Geometry test for $\mathrm{IX}^{th}$ class.

P0. $\ \boxed{\ \mathrm{Let\ an\ acute}\ \triangle\ ABC,\ \mathrm{its\ circumcircle}\ w=\mathbb C(O,R)\ \mathrm{and\ the\ intersections}\ \left\{\begin{array}{c} D\in BC\cap AO\\\ E\in CA\cap BO\\\ F\in AB\cap CO\end{array}\right\|\ .\ \mathrm{Prove\ that}\ \frac {1}{AD}+\frac {1}{BE}+\frac{1}{CF}=\frac 2R\ .\ }$

Proof. Observe that $\frac {AD}{\sin B}=\frac c{\cos (B-C)}\iff$ $AD=\frac {c\sin B}{\cos (B-C)}\iff$ $AD=\frac {2R\sin B\sin C}{\cos (B-C)}\iff$ $\frac 1{AD}=\frac {\cos (B-C)}{2R\sin B\cdot \sin C}\implies$ $\underline{\underline{\sum\frac 1{AD}}}=\frac 1{2R}\cdot\sum\frac {\cos B\cos C+\sin B\sin C}{\sin B\sin C}=$

$\frac 1{2R}\cdot \sum\left(\cot B\cot C+1\right)=\frac 1{2R}\cdot (1+3)=\frac 4{2R}=\underline{\underline{\frac 2R}}\implies$ $\frac {1}{AD}+\frac {1}{BE}+\frac{1}{CF}=\frac 2R\ .$ Remark. $c\cdot\sin B=h_a=AD\cos (B-C)\implies$ $\frac 1{AD}=$ $\frac {\cos (B-C)}{c\sin B}=$ $\frac 1{2R}\cdot \frac{\cos (B-C)}{\sin B\sin C}$ a.s.o.

P1 (Thanos Kalogerakis). Let an $A$ -isosceles $\triangle\ ABC\ ,$ where denote the midpoint $M$ of $[AB]\ ,$ the midpoint $N$ of $[CM]$ and $D\in BN\cap AC\ ,$ where $DN=DC\ .$ Find the ratio $\frac {AB}{BC}\ .$

Proof.

P2 (Matu BARCENA). Let an $B$ -isosceles $\triangle\ ABC$ and $D\in BC\ .$ Construct an $B$ -isosceles $\triangle\ BDE$ so that $BC$ separates $\{A,E\}$ and $\widehat{BAC}\equiv\widehat{BDE}\ .$ Prove that $AD=CE\ .$

Proof 1 (synthetic). Observe that $\widehat{ABD}\equiv \widehat {CBE}$ and $\left\{\begin{array}{ccc} BA & = & BC\\\ BD & = & BE\end{array}\right\|\implies$ $\triangle ABD\equiv\triangle CBE\implies AD=CE\ .$

Proof 2 (metric). Denote $BA=BC=a\ ,$ $BD=BE=b$ and $m\left(\widehat{ABC}\right)=m\left(\widehat{CBE}\right)=\phi\ .$ Observe that $DC=a-b$ and $AC^2=2a^2-2a^2\cos\phi =4a^2\sin^2\frac {\phi}2\ .$ Apply the Stewart's

theorem to the cevian $AD/\triangle ABC\ :\ a\cdot AD^2+ab(a-b)=$ $a^2(a-b)+b\cdot AC^2\iff$ $\cancel a\cdot AD^2+\cancel ab(a-b)=$ $a\cancel{^2}(a-b)+4a\cancel{^2}b\sin^2\frac {\phi}2\iff$ $\boxed{\ AD^2=(a-b)^2+4ab\sin\frac{\phi}2\ }\ (1)\ .$

Apply the generalized Pythagoras' theorem to the side $[CE]/\triangle ABC\ :\ CE^2=$ $a^2+b^2-2ab\cos\phi =$ $a^2+b^2-2ab\left(1-2\sin^2\frac {\phi}2\right)=$ $(a-b)^2+4ab\sin^2\frac {\phi}2\implies$

$\boxed{\ CE^2=(a-b)^2+4ab\sin^2\frac {\phi}2\ }\ (2)\ .$ From the relations $(1\wedge 2)$ obtain that $AD=CE=\sqrt{(a-b)^2+4ab\sin^2\frac {\phi}2}\ .$ Remark. $\phi =60^{\circ}\implies$ $AD=CE=\sqrt {a^2-ab+b^2}\ .$

P3 (Matu BARCENA). Let $\triangle\ ABC$ and $D\in (AC)$ so that $m\left(\widehat{ADB}\right)=3x\ ,$ $m\left(\widehat{ABD}\right)=6x$ and $m\left(\widehat{DBC}\right)=x\ .$ Prove that $C=30^{\circ}\ .$

Proof 1. Observe that $C=3x-x\ ,$ i.e. $\boxed{\ C=2x\ }\ (*)\ .$ $\frac {\sin 3x}{\sin 2x}=\frac {BC}{BD}=\frac {AD}{BD}=\frac {\sin 6x}{\sin 9x}\implies$ $\frac {\sin 3x}{\sin 2x}=\frac {\sin 6x}{\sin 9x}\implies$ $\frac 1{\sin 2x}=\frac {2\cos 3x}{\sin 9x}\iff$

$\sin 9x=\sin 5x-\sin x\iff$ $\sin 9x+\sin x=\sin 5x\iff$ $2\sin 5x\cos 4x=\sin 5x\iff$ $\cos 4x=\frac 12\iff$ $4x=\frac {\pi}3\iff$ $x=\frac {\pi}{12}\ \stackrel{(*)}{=}\ \boxed{\ C=30^{\circ}\ }\ .$

Proof 2.

P4 (Iulian Cristi). $(\forall )\ \{a,b,c,d\}\subset \mathbb R^*_+$ there is the inequality $\frac a{b+c+d}+\frac b{a+c+d}+\frac c{a+b+d}+\frac d{a+b+c}\ge \frac 43\ .$

Proof 1. $\sum\frac a{b+c+d}=$ $\sum\left(\frac {a+b+c+d}{b+c+d}-1\right)=$ $\sum a\cdot\sum\frac 1{b+c+d}-4\ge$ $\sum a\cdot\frac {4^2}{\sum (b+c+d)}-4=$ $\cancel{\sum a}\cdot\frac {16}{3\cancel{\sum a}}-4=$ $\frac {16}3-4=\frac 43\implies$ $\boxed{\ \sum\frac a{b+c+d}\ge \frac 43\ }\ .$

Proof 2. Let $\boxed{\ a+b+c+d=s\ }\ .$ Apply the Chebyshev's inequality $(a+b+c+d)^2\le 4\cdot\left(a^2+b^2+c^2+d^2\right)\ ,$ i.e. $\boxed{\ \left(\sum a\right)^2\le 4\cdot\sum a^2\ }\ (*)$

and the C.B.S - inequality $:\ \sum\frac a{b+c+d}=$ $\sum\frac {a^2}{a(s-a)}\ \stackrel{(C.B.S)}{\ge}\ \frac {\left(\sum a\right)^2}{\sum a(s-a)}=$ $\frac {s^2}{s^2-\sum a^2}\ \stackrel{(*)}{\ge}\ \frac {s^2}{s^2-\frac{s^2}4}=\frac 43\implies$ $\boxed{\ \sum\frac a{b+c+d}\ge \frac 43\ }\ .$

P5 (2003 China Girls Mathematics Olympiad) (<= click). Let $\triangle ABC\ ,\ b\ne c$ and its incircle $w=\mathbb C(I,r)\ .$ Denote $D\in AI\cap (BC)\ ,$ $E\in BI\cap (CA)$ and

$F\in CI\cap (AB)\ .$ Prove that $:\ DE=DF\iff$ $\frac a{b+c}=\frac b{c+a}+\frac c{a+b}\ ;\ m\left(\widehat{BAC}\right)>90^{\circ}\ ;$ find the value of the angle $\widehat{ACB}$ if we know that $B=2C\ .$

Proof. $\left\{\begin{array}{ccccc} D\in (BC) & \implies & \frac {DB}c=\frac {DC}b=\frac a{b+c}\\\\ E\in (CA) & \implies & \frac {EC}a=\frac {EA}c=\frac b{c+a}\\\\ F\in (AB) & \implies & \frac {FA}b=\frac {FB}a=\frac c{a+b}\end{array}\right\|\ (*).$ Applly Stewart's relation to the cevians $:\ \left\{\begin{array}{ccccc} F/\triangle BFC\ : & a\cdot FD^2+a\cdot DB\cdot DC & = & DC\cdot FB^2+DB\cdot FC^2\\\\ E/\triangle BEC\ : & a\cdot ED^2+a\cdot DB\cdot DC & = & DB\cdot EC^2+DC\cdot EB^2\end{array}\right\|\ominus\implies$

$a\cdot\left(FD^2-ED^2\right)=\left(DC\cdot FB^2+DB\cdot FC^2\right)-\left(DB\cdot EC^2+DC\cdot EB^2\right)\ .$ Therefore, $DF=DE\iff$ $DC\cdot FB^2+DB\cdot FC^2=DB\cdot EC^2+DC\cdot EB^2\iff$

$\cancel b\cdot \frac {a\cancel{^2}c\cancel{^2}}{(a+b)^2}+\cancel c\cdot\frac {4\cancel{ab}s(s-c)}{(a+b)^2}=\cancel c\cdot\frac {a\cancel{^2}b\cancel{^2}}{(a+c)^2}+\cancel b\cdot\frac {4\cancel{ac}s(s-b)}{(a+c)^2}\iff$ $\frac {ac+4s(s-c)}{(a+b)^2}=\frac {ab+4s(s-b)}{(a+c)^2}\iff$ $\frac {ac+(a+b)^2-c^2}{(a+b)^2}=\frac {ab+(a+c)^2-b^2}{(a+c)^2}\iff$

$\cancel 1+\frac {ac-c^2}{(a+b)^2}=\cancel 1+\frac {ab-b^2}{(a+c)^2}\iff$ $\frac {c(a-c)}{(a+b)^2}=\frac {b(a-b)}{(a+c)^2}\iff$ $c(a+c)\left(a^2-c^2\right)=b(a+b)\left(a^2-b^2\right)\iff$ $c\left(a^3+ca^2-c^2a-c^3\right)=b\left(a^3+ba^2-b^2a-b^3\right)\iff$

$a^3(b-c)+a^2\left(b^2-c^2\right)=$ $a\left(b^3-c^3\right)+\left(b^4-c^4\right)\iff$ $a^3+a^2(b+c)=a\left(b^2+bc+c^2\right)+(b+c)\left(b^2+c^2\right)\iff$ $a^2(a+b+c)=\left(b^2+c^2\right)(a+b+c)+(b+c)+abc\iff$

$(a+b+c)\left(a^2-b^2-c^2\right)=abc\ .$ In conclusion, $\boxed{\ DF=DE\iff (a+b+c)\left(a^2-b^2-c^2\right)=abc\ }$ and $a^2-b^2-c^2\ge 0\implies$ $m\left(\widehat{BAC}\right)>90^{\circ}\ .$ Prove easily that

$\boxed{\ \frac a{b+c}=\frac b{c+a}+\frac c{a+b}\iff (a+b+c)\left(a^2-b^2-c^2\right)=abc\ }\ (1)\ .$ Remark. $\left\{\begin{array}{cccc} \triangle AFD\ : & \frac {DF}{\sin \frac A2} & = & \frac {AD}{\sin\widehat{AFD}}\\\\ \triangle AED\ : & \frac {DE}{\sin\frac A2} & = & \frac {AD}{\sin\widehat{AED}}\end{array}\right\|\ \stackrel{DF=DE}{\iff}\ \sin\widehat{AFD}\equiv \sin\widehat{AED}\iff$

$\widehat{AFD}=\widehat{AED}\ \vee\ \widehat{AFD}+\widehat{AED}=180\ ,$ i.e. $AEDF$ is cyclic a.s.o. Is well-known that $B=2C\iff b^2=c(c+a)\ ,$ i.e. $\frac a{\sin 3C}=\frac b{\sin 2C}=\frac c{\sin C}.$ Can use $(1)$ a.s.o. Very nice problem!

P6 (Ashok Kumar Govsa). Find the product $abc\ ,$ if $\frac {\log a}{b-c}=\frac {\log b}{c-a}=\frac {\log c}{a-b}\ .$

Proof. Exists $k\in\mathbb R^*$ so that $\frac {\log a}{b-c}=\frac {\log b}{c-a}=\frac {\log c}{a-b}=k\ ,$ i.e. $\log (abc)=\log a+\log b+\log c=k[(b-c)+(c-a)+(a-b)]=0$ $\implies$ $\log (abc)=0\implies$ $abc=10^0\implies$ $\boxed{\ abc=1\ }\ .$

P7 (Van Khea, Cambodia). Let an acute $\triangle ABC$ with circumcircle $w=\mathbb C(O,R)$ and $\{D,E,F\}\subset w$ so that $O\in AD\cap BE\cap CF\ .$ Prove $\boxed{\frac {DB\cdot DC}{AB\cdot AC}+\frac {EC\cdot EA}{BC\cdot BA}+\frac {FA\cdot FB}{CA\cdot CB}=1}\ (*)\ .$

Proof. Is well-known that the quadrilaterals $HBDC\ ,$ $HCEA$ and $HAFB$ are parallelograms, where $H$ is the orthocenter of $\triangle ABC\ .$ Therefore, $\frac {DB\cdot DC}{AB\cdot AC}=$ $\frac {DB\cdot DC\cdot\sin\widehat{BDC}}{AB\cdot AC\cdot\sin\widehat{BAC}}=$

$\frac {[BDC]}{[BAC]}=$ $\frac {[BHC]}{[BAC]}=$ $\frac {[BHC]}S=$ $\frac 1S\cdot [BHC]\implies$ $\frac {DB\cdot DC}{AB\cdot AC}=$ $\frac 1S\cdot [BHC]\implies$ $\sum \frac {DB\cdot DC}{AB\cdot AC}=$ $\frac 1S\cdot\sum [BHC]=$ $\frac 1S\cdot S=1\ .$ Remark. Can apply the well-known property from here.

P8 (Emil Stoyanov). Let $P$ be the interior point of $\triangle ABC$ for which $\widehat{AFE}\equiv\widehat{BFD}$ and $\widehat{AEF}\equiv\widehat{CED}\ .$ Prove that $P\equiv H$ what is the orthocenter of $\triangle ABC\ .$

Proof 1. Construct the points $:\ \left|\begin{array}{ccc} X\in EF\ ,\ Y\in ED & ; & B\in (XY)\ ,\ XY\parallel AC\\\\ U\in FE\ ,\ V\in FD & ; & C\in (UV)\ ,\ UV\parallel AB\end{array}\right|\ .$ Observe that $\left\{\begin{array}{ccc} \widehat{AFE}\equiv\widehat{BFD}\implies\widehat{FUV}\equiv\widehat{FVU} & \implies & \triangle\ UFV\ \mathrm{is}\ F-\mathrm{isosceles}\\\\ \widehat{AEF}\equiv\widehat{CED}\implies\widehat{EXY}\equiv\widehat{EYX} & \implies & \triangle\ XEY\ \mathrm{is}\ E-\mathrm{isosceles}\end{array}\right\|\ .$

Therefore, $P\in AD\cap BE\cap CF\iff$ $\frac {DB}{DC}\cdot \frac {EC}{EA}\cdot\frac {FA}{FB}=1\implies$ $\left\{\begin{array}{ccccc} \frac {BY}{\cancel{CE}}\cdot \frac {\cancel{EC}}{\cancel{EA}}\cdot \frac {\cancel{AE}}{BX}=1 \implies & BY=BX \implies EB\perp XY & \implies & BE\perp AC \implies P\in BH\\\\ \frac {\cancel{FB}}{CV}\cdot \frac {CU}{\cancel{FA}}\cdot \frac {\cancel{FA}}{\cancel{FB}}=1 \implies & CU=CV\implies FC\perp UV & \implies & CF\perp AB \implies P\in CH\end{array}\right\|\implies P\equiv H\ .$

Proof 2. Denote $\left\{\begin{array}{ccccc} m\left(\widehat{CED}\right) & = & m\left(\widehat{AEF}\right) & = & y\\\\ m\left(\widehat{AFE}\right) & = & m\left(\widehat{BFD}\right) & = & z\end{array}\right\|$ and apply the theorem of SINES in the triangles $:\ \left\{\begin{array}{cccc} \triangle BDF\ : & \frac {BD}{BF} & = & \frac {\sin z}{\sin (B+z)}\\\\ \triangle CED\ : & \frac {CE}{CD} & = & \frac {\sin (C+y)}{\sin y}\\\\ \triangle AFE\ : & \frac {AF}{AE} & = & \frac {\sin y}{\sin z}\end{array}\right\|\bigodot\implies$ $\frac {BD}{BF}\cdot\frac {CE}{CD}\cdot\frac {AF}{AE}=$ $\frac {\sin (C+y)}{\sin (B+z)}\ .$

Observe that $P\in AD\cap BE\cap CF\iff$ $1=\frac {DB}{DC}\cdot \frac {EC}{EA}\cdot \frac {FA}{FB}=$ $\frac {BD}{BF}\cdot\frac {CE}{CD}\cdot\frac {AF}{AE}=$ $\frac {\sin (C+y)}{\sin (B+z)}\iff$ $\frac {\sin (C+y)}{\sin (B+z)}=1\iff$ $\sin (C+y)=\sin (B+z)\iff$ $\boxed{C+y=B+z=x}\ (*)\ .$

Let $U\in DF$ and $V\in DE$ so that $A\in UV$ and $UV\parallel BC\ .$ Prove easily $\triangle UDV$ is $D$ -isosceles and $1=\frac {DB}{DC}\cdot \frac {EC}{EA}\cdot \frac {FA}{FB}=$ $\frac {\cancel{DB}}{\cancel{DC}}\cdot \frac {\cancel{DC}}{AV}\cdot\frac {AU}{\cancel{DB}}=$ $\frac {AU}{AV}\iff$ $AU=AV$ $\iff$ $A$ is the midpoint

of $[UV]$ in $D$ -isosceles $\triangle UDV\iff$ $AD\perp UV\parallel BC\implies AD\perp BC\ .$ In conclusion, $A+y+z=B+z+x=C+x+y=x+y+z=180^{\circ}\iff$ $x=A\ ,\ y=B\ ,\ z=C$ a.s.o.

P9. Prove that $(\forall )\ \triangle ABC$ there is the following inequality $:\ \frac {a(b+c)}{s-a}+\frac {b(c+a)}{s-b}+\frac {c(a+b)}{s-c}\ge 12R\sqrt 3$ (standard notations).

Proof. $\sum\frac {a(b+c)}{s-a}=$ $\sum\frac {s^2-(s-a)^2}{s-a}=$ $s^2\cdot \sum\frac 1{s-a}-\sum (s-a)=$ $s^2\cdot\sum\frac {r_a}{r_a(s-a)}-s=$ $s^2\cdot\sum\frac {r_a}{sr}-s=$

$\frac sr\cdot \sum r_a-s=$ $\frac sr\cdot (4R+r)-s=$ $\frac {4R}r\cdot s\ge$ $\frac {4R}r\cdot 3r\sqrt 3=12R\sqrt 3$ $\implies$ $\sum\frac {a(b+c)}{s-a}\ge 12R\sqrt 3\ .$

P10. Let $ABCD$ be a tangential quadrilateral with the incircle $w=\mathbb C(I,r)$ what touches $w$ at $X\in (AB)\ ,$ $Y\in (BC)\ ,$ $Z\in (CD)$ and $T\in (DA)\ .$ Denote $\left\{\begin{array}{ccccc} AX & = & AT & = & a\\\ BY & = & BX & = & b\\\ CZ & = & CY & = & c\\\ DT & = & DZ & = & d\end{array}\right\|\ .$

Prove that $\boxed{\ r^2=\frac {ab(c+d)+cd(a+b)}{(a+b)+(c+d)}\ }\ (*)$ and the equivalence $(IA+ID)^2+(IB+IC)^2=(AB+CD)^2\iff ABCD\ \mathrm{is\ an\ isosceles\ trapezoid}\ .$

Proof. Observe that $\left\{\begin{array}{ccc} AB=a+b & ; & BC=b+c\\\\ CD=c+d & ; & DA=d+a\end{array}\right\|$ and $\left\{\begin{array}{ccc} \tan\frac A2=\frac ra & ; & \tan\frac B2=\frac rb\\\\ \tan\frac C2=\frac rc & ; & \tan \frac D2=\frac rd\end{array}\right\|\ .$ Hence $A+B+C+D=360^{\circ}\iff$ $\frac {A+B}2+\frac {C+D}2=180^{\circ}\iff$

$\tan\frac {A+B}2+\tan\frac {C+D}2=0\iff$ $\frac {\frac {\cancel r}a+\frac {\cancel r}b}{1-\frac {r^2}{ab}}+\frac {\frac {\cancel r}c+\frac {\cancel r}d}{1-\frac {r^2}{cd}}=0\iff$ $\frac {a+b}{ab-r^2}+\frac {c+d}{cd-r^2}=0\iff$ $(a+b)\left(cd-r^2\right)+(c+d)\left(ab-r^2\right)=0\iff$ $r^2=\frac {ab(c+d)+cd(a+b)}{(a+b)+(c+d)}\ .$

P11. Prove that $(\forall )\ \triangle ABC$ there is the following identity: $\boxed{\ \frac {a^2}{r_a}+\frac {b^2}{r_b}+\frac {c^2}{r_c}=4(R+r)\ }$ (standard notation). Remark. $12r\le \frac {a^2}{r_a}+\frac {b^2}{r_b}+\frac {c^2}{r_c}\le 6R\ .$

Proof 1. $\left\{\begin{array}{ccccc} \sum a^2=2\left(s^2-r^2-4Rr\right) & (1)\\\\ \sum a^3=2s\left(s^2-3r^2-6Rr\right) & (2)\end{array}\right\|$ $\implies$ $\sum \frac {a^2}{r_a}=$ $\sum\frac{a^2(s-a)}{sr}=$ $\frac 1{sr}\cdot \left(s\cdot\sum a^2-\sum a^3\right)\ \stackrel{1\wedge 2}{=}$ $\frac 1{sr}\cdot \left[2s\left(s^2-r^2-4Rr\right)-2s\left(s^2-3r^2-6Rr\right)\right]=$

$\frac 2r\cdot\left[\left(\cancel{s^2}-r^2-4Rr\right)-\left(\cancel{s^2}-3r^2-6Rr\right)\right]=$ $\frac 2r\cdot \left(2r^2+2Rr\right)=4(R+r)\implies$ $\boxed{\sum\frac {a^2}{r_a}=4(R+r)}\ .$ Remark. Observe that $\sum\frac {a^2}{6r_a}=\frac{2(R+r)}3\in [2r,R]\ .$

Proof 2. $\left\{\begin{array}{ccccc} x=s-a & \implies & a=y+z & \implies & x+y+z=s\\\\ y=s-b & \implies & b=z+x & \implies & xy+yz+zx=r(4R+r)\\\\ z=s-c & \implies & c=x+y & \implies & xyz=sr^2\end{array}\right\|\ (*)\ \implies$ $S\cdot \sum \frac {a^2}{r_a}=$ $\sum a^2(s-a)\ \stackrel{(*)}{=}\ \sum x(y+z)^2=$ $\sum \left[x\left(y^2+z^2\right)+2xyz\right]=$

$6xyz+\sum yz[(y+z)=$ $6xyz+\sum yz[(x+y+z)-x]=$ $\sum x\cdot\sum yz+3xyz\ \stackrel{(*)}{=}\ s\cdot r(4R+r)+3sr^2=$ $4sr(R+r)$ $\implies$ $\cancel S\cdot \sum\frac{a^2}{r_a}=$ $4\cancel{sr}(R+r)$ $\implies$ $\sum\frac {a^2}{r_a}=4(R+r)\ .$

P12. Let $ABC$ be an $A$ -rightangled triangle with the inradius $r,$ $B$ -exradius $r_b$ and $C$ -exradius $r_c.$ Prove that $\left(cr+br_b\right)\cdot\left(br+cr_c\right)=a^2\cdot S\ ,$ where $S$ is the area of $\triangle ABC\ .$

Proof. $\left(cr+br_b\right)\left(br+cr_c\right)=$ $bc\left(r^2+r_br_c\right)+r\left(b^2r_b+c^2r_c\right)=$ $bc\left(r^2+rr_a\right)+r\left[b^2(s-c)+c^2(s-b)\right]=$ $r\left[bc\left(r+r_a\right)+b^2(\underline s-c)+c^2(\underline s-b)\right]=$

$r\left[\cancel{bc\left(b+c\right)}+sa^2-\cancel{bc(b+c)}\right]=$ $a^2sr=a^2S.$ I used the following identities in any $A$ -right $\triangle ABC\ :\ \left\{\begin{array}{ccc} r=s-a & ; & r_a=s\\\ r_b=s-c & ; & r_c=s-b\end{array}\right\|\ .$

P13. Let an acute $\triangle ABC,$ its circumcircle $w=\mathbb C(O,R)$ and the intersections $\left\{\begin{array}{c} D\in BC\cap AO\\\ E\in CA\cap BO\\\ F\in AB\cap CO\end{array}\right\|.$ Prove that $\frac {1}{AD}+\frac {1}{BE}+\frac{1}{CF}=\frac 2R$

Proof 1. Observe that $h_a=b\sin C=2R\sin B\sin C$ and $m\left(\widehat{PAD}\right)=|B-C|.$ Thus, $h_a=AD\cdot \cos (B-C)\implies$ $2R\sin B\sin C=$

$AD\cdot \cos (B-C)\implies$ $\frac {2R}{AD}=\frac {\cos (B-C)}{\sin B\sin C}=$ $1+\cot B\cot C\implies$ $2R\cdot\sum \frac 1{AD}=\sum (1+\cot B\cot C)=4$ $\implies$ $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Proof 2. Apply the theorem of Sines in $\triangle ADC\ :\ \frac {AD}{\sin C}=\frac {AC}{\sin \widehat{ADC}}\iff$ $\frac {AD}{\sin C}=$ $\frac b{\cos(B-C)}\iff$ $\frac {2R}{AD}=\frac {\cos(B-C)}{\sin B\sin C}$ a.s.o. In conclusion, $2R\cdot \sum \frac 1{AD}=$

$\sum \frac {\cos(B-C)}{\sin B\sin C}=$ $\sum\frac {\cos B\cos C+\sin B\sin C}{\sin B\sin C}=$ $\sum\left(\cot B\cot C+1\right)=4\implies$ $2R\cdot \sum \frac 1{AD}=4\implies$ $\sum\frac 1{AD}=\frac 2R\ ,$ i.e. $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Proof 3. Let circumcircle $w=\mathbb C(O,R)$ and $\left\{\begin{array}{ccc} \{A,X\}=\{A,O\}\cap w\\\\ \{B,Y\}=\{B,O\}\cap w\\\\ \{C,Z\}=\{C,O\}\cap w\end{array}\right\|\ .$ Thus, $\frac {2R}{AD}=\frac {AX}{AD}=$ $\frac{AD+DX}{AD}=$ $1+\frac {DX}{DA}=1+\frac {XB\cdot XC}{AB\cdot AC}=$

$1+\cot C\cot B\implies$ $\sum\frac {2R}{AD}=\sum\left(1+\cot C\cot B\right)=3+\sum \cot C\cot B=4\implies$ $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Remark. I used two well-known identities $:\ \boxed{\sum\cot B\cot C=1}\ (1)$ and the remarkable property PP1 from here.

Proof 4. $2S=BC\cdot AD\cdot\sin \widehat{ADB}=$ $a\cdot AD\cdot \cos (B-C)=$ $AD\cdot 2R\sin A\cos (B-C)=$ $AD\cdot 2R\sin (B+C)\cos (B-C)=$ $R\cdot AD\cdot (\sin 2B+\sin 2C)\implies$

$2S=R\cdot AD\cdot (\sin 2B+\sin 2C)\implies$ $\frac 1{AD}=\frac {R(\sin 2B+\sin 2C)}{2S}\implies$ $\sum\frac 1{AD}=\frac {\cancel 2R\cdot\sum\sin 2A}{\cancel 2S}=$ $\frac {4\cancel R\prod\sin A}{2R\cancel{^2}\prod\sin A}=\frac 2R\implies$ $\boxed{\frac 1{AD}+\frac 1{BE}+\frac 1{CF}=\frac 2R}\ (*)\ .$

Proof 5. Denote the orthocenter $H$ of $\triangle\ ABC\ ,$ $P\in AH\cap BC\ ,$ $\{A,S\}=w\cap AO$ and $\{A,Q\}=w\cap AH\ .$ Thus, $\frac {2R}{AD}=\frac {AS}{AD}=\frac {AQ}{AP}=$ $\frac {h_a+HP}{h_a}=$

$1+\frac {HP}{h_a}\implies$ $2R\cdot\sum\frac 1{AD}=3+\sum\frac {a\cdot HP}{a\cdot h_a}=$ $3+\sum\frac {[BHC]}{[BAC]}=4\implies$ $2R\cdot\sum\frac 1{AD}=4\implies$ $\sum\frac 1{AD}=\frac 4{2R}\implies$ $\sum \frac 1{AD}=\frac 2R\ .$

P14 (George APOSTOLOPOULOS, Greece). Prove that $(\forall )\ \triangle\ ABC$ there is the following inequality $\frac 1{r^3}\ge \frac 1{r_a^3}+\frac 1{r_b^3}+\frac 1{r_c^3}+\frac {64}{9R^3}$ (standard notations).

Proof. $\left\{\begin{array}{ccc} 3R\sqrt 3 & \ge & 2s\\\\ R & \ge & 2r\end{array}\right\|\bigodot\implies$ $3R^2\sqrt 3\ge 4S\implies$ $27R^4\ge 16S^2\implies$ $\frac {3R}{S^2}\ge \frac {16}{9R^3}\implies$ $\boxed{\ \frac {12R}{S^2}\ge \frac {64}{9R^3}}\ (*)\ .$ I"ll use an well-known identities $:\ \left\{\begin{array}{ccc} r_a+r_b+r_c & = & 4R+r\\\\ r_ar_b+r_br_c+r_cr_a & = & s^2\\\\ rr_ar_br_c & = & S^2\end{array}\right|\ ;$

$\left\{\begin{array}{ccc} \frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c} & = & \frac 1r\\\\ \frac 1{r_ar_b}+\frac 1{r_br_c}+\frac 1{r_cr_a} & = & \frac{r(4R+r)}{S^2}\end{array}\right|\ ;\ \boxed{\ (x+y+z)^3-\left(x^3+y^3+z^3\right)=3(x+y+z)(xy+yz+zx)-3xyz\ }\ ,$ where $\{x,y,z\}=\left\{\frac 1{r_a},\frac 1{r_b},\frac 1{r_c}\right\}\ .$ Therefore,

$\left(\frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c}\right)^3-\sum\frac 1{r_a^3}=$ $3\cdot\sum\frac 1{r_a}\cdot\sum\frac 1{r_br_c}-3\cdot\prod\frac 1{r_a}\implies$ $\frac 1{r^3}-\sum\frac 1{r_a^3}=3\cdot\frac 1{\cancel{r}}\cdot\frac{\cancel r(4R+r)}{S^2}-\frac{3r}{S^2}=\frac {12R}{S^2}\ \stackrel{(*)}{\ge}\ \frac {64}{9R^3}\implies$ $\frac 1{r^3}-\sum\frac 1{r_a^3}\ge \frac {64}{9R^3}\ ,$ i.e. $\frac 1{r^3}\ge \sum\frac 1{r_a^3}+\frac {64}{9R^3}\ .$

P15 (Adil Abdullayev, Azerbaijan) (<= click =>) here.

Proof 1. Denote $\{A,S\}=AI\cap\Omega\ ,$ where $\Omega$ is the circumcircle of $\triangle\ ABC\ .$ Thus, $\frac a{R_a}=\frac a{SI}=\frac {a\cdot IA}{2Rr}\implies$ $\frac {a^2}{R_a^2}=$ $\frac {a^{\cancel 2}}{\cancel 4R^{\cancel 2}r^{\cancel 2}}\cdot \frac {\cancel{bc}(s-a)}{\cancel s}=$ $\frac {a(s-a)}{Rr}$ $\implies$

$\sum\frac {a^2}{R_a^2}=$ $\frac 1{Rr}\cdot \sum a(s-a)=$ $\frac 1{Rr}\cdot\left(s\sum a-\sum a^2\right)=$ $\frac 1{Rr}\cdot\left[\cancel{2s^2}-2\left(\cancel{s^2}-r^2-4Rr\right)\right]=$ $\frac {2r(4R+r)}{Rr}=8+\frac {2r}R=8+\frac {2r}R\implies$ $\sum\frac {a^2}{R_a^2}=8+\frac {2r}R\ .$

Proof 2. $\frac {a^2}{R_a^2}=\left(\frac {a}{SB}\right)^2\ \stackrel{\triangle BSC}{=}\ \left(\frac {\sin A}{\sin\frac A2}\right)^2=\left(2\cos\frac A2\right)^2=$ $2\cdot 2\cos^2\frac A2=$ $2(1+\cos A)\implies$ $\sum\frac {a^2}{R_a^2}=\sum2(1+\cos A)=6+2\left(1+\frac rR\right)=8+\frac {2r}R\implies$ $\sum\frac {a^2}{R_a^2}=8+\frac {2r}R\ .$

P16 (Adil Abdullayev, Azerbaijan) $\boxed{\ \mathrm{Prove\ that\ for}\ (\forall\ )\ \triangle\ ABC\ \mathrm{there\ is\ the\ inequality}\ a^4+b^4+c^4\ge 16S^2\ .}$

Proof. $\ 3\cdot\sum a^4 \ge \left(\sum a^2\right)^2\ \stackrel{Weitzenbock}{\ge}\ \left(4S\sqrt 3\right)^2=48S^2\implies 3\left(a^4+b^4+c^4\right)\ge 48S^2\implies a^4+b^4+c^4\ge 16S\ .$

P17 (Adil Abdullayev, Azerbaijan) $\boxed{\ \mathrm{Let}\ \triangle ABC\ \mathrm{with\ the\ orthocenter}\ H\ \mathrm{and\ the\ incircle}\ w=\mathbb C(I,r).\ \mathrm{Prove\ that\ the\ equivalence\ :}\ H\in w\iff\tan A+\tan B+\tan C=\frac {2s}r\ }$

Proof. I'll apply the well known identities $\odot\begin{array}{cccccc} \nearrow & 2R^2\cdot \prod \sin A & = & S & (1) & \searrow\\\\ \rightarrow & 1+\prod\cos A & = & \frac {a^2+b^2+c^2}{8R^2} & (2) & \rightarrow\\\\ \searrow & 4R^2+4Rr+3r^2-s^2 & = & HI^2 & (3) & \nearrow\end{array}\odot$ Indeed, $\underline{\underline{H\in w}}\iff$ $HI=r\iff$ $HI^2=r^2\ \stackrel{(3)}{\iff}\ 4R^2+$ $4Rr+3r^2-s^2=r^2\iff$

$\underline{\underline{s^2=2\left(2R^2+2Rr+r^2\right)}}\ (4)\ .$ On other hand $\sum\tan A=\prod\tan A=\frac {\prod\sin A}{\prod\cos A}\ \stackrel{(1\wedge 2)}{=}\ \frac {\frac S{2R^2}}{\frac {a^2+b^2+c^2}{8R^2}-1}=$ $\frac {4S}{a^2+b^2+c^2-8R^2}=$ $\frac {4sr}{2\left(s^2-r^2-4Rr\right)-8R^2}\ \stackrel{(4)}{=}\ \frac {2sr}{s^2-r^2-4Rr-4R^2}=$

$\frac {2sr}{\cancel{4R^2}+\cancel{4Rr}+2r^2-r^2-\cancel{4Rr}-\cancel{4R^2}}=$ $\frac {2sr}{r^2}=\frac {2s}r\ .$ In conclusion, $\underline{\underline{\tan A+\tan B+\tan C=\frac {2s}r}}\iff$ $\underline{\underline{s^2=2\left(2R^2+2Rr+r^2\right)}}\ .$ Hence $\boxed{H\in w\iff \tan A+\tan B+\tan C=\frac {2s}r}\ .$

Remarks..

$\boxed{(1)}\blacktriangleright\ 4RS=abc=\prod (2R\sin A)=8R^3\prod \sin A\implies$ $4RS=8R^3\prod\sin A\iff$ $S=\frac {8R^3\prod\sin A}{4R}\iff$ $\boxed{\ 2R^2\cdot\prod\sin A=S\ }\ (1)\ .$

$\boxed{(2)}\blacktriangleright\ 4\cdot\prod\cos A=2\cos A\cdot 2\cos B\cos C=$ $2\cos A\left[\cos (B+C)+\cos (B-C)\right]=$ $2\cos A[\cos (B-C)-\cos A]=$ $-2\cos^2A-2\cos (B+C)\cos (B-C)]=$

$-2\cos^2A-\cos 2B-\cos 2C=$ $-2\left(1-\sin^2A\right)-\left(1-2\sin^2B\right)-\left(1-2\sin^2C\right)=$ $-4+2\left(\sin^2A+\sin^2B+\sin^2C\right)=$ $-4+2\cdot \sum\left(\frac a{2R}\right)^2=$ $-4+\frac {\sum a^2}{2R^2}\implies$

$4\cdot\prod\cos A=-4+\frac {\sum a^2}{2R^2}\iff$ $4\cdot \left(1+\prod\cos A\right)=\frac 1{2R^2}\cdot\sum a^2\iff$ $\boxed{\ 1+\prod\cos A=\frac {a^2+b^2+c^2}{8R^2}\ }\ (2)\ ,$ where $\boxed{\ a^2+b^2+c^2=2\left(s^2-r^2-4Rr\right)\ }\ .$

$\boxed{(3)}\blacktriangleright$ I"ll apply the distancies between two remarkable points from here: [1] $\wedge$ [2] $\wedge$ [3] and Stewart's relation to $IG/\triangle HIO\ :\ IH^2\cdot GO+IO^2\cdot GH=IG^2\cdot HO+HO\cdot GO\cdot GH\iff$

$3\cdot IH^2+6\cdot IO^2=9\cdot IG^2+2\cdot HO^2\iff$ $3\cdot IH^2+6\left(R^2-2Rr\right)=$ $\left(s^2+5r^2-16Rr\right)+$ $2\cdot 9\cdot\left[R^2-\frac {2\left(s^2-r^2-4Rr\right)}9\right]\iff$ $3\cdot IH^2+6R^2-12Rr=$ $s^2+5r^2-\cancel{16Rr}+$

$18R^2-$ $4\left(s^2-r^2-\cancel{4Rr}\right)\iff$ $3IH^2+6R^2-12Rr=$ $-3s^2+9r^2+18R^2\iff$ $3\cdot IH^2=12R^2+12Rr+9r^2-3s^2\iff$ $\boxed{\ IH^2=4R^2+4Rr+3r^2-s^2\ }\ (3)\ .$

P18 (Mehmet SAHIN, Turkey) (<= click).

Proof. $[BI_aC]=\frac 12\cdot BC\cdot I_aE=\frac {ar_a}2\iff$ $\boxed{\ [BI_aC]=\frac {ar_a}2\ }\ (1)\ .$ Find analogously $[CI_bA]=\frac {br_b}2$ and $[AI_cB]=\frac {cr_c}2\ .$ On other hand $[DEF]=[DI_aE]+[FI_aE]-[DI_aF]=$

$\frac 12\cdot \left(r^2\sin B+r_a^2\sin C-r_a^2\sin A\right)=$ $\frac {r_a^2}2\cdot\frac {b+c-a}{2R}=$ $\frac {2r_a^2(s-a)}{4R}\implies$ $\boxed{\ [DEF]=\frac {r_a^2(s-a)}{2R}\ }\ (2)\ .$ Find analogously $[LKM]=\frac {r_b^2(s-b)}{2R}$ and $[PNR]=\frac {r_c^2(s-c)}{2R}\ .$ Therefore,

$\frac {[BI_aC]}{[DEF]}\ \stackrel{(1\wedge 2)}{=}\ \frac {a\cancel{r_a}}{\cancel 2}\cdot\frac {\cancel 2R}{r_a^{\cancel 2}(s-a)}=\frac {aR}{r_a(s-a)}=\frac {aR}S\implies$ $\boxed{\ \frac {[BI_aC]}{[DEF]}=\frac {aR}S\ }\ (3)\ .$ In conclusion, $\sum\frac {[BI_aC]}{[DEF]}\ \stackrel{(3)}{=}\ \sum\frac {aR}S=\frac RS\cdot \sum a=\frac R{sr}\cdot 2s=\frac {2R}r\implies$ $\sum\frac {[BI_aC]}{[DEF]}=\frac {2R}r\ .$

P19 (sqing). $\boxed{\ \mathrm{Prove\ that }\ (\forall )\ \{x,a,b,c\}\subset\mathbb R^*_+\ \mathrm{there\ is\ the\ inequality}\ (x+a+b)(x+b+c)(x+c+a)\ge (x+2a)(x+2b)(x+2c)\ }$

Proof. Denote $P(x) =(x+a+b)(x+b+c)(x+c+a)- (x+2a)(x+2b)(x+2c)\ .$ Observe that $P(x)=x\cdot\sum\left(a^2-bc\right)+\prod (b+c)-8abc\ge 0$ for any $x\ge 0\ .$

P20 (M.O. Sanchez) - Olimpiada de Centre America y el Caribe. Let a line $d ,$ the points $\{A,B,C,D\}\subset d$ in this order and the equilateral triangles $APB\ ,$ $BQC$ and $CRD$ so that the line

$d$ doesn't separate the points $P\ ,$ $Q$ and $R\ .$ Denote $AB=a\ ,$ $BC=b$ and $CD=c\ .$ Prove that $\frac 1a+\frac 1c=\frac 1b$ and the quadrilateral $PBCR$ is circumscriptible, i.e. $PB+RC=PR+BC\ .$

Proof. Let $\left\{\begin{array}{ccc} m\left(\widehat{PQB}\right) & = & x\\\ m\left(\widehat{RQC}\right) & = & y\end{array}\right\|$ where $\boxed{\ x+y =180^{\circ}\ }\ (*)$ and apply theorem of Sines in $:\ \left\{\begin{array}{cc} \triangle PQB\ : & \frac ba=\frac {\sin (60+x)}{\sin x}\\\\ \triangle RQC\ : & \frac bc=\frac {\sin (60+y)}{\sin y}\end{array}\right\|$ $\implies$ $\frac ba+\frac bc=$ $\frac {\sin (60+x)}{\sin x}+\frac {\sin (60+y)}{\sin y}=$

$\frac {\sin 60\cos x+\sin x\cos 60}{\sin x}+\frac {\sin 60\cos y+\cos 60\sin y}{\sin y}=$ $\left(\frac {\sqrt 3}{2}\cdot\cot x+\frac 12\right)+\left(\frac {\sqrt 3}{2}\cdot\cot y+\frac 12\right)=$ $\frac {\sqrt 3}{2}\cdot \left(\cot x+\cot y\right)+1=1\implies$ $\frac ba+\frac bc=1\implies\ ,$ i.e. $\boxed{\ b(a+c)=ac\ }\ (1)\ .$

Let $S$ so that $\triangle BSC$ is equilateral and the line $d$ separates $Q$ and $S\ .$ Observe that $B\in (PS)$ and $C\in (RS)\ .$ Thus, $PR^2=PS^2+RS^2-PS\cdot RS=(a+b)^2+(c+b)^2-(a+b)(c+b)=$

$\left(b^2+a^2+2ba\right)+\left(\cancel{b^2}+c^2+2bc\right)-\left[\cancel{b^2}+b(a+c)+ac\right]=$ $a^2+b^2+c^2+b(a+c)-ac\ \stackrel{(*)}{=}\ a^2+b^2+c^2\implies$ $\boxed{\ PR^2=a^2+b^2+c^2\ }\ (2)\ .$ Thus, $PBCR$ is circumscriptible $\iff$

$PR+b=a+c\iff$ $PR=a+c-b\iff$ $PR^2=(a+c-b)^2\ \stackrel{(2)}{\iff}\ a^2+b^2+c^2=a^2+b^2+c^2-2b(a+c)+2ac\iff b(a+c)\ \stackrel{(1)}{=}\ ac\ ,$ what is true. Very nice problem!

P21 (M.O. Sanchez). $A$ -isosceles $\triangle ABC$ with $D\in (BC)$ and $E\in (AC)$ so that $m\left(\widehat{ACB}\right)=20^{\circ}$ and $m\left(\widehat{ABE}\right)=m\left(\widehat{CBE}\right)= m\left(\widehat{BAD}\right)=10^{\circ}.$ Prove that $BE^2+CD^2=(b+c)^2\ .$

Proof. The theorem of Sines in $\triangle\ BEC\ :\ \frac {BE}{BC}=\frac {\sin 20^{\circ}}{\sin 150^{\circ}}=$ $\frac {\sin 20^{\circ}}{\sin 30^{\circ}}\implies$ $\boxed{\ BE=2a\cdot\sin 20^{\circ}\ }\ .$ The theorem of Sines in $ADC$ and $ABC\ :\ \frac {DC}{BC}=$ $\frac {DC}{AC}\cdot\frac {AC}{BC}=$ $\frac {\sin 130^{\circ}}{\sin 30^{\circ}}\cdot \frac {\sin 20^{\circ}}{\sin 140^{\circ}}=$

$2\sin 50^{\circ}\cdot \frac {\sin 20^{\circ}}{\sin 40^{\circ}}=$ $\frac {\cos 40^{\circ}}{\cos 20^{\circ}}\implies$ $\boxed{\ DC=\frac {a\cos 40}{\cos 20}\ }\ .$ Hence $BE^2+DC^2=a^2\cdot\left(4\sin ^2 20+\frac {\cos ^240}{\cos ^2 20}\right)=$ $a^2\cdot\left(\frac {\sin ^240+\cos^240}{\cos ^2 20}\right)=$ $\frac {a^2}{\cos^220}=$ $\left(\frac a{\cos 20}\right)^2=(2b)^2=(b+c)^2\ .$

P22 (Tran Quang Hung, Vietnam).

Proof. Observe that $\frac 1c=\frac 1a+\frac 1b\iff$ $\boxed{\ bc=ab-ac\ }\ (*)\ .$ Define $\ :\ \left\{\begin{array}{ccccc} K\in EF\cap AB & \mathrm{and} & m\left(\widehat{BEK}\right)=x & \implies & \tan x=\frac {KB}{KE}=\frac {a-c}a\\\\ L\in EF\cap IJ & \mathrm{and} & m\left(\widehat{IEL}\right)=y & \implies & \tan y=\frac {LI}{LE}=\frac {b-c}{b+c}\end{array}\right\|\implies$ $\tan (x+y)=$ $\frac {\tan x+\tan y}{1-\tan x\tan y}=$

$\frac {\frac {a-c}a+\frac {b-c}{b+c}}{1-\frac {a-c}a\cdot \frac {b-c}{b+c}}=\frac {(a-c)(b+c)+a(b-c)}{a(b+c)-(a-c)(b-c)}=$ $\frac {2ab-bc-c^2}{2ac+bc-c^2}\ \stackrel{(*)}{=}\ \frac {2ab-(ab-ac)-c^2}{2ac+(ab-ac)-c^2}=$ $\frac {ab+ac-c^2}{ac+ab-c^2}=1\implies$ $x+y=45^{\circ}\implies$ $m\left(\widehat{BEI}\right)= 135^{\circ}$ (Dũng Nguyễn Tiến's proof).

P23 (Mehmet Şahin). The ncircle $w=\mathbb C(I,r)$ of $\triangle ABC$ touches $[BC]\ ,$ $[CA]\ ,$ $[AB]$ at $D\ ,$ $E\ ,$ $F\ .$ Prove that $\boxed{\ DE+EF+FD\le 3r\sqrt 3\le s\sqrt{\frac {2r}R}\le s\le\frac {3R\sqrt 3}2\ }$ (standard notations).

Proof. $EF=IA\cdot \sin A\iff EF^2=\frac {bc(s-a)}s\cdot \left(\frac {a}{2R}\right)^2=$ $\frac{abc}{4sR^2}\cdot a(s-a)=$ $\frac rR\cdot a(s-a)\implies$ $\sum EF^2=\frac {r}R\cdot \sum a(s-a)=\frac rR\cdot 2r(4R+r)\implies$ $\boxed{\ \sum EF^2=\frac {2r^2}R\cdot (4R+r)\ }\ .$

In conclusion, $\left(\sum EF\right)^2\le 3\cdot \sum EF^2=$ $\frac {2r}R\cdot 3r(4R+r)\ \stackrel{(*)}{\le}\ \frac {2r}R\cdot s^2\implies$ $DE+EF+FD\le s\cdot\sqrt {\frac {2r}R}\le s\le\frac {3R\sqrt 3}2\ .$ I used the well known inequality $\boxed{\ s^2\ge 3r(4R+r)\ }\ (*)\ .$ Thus,

$EF=2r\cos\frac A2$ a.s.o. $\implies$ $\sum EF=2r\cdot\sum\cos\frac A2\le 2r\cdot 3\cos 30^{\circ}=3r\sqrt 3\implies$ $\boxed{\sum EF\le 3r\sqrt 3}\ .$ Prove easily $\boxed{3r\sqrt 3\le s\sqrt{\frac {2r}R}}$ using only the well known inequality $s^2\ge 16Rr-5r^2\ .$

• P24 <= click => Proof (Daniel DAN) •

P25 (Miquel Ochoa SANCHEZ) <= click.

Proof. Denote the midpoint $M$ of $[DE]\ .$ Observe that $\widehat{BAH}\equiv\widehat{ACH}\iff \triangle HBA\sim\triangle HAC\iff$ $\frac {HB}{HA}=\frac cb=\frac {HA}{HC}\implies$ $HA^2=HB\cdot HC\iff$ $b^2-HC^2=HB\cdot HC\iff$

$b^2=HC(HB+HC)\iff$ $\boxed{\ b^2=HC\cdot (a+2\cdot HB)\ }\ (*)\ .$ Apply generalized Phytagoras' theorem in $\triangle ABC$ to $:\ \left\{\begin{array}{c} AC=b\ :\ b^2=a^2+c^2+2a\cdot HB\iff HB=\frac {b^2-a^2-c^2}{2a}\\\\ AB=c\ :\ c^2=a^2+b^2-2a\cdot HC\iff HC=\frac {a^2+b^2-c^2}{2a}\end{array}\right\|\ \stackrel{(*)}{\implies}$

$\ b^2=\frac {a^2+b^2-c^2}{2a}\cdot \left(a+\frac {b^2-a^2-c^2}a\right)\iff$ $2a^2b^2=\left(a^2+b^2-c^2\right)\left(b^2-c^2\right)\iff \cancel 2a^2b^2=a^2\left(\cancel{b^2}-c^2\right)+\left(b^2-c^2\right)^2\iff$ $\boxed{\ a^2\left(b^2+c^2\right)=\left(b^2-c^2\right)^2\ }\ .$

P26 (Adil Abdullayev, BAKU) <= click.

Proof. I"ll apply the identity $\boxed{\ bc=2Rh_a\ }$ a.s.o. and the well known inequality $\boxed{\ m_a^2\ge s(s-a)\ }\ (*)$ a.s.o. Thus, $\frac {m_a}{h_a}\ \stackrel{(*)}{\ge}\ 2R\cdot \frac {\sqrt{s(s-a)}}{bc}$ and $\prod\frac {m_a}{h_a}\ge 8R^3\cdot \frac {\sqrt{s^3(s-a)(s-b)(s-c)}}{(abc)^2}=$

$\frac {8sR\cancel{^3}\cancel S}{16\cancel{R^2}S\cancel{^2}}=\frac {sR}{2S}=\frac R{2r}\implies$ $\prod\frac {m_a}{h_a}\ge \frac R{2r}\implies$ $\boxed{\ 4\left(\prod\frac {m_a}{h_a}-1\right)\ge 4\left(\frac R{2r}-1\right)\ }\ (1)\ .$ Remain to prove $\boxed{\ 4\left(\frac R{2r}-1\right)\ge \frac sr-3\sqrt 3\ }\ (2)\ .$ Indeed, $\underline{\underline{4\left(\frac R{2r}-1\right)\ge\frac sr-3\sqrt 3}}\iff$

$\frac {2R}r\ge \frac sr+4-3\sqrt 3\iff$ $s\le 2R+\left(3\sqrt 3-4\right)r\iff$ $\underline{\underline{s^2\le \left[2R+\left(3\sqrt 3-4\right)r\right]^2}}\ .$ I"ll apply now the well known inequality $\boxed{ s^2\le 4R^2+4Rr+3r^2\ }\ (3)\ .$ Remain to prove only

$4R^2+4Rr+3r^2\le \left[2R+\left(3\sqrt 3-4\right)r\right]^2\ ,$ i.e. $\cancel{4R^2}+4Rr+3r^2\le \cancel{4R^2}+4Rr\left(3\sqrt 3-4\right)+\left(3\sqrt 3-4\right)^2r^2\iff$ $4R+3r\le 4R\left(3\sqrt 3-4\right)+\left(43-24\sqrt 3\right)r\iff$

$4R\le 4R\left(3\sqrt 3-4\right)+\left(40-24\sqrt 3\right)r\iff$ $R\le R\left(3\sqrt 3-4\right)+\left(10-6\sqrt 3\right)r\iff$ $0\le R\cancel{\left(3\sqrt 3-5\right)}-2\cancel{\left(3\sqrt 3-5\right)}r\iff$ $R\ge 0\ ,$ what is true (Gabi Cuc Cucoanes' proof).

P27 (own). $\boxed{\ \mathrm{Prove\ that}\ \forall\ \mathrm{a\ nonobtuse\ \triangle\ ABC\ there\ is\ the\ inequality}\ \sum am_a\le (R+r)^2\cdot\sum \frac a{m_a}\ \mathrm{(standard\ notations).}\ }$

Proof. Let $M$ be the midpoint of $[BC]$ and the circumcircle $w=\mathbb C(O,R)\ .$ Thus, $AM\le OA+OM\iff m_a\le R(1+\cos A)\iff$ $am_a\le R(a+a\cdot \cos A)\implies$

$\sum am_a\le R\cdot\left(2s+\sum a\cos A\right)=$ $2sR+2R^2\sum\sin A\cos A=$ $2sR+R^2\sum \sin 2A=$ $2sR+4R^2\prod\sin A=$ $2sR+2S=2sR+2sr=2s(R+r)\implies$

$\boxed{\ am_a+bm_b+cm_c\le 2s(R+r)\ }\ (1)\ .$ On other hand $\sum\frac a{m_a}=\sum\frac {a^2}{am_a}\ge \frac {(a+b+c)^2}{am_a+bm_b+cm_c}\ \stackrel{(1)}{\ge}\ \frac {4s^2}{2s(R+r)}=\frac {2s}{R+r}\implies$ $\boxed{\ (R+r)\sum\frac a{m_a}\ge a+b+c\ }\ (2)\ .$

In conclusion, $\left\{\begin{array}{cccc} (1) & am_a+bm_b+cm_c & \le & (R+r)(a+b+c)\\\\ (2) & a+b+c & \le & (R+r)\left(\frac a{m_a}+\frac b{m_b}+\frac c{m_c}\right)\end{array}\right\|\ \bigodot\ \implies$ $\boxed{\ am_a+bm_b+cm_c\le (R+r)^2\cdot \left(\frac a{m_a}+\frac b{m_b}+\frac c{m_c}\right)\ }\ (3)\ .$

P28 (Cristian TELLO) <= click.

Proof. Observe that $BH=2R\ ,\ BE=BF=EF=R\sqrt 3\ ,\ PH^2=4R^2-PB^2$ and the required relation $PB^2+3PH^2=2\left(PE^2+PF^2\right)$ is equivalent with the relation

$PB^2+3\cdot\left(4R^2-PB^2\right)=2\left(PE^2+PF^2\right)\iff$ $\boxed{\ PB^2+PE^2+PF^2=6R^2\ }\ (*)\ .$ The generalized Phytagoras' theorem in $\triangle PBF\ :\ 3R^2=BF^2=PB^2+PF^2+PB\cdot PF$

implies $\boxed{\ PB^2+PF^2+PB\cdot PF=3R^2\ }\ (1)\ .$ Using the Ptolemy's theorem to the quadrilateral $PBEF$ obtain that $\boxed{\ PE=PB+PF\ }\ (1)\ .$ Hence the relation $(*)$ becomes

$PB^2+\left(PB+PF\right)^2+PF^2=6R^2\ ,$ i.e. $PB^2+PF^2+PB\cdot PF=3R^2\ ,$ what is the true relation $(1)\ .$ .