\underline{\sin 40}^{\circ}\cancel{\sin 10^{\circ}}\sin x=\underline{\sin 20}^{\circ}\sin \left(100^{\circ}-x\right)\cancel{\sin 10^{\circ}}\iff$ $2\cos 20\sin x=\sin \left(80^{\circ}+x\right)\iff$ $\sin \left(x+20^{\circ}\right)+\sin \left(x-20^{\circ}\right)=$ $\sin \left(80^{\circ}+x\right)\iff$ $\sin \left(80^{\circ}+x\right)-\sin \left(x+20^{\circ}\right)=$

$\sin \left(x-20^{\circ}\right)\iff$ $\cancel{2\sin 30^{\circ}}\cos \left(50^{\circ}+x\right)=\cos\left(110^{\circ}-x\right)\iff$ $\cos \left(50^{\circ}+x\right)=\cos\left(110^{\circ}-x\right)\iff$ $50^{\circ}+x=110^{\circ}-x\iff x=30^{\circ},$ i.e. $m\left(\widehat{DCA}\right)=30^{\circ}\ ,\ m\left(\widehat{DCB}\right)=70^{\circ}\ .$

Thus $m\left(\widehat{BAC}\right)+ m\left(\widehat{ACD}\right)=60^{\circ}+30^{\circ}=90^{\circ},$ i.e. $CD\perp AB\ .$ Hence $:\ CD\perp AB\iff$ $CA^2-CB^2=DA^2-DB^2\iff$ $b^2-d^2=c^2-a^2\iff$ $\boxed{a^2+b^2=c^2+d^2}\ .$

P13. In $\triangle ABC$ , the $A$ -exincircle $w_a$ is tangent to $AB$ and $AC$ in $P$ and $Q$ respectively. The $B$ -exincircle $w_b$ is tangent to $BA$ and $BC$

in $M$ and $N$ respectively. Denote the projection $K$ of $C$ on $MN$ and the projection $L$ of $C$ on $PQ$ . Prove that $MKLP$ is cyclically.

Proof. Denote $S\in KM\cap LP$ . Observe that $AM=BP=s-c$ , $MP=a+b$ and $\left\{\begin{array}{c} m\left(\widehat{PMS}\right)=90^{\circ}-\frac B2\\\\ m\left(\widehat{MPS}\right)=90^{\circ}-\frac A2\\\\ m\left(\widehat{MSP}\right)=90^{\circ}-\frac C2\end{array}\right\|$ . Apply the theorem of Sines in $\triangle MPS\ :\ \frac {a+b}{\cos\frac C2}=$ $\frac {SM}{\cos \frac A2}=\frac {SP}{\cos \frac B2}\implies$

$SM^2-SP^2=\frac {(a+b)^2}{\cos^2\frac C2}\cdot\left(\cos ^2\frac A2-\cos^2\frac B2\right)=$ $\frac {ab(a+b)^2}{2s(s-c)}\cdot(\cos A-\cos B)=$ $\frac {ab(a+b)^2}{2s(s-c)}\cdot\frac {2(b-a)s(s-c)}{abc}\implies$ $\boxed{SM^2-SP^2=\frac {(a+b)(b^2-a^2)}{c}}$ . Apply the theorem of Cosines in

the triangles $CAM$ and $CBM\ :\ \left\{\begin{array}{c} CM^2=(s-c)^2+b^2+2b(s-c)\cos A\\\\ CP^2=(s-c)^2+a^2+2a(s-c)\cos B\end{array}\right\|\implies$ $CM^2-CP^2=b^2-a^2+2(s-c)(b\cos A-a\cos B)$ . Observe that $b\cos A-a\cos B=$ $\frac {b^2-a^2}{c}$ and

$CM^2-CP^2=$ $b^2-a^2+2(s-c)\cdot \frac {b^2-a^2}{c}=\left(b^2-a^2\right)\cdot\left(1+\frac {a+b-c}{c}\right)\implies$ $\boxed{CM^2-CP^2=\frac {(a+b)(b^2-a^2)}{c}}$ . Thus, $SM^2-SP^2=CM^2-CP^2=\frac {(b-a)(a+b)^2}{c}\implies$

$SC\perp AB$ . Denote $R\in SC\cap AB$ . Since $RMKC$ and $RPLC$ are cyclically obtain that $\left\{\begin{array}{c} SC\cdot SR=SK\cdot SM\\\\ SC\cdot SR=SL\cdot SP\end{array}\right\|\implies$ $SK\cdot SM=SL\cdot SP\implies$ $MKLP$ is cyclically.

P14. Let $ABC$ be an acute triangle with the incircle $\omega =C(I,r)$ and the circumcircle $\rho =C(O,R)$ .The circles $C_{1}=C(P,r_{1})$ and $C_{2}=(Q,r_{2})$ are tangent internally to the circle $\rho$ in the same

point $A\ .$ The circle $w$ is tangent externally to the circle $C_{1}$ and is tangent internally to the circle $C_{2}\ .$ Prove that $\boxed{PQ=\frac{a^{2}(p-a)}{4S}}$ , where $2p=a+b+c$ and $S\equiv [ABC]$ - the area of $\triangle ABC\ .$

Proof. Prove easily that $\left\{\begin{array}{ccccc}IP=r+r_{1}& ; & IQ=r_{2}-r\\\\ PO=R-r_{1}& ; & PA=r_{1}\\\\ QO=R-r_{2}& ; & QA=r_{2}\\\\ OA=R & ; & \boxed{PQ=r_{2}-r_{1}}\end{array}\right\|$ . The relations $p(p-a)+(p-b)(p-c)=bc$ and $\left\{\begin{array}{ccc}IO^{2}=R(R-2r) & ; & IA^{2}=\frac{bc(p-a)}{p}\\\\ IA^{2}-r^{2}=(p-a)^{2}& ; & IA^{2}+4Rr=bc\end{array}\right\|$ are well-known. Denote

$IO=m\ ,$ $IA=n$ and apply Stewart's theorem in $\triangle OIA$ for $IP$ and $IQ\ :\ \left\{\begin{array}{c}m^{2}r_{1}^{2}+n^{2}(R-r_{1})=R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\ m^{2}r_{2}+n^{2}(R-r_{2})=R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\right\|$ $\implies$ $r_{1}(R^{2}+2Rr+n^{2}-m^{2})=R(n^{2}-r^{2})=$

$r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $\implies$ $r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr)=$ $R(p-a)^{2}=$ $r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $\implies$ $r_{1}bc=R(p-a)^{2}=r_{2}(bc-4Rr)$ $\implies$

$PQ=r_{2}-r_{1}=$ $\frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $\frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $\frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $\implies$ $\boxed{\ PQ=\frac{a^{2}(p-a)}{4S}\ }\ .$

Remark. $8\cdot \prod PQ=\frac{8(abc)^{2}(p-a)(p-b)(p-c)}{(4S)^{3}}=\frac{8(4RS)^{2}Sr}{(4S)^{3}}=$ $R^{2}\cdot 2r\le R^3$ $\implies$ $\boxed{8\cdot \prod PQ\le R^3}\implies$ $\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\| .$

P15. Let an equilateral $\triangle ABC$ and an interior $P$ so that $:\ \left\{\begin{array}{ccccc} PA=x & : & D\in BC & ; & PD\perp BC\\\ PB=y & : & E\in CA & ; & PE\perp CA\\\ PC=z & : & F\in AB & ; & PF\perp AB\end{array}\right\|\ .$ Prove that $[DEF]=\frac 34\cdot \sqrt{p(p-x)(p-y)(p-z)}\ ,$ where $2p=x+y+z\ .$

Proof. $PEAF\ ,$ $PFBD\ ,$ $PDCE$ are cyclic with the diameters $PA\ ,$ $PB\ ,$ $PC$ respectively. So $EF=x\cdot\sin A\ ,$ $FD=y\cdot\sin B\ ,$ $DE=z\cdot\sin C\iff$ $\frac {EF}x=\frac {FD}y=\frac {DE}z=\frac {\sqrt 3}2\iff$

$\triangle DEF\sim\triangle XYZ\ ,$ where $XY=z\ ,$ $YZ=x\ ,$ $ZX=y$ and $[XYZ]=\sqrt {p(p-x)(p-y)(p-z)}\ .$ Thus, $\frac {[DEF]}{[XYZ]}=\left(\frac {\sqrt 3}2\right)^2=\frac 34\implies$ $[DEF]=\frac 34\cdot \sqrt{p(p-x)(p-y)(p-z)}\ .$

P16 (PAN African 2017). Let $\triangle ABC$ with orthocenter $H\ .$ The circle with diameter $[AC]$ cuts again the

circumcircle of $\triangle ABH$ at $K$ . Prove that the point $M\in CK\cap BH$ is the midpoint of the segment $[BH]\ .$

Proof.

$\blacktriangleright\ \left\{\begin{array}{ccccc} BAHK\ \mathrm{is\ cyclic} & \iff & \widehat{BAK}\equiv \widehat{BHK}\equiv\underline{\widehat{MHK}}\\\\ ACKE\ \mathrm{is\ cyclic} & \iff & \widehat{KAE}\equiv\widehat {KCE}\equiv\underline{\widehat {KCH}}\end{array}\right\|\implies$ $\widehat{MHK}\equiv\widehat {KCH}\implies \boxed{MH^2=MK\cdot MC}\ (1)\ .$

$\blacktriangleright\ \left\{\begin{array}{ccccc} BAHK\ \mathrm{is\ cyclic} & \iff & \widehat{KAH}\equiv \widehat{KBH}\equiv\underline{\widehat{KBM}}\\\\ ACFK\ \mathrm{is\ cyclic} & \iff & \widehat{KAF}\equiv\widehat {KCF}\equiv\underline{\widehat {KCB}}\end{array}\right\|\implies$ $\widehat{KBM}\equiv\widehat {KCB}\implies \boxed{MB^2=MK\cdot MC}\ (2)\ .$

In conclusions, from the relations $(1)$ and $(2)$ obtain that $MB=MH\ ,$ i.e. the point $M$ is the midpoint of the segment $[BH]\ .$

P17 (Adil Abdullayev). Prove that $(\forall )\ \triangle\ ABC$ there is the bilateral inequality $18r\le\sum \frac {a^2}{r_a-r}\le 9R\ .$

Proof. $S=rs=r_a(s-a)\iff$ $s\left(r_a-r\right)=ar_a\iff$ $\boxed{\frac a{r_a-r}=\frac s{r_a}=\frac {s-a}r}\ (*)\ .$ Therefore, $\sum \frac {a^2}{r_a-r}\ \stackrel{(*)}{=}\ \sum \left(a\cdot\frac {s-a}r\right)=\frac 1r\cdot\sum \left(as-a^2\right)=\frac 1r\cdot \left[\cancel{2s^2}-2\left(\cancel{s^2}-r^2-4Rr\right)\right]=$

$\frac {2r(4R+r)}r=2(4R+r)\implies$ $\boxed{\sum \frac {a^2}{r_a-r}=2(4R+r)}\ (1)\ .$ Hence $18r\le\sum \frac {a^2}{r_a-r}\le 9R\ \stackrel{(1)}{\iff}\ \odot \begin{array}{ccc} \nearrow & 18r\le 2(4R+r) \iff 18r-2r\le 8R & \searrow\\\\ \searrow & 2(4R+r)\le 9R \iff 2r\le 9R-8R & \nearrow\end{array}\odot\iff 2r\le R$ what is true.

P18 (Kadir ALTINTAS). Let $\triangle ABC$ with $a\ne c\ ,$ the incircle $w=\mathbb C(I,r)\ ,$ $T\in AC\cap w$ and the Lemoine's point $K\ .$ Prove that $\boxed{K\in BT\iff b(a+c)=a^2+c^2\iff IK\parallel AC}\ .$

Proof. Denote $\{T,L,S,H\}\subset AC$ so that $:\ T\in w\ ;\ L\in BK\ ;\ KS\perp AC\ ;\ BH\perp AC\ .$ Is well-known that $:\ IT=r\ ;\ \frac {LA}{LC}=\frac {c^2}{a^2}$ and $\boxed{\frac {KB}{a^2+c^2}=\frac{KL}{b^2}=\frac {BL}{a^2+b^2+c^2}}\ (1)\ \ ;$

$SK\parallel HB\iff\frac{SK}{HB}=\frac{LK}{LB}\ \stackrel{(1)}{\iff}\ \frac {SK}{h_b}=$ $\frac {b^2}{a^2+b^2+c^2}\iff$ $SK=\frac {b\cdot \left(bh_b\right)}{a^2+b^2+c^2}=\frac {b\cdot (2sr)}{a^2+b^2+c^2}\iff$ $\boxed{SK=\frac {rb(a+b+c)}{a^2+b^2+c^2}}\ (2)\ .$

$\blacktriangleright\boxed{K\in BT}\iff$ $L\equiv T\iff$ $\frac {LA}{LC}=\frac {TA}{TC}\iff$ $\frac {c^2}{a^2}=\frac {s-a}{s-c}\iff$ $c^2(s-c)=a^2(s-a)\iff$ $a^3-c^3=s\left(a^2-c^2\right)\iff$ $\cancel{(a-c)}\left(a^2+ac+c^2\right)=s\cancel{(a-c)}(a+c)\iff$

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

$\blacktriangleright\ \boxed{IK\parallel AC}\iff$ $\delta_{AC}(I)=\delta_{AC}(K)\iff$ $IT=KS\iff\ \stackrel{(2)}{\iff}\ \cancel r=\frac {\cancel rb(a+b+c)}{a^2+b^2+c^2}\iff$ $b(a+\cancel b+c)=a^2+\cancel{b^2}+c^2\iff$ $\boxed{b(a+c)=b^2+c^2}\ .$

P19 (Kadir ALTINTAS). Let $\triangle ABC$ with $\overline{\overline{\{a,b,c\}}}=3\ ,$ the circumcircle $w=\mathbb C(O,R)\ ,$ the midpoint $M$ of $[BC]\ ,$ $\{A,D\}=AM\cap w\ .$ Prove that $\boxed{a(b+c)=b^2+c^2\iff DA=DB+DC}\ .$

Proof. Apply the lemma (<= click) to the $M$ in the cyclic quadrilateral $ABDC\ :\ \frac {MB}{AB\cdot DB}=\frac {MC}{AC\cdot DC}\iff$ $c\cdot DB=b\cdot DC\iff$ $\frac {DB}b=\frac {DC}c=$

$\boxed{\frac {DB+DC}{b+c}}=$ $\frac {b\cdot DB}{b^2}=\frac {c\cdot DC}{c^2}=\boxed{\frac {a\cdot DA}{b^2+c^2}}\iff$ $\boxed{\frac {DB+DC}{a(b+c)}=\frac {DA}{b^2+c^2}}\ (*)\ .$ In conclusion, $a(b+c)=b^2+c^2\ \stackrel{(*)}{\iff}\ DB+DC=DA\ .$

P20 (extension). Let $\triangle ABC$ with $\overline{\overline{\{a,b,c\}}}=3\ ,$ the circumcircle $w\ ,$ the point $M\in (BC)$ and $\{A,D\}=AM\cap w\ .$ Prove that $\boxed{\frac {MB}{MC}=\frac {c(c-a)}{b(a-b)}\iff DA=DB+DC}\ .$ Superbly !

Proof. Let $\frac {MB}{MC}=m$ and apply the lemma (<= click) to the $M$ in the cyclic quadrilateral $ABDC\ :\ \frac {MB}{AB\cdot DB}=\frac {MC}{AC\cdot DC}\iff$ $\frac m{c\cdot DB}=\frac 1{b\cdot DC}\iff$ $\frac {DB}{mb}=\frac {DC}c=\boxed{\frac {DB+DC}{mb+c}}=$

$\frac {b\cdot DB}{mb^2}=\frac {c\cdot DC}{c^2}=$ $\frac {b\cdot DB+c\cdot DC}{mb^2+c^2}\ \stackrel{\mathrm{Ptolemy}}{=}\ \boxed{\frac {a\cdot AD}{mb^2+c^2}}\ .$ Thus, $\boxed{\frac {DB+DC}{mb+c}=\frac {a\cdot AD}{mb^2+c^2}}\ (1)\ .$ Hence $m=\frac {c(c-a)}{b(a-b)}\iff a(mb+c)=mb^2+c^2\ \stackrel{(1)}{\iff}\ DB+DC=DA\ .$

Particular cases.

$\mathrm{PC1.}\ \blacktriangleright\ m=\frac {MB}{MC}=1\ ,$ i.e. $AM$ is the $A$ -median of $\triangle ABC\ :\ \boxed{DB+DC=DA\iff a(b+c)=b^2+c^2}\ .$

Indeed $:\ 1=\frac {c(c-a)}{b(a-b)}\iff$ $b(a-b)=c(c-a)\iff$ $ba-b^2=c^2-ac\iff$ $a(b+c)=b^2+c^2\ .$

$\mathrm{PC2.}\ \blacktriangleright\ m=\frac {MB}{MC}=\frac {c^2}{b^2}\ ,$ i.e. $AM$ is the $A$ -symmedian of $\triangle ABC\ :\ \boxed{DB+DC=DA\iff \frac 1b+\frac 1c=\frac 2a}\ .$

Indeed $:\ \frac {c^2}{b^2}=\frac {c(c-a)}{b(a-b)}\iff$ $\frac cb=\frac {c-a}{a-b}\iff$ $ac-bc=bc-ab\iff$ $a(b+c)=2bc\iff$ $\frac 1b+\frac 1c=\frac 2a\ .$

$\mathrm{PC3.}\ \blacktriangleright\ m=\frac {MB}{MC}=\frac {s-c}{s-b}\ ,$ i.e. $AM$ is the $A$ -Nagel cevian of $\triangle ABC\ :\ \boxed{DB+DC=DA\iff \frac rR+\frac a{b+c}=1}\ .$

Indeed $:\ \frac {s-c}{s-b}=\frac {c(c-a)}{b(a-b)}\iff$ $b(a-b)(a+b-c)=c(c-a)(a+c-b)\iff$ $(b+c)\left(a^2+bc\right)=2abc+\left(b^3+c^3\right)\iff$ $\left(a^2+bc\right)(b+c)=2abc+(b+c)\left(b^2+c^2-bc\right)\iff$

$(b+c)\left[a^2-(b-c)^2\right]=2abc\iff$ $4(b+c)(s-b)(s-c)=2abc\iff$ $4(b+c)sr^2=8Rsr(s-a)\iff$ $r(b+c)=R(b+c-a)\iff\frac rR=\frac {b+c-a}{b+c}\iff$ $\frac rR+\frac a{b+c}=1\ .$

$\mathrm{PC4.}\ \blacktriangleright\ m=\frac {MB}{MC}=\frac {c\cdot\cos C}{b\cdot \cos B}\ ,$ i.e. $[AD]$ is the diameter of the circumcircle $w\ :\ \boxed{DB+DC=DA\iff \cos B+\cos C=1\iff \frac rR+\frac a{b+c}=1}\ .$

Indeed $:\ \frac {\cancel c\cdot \cos C}{\cancel b\cdot\cos B}=\frac {\cancel c(c-a)}{\cancel b(a-b)}\iff$ $a(\cos C+\cos B)=c\cdot\cos B+b\cdot\cos C=a\iff$ $\boxed{\cos B+\cos C=1}\iff$ $b\left(a^2+c^2-b^2\right)+c\left(a^2+b^2-c^2\right)=2abc\iff$

$a^2(b+c)+bc(b+c)-(b+c)\left(b^2+c^2-2bc\right)=2abc\iff$ $(b+c)\left(a^2+2bc-b^2-c^2\right)=2abc\iff$ $(b+c)\left[a^2-(b-c)^2\right]=2abc\iff$ $4(b+c)(s-b)(s-c)=2abc\iff$

$\cancel 4(b+c)\cancel sr^{\cancel 2}=2\cdot \cancel 4R\cancel{rs}(s-a)\iff$ $r(b+c)=2R(s-a)\iff$ $r(b+c)=R(b+c-a)\iff$ $\frac rR=\frac {b+c-a}{b+c}\iff$ $\boxed{\frac rR+\frac a{b+c}=1}\ .$

Remark. $\cos B+\cos C=1\iff$ $\sum\cos A=1+\cos A\iff$ $\cancel 1+\frac rR=\cancel 1+\cos A\iff$ $R\cos A=r\iff$ a.s.o.

P21 (Adil Abdullayev). Prove that $(\forall )\ \triangle\ ABC$ there is the bilateral inequality $\frac {4(2R-r)}{3Rr\sqrt 3}\le\sum \frac a{r_a^2}\le \frac {2(2R-r)}{3r^2\sqrt 3}\ .$

Proof. Observe that $\boxed{\sum\frac a{r_a^2}=\frac 1{S^2}\cdot \sum a(s-a)^2}\ (1)\ ,$ where $S=sr$ and it is the area of $\triangle ABC\ .$ I"ll use only well-known identity $\boxed{s(s-a)+(s-b)(s-c)=bc}\ (2)\ .$ Hence $\sum a(s-a)^2=$

$\frac 1s\cdot\sum \left[a(s-a)\cdot s(s-a)\right]\ \stackrel{(2)}{=}\ \frac 1s\cdot \sum a(s-a)[bc-(s-b)(s-c)]=$ $\frac {abc}s\cdot\sum (s-a)-\frac 1s\cdot \sum a(s-a)(s-b)(s-c)=$ $abc-\frac 1s\cdot sr^2\cdot \sum a=$ $4RS-r^2\cdot 2s=$

$4RS-2rS=2S(2R-r)\implies$ $\boxed{\sum a(s-a)^2=2S(2R-r)}\ (2)\ .$ Hence $\sum\frac a{r_a^2}\ \stackrel{(1)}{=}\ \frac 1{S^2}\cdot \sum a(s-a)^2\ \stackrel {(2)}{=}\ \frac 1{S^2}\cdot 2S(2R-r)=\frac {2(2R-r)}S\implies$ $\boxed{\sum \frac a{r_a^2}=\frac {2(2R-r)}S}\ (3)\ .$

In conclusion, the required inequality becomes $\frac {4\cancel{(2R-r)}}{3R\cancel r\sqrt 3}\le\frac {2\cancel{(2R-r)}}{s\cancel r}\le \frac {2\cancel{(2R-r)}}{3r\cancel{^2}\sqrt 3}\iff$ $\frac 2{3R\sqrt 3}\le \frac 1s\le \frac 1{3r\sqrt 3}\iff$ $\boxed{3r\sqrt 3\le s\le\frac {3R\sqrt 3}2}\ ,$ what is true.

This post has been edited 501 times. Last edited by Virgil Nicula, Mar 30, 2018, 7:36 AM

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372554 views!
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369302 views!

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You seem to have solved every existing problem.
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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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Thank you a lot dear Virgil for an interesting and instructive blog.
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Thanks to all !
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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.
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Thanks Thomas_.
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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!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
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72 shouts

My (math)blog

453. Working page III.

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