420. Some geometry problems for high school II.

by Virgil Nicula, Mar 28, 2015, 4:35 PM

P1 (Miguel Ochoa Sanchez, Peru). Let an isosceles $\triangle ABC$ with $B=C$ . Its incircle $w$ touches $AB$ , $AC$ in $T$ , $S$ respectively. Choose an arbitrary

point $P$ which belongs to the minor arc $\overarc[]{TS}$ of $w$ . The tangent at $w$ in $P$ cuts $AB$ , $AC$ In $D$ , $E$ respectively. Prove that $\frac {DA}{DB}+\frac {EA}{EC}=\frac {1-\cos B}{\cos B}$ (constant).

Proof. Denote $N\in BE\cap CD$ and $R\in AN\cap BC$ . Apply the Newton's theorem to $BDEC$ , i.e. $N\in TS$ . Now I"ll use the van Aubel's

theorem to $N$ in $\triangle ABC\ :\ \frac {DA}{DC}+\frac {EA}{EC}=$ $\frac {NA}{NR}\ \stackrel{(TS\parallel BC)}{\implies}$ $\frac {NA}{NR}=\frac {TA}{TB}\implies$ $\boxed{\frac {DA}{DC}+\frac {EA}{EC}=\frac {TA}{TB}}\ (*)$ . Thus, $\frac {TA}{TB}+1=\frac {AB}{TB}=$

$\frac {AB}{BM}=\frac 1{\cos B}\implies$ $\frac {TA}{TB}=\frac 1{\cos B}-1=$ $\frac {1-\cos B}{\cos B}\implies$ $\frac {TA}{TB}=\frac {1-\cos B}{\cos B}\ \stackrel{(*)}{\implies}\ \frac {DA}{DB}+$ $\frac {EA}{EC}=\frac {1-\cos B}{\cos B}$ (constant).

Generalization. Let $\triangle ABC$ with the incircle $w$ and two points $M\in (AB)$ , $N\in (AC)$ . Prove that $MN$ is tangent to $w\implies$ $(s-b)\cdot \frac {MA}{MB}+(s-c)\cdot\frac {NA}{NC}=(s-a)$ (constant).

Proof 1. Let: $w=C(I,r)$ , $T\in MN\cap w$ ; $AM=x$ , $AN=y$ ; $E\in (AC)\cap w$ , $F\in AB\cap w$ . Therefore, $(s-b)\cdot \frac {MA}{MB}+(s-c)\cdot\frac {NA}{NC}=(s-a)\iff$

$\frac {x(s-b)}{c-x}+\frac {y(s-c)}{b-y}=(s-a)\iff$ $x(s-b)(b-y)+y(s-c)(c-x)=(s-a)(c-x)(b-y)\iff$ $b(s-b)x+c(s-c)y-axy=$

$bc(s-a)-c(s-a)y-b(s-a)x+(s-a)xy\iff$ $bc(x+y)=bc(s-a)+sxy$ . In conclusion, $(s-b)\cdot \frac {MA}{MB}+(s-c)\cdot\frac {NA}{NC}=(s-a)\iff$

$x+y=(s-a)+\frac {sxy}{bc}\ (*)$ . Observe that $MN=PM+PN=MF+NE=$ $(s-a-x)+(s-a-y)\implies$ $\boxed{MN=2(s-a)-(x+y)}\ (1)$ . Hence

$[MAN]=[AEIF]-[IFMNE]=$ $2[IEA]-2[MIN]=$ $r(s-a)-r\cdot MN\ \stackrel{(1)}{=}\ r(s-a)-$ $r[2(s-a)-(x+y)]=$ $r[(x+y)-(s-a)]\implies$

$[MAN]=r[(x+y)-(s-a)]\ (2)\implies$ $\frac {[MAN]}{[ABC]}=\frac {[MAN]}{S}=\frac {xy}{bc}\implies$ $[MAN]=\frac {xyrs}{bc}\stackrel{(2)}{\implies}\frac {xyrs}{bc}=$ $r[(x+y)-(s-a)]\implies$ $x+y=(s-a)+\frac {sxy}{bc}$ , i.e. $(*)$ .

Proof 2. Let the incircle $w=C(I,r)$ , $T\in MN\cap w$ and $\left\{\begin{array}{cccc} TM=u\ ; & m\left(\widehat{TIM}\right)=\alpha & \implies & \tan\alpha =\frac ur\\\\ TN=v\ ; & m\left(\widehat{TIN}\right)=\beta & \implies & \tan\beta =\frac vr\end{array}\right\|$ .

Thus, $\boxed{\alpha +\beta =90^{\circ}-\frac A2}$ and $\cot\frac A2=\tan\left(90^{\circ}-\frac A2\right)=\tan\left(\alpha +\beta\right)$ . Hence:

$\blacktriangleright\ \tan (\alpha +\beta )=\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=$ $\frac {\frac ur+\frac vr}{1-\frac {uv}{r^2}}\iff$ $\frac {s-a}r=\frac {r(u+v)}{r^2-uv}\iff$ $\frac {r^2(u+v)}{s-a}+uv=r^2\iff$

$\frac {sr^2(u+v)}{s-a}+suv=sr^2\iff$ $\boxed{(s-b)(s-c)(u+v)+suv=(s-a)(s-b)(s-c)}$ .

$\blacktriangleright\ (s-b)+(s-c)=a$ and $(s-b)\cdot \frac {MA}{MB}+(s-c)\cdot\frac {NA}{NC}=(s-a)\implies$ $(s-b)\cdot\left(1+\frac {MA}{MB}\right)+(s-c)\cdot\left(1+\frac {NA}{NC}\right)=(s-a)+a\iff$

$(s-b)\cdot \frac {AB}{MB}+(s-c)\cdot \frac {AC}{NC}=s\iff$ $\frac {c(s-b)}{s-b+u}+\frac {b(s-c)}{s-c+v}=s\iff$ $c(s-b)(s-c+v)+b(s-c)(s-b+u)=$ $s(s-b+u)(s-c+v)\iff$

$c(s-b)(s-c)+c(s-b)v+b(s-b)(s-c)+b(s-c)u=$ $s(s-b)(s-c)+s(s-b)v+s(s-c)u+suv\iff$ $(b+c-s)(s-b)(s-c)=$

$(s-b)(s-c)v+(s-b)(s-c)u+suv\iff$ $\boxed{(s-b)(s-c)(u+v)+suv=(s-a)(s-b)(s-c)}$

Particular cases.

$1\blacktriangleright$ If $\triangle ABC$ is equilateral, then $\boxed{\frac {MA}{MB}+\frac {NA}{NC}=1}$ .

$2\blacktriangleright$ If $\triangle ABC$ is $A$ -isosceles, i.e. $b=c$ , then $\boxed{\frac {MA}{MB}+\frac {NA}{NC}=\frac {s-a}{s-b}}$ . So $\frac {s-a}{s-b}=\frac {b+c-a}{a-b+c}=$ $\frac {2b-a}{a}=\frac {b}{\frac a2}-1=$ $\frac 1{\cos B}-1\implies$ $\boxed{\frac {MA}{MB}+\frac {NA}{NC}=\frac {1-\cos B}{\cos B}}$ .

P2. Let $\triangle ABC$ with incircle $w=C(I,r)$ and $D\in (BC)$ . Let incenters $E$ , $F$ of $\triangle ADB$ , $\triangle ADC$ . Prove that $BEFC$ is cyclic $\iff \frac {AD+DB}{AD+DC}=\frac {AB}{AC}$ .

Proof. $I\in BE\cap CF$ and the projections $(X,K,Y)$ of $(E,I,F)\implies$

$\odot\begin{array}{ccccc} \nearrow & KX=KB-XB=(s-b)-\frac 12\cdot (c+DB-AD) & \implies & KX=a-b+AD-DB=AD+DC-b & \searrow\\\\ \searrow & KY=KC-YC=(s-c)-\frac 12\cdot (b+DC-AD) & \implies & KY=a-c+AD-DC=AD+DB-c & \nearrow\end{array}\odot$ . Thus, $\left\{\begin{array}{c} KX=IE\cdot\cos\frac B2\\\\ KY=IF\cdot\cos\frac C2\end{array}\right\|\implies$

$\frac {IE}{IF}=$ $\frac {KX}{KY}\cdot\frac {\cos \frac C2}{\cos\frac B2}=$ $\frac {AD+DC-b}{AD+DB-c}\cdot\sqrt{\frac {c(s-c)}{b(s-b)}}$ . So $BEFC$ is cyclic $\iff IE\cdot IB=IF\cdot IC\iff$ $\frac {IE}{IF}=\frac {IC}{IB}\iff$ $\frac {AD+DC-b}{AD+DB-c}\cdot\sqrt{\frac {c(s-c)}{b(s-b)}}=$

$\sqrt{\frac {b(s-c)}{c(s-b)}}\iff$ $\frac {AD+DC-b}{AD+DB-c}=\frac bc\iff$ $\frac {AD+DC-b}{b}=\frac {AD+DB-c}c\iff$ $\frac {AD+DC}{b}-1=\frac {AD+DB}c-1\iff$ $\frac {AD+DB}{AD+DC}=\frac {AB}{AC}$ .

Generalization. Let $\triangle ABC$ , its interior $P$ , $\left\{\begin{array}{c} E\in (PB)\\\\ F\in (PC)\end{array}\right\|$ and $\left\{\begin{array}{ccc} R\in (BC) & , & PR\perp BC\\\\ X\in BC & , & EX\perp BC\\\\ Y\in BC & , & FY\perp BC\end{array}\right\|$ . Prove that $BEFC$ is cyclic $\iff \frac {RB}{RX}=\frac {RC}{RY}\cdot\left(\frac {PB}{PC}\right)^2$ .

Proof. $BEFC$ is cyclically $\iff PE\cdot PB=PF\cdot PC\iff$ $\frac {PE}{PB}\cdot PB^2=\frac {PF}{PC}\cdot PC^2\iff$ $\frac {RX}{RB}\cdot PB^2=\frac {RY}{RC}\cdot PC^2\iff$ $\boxed{\frac {RB}{RX}=\frac {RC}{RY}\cdot\left(\frac {PB}{PC}\right)^2}\ (*)$ .

Particular case. If $P$ is the incenter of $\triangle ABC$ and $E$ , $F$ are the incenters of $\triangle ABD$ , $\triangle ACD$ respectively , where $D$ ia a mobile point of $(BC)$

obtain that $\left\{\begin{array}{ccccc} 2\cdot RB=a+c-b & ; & 2\cdot BX=c+DB-DA & \implies & RX=RB-XB=\frac 12\cdot (DC+DA-b)\\\\ 2\cdot RC=a+b-c & ; & 2\cdot CY=b+DB-DA & \implies & RY=RC-YC=\frac 12\cdot (DB+DA-c)\end{array}\right\|\stackrel{(*)}{\implies}$

$\frac {s-b}{DC+DA-b}=\frac {s-c}{DB+DA-c}\cdot\frac {c(s-b)}{b(s-c)}\iff$ $\frac {DB+DA-c}{DC+DA-b}=\frac cb=\frac {DB+DA}{DC+DA}\implies$ $\frac {DA+DB}{DA+DC}=\frac cb$ (constant).

P3. Let an $A$ -isosceles $\triangle ABC$ and $\left\{\begin{array}{c} D\in (BC)\\\ E\in (CA)\\\ F\in (AB)\end{array}\right\|$ so that $\left\{\begin{array}{c} ED=EA\\\ FD=FA\\\ EF\perp AD\end{array}\right\|$ . i.e. $AEDF$ is a deltoid.

Denote $AB=l$ , $DB=m$ and $DC=n$ . Prove that $EF=\frac {\sin A\left(l^2-mn\right)\sqrt{l^2-mn}}{2\left(l-m\sin\frac A2\right)\left(l-n\sin\frac A2\right)}$ .

Proof. Denote: the midpoint $M$ of $[AD]$ , i.e. $M\in EF$ ; the projections $U$ , $V$ of $D$ on $AB$ , $AC$ respectively. Observe that $\left\{\begin{array}{ccc} AU & = & l-m\sin\frac A2\\\\ AV & = & l-n\sin\frac A2\end{array}\right\|$ and prove easily that (Stewart)

$AD^2=l^2-mn$ . Thus, $EF=ME+MF=$ $AM\cdot\left(\tan\widehat{DAB}+\tan\widehat{DAC}\right)=$ $\frac {AM\cdot \sin A}{\cos\widehat{DAB}\cos\widehat{DAC}}=$ $\frac {AD\cdot\sin A}{2\cdot \frac{AU}{AD}\cdot\frac {AV}{AD}}\implies$ $EF=\frac {\sin A\left(l^2-mn\right)\sqrt{l^2-mn}}{2\left(l-m\sin \frac A2\right)\left(l-n\sin \frac A2\right)}$ .

Particular case. $A=60^{\circ}\ ,\ l=30\ ,\ m=6\ ,\ n=24\implies$ $EF=\frac {\frac {\sqrt 3}2\cdot \left(30^2-6\cdot 24\right)\cdot\sqrt{30^2-6\cdot 24}}{2\cdot (30-3)(30-12)}=$ $\frac {\frac {\sqrt 3}2\cdot 756\cdot\sqrt{756}}{2\cdot 27\cdot 18}=$ $\frac {\frac {\sqrt 3}2\cdot 7\cdot 9\cdot 12\cdot\sqrt{7\cdot 9\cdot 12}}{2\cdot 27\cdot 18}\implies EF=7\sqrt 7$ .

Lemma. Let $\triangle ABC$ and the points $\left\{\begin{array}{ccc} E\in (AC) & ; & F\in (AB)\\\\ P\in (EF) & ; & D\in AP\cap BC\end{array}\right\|$ . Prove that $\frac {FB}{FA}\cdot DC+\frac {EC}{EA}\cdot DB=\frac {PD}{PA}\cdot BC$ .

P4. Let $\triangle ABC$ with the incircle $w=C(I,r)$ and the Gergonne's point $\Gamma$ . Denote $P\in I\Gamma\cap BC$ . Prove that $\frac {PB}{PC}=\frac {(a-c)(s-b)^2}{(a-b)(s-c)^2}$ .

Proof. Denote $\left\{\begin{array}{ccc} D\in BC\cap w & ; & E\in CA\cap w\\\\ F\in AB\cap w & ; & S\in BI\cap AC\end{array}\right\|$ . Is well-known that $\Gamma\in AD\cap BE\cap CF$ , where $\left\{\begin{array}{c} AE=AF=s-a\\\\ BF=BD=s-b\\\\ CD=CE=s-c\end{array}\right\|$ . Therefore, that $\frac {SA}c=\frac {SC}a=\frac b{a+c}\implies$

$ES=AS-AE=$ $\frac {bc}{a+c}-(s-a)\implies$ $\boxed{ES=\frac {(a-c)(s-b)}{a+c}}$ . From the van Aubel's relation obtain that $\left\{\begin{array}{ccc} \frac {IB}{IS} & = & \frac {a+c}b\\\\ \frac {\Gamma B}{\Gamma E} & = & \frac {b(s-b)}{(s-a)(s-c)}\end{array}\right\|$ . Apply the upper lemma in $\triangle CBE\ :$

$\frac {\Gamma E}{\Gamma B}\cdot SC+\frac {PC}{PB}\cdot SE=\frac {IS}{IB}\cdot CE\iff$ $\frac {(s-a)(s-c)}{b(s-b)}\cdot \frac {ba}{a+c}+\frac {PC}{PB}\cdot\frac {(a-c)(s-b)}{a+c}=\frac b{a+c}\cdot (s-c)\iff$ $\frac {PC}{PB}=\frac {(s-c)[b(s-b)-a(s-a)]}{(a-c)(s-b)^2}=$

$\frac {(a-b)(s-c)^2}{(a-c)(s-b)^2}\implies$ $\boxed{\frac {PB}{PC}=\frac {(a-c)(s-b)^2}{(a-b)(s-c)^2}}$ . Particular case. $AB\perp AC\implies$ $\left\{\begin{array}{cccc} s-b=r_c\ ; & (a-c)(s-b) & = & b(s-a)\\\\ s-c=r_b\ ; & (a-b)(s-c) & = & c(s-a)\end{array}\right\|$ $\implies$ $\frac {PB}{PC}=\frac {b(s-b)}{c(s-c)}=\frac {br_c}{cr_b}$ .

Remark. Denote $R\in IP\cap AB$ and apply the upper lemma to the triangle $ABC\ :\ \frac {RA}{RB}\cdot SC+\frac {PC}{PB}\cdot SA=$ $\frac {IS}{IB}\cdot AC\implies\ \ldots\ \implies$ $\boxed{\frac {RA}{RB}=\frac {(c-b)(s-a)^2}{(c-a)(s-b)^2}}$ .

P5. Let $ABC$ be a triangle. Consider the points $X\in (BC)$ , $Y\in (CA)$ , $Z\in (AB)$ , $P\in AX\cap YZ$ . Prove easily that (for example, with areas)

$\boxed {\ \frac {XB}{XC}=\frac {AB}{AC}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\ (1)\ \wedge\ \ \frac {PZ}{PY}=\frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot \frac {AC}{AB}\ (2)\ \wedge\ \ \frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\ (3)\ }$ .

Proof.

$1\blacktriangleright$ Apply the theorem of Sinus in the mentioned triangles: $\left\{\begin{array}{cccc} \triangle XAB\ : & \frac {XB}{\sin\widehat{XAB}} & = & \frac {XA}{\sin B}\\\\ \triangle XAC\ : & \frac {XA}{\sin C} & = & \frac {XC}{\sin\widehat{XAC}}\end{array}\right|\ \bigodot \implies$ $\ \frac {XB}{XC}=\frac {\sin C}{\sin B}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\implies$

$\frac {XB}{XC}=\frac {AB}{AC}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\ (1)$ . Otherwise (with areas). $\frac {XB}{XC}=\frac {[XAB]}{[XAC]}=\frac {AX\cdot AB\cdot\sin \widehat{XAB}}{AX\cdot AC\cdot\sin\widehat{XAC}}\implies$ $\frac {XB}{XC}=\frac {AB}{AC}\cdot \frac {\sin \widehat{XAB}}{\sin\widehat{XAC}}\ (1)$ .

$2\blacktriangleright$ Apply the relation $(1)$ in the mentioned triangles: $\left\{\begin{array}{cc} \triangle AZY\ : & \frac {PZ}{PY}=\frac {AZ}{AY}\cdot\frac {\sin\widehat{PAZ}}{\sin\widehat{PAY}}\\\\ \triangle ABC\ : & \frac {XB}{XC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{XAB}}{\sin\widehat{XAC}}\end{array}\right|\implies$ $\frac {PZ}{PY}:\frac {XB}{XC}=\frac {AZ}{AY}:\frac {AB}{AC}\implies$ $\frac {PZ}{PY}=\frac {XB}{XC}\cdot\frac {AZ}{AY}\cdot \frac {AC}{AB}\ (2)$ .

$3\blacktriangleright$ First method. Denote $\left\{\begin{array}{c} U\in BC\cap YZ\\\ V\in YZ\ ,\ VA\parallel BC\end{array}\right\|$ . Thus, $\frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\iff$ $\frac {UX}{VA}\cdot BC=\frac {UC}{VA}\cdot BX+\frac {UB}{VA}\cdot XC\iff$

$\boxed{UX\cdot BC=UC\cdot BX+UB\cdot XC} \ (*)\ ,$ what is true because $UC\cdot BX+UB\cdot CX=$ $XB\cdot (UX+XC)+XC\cdot (UX-XB)=$

$UX\cdot (BX+CX)=UX\cdot BC$ . In conclusion, $\frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\ (3)$ .

Second method. Denote $U\in YZ\cap BC$ . Apply the Menelaus' theorem to the transversals $:$

$\left\{\begin{array}{cccccc} \overline{PZU}/\triangle ABX\ : & \frac {UB}{UX}\cdot\frac {PX}{PA}\cdot\frac {ZA}{ZB}=1 & \implies & \frac {ZB}{ZA} & = & \frac {UB}{UX}\cdot \frac {PX}{PA}\\\\ \overline{PYU}/\triangle ACX\ : & \frac {UX}{UC}\cdot\frac {YC}{YA}\cdot\frac {PA}{PX}=1 & \implies & \frac {YC}{YA} & = & \frac {UC}{UX}\cdot\frac {PX}{PA}\end{array}\right\|$ $\begin{array}{cc} \odot\ XC & \searrow\\\\ \odot\ XB & \nearrow\end{array}$ $\bigoplus$ $\implies$ $\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC=$

$\frac {PX}{PA}\cdot \frac {UB\cdot XC+UC\cdot XB}{UX}\ \stackrel{(*)}{=}\ \frac {PX}{PA}$ $\cdot \frac {BC\cdot UX}{UX}=$ $\frac {PX}{PA}\cdot BC$ $\implies$ $\frac {PX}{PA}\cdot BC=\frac {YC}{YA}\cdot BX+\frac {ZB}{ZA}\cdot XC\ (3)$ .

Remark. For third relation can use an equivalent form : $\boxed {\ \frac {AX}{AP}\cdot BC=\frac {AC}{AY}\cdot BX+\frac {AB}{AZ}\cdot XC\ }$ .

Application 1. Let $ABC$ be a triangle, let $D\in BC$ , $E\in CA$ , $F\in AB$ so that $AD\cap BE\cap CF\ne\emptyset$

and let $P\in EF$ , $Q\in FD$ , $R\in DE$ so that $DP\cap EQ\cap FR\ne\emptyset$ . Prove that $AP\cap BQ\cap CR\ne\emptyset$ .

Proof. Denote $\left\{\begin{array}{c} X\in BC\cap AP\\\ Y\in CA\cap BQ\\\ Z\in AB\cap CR\end{array}\right\|$ . Apply the Ceva's theorem to the concurrent lines $\left\{\begin{array}{ccc} AD\cap BE\cap CF\ne\emptyset & \implies & \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1\\\\ DP\cap EQ\cap FR\ne\emptyset & \implies & \frac {PE}{PF}\cdot\frac {QF}{QD}\cdot\frac {RD}{RE}=1\end{array}\right\|\ (*)$ .

Apply upper relation $(2)\ :\ \left\{\begin{array}{c} \frac {XB}{XC}=\frac {PF}{PE}\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}\\\\ \frac {YC}{YA}=\frac {QD}{QF}\cdot\frac {BD}{BF}\cdot\frac {BA}{BC}\\\\ \frac {ZA}{ZB}=\frac {RE}{RD}\cdot\frac {CE}{CD}\cdot\frac {CB}{CA}\end{array}\right\|\bigodot\ \stackrel{(*)}{\implies}\ \frac {XB}{XC}$ $\cdot\frac {YC}{YA}\cdot\frac {ZA}{ZB}=1\implies$ $AX\cap BY\cap CZ\ne\emptyset\implies$ $AP\cap BQ\cap CR\ne\emptyset$ .

P6. In acute $\triangle ABC$ , an arbitrary $P$ is chosen on altitude $AH\ ,\ H\in BC$ . Let $E$ , $F$ be the midpoints of $[CA]$ ,

$[AB]$ respectively. The perpendiculars from $E$ to $CP$ and from $F$ to $BP$ meet at $K$ . Prove that $KB = KC$ .

Proof. From $\left\{\begin{array}{ccc} KE\perp PC & \implies KP^2-KC^2=EP^2-EC^2\\\\ KF\perp BP & \implies KB^2-KP^2=FB^2-FP^2\end{array}\right\|\bigoplus$ $\implies KB^2-KC^2=\left(PE^2-PF^2\right)+\left(FB^2-EC^2\right)$ . Since $AP \perp EF\implies$ $PE^2-PF^2=AE^2-AF^2$ obtain that $KB^2-KC^2=\left(AE^2-AF^2\right)+\left(FB^2-EC^2\right)\ .$ Since $\left\{\begin{array}{ccc} EA & = & EC\\\\ FA & = & FB\end{array}\right\|$ obtain that $KB^2-KC^2=0$ , i.e. $KB=KC$ .

P7 (E. Vargas Diaz). Semicircle $w=\mathbb C(O,r)$ with diameter $[AC]$ and $B\in (AC)$ $;$ semicircle $\alpha =\mathbb C(X,x)$ with diameter $[AB]$ and semicircle $\beta =\mathbb C(Y,y)$ with diameter $[BC]$ which

are interior to $w$ $;$ the circle $\gamma=\mathbb C(Z,z)$ which is interior tangent to $w$ at $D$ and is exterior tangent to $\alpha$ , $\beta$ at $E$ , $F$ respectively. Let $T\in EF\cap AC$ . Prove that $DT$ is tangent to $\gamma$ .

Proof 1 (synthetic). Denote radical centers $\left\{\begin{array}{ccc} U & \mathrm{for\ :} & \{w ,\alpha , \gamma \}\\\\ V & \mathrm{for\ :} & \{w ,\beta , \gamma \}\\\\ W & \mathrm{for\ :} & \{\alpha , \beta , \gamma \}\end{array}\right\|$ . Thus, $\gamma$ is incircle of $\triangle UVW$ and $\left\{\begin{array}{c} UA=UE=UD\\\\ VC=VF=VD\\\\ WB=WE=WF\end{array}\right\|\ (*)$ . Let $T\in AC\cap UV\implies$

$\frac {TC}{TA}=\frac {TV}{TU}=\frac {VC}{UA}$ . Apply the Menelaus' theorem to $\{EFT\}$ and $\triangle UVW\ :\ \frac {TV}{TU}\cdot\frac {EU}{EW}\cdot\frac {FW}{FV}=\frac {VC}{UA}\cdot\frac {UE}{WE}\cdot\frac {WF}{VF}\ \stackrel{(*)}{=}\ \frac {VF}{UE}\cdot\frac {UE}{WF}\cdot\frac {WF}{VF}=1\iff T\in EF$

Remark.

$\frac {UA}{x^2}=\frac {VC}{y^2}=\frac 1{WB}=\frac {UV}{x^2+y^2}=\sqrt{\frac {x+y+z}{xyz}}=\frac {x+y}{xy}\implies$ $z=\frac {xy(x+y)}{x^2+xy+y^2}\implies \frac 1z+\frac 1{x+y}=\frac 1x+\frac 1y$ .
.

$\frac {TC}{TA}=\frac {y^2}{x^2}\implies$ $\frac {TA}{x^2}=\frac {TC}{y^2}=$ $\frac {2(x+y)}{x^2-y^2}=$ $\frac 2{x-y}\implies\odot$ $\begin{array}{ccc} \nearrow & TA=\frac {2x^2}{x-y} & \searrow\\\\ \searrow & TC=\frac {2y^2}{x-y} & \nearrow\end{array}$ $\odot\implies$ $\left\{\begin{array}{ccc} 2\cdot TO=TA+TC & \implies & TO=\frac {x^2+y^2}{x-y}\\\\ TD^2=TA\cdot TC & \implies & TD=\frac {2xy}{x-y}\end{array}\right\|$ .

$\odot\begin{array}{ccc} \nearrow & TX=TA-AX=\frac {2x^2}{x-y}-x=\frac {x(x+y)}{x-y} & \searrow\\\\ \searrow & TY=TC+BY=\frac {2y^2}{x-y}+y=\frac {y(x+y)}{x-y} & \nearrow\end{array}$ $\odot\implies \frac {TX}x=\frac {TY}y=\frac {x+y}{x-y}$ .

Proof 2 (metric). Observe that $\left\{\begin{array}{ccc} OX=y & ; & OY=x\\\\ BO=x-y & ; & x+y=r\end{array}\right\|$ and $\left\{\begin{array}{ccc} XY & = & x+y\\\\ XZ & = & x+z\\\\ YZ & = & y+z\end{array}\right\|$ . Apply the Stewart's relation to the cevian $[ZO$ and $\triangle XYZ\ :$

$XZ^2\cdot OY+YZ^2\cdot OX=$ $OZ^2\cdot XY+OX\cdot OY\cdot XY\iff$ $x(x+z)^2+y(y+z)^2=$ $(x+y)(x+y-z)^2+xy(x+y)\iff$ $z^2(x+y)+\left(x^3+y^3\right)+$

$2z\left(x^2+y^2\right)=$ $z^2(x+y)-2z(x+y)^2+$ $(x+y)^3+xy(x+y)\iff$ $4z\left(x^2+xy+y^2\right)=4xy(x+y)\iff$ $\boxed{z=\frac {xy(x+y)}{x^2+xy+y^2}}\ (*)$ , i.e. $\boxed{\frac 1z+\frac 1{x+y}=\frac 1x+\frac 1y}$ .

Apply Menelaus' theorem to $\overline {EFT}$ and $\triangle XYZ\ :\ \frac {TY}{TX}\cdot$ $\frac {EX}{EZ}\cdot\frac {FZ}{FY}=1\iff$ $\frac {TY}{TX}\cdot \frac xz\cdot\frac zy=1\iff$ $\boxed{\frac {TX}x=\frac {TY}y=\frac {x+y}{x-y}}\ (1)$ . Thus, $OT=OY+YT\ \stackrel{(1)}{=}\ x+\frac {y(x+y)}{x-y}$

Thus, $\boxed{OT=\frac {x^2+y^2}{x-y}}\ (2)$ . Apply Stewart's relation to $[ZT$ and $\triangle XYZ\ :\ ZX^2\cdot YT+ZT^2\cdot XY=ZY^2\cdot XT+YX\cdot YT\cdot XT\iff$ $(x+z)^2\cdot\frac {y(x+y)}{x-y}+$

$ZT^2\cdot (x+y)=(y+z)^2\cdot\frac {x(x+y)}{x-y}+(x+y)\cdot\frac {y(x+y)}{x-y}\cdot\frac {x(x+y)}{x-y}\iff$ $\boxed{y(x+z)^2(x-y)+(x-y)^2\cdot ZT^2=x(x-y)(y+z)^2+xy(x+y)^2}\ (3)$ . Therefore,

$TD$ is tan. to $w\iff$ $ZO\perp TD\iff$ $ZT^2-ZD^2=OT^2-OD^2\ \stackrel{(2)}{\iff}\ ZT^2-z^2=$ $\left(\frac {x^2+y^2}{x-y}\right)^2-(x+y)^2\iff$ $ZT^2(x-y)^2=z^2(x-y)^2+\left(x^2+y^2\right)^2-$

$\left(x^2-y^2\right)^2\ \stackrel{(3)}{\iff}\ z^2(x-y)^2+$ $\left(x^2+y^2\right)^2-\left(x^2-y^2\right)^2=$ $x(x-y)(y+z)^2+xy(x+y)^2-y(x+z)^2(x-y)\iff$ $z^2(x-y)^2+4x^2y^2=$ $xy^2(x-y)+$

$2xy(x-y)\cdot z+x(x-y)\cdot z^2+$ $xy(x+y)^2-x^2y(x-y)-2xy(x-y)\cdot z-y(x-y)\cdot z^2\iff$ $z^2(x-y)^2+4x^2y^2=(x-y)^2z^2-xy(x-y)^2+$

$xy(x+y)^2\iff$ $4x^2y^2=xy\left[(x+y)^2-(x-y)^2\right]\iff$ $4xy=(x+y)^2-(x-y)^2$ what is truly. In conclusion, $TD$ is tangent to $w$ .

P8 (Cristian Tello). Let the circle $w=\mathbb C(O,R)$ with the diameter $[AB]$ and the circle $\alpha =\mathbb C(I,r)$ what is interior tangent

to the circle $w$ at the point $T$ and is tangent to $AB$ at $L\in (AO)$ so that $AI\perp TO$ . Prove that $\left(\frac R{R-r}\right)^2=\frac {2R-r}r$ .

Proof. Let the diameter $[TS]$ of $w$ . Thus, $\left\{\begin{array}{ccccccc} \mathrm{formulas\ of\ area\ in} & \triangle AIO\ : & IL\cdot AO=IA\cdot IO & \implies & r^2R^2 & = & IA^2\cdot (R-r)^2\\\\ \mathrm{theorem\ of\ height\ in} & \triangle ATS\ : & IA^2=IT\cdot IS & \implies & IA^2 & = & r(2R-r)\end{array}\right\|\bigodot$ $\implies$ $\left(\frac R{R-r}\right)^2=\frac {2R-r}r$ .

Lemma. Let $\widehat{XOY}$ and fixed $\{A,B\}\subset [OX$ , $\{C,D\}\subset [OY$ . Prove that the geometric locus of mobile interior $L$ w.r.t. given angle for which $[LAB]+[LCD]=k$ (constant) is a line.

Particular cases.

$1\blacktriangleright$ Let convex $ABCD$ so $E\in AD\cap BC$ and $F\in AB\cap CD\ (ABCDEF$ is a "complete quadrilateral" $)$ . Prove that the midpoints of diagonals $[AC]$ , $[BD]$ , $[EF]$ are collinear.

$2\blacktriangleright$ If $ABCD$ is a tangential (circumscribed) quadrilateral with the incenter $I$ , then $I\in MN$ where $M$ , $N$ are the midpoints of the diagonals (Gauss ' line).

Application. Let the tangential $ABCD$ with the incenter $I$ and the midpoint $M$ of $[BD]$ . Denote $E\in IM\cap AD$ . Prove that $2\cdot [EBC]=[ABCD]$ .

P9. Let $\triangle ABC$ and $\{D,E\}\subset \in (BC)$ so that $m\left(\widehat{BAD}\right)=\left(\widehat {CAE}\right) = \phi$ . Let $\left\|\begin{array}{ccc} M\in (BC) & , & MB = MC \\\\ \{X,T\}\subset AD & , & \left|\begin{array}{c} BX\perp AD\\\\ CT\perp AD\end{array}\right|\\\\ \{Y,Z\}\subset AE\ & , &\left| \begin{array}{c} BZ\perp AE\\\\ CY\perp AE\end{array}\right|\end{array}\right\|$ . Prove that $\left\{\begin{array}{c} MX = MY = MZ = MT \\\\ m(\widehat {XMY}) = 180^{\circ} - 2\phi \\\\ m(\widehat {ZMT}) = 180^{\circ} - 2(A - \phi )\end{array}\right\|$ .

Proof. Let $K\in XZ\cap AC$ and let the midpoint $P$ of $AB$ . Since $\angle AZB = 90 = \angle AXB$ , we have that $AZXB$ is cyclic. Since $\angle AZB = 90$ and $AP = BP$ , we see

that $P$ is the center of the circumcircle of $AZXB$ . Thus, $PZ = PX$ . Now, $\angle AXK = \angle AXZ =$ $\angle ABZ =$ $90 - \angle BAZ =$ $90 - \angle XAK$ . It follows that $\angle XKA =$

$180 - \angle AXK - \angle XAK = 90$ , so $XZ\perp AC$ . Yet, $MP$ is a midline of $\triangle ABC$ , so $MP\parallel AC$ , which means that $MP\perp XZ$ . Yet, $P$ is on the perpendicular bisector of $XZ$

since $PZ=PX$ , so $M$ is on the perpendicular bisector of $XZ$ , so $MX = MZ$ . Similarly, $MT = MY$ . Notice that $\angle XAB = \angle YAC$ and $\angle AXB = 90 = \angle AYC$ , so

$\triangle AXB\sim \triangle AYC$ . So $\frac {AX}{AY} = \frac {AB}{AC}$ . Similarly, $\triangle AZB\sim \triangle ATC$ , so $\frac {AZ}{AT} = \frac {AB}{AC} =$ $\frac {AX}{AY}\implies$ $AZ\cdot AY = AT\cdot AX$ , so $XZYT$ is cyclic. Since $MX = MZ$ , we have

that $M$ on the perpendicular bisector of $XZ$ . Similarly, it lies on the perpendicular bisector of $TY$ , so $M$ is the circumcenter of $XZYT$ . It follows that $MX = MZ = MY = MT$ .

P10. Let $ABC$ be a triangle for which $a > b$ , $a > c$ . Consider the points $\{P,Q\}\subset [BC]$ so that $PC = b$ , $BQ = c$ . The line $AQ$ cut again the circumcircle

of $\triangle ABC$ in $D$ and the line $BC$ cut again the circumcircle of $\triangle APD$ in $E$ . Prove that $\boxed {\ \frac {PQ}{CE} = 2 + \frac {b^2 + c^2 - a^2}{(a - b)(a - c)}\ }$ and $A = 90^{\circ}\ \Longleftrightarrow\ PQ = 2\cdot CE$ .

Proof. Using the power of a point w.r.t. a circle obtain : $QB\cdot QC = QA\cdot QD = QP\cdot QE$ . Since $QB = c$ , $QC = a - c$ , $QP = b + c - a$ results

$QE = \frac {c(a - c)}{b + c - a}$ and $CE = QE - QC$ $\implies$ $CE = \frac {(a - b)(a - c)}{b + c - a}$ . Therefore, $\frac {PQ}{CE} = \frac {(b + c - a)^2}{(a - b)(a - c)} = 2 + \frac {b^2 + c^2 - a^2}{(a - b)(a - c)}$ a.s.o.

Remarks. If $A = 120^{\circ}$ , then $PQ < CE$ . Here is an identity with many applications $:\ \boxed {\ \sum\frac {b^2 + c^2 - a^2}{(a - b)(a - c)}+2=0\ }$ , where $\{a,b,c\}\subset\mathbb C$ and $a\ne b\ne c\ne a$ . This

identity is equivalently with the relation $\left(b^2+c^2-a^2\right)(b-c)+\left(c^2+a^2-b^2\right)(c-a)+\left(a^2+b^2-c^2\right)(a-b)=2(a-b)(b-c)(c-a)$ . Prove similarly that $:$

$1\blacktriangleright$ Let $\triangle ABC$ for which $a \ne b\ne c\ne a$ . Consider the points $P\in (CB$ , $Q\in (BC$ so that $CP = b$ , $BQ = c$ . The line $AQ$ cut again the circumcircle of $\triangle ABC$ in $D$ and

$BC$ cut again the circumcircle of $\triangle APD$ in $E$ . Prove that $\boxed {\ \frac {\overline {PQ}}{\overline {CE}} = 2 + \frac {b^2 + c^2 - a^2}{(a - b)(a - c)}\ }$ , where $X(x)$ , $\overline {XY} = y - x$ - an analytical geometry on the orientated line $d = BC$ .

$2\blacktriangleright$ Let $ABC$ be a triangle . Consider the points $\{P,Q\}\subset BC$ so that $C\in (BP)$ , $B\in (CQ)$ and $CP = b$ , $BQ = c$ . The line $AQ$ cut

again the circumcircle of $\triangle ABC$ in $D$ and the line $BC$ cut again the circumcircle of $\triangle APD$ in $E$ . Prove that $\boxed {\ \frac {PQ}{CE} = 2 + \frac {b^2 + c^2 - a^2}{(a + b)(a + c)}\ }\ .$

$3\blacktriangleright$ Let $\triangle ABC$ and $\{P,Q\}\subset (BC)$ so that $BQ^2 + CP^2 = BC^2$ . The lines $\left\{\begin{array}{ccccc} AQ & \mathrm{cut\ again\ circumcircle\ of} & \triangle ABC & \mathrm{in} & D\\\\ BC & \mathrm{cut\ again\ circumcircle\ of} & \triangle APD & \mathrm{in} & E\end{array}\right\|$ . Prove that $PQ = 2\cdot CE$ .

$4\blacktriangleright$ Let an $A$ -right $\triangle ABC$ and $\{X,Y,Z,T\}\subset BC$ (line !) so that $\left\|\begin{array}{cccccc} B\in (XY) & , & Y\in (BC) & , & BX = BY = AB \\ \\ C\in (ZT) & , & Z\in (BC) & , & CZ = CT = AC\end{array}\right\|$ . Prove that $\widehat {XAZ}\equiv\widehat {ZAY}\equiv\widehat {YAT}$ .

P11 (E. Vargas Diaz). Let a convex $ABCD$ which is inscribed in $w=\mathbb C(O,a)$ with the diameter $[AB]$ . Denote $\left\{\begin{array}{ccc} P\in AC\cap BD & ; & OP=p\\\\ Q\in OP\cap CD & ; & PQ=x\end{array}\right\|$ . Prove that $\boxed{x=\frac {p\left(a^2-p^2\right)}{a^2+p^2}}\ .$

Proof. $\left\{\begin{array}{ccc} \mathrm{Median}\ PO\ \mathrm{in}\ \triangle APB\ : & 2\left(PA^2+PB^2\right)=AB^2+4PO^2=4a^2+4p^2\iff \boxed{PA^2+PB^2=2\left(a^2+p^2\right)} & (1)\\\\ \mathrm{Power\ of}\ P\ \mathrm{w.r.t.}\ w\ : & PA\cdot PC=PB\cdot PD=OA^2-PO^2=a^2-p^2\implies\boxed{PA\cdot PC=PB\cdot PD=a^2-p^2} & (2)\end{array}\right\|$ . Let $\left\{\begin{array}{ccc} X\in OD\cap AC\\\\ Y\in OC\cap BD\end{array}\right\|$ and

Aubel's relation to $\overline{OPQ}$ in $\triangle COD\ :\ \boxed{\frac {PO}{PQ}=\frac {XO}{XD}+\frac {YO}{YC}}\ (*)$ . Menelaus' theorem $:\ \left\{\begin{array}{cc} \overline{AXP}/\triangle OBD\ : & \frac {AO}{AB}\cdot\frac {PB}{PD}\cdot \frac {XD}{XO}=1\implies\frac {XO}{XD}=\frac {PB}{2\cdot PD}=\frac {PB^2}{2\cdot PB\cdot PD}\\\\ \overline{BYP}/\triangle OAC\ : & \frac {BO}{BA}\cdot\frac {PA}{PC}\cdot \frac {YC}{YO}=1\implies\frac {YO}{YC}=\frac {PA}{2\cdot PC}=\frac {PA^2}{2\cdot PA\cdot PC}\end{array}\right\|$ $\bigoplus$ $\stackrel{(*)}{\implies}$

$\frac px=$ $\frac {PO}{PQ}\ \stackrel{*}{=}\ \frac {PB^2}{2\cdot PB\cdot PD}+$ $\frac {PA^2}{2\cdot PA\cdot PC}\ \stackrel{2}{=}$ $\frac {PA^2+PB^2}{2\cdot PA\cdot PC}\ \stackrel{1\wedge 2}{=}\ \frac {2\left(a^2+p^2\right)}{2\left(a^2-p^2\right)}=\frac {a^2+p^2}{a^2-p^2}$ $\implies$ $\frac px=\frac {a^2+p^2}{a^2-p^2}\implies$ $x=\frac {p\left(a^2-p^2\right)}{a^2+p^2}\ .$ Very nice problem!

An easy extension. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\frac {PM}{PN}=\frac {PM^2+MA\cdot MB}{p_w(P)}}\ ,$ where $PA\cdot PC=p_w=PB\cdot PD$ .

Proof 1. Let $\left\{\begin{array}{ccc} X\in MD\cap AC\\\\ Y\in MC\cap BD\end{array}\right\|$ and apply Aubel's relation to $\overline{MPN}$ in $\triangle CMD\ :\ \boxed{\frac {PM}{PN}=\frac {XM}{XD}+\frac {YM}{YC}}\ (1)$ . Apply the Stewart's relation to the cevian $PM$ in $\triangle APB\ :$

$\boxed{AB\cdot\left(PM^2+MA\cdot MB\right)=PA^2\cdot MB+PB^2\cdot MA}\ (*)$ . Apply the Menelaus' theorem to the transversals $\overline{AXP}/\triangle MBD$ and $\overline{BYP}/\triangle MAC$ respectively $:$

$\left\{\begin{array}{c} \frac {AM}{AB}\cdot\frac {PB}{PD}\cdot \frac {XD}{XM}=1\implies\frac {XM}{XD}=\frac {MA}{AB}\cdot \frac {PB}{PD}\\\\ \frac {BM}{BA}\cdot\frac {PA}{PC}\cdot \frac {YC}{YM}=1\implies\frac {YM}{YC}=\frac {MB}{AB}\cdot \frac {PA}{PC}\end{array}\right\|$ $\bigoplus$ $\implies$ $\frac {PM}{PN}\ \stackrel{(1)}{=}\ \frac {MA}{AB}\cdot \frac {PB^2}{p_w(P)}+$ $\frac {MB}{AB}\cdot \frac {PA^2}{p_w(P)}=$ $\frac {MA\cdot PB^2+MB\cdot PA^2}{AB\cdot p_w(P)}\ \stackrel{(*)}{=}\ \boxed{\frac {PM}{PN}=\frac {PM^2+MA\cdot MB}{p_w(P)}}\ (2)$ .

Remark. Get analogously $\boxed{\frac {PN}{PM}=\frac {PN^2+NC\cdot ND}{p_w(P)}}\ (3)$ . From the product of $(2)$ and $(3)\ :\ \boxed{\left(PM^2+MA\cdot MB\right)\left(PN^2+NC\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .

Proof 2 (Suli). Let the circumcircles of $DPC$ , $APB$ intersect $MN$ at $R$ , $S$ respectively. Observe that $\left\{\begin{array}{ccc} MA\cdot MB & = & MP\cdot MS\\\\ NC\cdot ND & = & NP\cdot NR\end{array}\right\|\ (4)$ . Therefore,

$\left\{\begin{array}{ccccccc} \widehat{DRP}\equiv\widehat{DCP}\equiv \widehat{DBM} & \implies & \widehat{DRP}\equiv \widehat{DBM} & \implies & DRBM\ \mathrm{is\ cyclic} & \implies & PB\cdot PD=\cdot PM\cdot PR\\\\ \widehat{ASP}\equiv\widehat{ABP}\equiv \widehat{ACD} & \implies & \widehat{ASN}\equiv \widehat{ACN} & \implies & ASCN\ \mathrm{is\ cyclic} & \implies & PA\cdot PC=\cdot PN\cdot PS\end{array}\right\|$ . In conclusion, $PA \cdot PB \cdot PC \cdot PD =$

$PN\cdot PR\cdot PM \cdot PS=$ $PN(PN+NR)\cdot PM(PM+MS)=$ $\left(PN^2 + PN \cdot NR\right)\left(PM^2 + PM \cdot MS\right)\ \stackrel{(4)}{=}\ \left(PN^2 + ND \cdot NC\right)\left(PM^2 + MA \cdot MB\right)$ .

P12. Let $\triangle ABC$ such that $C<A<\frac{\pi}{2}$ . Let $D\in [AC]$ such that $BD=BA$ . The incircle $w=C(I,r)$ of $\triangle ABC$ touches its

sides in $K\in [AB]$ and $L\in [AC]$ . Let $w_{1}=C(J,r_{1})$ be the incircle of $\triangle BCD$ . Denote $T\in KL\cap AJ$ . Prove that $TA=TJ$ .

Proof.. The incircle $w_{1}$ touches $\triangle BCD$ in $U\in BD$ , $V\in BC$ and $W\in BC$ . Observe that $DC=AC-AD=b-2c\cdot\cos A=$ $\frac{b^{2}-(b^{2}+c^{2}-a^{2})}{b}$ $\implies$ $DC=\frac{a^{2}-c^{2}}{b}$ .

Thus, $VC=\frac{1}{2}\cdot (BC+CD-BD)=\frac{1}{2}\cdot(a+\frac{a^{2}-c^{2}}{b}-c)\implies VC=\frac{p(a-c)}{b}$ . Since $VC=JC\cdot\cos \frac{C}{2}$ obtain that $\boxed{JC=\frac{p(a-c)}{b\cdot\cos\frac{C}2}}$ . Denote $P\in KL\cap CI$ . The

property $PC\perp PB$ is well-known (see the lower remark). Thus, $APJW$ is cyclically. Analogously $P\in UV\cap CJ$ , i.e. $\boxed{\ P\in KL\cap UV\ }$ . Observe that $\boxed{PC=a\cdot \cos \frac{C}{2}}\implies$

$\frac{PC}{JC}=\frac{ab\cdot\cos^{2}\frac{C}{2}}{p(a-c)}=\frac{p-c}{a-c}$ $\implies$ $\frac{PC}{p-c}=\frac{JC}{a-c}=\frac{PJ}{p-a}$ $\implies$ $\boxed{\ \frac{PJ}{PC}=\frac{p-a}{p-c}\ }$ . Apply the Menelaus' theorem to transversal $\overline{PTL}$ in $\triangle AJC$ : $\frac{PJ}{PC}\cdot\frac{LC}{LA}\cdot\frac{TA}{TJ}=1$ $\implies$

$\frac{p-a}{p-c}\cdot\frac{p-c}{p-a}\cdot\frac{TA}{TJ}=1$ $\implies TA=TJ$ . So $PC\perp PB$ $\Longleftrightarrow$ the quad. $PKIB$ is cyclically $\Longleftrightarrow$ $m(\widehat{PKB})=m(\widehat{PIB})=\frac{B+C}{2}$ or $m(\widehat{PKA})=m(\widehat{PIB})=\frac{B+C}{2}$

P13 (M. O. Sanchez). Let an equilateral $\triangle ABC$ with the circumcircle $\mathbb C(O,R)$ and a mobil point $D\in w$ so that $BC$ separates $A$ and $D\ .$ Denote $:$ the midpoints $M\ ,$ $N$ of the

arcs $\overarc{AB}\ ,$ $\overarc{AC}$ respectively so that $AB$ separates $D\ ,$ $M$ and $AC$ separates $D\ ,$ $N\ ;$ the intersections $E\in AC\cap DN$ and $F\in AB\cap DM\ .$ Prove that $EF=EN+FM\ .$

Proof. Is well-known the Pompeiu's theorem $\boxed{DB+DC=DA}\ (*)$ (is a particular case of the Ptolemy's theorem). The rays $[DM\ ,$ $[DN$ are the bisectors of the angles $\widehat{ADB}\ ,$ $\widehat{ADC}$

and the points $M\ ,$ $N$ are the symmetrical points of $O$ w.r.t. the lines $AB\ ,$ $AC$ respectively. Hence $\boxed{EN=EO\ ,\ FM=FO}\ (1)\ .$ Therefore, $\frac {FB}{FA}=\frac {DB}{DA}$ and $\frac {EC}{EA}=\frac{DC}{DA}\implies$

$\frac {FB}{FA}+\frac {EB}{EA}=\frac {DB}{DA}+\frac {DC}{DA}=\frac{DB+DC}{DA}\ \stackrel{(*)}{=}\ 1\implies$ $\frac {FB}{FA}+\frac {EB}{EA}=1$ $\iff O\in EF\ .$ In conclusion, $EF=OE+OF\ \stackrel{(1)}{\iff}\ EF=EN+FM\ .$ Very nice problem!

Remark. Apply the Pascal's theorem to the cyclical hexagon $ACMDNB\ \odot\begin{array}{ccc} \nearrow & E\in AC\cap DN & \searrow\\\ \rightarrow & O\in CM\cap NB & \rightarrow\\\ \searrow & F\in BA\cap MD & \nearrow\end{array}\odot\implies O\in EF\ .$

P14 (Thanasis Gakopoulos, Greece). Prove that $(\forall )\ \triangle ABC$ there is the inequality $\boxed{\ \frac 1{II_a^2}+\frac 1{II_b^2}+\frac 1{II_c^2}\ge \frac {3s}{8RS}\ }$ (standard notations).

Proof 1. Observe that $\triangle ABI_a\sim \triangle AIC\iff$ $\frac {AB}{AI}=\frac {AI_a}{AC}\iff$ $\frac c{AI}=\frac {AI_a}b\iff$ $\boxed{\ AI_a\cdot AI=bc\ }\ (1)\ .$ Denote the projections $\mathrm{pr}_{\mathrm{AB}}(I)=P$ and $\mathrm{pr}_{\mathrm{AB}}(I_a)=Q\ ,$

where is well known that $\left\{\begin{array}{ccc} AP=s-a & ; & IP=r\\\\ AQ=s & ; & I_aQ=r_a\end{array}\right\|\ .$ Hence $IP\parallel I_aQ\iff$ $\triangle AIP\sim AI_aQ\iff$ $\frac{AI}{AP}=\frac {AI_a}{AQ}\iff$ $\frac {AI}{s-a}=\frac {AI_a}s=\frac {II_a}a=$ $\sqrt{\frac {AI\cdot AI_a}{s(s-a)}}\ \stackrel{(1)}{=}$

$\sqrt{\frac {bc}{s(s-a)}}\implies$ $\frac 1{II_a^2}=\frac {s(s-a)}{a^2bc}=\frac {s(s-a)}{4aRS}\implies$ $\boxed{\ \frac 1{II_a^2}=\frac {s(s-a)}{4aRS}\ }\ (2)\ .$ I"ll apply now the well known inequality $\sum a\cdot\sum\frac 1a\ge 9\ ,$ i.e. $\boxed{\ \sum \frac 1a\ge \frac 9{2s}\ }\ (3)\ .$

In conclusion, $\sum\frac 1{II_a^2}\ \stackrel{(2)}{=}\ \frac 1{4RS}\cdot \sum \frac {s(s-a)}a=$ $\frac 1{4RS}\cdot \left(s^2\cdot\sum\frac 1a-\sum s\right)\ \stackrel{(3)}{\ge}$ $\frac 1{4RS}\cdot \left(s^{\cancel 2}\cdot\frac 9{2\cancel s}-3s\right)=$ $\frac s{4RS}\cdot\frac 32=\frac {3s}{8RS}\implies$ $\frac 1{II_a^2}+\frac 1{II_b^2}+\frac 1{II_c^2}\ge \frac {3s}{8RS}\ .$

Proof 2. $IBI_aC$ is inscribed in the circle with the diameter $II_a\implies$ $BC=II_a\cdot\sin\widehat{BIC}\implies$ $II_a=\frac a{\sin\left(\frac {\pi}2+\frac A2\right)}=$ $\frac {2R\sin A}{\cos\frac A2}=$ $4R\sin\frac A2\implies$ $\boxed{\ II_a=4R\sin\frac A2\ }$ a.s.o.

P15 (Thanasis Gakopoulos, Greece). Prove that $(\forall )\ \triangle ABC$ there is the identity $\boxed{\ \frac 1{a\cdot AI_a^2}+\frac 1{b\cdot BI_b^2}+\frac 1{c\cdot CI_c^2}=\frac 1{a\cdot AI^2+b\cdot BI^2+c\cdot CI^2}=\frac 1{4RS}\ }$ (standard notations).

Proof. Have from the upper problem $\boxed{\ AI\cdot AI_a=bc\ }\ (1)$ and $\boxed{\ AI^2=\frac {bc(s-a)}s\ }\ (2)\ .$ Thus, $\sum a\cdot AI^2=\sum a\cdot \frac {bc(s-a)}s=$ $\frac {abc}s\cdot s=abc=4RS\implies$ $\boxed{\ \sum a\cdot AI^2=4RS\ }\ (3)\ .$

On other hand, $\frac 1{a\cdot AI_a^2}=$ $\frac {AI^2}{a\cdot \left(AI\cdot AI_a\right)^2}\ \stackrel{(1\wedge 2)}{=}\ \frac {\cancel{bc}(s-a)}s\cdot\frac 1{a(bc)^{\cancel 2}}\implies$ $\boxed{\ \frac 1{a\cdot AI_a^2}=\frac{s-a}{s\cdot abc}\ }\ (4)\implies$ $\sum \frac 1{a\cdot AI_a^2}=\frac 1{s\cdot 4RS}\cdot s=\frac 1{4RS}\implies$ $\boxed{\ \sum \frac 1{a\cdot AI_a^2}=\frac 1{4RS}\ }\ (5)\ .$

This post has been edited 353 times. Last edited by Virgil Nicula, Jan 27, 2018, 4:33 PM

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420. Some geometry problems for high school II.

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