337. Nice metrical problems.

by Virgil Nicula, Feb 26, 2012, 1:37 PM

PP1. Let $\triangle ABC$ for which $A= \frac {\pi}{4}$ and exists $D\in (AC)$ so that $AD = a$ and $m\left(\widehat {DBC}\right) = \frac {\pi}{8}$ . Find the value of $C$ .

Proof 1 (metric). Let $E\in AC$ so that $A\in (EC)$ and $AE=c$ . Since $\widehat {CBD}\equiv\widehat{CEB}$ obtain $\triangle CBD\sim \triangle CEB$ , i.e. $\frac {CB}{CE}=\frac {CD}{CB}\iff$ $CB^2=CD\cdot CE\iff$

$\boxed{a^2=(b-a)(b+c)}\ (1)$ . The generalized Pythagoras' in $ABC\ :$ $\boxed{a^2=b^2+c^2-bc\sqrt 2}\ (2)$ . We"ll eliminate $a$ between $(1)$ and $(2)$ , i.e. $\left\{\begin{array}{c} a^2=(b-a)(b+c)\\\\ a^2=b^2+c^2-bc\sqrt 2\end{array}\right\|$ $\implies$

$\boxed{a=\frac {\left(1+\sqrt 2\right)\cdot bc-c^2}{b+c}}$ , $(b+c)^2\left(b^2+c^2-bc\sqrt 2\right)=$ $c^2\left[b\left(1+\sqrt 2\right)-c\right]^2$ . Let $\boxed{\frac bc=\lambda}$ . Relation becomes $\left(\lambda +1\right)^2\left(\lambda ^2-\lambda \sqrt 2+1\right)=$ $\left[\left(1+\sqrt 2\right)\lambda -1\right]^2\stackrel{(\lambda \ne 0)}{\iff}$

$\lambda ^3+\left(2-\sqrt 2\right)\cdot \lambda ^2-$ $\left(1+4\sqrt 2\right)\cdot \lambda+$ $\left(4+\sqrt 2\right)=0$ $\stackrel{(\lambda >0)}{\iff}$ $\lambda\in\left\{\ \sqrt 2\ ,\ -1+\sqrt {2\left(1+\sqrt 2\right)}\ \right\}\ (3)$ . Using $(2)$ obtain $a=c\cdot \sqrt{\lambda^2-\lambda\sqrt 2+1}$ . i.e. $\frac {a}{\sqrt{\lambda^2-\lambda\sqrt 2+1}}=$ $\frac {b}{\lambda}=\frac c1$ . Hence $\boxed{\cos C=\frac {2\lambda -\sqrt 2}{2\sqrt{\lambda^2-\lambda\sqrt 2+1}}}$ , where $C$ appears in $(3)$ . Remark that for $\lambda =\sqrt 2$ obtain $C=\frac {\pi}{4}$ , i.e. $\triangle ABC$ is $B$ -isosceles and $B$ -right.

PP2. Let $\triangle ABC$ with circumcircle $w=C(O,R)$ and $\left\{\begin{array}{c} B=80^{\circ}\\\\ C=40^{\circ}\end{array}\right\|$ . Denote $\left\{\begin{array}{c} D\in BC\ ,\ AD\perp BC\\\\ E\in AO\cap BC\end{array}\right\|$ . Prove that $CE=2\cdot BD$ .

Proof 1. Let $\{A,E\}\cap w=\{A,S\}$ , $F\in AB$ for which $DF\parallel AE$ . Thus $E$ , $F$ are homologously in the similarity $\triangle ACS\sim ADB$ , $\triangle ECS$ is $S$ -isosceles and $\triangle DBF$ ,

$\triangle EBA$ are $B$ -isosceles , i.e. $SE=SC$ , $BF=BD$ , $BE=BA$ . Hence $\frac {CE}{BD}=$ $\frac {CE}{DF}\cdot \frac{DF}{DB}=$ $\frac {AS}{AB}\cdot \frac{AE}{BE}=$ $\frac {1}{\sin 40^{\circ}}\cdot \frac {EC}{ES}=2\implies$ $CE=2\cdot BD$ .

Proof 2. Evaluate the areas of the triangles $DAB$ and $EAC\ :\ \left\{\begin{array}{c} DB\cdot\delta_{BC}(A)=AB\cdot AD\cdot\sin \widehat {DAB}\\\\ EC\cdot \delta_{BC}(A)=AC\cdot AE\cdot \sin\widehat{EAC}\end{array}\right\|$ $\implies$ $\frac {EC}{BD}=\frac {AC}{AB}\cdot \frac {AE}{AD}=$ $\frac {\sin 80}{\sin 40}\cdot \frac {1}{\cos 40}=2\implies$ $CE=2\cdot BD$ .

Generalization. Consider in $\triangle ABC$ two points $\{D,E\}\subset (BC)$ so that $D\in (BE)$ and $m\left(\widehat{DAB}\right)=x$ , $m\left(\widehat{EAC}\right)=y$ . Prove that $\boxed{\frac {BD}{CE}=\frac {AB}{AC}\cdot\frac {AD}{AE}\cdot\frac {\sin x}{\sin y}}$ .

Proof 1. $\frac {BD}{CE}=\frac {AB}{AC}\cdot\frac {AD}{AE}\cdot\frac {\sin x}{\sin y}\iff$ $\frac {BD}{AB}\cdot\frac {AC}{CE}\cdot\frac {AE}{AD}\cdot\frac {\sin y}{\sin x}=1\iff$ $\frac {\sin x}{\sin (B+x)}\cdot\frac {\sin (C+y)}{\sin y}\cdot\frac {\sin (B+x)}{\sin (C+y)}\cdot \frac {\sin y}{\sin x}=1$ , what is truly.

Proof 2. Evaluate the areas of the triangles $DAB$ and $EAC\ :\ \left\{\begin{array}{c} DB\cdot\delta_{BC}(A)=AB\cdot AD\cdot\sin x\\\\ EC\cdot \delta_{BC}(A)=AC\cdot AE\cdot \sin y\end{array}\right\|$ $\implies\ \frac {BD}{CE}=\frac {AB}{AC}\cdot \frac {AD}{AE}\cdot \frac {\sin x}{\sin y}$ .

Proof 3. Apply the well-known relations $\left\{\begin{array}{c} \frac {BD}{BC}=\frac {AD}{AC}\cdot\frac {\sin\widehat{BAD}}{\sin\widehat{BAC}}=\frac {AD}{AC}\cdot\frac {\sin x}{\sin A}\\\\ \frac {CE}{CB}=\frac {AE}{AB}\cdot\frac {\sin\widehat{CAE}}{\sin\widehat{CAB}}=\frac {AE}{AB}\cdot\frac {\sin y}{\sin A}\end{array}\right\|\implies\ \frac {BD}{CE}=\frac {AB}{AC}\cdot \frac {AD}{AE}\cdot \frac {\sin x}{\sin y}$ .

PP3. Let $G$ be the centroid of $\triangle ABC$ . Prove that $\boxed{\ PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3\cdot PG^2\ ,\ (\forall )\ P\ }$ .

Proof 1. Prove easily that $BC^2+3\cdot GA^2=2\cdot\left(GA^2+GB^2+GC^2\right)\ (*)$ . Let the midpoint $M$ of $[BC]$ and $S\in AM$ so

that $M$ is midpoint of $[GS]$ . Observe that $GS=GA$ and $AS=2\cdot GA$ . Apply Appolonius' theorem to the $P$ -medians in triangles:

$\left\|\begin{array}{cccc} \triangle BPC\ : & 4\cdot PM^2=2\left(PB^2+PC^2\right)-BC^2\\\\ \triangle GPS\ : & 4\cdot PM^2=2\left(PG^2+PS^2\right)-GS^2\\\\ \triangle APS\ : & 4\cdot PG^2=2\left(PA^2+PS^2\right)-AS^2\end{array}\right\|\ \implies$ $\left\|\begin{array}{c} 2\left(PB^2+PC^2\right)-BC^2=2\left(PG^2+\underline {PS}^2\right)-GS^2\\\\ 4\cdot PG^2=2\left(PA^2+\underline{PS}^2\right)-AS^2\end{array}\right\|$ .

Eliminate $PS$ between upper relations: $2\cdot PB^2+2\cdot PC^2-BC^2=2\cdot PG^2+\left[4\cdot PG^2 -2\cdot PA^2+4\cdot GA^2\right]-GA^2\iff$

$2\left(PA^2+PB^2+PC^2\right)=$ $6\cdot PG^2+$ $\left(BC^2+3\cdot GA^2\right)\stackrel{(*)}{\iff}$ $PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3\cdot PG^2$ .

Remark. I used only the Appolonius' theorem (the "relation of the median in a triangle" or the "law of parallelogram"). $AB^2+AC^2=2\left(AM^2+BM^2\right)$ .

Proof 2. Apply the Stewart's theorem to the cevian $PG$ in $\triangle APM\ :\ PG^2\cdot AM+AG\cdot GM\cdot AM=PA^2\cdot GM+PM^2\cdot GA\iff$

$PG^2\cdot \frac 32\cdot GA+\frac 34\cdot GA^3=PA^2\cdot \frac 12\cdot GA+PM^2\cdot GA\iff$ $6\cdot PG^2+3\cdot GA^2=2\cdot PA^2+4\cdot PM^2\ (1)$ . Apply relation of median

in $\triangle BPC\ :\ 4\cdot PM^2=2\left(PB^2+PC^2\right)-BC^2$ . The relation $(1)$ becomes $6\cdot PG^2+3\cdot GA^2=2\cdot PA^2+\left[2\cdot \left(PB^2+PC^2\right)-BC^2\right]$

$\iff 6\cdot PG^2+\left(BC^2+3\cdot GA^2\right)=$ $2\cdot\left(PA^2+PB^2+PC^2\right)\stackrel{(*)}{\iff}$ $PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3\cdot PG^2$ .

Remark. $(\forall )\ P$ there is $\boxed{x\cdot PA^2+y\cdot PB^2+z\cdot PC^2=PS^2-p_w(S)}\ (*)$ , where $(x,y,z)$ are normal barycentrical coordinates

$(x+y+z=1)$ of given $S$ and $p_w(S)$ is power of $S$ w.r.t. circumcircle of given $\triangle ABC$ . In particular case $S:=G$ we have $S\left(\frac 13,\frac 13,\frac 13\right)\ ,$

$p_w(G)=-\frac {a^2+b^2+c^2}{9}=-\frac 13\cdot\sum GA^2$ and $(*)$ becomes our identity $\sum PA^2=\sum GA^2+3\cdot PG^2$ .

PP4. If $a,b,c$ and $a_1 , b_1 , c_1$ are lengths of sides in $\triangle ABC$ , $\triangle A_1B_1C_1$ respectively while $\alpha , \beta , \gamma$ and $\alpha_1 , \beta_1 , \gamma_1$ their angles so that $\alpha +\alpha_1 =\pi$ , $\beta=\beta_1$ . Prove that $aa_1=bb_1+cc_1$

Proof 1. I'll use the well-known remarkable identity $\boxed{\ \sin^2x-\sin^2y=\sin (x+y)\sin (x-y)\ }\ (*)$ . Indeed, $\sin^2x-\sin^2y=$ $\sin (x+y)\sin (x-y)\iff$ $2\sin^2x-2\sin^2y=$

$2\sin (x+y)\sin (x-y)\iff$ $(1-\cos 2x)-(1-\cos 2y)=$ $\cos 2y-\cos 2x$ , what is truly. Denote the length $R$ of the circumradius for $\triangle ABC$ and the length $R_1$ of the

circumradius for $\triangle A_1B_1C_1$ . Thus, $\left\{\begin{array}{ccc} A_1 & = & 180^{\circ}-A\\\\ B_1 & = & B\\\\ C_1 & = & A-B\end{array}\right\|$ and $aa_1=bb_1+cc_1$ $\iff$ $4RR_1\sin^2A=$ $4RR_1\sin^2 B+4RR_1\sin C\sin (A-B)\iff$ $\boxed{\sin^2A-\sin^2 B=\sin (A+B)\sin (A-B)}$ .

$(*)$ Remark. $\sin^2x-\sin^2y=\sin (x+y)\sin (x-y)\iff$ $2\sin^2x-2\sin^2y=2\sin (x+y)\sin (x-y)\iff$ $(1-\cos 2x)-(1-\cos 2y)=\cos 2y-\cos 2x$ , what is truly.

Proof 2. Denote the point $B_1\in AB$ so that $A\in \left(BB_1\right)\ ,\ AB_1=c_1$ and the point $C_1\in AC$ so that $C\in \left(AC_1\right)\ ,\ AC_1=b_1$ . Observe that $A\equiv A_1$ . Finally denote $B_2\in BB_1$ so that $CB_2\parallel B_1C_1$ . Obviously, the $\triangle BCB_2$ is isosceles with apex $C$ , hence $CB_2=a$ . Following equality holds: $\frac{b}{b_1}=\frac{a}{a_1}=\frac{c_2}{c_1}\ (*)$ , where $AB_2=c_2$ . Apply the Stewart's relation to $\triangle BCB_2$ with cevian $AC$ and obtain that $BC^2\cdot \left(AB_2+AB\right)=$ $AC^2\cdot BB_2+AB\cdot AB_2\cdot BB_2\iff$ $a^2=b^2+c\cdot c_2\stackrel{(*)}{\iff} a^2=b^2+\frac{a\cdot c\cdot c_1}{a_1}\iff$ $a^2\cdot a_1=a_1\cdot b^2+a\cdot c\cdot c_1\stackrel{(*)}{\iff}$ $aa_1=bb_1+cc_1$ .

PP5. $ABCD$ is a square. A semicircle $smc$ is drawn on $AB$ with $AB$ as diameter. A quartes-circle $qs$ is drawn

with $C$ as center and $CD$ as radius. Denote $\{B,P\}=smc\cap qs$ . Find the area of their intersection.

Proof. Let $[XY]$ be a chord of $w=C(I,r)$ so that $m\left(\widehat{XOY}\right)=\phi<\pi$ . Then is well-known that $[XOY]=\frac {r^2\sin \phi}{2}$ , the length of the arc $\stackrel{\frown}{XY}$ is $l(\phi )=r\phi$ , the area of sector

$\stackrel{\frown}{\vee}(XOY)$ is $sc(r,\phi )=\frac {r\cdot l(\phi )}{2}=\frac{ r^2\phi}{2}$ and area of the segment $\stackrel{\frown}{-}(XY)$ w.r.t. $w$ is $\boxed{sg(r,\phi )=sc(r,\phi )-[XOY]=\frac {r^2}{2}\cdot (\phi -\sin\phi)}$ . Let $AB=r\ ,\ \alpha =m\left(\widehat{BNP}\right)$ . Thus

$\tan\frac {\alpha}{2} =2\ ,\ \sin\alpha =\frac {2\tan\frac {\alpha}{2}}{1+\tan^2\frac {\alpha}{2}}=\frac 45$ and $\tan\alpha =\frac {2\tan\frac {\alpha}{2}}{1-\tan^2\frac {\alpha}{2}}=-\frac 43\iff$ $\tan(\pi -\alpha)=\frac 43\iff$ $\boxed{\alpha =\pi -\arctan\frac 43}$ . The required area is $sg\left(\frac r2,\alpha\right)+sg\left(r, \pi -\alpha\right)=$

$\frac {r^2}{8}\cdot (\alpha -\sin \alpha )+\frac{r^2}{2}\cdot (\pi -\alpha-\sin\alpha )=$ $\frac {r^2}{8}\cdot (\alpha -\sin \alpha +4\pi -4\alpha -4\sin\alpha)=$ $\frac {r^2}{8}\cdot (4\pi -3\alpha -5\sin\alpha )=$ $\frac {r^2}{8}\cdot \left(4\pi -3\pi +3\arctan\frac 43-5\cdot\frac 45\right)=$

$\frac {r^2}{8}\cdot\left(\pi +3\arctan\frac 43-4\right)$ . In conclusion, the required area is $\boxed{\frac {r^2}{8}\cdot\left(2\pi -4-\arctan\frac {44}{117}\right)}$ .

PP6 (Euler). Let $\triangle ABC$ and $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ so that $P\in AD\cap BE\cap CF$ . Prove that $\boxed{\frac{PA}{PD}\cdot\frac{PB}{PE}\cdot \frac{PC}{PF}=\frac{PA}{PD}+\frac{PB}{PE}+\frac{PC}{PF}+2}\ (*)$ .

Proof 1. Denote the areas $[BPC]=x$ , $[CPA]=y$ and $[APB]=z$ . Prove easily that $\left\{\begin{array}{c} \frac {DB}{DC}=\frac zy\\\\ \frac {EC}{EA}=\frac xz\\\\ \frac {FA}{FB}=\frac yx\end{array}\right\|$ . Using the Van Aubel's relations obtain

$\left\{\begin{array}{c} \frac {PA}{PD}=\frac {EA}{EC}+\frac {FA}{FC}=\frac {z+y}{x}\\\\ \frac {PB}{PE}=\frac {FB}{FA}+\frac {DB}{DC}=\frac {x+z}{y}\\\\ \frac {PC}{PF}=\frac {DC}{DB}+\frac {EC}{EA}=\frac {y+x}{z}\end{array}\right\|$ . The relation $(*)$ is equivalently with $\frac {(x+y)(y+z)(z+x)}{xyz}=\frac {x+y}{z}+\frac {y+z}{x}+\frac {z+x}{y}+2\iff$

$(x+y)(y+z)(z+x)=xy(x+y)+yz(y+z)+zx(z+x)+2xyz$ , what is evidently or well-known.

Proof 2 (without Van Aubel's relation).Denote $P\in AD\cap BE\cap CF$ and $\left\{\begin{array}{c} \frac {PA}{PD}=u\\\\ \frac {PB}{PE}=v\\\\ \frac {PC}{PF}=w\end{array}\right\|$ . Therefore, $\sum \frac {1}{1+u}=\sum \frac {PD}{AD}=\sum \frac {[BPC]}{[BAC]}=1$ . Thus, $\sum \frac {1}{1+u}=1\iff$

$\sum (1+v)(1+w)=\prod (1+u)\iff$ $3+2\cdot \sum u+\sum vw=1+\sum u+\sum vw+uvw\iff$ $2+\sum u=uvw\iff$ $2+\sum \frac{PA}{PD}=\prod \frac{PA}{PD}$ .

PP7 (Iran, 2011). In $\triangle ABC$ let $X$ and $Y$ be the tangent points of incircle $w=C(I,r)$ with $AB$ and $AC$ respectively. The tangent at $A$ to the circumcircle $C(O,R)$

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

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

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

$\blacktriangleright\ \boxed{D\in XY}\iff$ $\frac {DB}{DC}=\frac{ZB}{ZC}\iff$ $\frac {c^2}{b^2}=\frac {s-b}{s-c}\iff$ $s\left(b^2-c^2\right)=b^3-c^3\iff$ $(a+b+c)(b+c)=2\left(b^2+bc+c^2\right)\iff$ $\boxed{a(b+c)=b^2+c^2}$ .

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

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

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

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

PP8. Let $ABCD$ be a square and $E$ , $F$ be two interior points for which $m(\widehat{EBF})=m(\widehat{EDF})=45^{\circ}$ . Prove that $EF^2=AE^2+CF^2$ .

Proof 1. I"ll work in the analytical system $xOy$ , where $A(0,0)$ , $B(1,0)$ , $C(1,1)$ , $D(0,1)$ . Denote: $p(d)$ - the slope of the line $d\ ;\ m(\widehat{ADE})=x<45^{\circ}$ , $m(\widehat{ABE})=y<45^{\circ}$ ;

$\tan x=m<1$ and $\tan y=n<1$ . Thus, $m(\widehat{CDF})=45^{\circ}-x$ , $m(\widehat{CBF})=45^{\circ}-y$ and $p(DE)=-\frac{1}{m}$ ; $p(BE)=-n$ ; $p(DF)=\frac{m-1}{m+1}$ ; $p(BF)=\frac{n+1}{n-1}$ . Therefore $:$

$\left\|\begin{array}{cccc}DE\ : & x+my & = & m\\\\ BE\ : & nx+y & = & n\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c} x_{e}=\frac{m(1-n)}{1-mn}\\\\ y_{e}=\frac{n(1-m)}{1-mn}\end{array}\right\|$ $\boxed{\begin{array}{c} \mathrm{coordinates\ of}\\\\ E\ \blacktriangleleft\ \mathrm{and}\ \blacktriangleright\ F\end{array}}$ $\left\|\begin{array}{cccc} DF\ : & (1-m)x+(1+m)y & = & 1+m\\\\ BF\ : & (1+n)x+(1-n)y & = & 1+n\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}x_{f}=\frac{n(1+m)}{m+n}\\\\ y_{f}=\frac{m(1+n)}{m+n}\end{array}\right\|$ .

Thus, $AE^{2}+CF^{2}=EF^{2}$ $\Longleftrightarrow$ $x^{2}_{e}+y^{2}_{e}+(1-x_{f})^{2}+(1-y_{f})^{2}=(x_{e}-x_{f})^{2}+(y_{e}-y_{f})^{2}$ $\Longleftrightarrow$ $1+x_{e}x_{f}+y_{e}y_{f}=x_{f}+y_{f}$ $\Longleftrightarrow$

$1+\frac{mn(1+m)(1-n)}{(m+n)(1-mn)}+\frac{mn(1+n)(1-m)}{(m+n)(1-mn)}=\frac{n(1+m)+m(1+n)}{m+n}\iff$ $(m+n)(1-mn)+mn[(1+m)(1-n)+(1+n)(1-m)]=$

$(1-mn)[n(1+m)+m(1+n)]$ , what is truly. Remark. If $M\in BE\cap AD$ and $N\in BF\cap CD$ , then $MN$ is tangent at $G$ to the circle $(B, BA)$ , similarly for $H$ .

Proof 2. Denote $\left\{\begin{array}{c} \theta =m\left(\widehat{DAE}\right)\\\\ \beta =m\left(\widehat {BCF}\right)\end{array}\right\|$ and construct $\left\{\begin{array}{cc} \triangle DEH\equiv\triangle DEA & (AE=EH\ ,\ AD=HD)\\\\ \triangle EGB\equiv\triangle EAB & (EG=EA\ ,\ BG=BA)\\\\ \triangle GFB\equiv\triangle CFB & (FC=FG\ ,\ BC=BG)\\\\ \triangle DHF\equiv\triangle DCF & (FH=FC\ ,\ DH=DC)\end{array}\right\|$ $\implies \triangle EGF\equiv\triangle EHF$ .

Since $90^{\circ}-\theta +\beta= \theta + 90^{\circ}-\beta\implies \beta =\theta\implies AE\parallel CF$ . In conclusion, the triangles $EGF$ and $EHF$ are right $\implies$ $EF^2=AE^2+CF^2$ .

PP9. Let $\triangle ABC$ and the feet $\{D,E\}\subset BC$ of $A$ -median and $A$ -bisector respectively. Circumcircle $w$ of $\triangle ADE$ cuts again $AB$ in $B'$ and $AC$ in $C'$ . Show that $BB'=CC'$ .

Proof 1. Observe that $\left\{\begin{array}{ccc} DB & = & DC\\\ b\cdot EB & = & c\cdot EC\end{array}\right\|\ (*)$ and from the powers of $B$ and $C$ w.r.t. the circle $w$ obtain that $\left\{\begin{array}{ccc} c\cdot BB' & = & BE\cdot BD\\\ CE\cdot CD & = & b\cdot CC'\end{array}\right\|\ \bigodot\stackrel{(*)}{\implies} BB'=CC'$ .

Proof 2. Denote the second intersection $M$ of $AE$ with the circumcircle of $\triangle ABC$ and its diameter $[MN]$ . Observe that $AEDN$

is a cyclical quadrilateral and $NB'=NC'$ . Prove easily that $\triangle BB'N=\triangle CC'N$ from where obtain that $BB'=CC'$ .

PP10. Let an equilateral $\triangle ABC$ with $AB=d$ and an internal point $P$ with $PA = a$ , $PB = b$ and $PC = c$ . Prove that $3\cdot \left(d^4+a^4+b^4+c^4\right) =\left (d^2+a^2+b^2+c^2\right)^2$ .

Proof. Suppose w.l.o.g. $d=1$ . So $\boxed{3\cdot\left(1+\sum a^4\right) =\left (1+\sum a^2\right)^2}\ (*)\iff$ $1+\sum a^4=\sum a^2+\sum b^2c^2$ . Denote $\left\{\begin{array}{c} x=m\left(\widehat{BPC}\right)\\\\ y=m\left(\widehat{CPA}\right)\\\\ z=m\left(\widehat{APB}\right)\end{array}\right\|$ . The generalized Pythagoras'
$:\ \left\{\begin{array}{c} \cos x=\frac {b^2+c^2-1}{2bc}\\\\ \cos y=\frac {c^2+a^2-1}{2ca}\\\\ \cos xz=\frac {a^2+b^2-1}{2ab}\end{array}\right\|$ . Show easily $\boxed{x+y+z=2\pi\implies\sum\cos^2x=1+2\cdot\prod\cos x}\ (1)$ . Indeed, $x+y+z=2\pi\implies$ $\cos (y+z)=\cos x\iff$

$\cos y\cos z-\cos x=\sin y\sin z\implies$ $(\cos y\cos z-\cos x)^2=\sin^2y\sin^2z\iff$ $\cos^2y\cos^2z+\cos^2x-2\cos x\cos y\cos z=\left(1-\cos^2y\right)\left(1-\cos^2z\right)\iff (1)$ .

Using the identity $(1)$ obtain easily that $\sum a^2\left(b^2+c^2-1\right)^2=4a^2b^2c^2+\prod\left(b^2+c^2-1\right)$ . Denote $\left\{\begin{array}{c} a^2=u\\\ b^2=v\\\ c^2=w\end{array}\right\|$ . Therefore, the previous relation becomes

$\sum u(v+w-1)^2=4uvw+\prod (v+w-1)\iff$ $1+\sum u^2=\sum u+\sum vw\iff$ $1+\sum a^4=\sum a^2+\sum b^2c^2\iff (*)$ .

PP11. Let $ABCD$ be a rectangle and let $E\in (BC)$ , $F\in (CD)$ be two points so that $[ABE]=8$ , $[ADF]=5$ , $[EFC]=9$ . Find the area $[AEF]$ of $\triangle AEF$ .

Proof. Let $\left\{\begin{array}{ccc} AB=a & BC=c\\\\ BE=e & CF=f\end{array}\right\|$ , where $0<e<c$ and $0<f<a$ . So $\left\{\begin{array}{cccc} [ABE]=8 & \implies & ea=16 & (1)\\\ [CEF]=9 & \implies & f(c-e)=18 & (2)\\\ [ADF]=5 & \implies & c(a-f)=10 & (3)\end{array}\right\|\implies$ $\boxed{f\stackrel{(3)}{=}\frac {ac-10}{c}}\ (4)$ and

$e\stackrel{(1)}{=}\frac {16}{a}\implies$ $f\stackrel{(1)\wedge (2)}{=}\frac {18}{c-\frac {16}{a}}\implies$ $f=\frac {18a}{ac-16}\ (5)$ . From $(4)$ and $(5)$ obtain $(ac)^2-44\cdot (ac)+160=0\iff$ $ac=40\iff$ $[ABCD]=40\iff [AEF]=18$ .

Generalization. Let $ABCD$ be a rectangle and let $E\in (BC)$ , $F\in (CD)$ be two points so that $\left\{\begin{array}{ccc} [ABE]=x & ; & [ADF]=y\\\ [EFC]=z & ; & [AEF]=t\end{array}\right\|$ . Prove that $\boxed{t^2+4xy=(x+y+z)^2}\ (*)$ .

Proof. Denote $\left\{\begin{array}{ccc} DC=a & BC=c\\\\ DF=f & BE=e\\\\ 0<f<a & 0<e<c\end{array}\right\|$ . Observe that $\left\{\begin{array}{c} CE=c-e\\\\ CF=a-f\end{array}\right\|$ and $\left\{\begin{array}{c} 2x=ae\\\\ 2y=cf\\\\ 2z= (c-e)(a-f)\end{array}\right\|$ . Therefore, $x+y+z+t=ac$

and $t^2+4xy=(x+y+z)^2\iff$ $[ac-(x+y+z)]^2+4xy=(x+y+z)^2\iff$ $(ac)^2+2x\cdot 2y=ac\cdot (2x+2y+2z)\iff$

$(ac)^2+ae\cdot cf=ac\cdot[ae+cf+(c-e)(a-f)]\iff$ $ac+ef=ae+cf+ac-cf-ae+ef$ , what is truly. Very nice relation $(*)\ !!$

PP12. Let $P\in (AB)$ , $M\in (BC)$ , $N\in (CD)$ , $Q\in (DA)$ be four points , where $ABCD$ is a square and $\widehat{ PMA}\equiv\widehat{MAN}\equiv\widehat{ANQ}$ . Prove that $AM^2+QN^2=AN^2+PM^2$

Proof. Prove easily that $\left\{\begin{array}{ccccc} \triangle BMP\sim\triangle DNA & \implies & \frac {BP}{DN}=\frac {BM}{DA} & \implies & BP=\frac {BM\cdot DN}{DA}\\\\ \triangle DQN\sim\triangle BMA & \implies & \frac {DQ}{BM}=\frac {DN}{BA} & \implies & DQ=\frac {BM\cdot DN}{BA}\end{array}\right\|$ $\implies\ BP=DQ\ (*)$ . Therefore,

$AM^2+QN^2=AN^2+PM^2\iff$ $\left(BA^2+BM^2\right)+\left(DQ^2+DN^2\right)=\left(DA^2+DN^2\right)+\left(BP^2+BM^2\right)\iff$ $DQ=BP$ , i.e. the relation $(*)$ .

PP13. Let $ABCD$ be a quadrilateral what is inscribed in the circle with the diameter $[AC]$ . Denote $\{M,N\}\subset (BD)$ so that $\left\{\begin{array}{c} AM\perp BD\\\\ CN\perp BD\end{array}\right\|$ . Show that $BM=DN$ .

Proof 1. $\left\{\begin{array}{ccccc} \widehat{BAM}\equiv\widehat{CAD} & \implies & \triangle ABM\sim\triangle ACD & \implies & \frac {BM}{CD}=\frac {AB}{AC}\\\\ \widehat{DCN}\equiv\widehat{ACB} & \implies & \triangle CDN\sim\triangle CAB & \implies & \frac {DN}{CD}=\frac {AB}{CA}\end{array}\right\|$ $\implies BM=DN$ .

Proof 2. $\left\{\begin{array}{cc} AB=a\\\ BC=b\\\ CD=c\\\ DA=d\end{array}\right\|\implies$ $\left\{\begin{array}{c} \frac {SB}{SD}= \frac {[ABC]}{[ADC]}=\frac {ab}{cd}\\\\ \frac {MB}{MD}\cdot \frac {SB}{SD}\stackrel{(\mathrm{Steiner's\ th}.)}{=}\left(\frac ad\right)^2\\\\ \frac {ND}{NB}\cdot\frac {SD}{SB}\stackrel{(\mathrm{Steiner's\ th}.)}{=}\left(\frac cb\right)^2\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} \frac {MB}{MD}\cdot \frac {ab}{cd} & = & \left(\frac ad\right)^2\\\\ \frac {ND}{NB}\cdot\frac {cd}{ab} & = & \left(\frac cb\right)^2\end{array}\right\|\implies$ $\frac {MB}{MD}=\frac {ND}{NB}=\frac {ac}{bd}$ .

An easy extension. Let $ABCD$ be a cyclical quadrilateral. Consider two points $\{M,N\}\subset (BD)$ so that $\left\{\begin{array}{c} \widehat{BAM}\equiv\widehat{CAD}\\\\ \widehat{DCN}\equiv\widehat{ACB}\end{array}\right\|$ ,

i.e. the rays $[AM\ ,\ [AC$ are isogonals in $\widehat {BAD}$ and the rays $[CN\ ,\ [CA$ are isogonals $\widehat{BCD}$ . Show that $BM=DN$ .

Proof 1. $\left\{\begin{array}{ccccc} \widehat{BAM}\equiv\widehat{CAD} & \implies & \triangle ABM\sim\triangle ACD & \implies & \frac {BM}{CD}=\frac {AB}{AC}\\\\ \widehat{DCN}\equiv\widehat{ACB} & \implies & \triangle CDN\sim\triangle CAB & \implies & \frac {DN}{CD}=\frac {AB}{CA}\end{array}\right\|$ $\implies BM=DN$ .

Proof 2. $\left\{\begin{array}{cc} AB=a\\\ BC=b\\\ CD=c\\\ DA=d\end{array}\right\|\implies$ $\left\{\begin{array}{c} \frac {SB}{SD}= \frac {[ABC]}{[ADC]}=\frac {ab}{cd}\\\\ \frac {MB}{MD}\cdot \frac {SB}{SD}\stackrel{(\mathrm{Steiner's\ th}.)}{=}\left(\frac ad\right)^2\\\\ \frac {ND}{NB}\cdot\frac {SD}{SB}\stackrel{(\mathrm{Steiner's\ th}.)}{=}\left(\frac cb\right)^2\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} \frac {MB}{MD}\cdot \frac {ab}{cd} & = & \left(\frac ad\right)^2\\\\ \frac {ND}{NB}\cdot\frac {cd}{ab} & = & \left(\frac cb\right)^2\end{array}\right\|\implies$ $\frac {MB}{MD}=\frac {ND}{NB}=\frac {ac}{bd}$ .

Proof 3. Denote the second intersections $X$ and $Y$ of the lines $AM$ and $CN$ with the circumcircle of $ABCD$ . Observe that

$\left\{\begin{array}{c} \widehat{BAM}\equiv\widehat{CAD}\implies CX\parallel BD\\\\ \widehat{DCN}\equiv\widehat{ACB} \implies AY\parallel BD\end{array}\right\|\implies$ $AY\parallel CX$ , i.e. $AYCX$ is isosceles trapezoid. By symmetry get that $MB=ND$ .

PP14. Circles $w_k=C\left(I_k,r_k\right)\ ,\ k\in\overline{1,3}$ are all exterior tangent to each other, are all interior tangent to the circle $w_e=C(O,R)$

and are all exterior tangent to the circle $w_i=C(I,r)$ . Denote $\left\{\begin{array}{c} r_1+r_2+r_3=m\\\ r_1r_2+r_2r_3+r_3r_1=n\\\ r_1r_2r_3=p\end{array}\right\|$ . Prove that $\left\{\begin{array}{c} R=\frac {p}{2\sqrt {mp}-n}\\\\ r=\frac {p}{2\sqrt {mp}+n}\end{array}\right\|$ .

Proof. Denote $\left\{\begin{array}{c} m\left(\widehat{I_2OI_3}\right)=x\\\ m\left(\widehat{I_3OI_1}\right)=y\\\ m\left(\widehat{I_1OI_2}\right)=z\end{array}\right\|$ . Prove easily that $x+y+z=2\pi\implies 1+\sum\cos x+4\prod\cos\frac x2=0\ (*)$ and

$\left\{\begin{array}{ccc} I_2I_3=r_2+r_3 & ; & OI_1=R-r_1\\\ I_3I_1=r_3+r_1 & ; & OI_2=R-r_2\\\ I_1I_2=r_1+r_2 & ; & OI_3=R-r_3\end{array}\right\|$ . Thus, $\cos x=\frac {\left(R-r_2\right)\left(R-r_3\right)-2r_2r_3}{\left(R-r_2\right)\left(R-r_3\right)}$ and $\cos\frac x2=\sqrt {\frac {R\left(R-r_2-r_3\right)}{\left(R-r_2\right)\left(R-r_3\right)}}$ .

Now I"ll use the well-known identity $(*)$ and get finally $2\cdot\prod \left(R-r_1\right)+2R\cdot\sqrt {R\cdot\prod \left(R-r_2-r_3\right)}=n\cdot R-3p\iff$

$nR-3p-2\left(R^3-mR^2+nR-p\right)=$ $2R\cdot\sqrt {R\left[R^3-2mR^2+\left(m^2+n\right)R-(mn-p)\right]}\iff$

$\left(2R^3-2mR^2+nR+p\right)^2=4R^3\left[R^3-2mR^2+\left(m^2+n\right)R-(mn-p)\right]\iff$ $4mpR^2=(nR+p)^2\iff$

$\boxed{R=\frac {p}{2\sqrt {mp}-n}}$ . I"ll find analogously and the second relation from the conclusion. Thus, $\left\{\begin{array}{c} II_1=r+r_1\\\ II_2=r+r_2\\\ II_3=r+r_3\end{array}\right\|$ .

I"ll use same notations $\left\{\begin{array}{c} m\left(\widehat{I_2II_3}\right)=x\\\ m\left(\widehat{I_3II_1}\right)=y\\\ m\left(\widehat{I_1II_2}\right)=z\end{array}\right\|$ . Hence $\cos x=\frac {\left(r+r_2\right)\left(r+r_3\right)-2r_2r_3}{\left(r+r_2\right)\left(r+r_3\right)}$ and $\cos\frac x2=\sqrt {\frac {r\left(r+r_2+r_3\right)}{\left(r+r_2\right)\left(r+r_3\right)}}$ .

Now I"ll use again the identity $(*)$ and get finally $2\cdot\prod \left(r+r_1\right)+2r\cdot\sqrt {r\cdot\prod \left(r+r_2+r_3\right)}=n\cdot r+3p\iff$

$nr+3p-2\left(r^3+mr^2+nr+p\right)=$ $2r\cdot\sqrt {r\left[r^3+2mr^2+\left(m^2+n\right)r+(mn-p)\right]}\iff$

$\left(-2r^3-2mr^2-nr+p\right)^2=4r^3\left[r^3+2mr^2+\left(m^2+n\right)r+(mn-p)\right]\iff$ $4mpr^2=(nr-p)^2\iff$

$\boxed{r=\frac {p}{2\sqrt {mp}+n}}$ . Observe that $\left(\frac 1r+\frac 1R\right)^2=16\cdot\frac {r_1+r_2+r_3}{r_1r_2r_3}$ and $\frac 1r+\frac 1R\ge \frac {12\sqrt 3}{r_1+r_2+r_3}$ .

Remark. Denote $\left\{\begin{array}{c} r_1r_2=w\\\ r_2r_3=u\\\ r_3r_1=v\end{array}\right\|$ . This problem has a real sense iff $4mp>n^2\iff$ $4(uv+vw+wu)>(u+v+w)^2\iff$ $u^2+v^2+w^2<2(uv+vw+wu)\iff$

$u^2-2(v+w)u+(v-w)^2<0\iff$ $v+w-2\sqrt {vw}<u<v+w+2\sqrt {vw}\iff$ $\left|\sqrt v-\sqrt w\right|<\sqrt u<\sqrt v+\sqrt w\iff$

$\left|\sqrt {r_1r_2}-\sqrt {r_1r_3}\right|<\sqrt {r_2r_3}<\sqrt {r_1r_2}+\sqrt {r_1r_3}\iff$ $\left|\frac {1}{\sqrt {r_2}}-\frac {1}{\sqrt {r_3}}\right|<\frac {1}{\sqrt{r_1}}<\frac {1}{\sqrt{r_2}}+\frac {1}{\sqrt {r_3}}$ , i.e. the values $\frac {1}{\sqrt{r_k}}\ ,\ k\in\overline{1,3}$ are the sidelengths of a triangle.

I used the well-known identities in $\triangle ABC\ :\ \cos A=\frac {b^2+c^2-a^2}{2bc}$ and $\cos \frac A2=\sqrt {\frac {s(s-a)}{bc}}$ , where $2s=a+b+c$ .

PP15. Let $P$ be an interior point of $\triangle ABC$ and denote $E\in BP\cap AC$ and $F\in CP\cap AB$ . Suppose that $[BPF] = 4$ , $[BPC]= 8$ and $[CPE] = 13$ . Find $[AFPE]\ .$

Proof. Denote $u=[AEF]\ ,\ v=[PEF]\ ,\ x=[AEPF]\ (u+v=x)$ . $(1)\ \ \boxed {\ [BPF]\cdot [CPE]=[PEF]\cdot [PBC]\ }\Longrightarrow 4\cdot 13=8\cdot v\Longrightarrow \boxed {\ v=\frac{13}{2}\ }\ .$

$\frac{[BAE]}{[BCE]}=\frac{EA}{EC}=\frac{[AEF]}{[CEF]}\Longrightarrow \frac{4+\frac{13}{2}+u}{8+13}=$ $\frac {u}{\frac{13}{2}+13}=\frac{4+\frac{13}{2}}{8-\frac{13}{2}}=$ $\frac{21}{3}=7\Longrightarrow$ $\boxed {\ u=21\cdot \frac{13}{2}}\Longrightarrow x=u+v=\frac{13}{2}\cdot (1+21)=13\cdot 11=143\Longrightarrow \boxed {\ x=143\ }\ .$

Remark. The relation $(1)$ is well-known in a convex quadrilateral $BCEF$ , where $P\in BE\cap CF\ !$

An easy extension. Let $P$ be an interior point of $\triangle ABC$ , $E\in BP\cap AC$ , $F\in CP\cap AB$ . Prove that $[BPF]=m\ \wedge\ [CPE]=$

$n\ \wedge\ [BPC]=p\ \implies\ [AFPE]=\frac {mn(2p+m+n)}{p^2-mn}\ .$ Particular case : $\left\{\begin{array}{c} m=2\\\ n=4\\\ p=8\end{array}\right\|\implies [APFE]=\frac {22}{7}\approx \pi$ .

Lemma. For the point $P$ which belongs to the inside of $\triangle ABC$ denote $\left\{\begin{array}{c} D\in AP\cap BC\\\\ E\in BP\cap CA\\\\ F\in CP\cap AB\end{array}\right|\implies$ $\boxed{\ \prod\frac {PA}{PD}=2+\sum\frac {PA}{PD}\ }\ (*)$ .

Proof. $(\exists ) \{x,y,z\}\subset \mathbb R^*_+$ so that $\left\{\begin{array}{c} \frac {DB}{DC}=\frac zy\\\\ \frac {EC}{EA}=\frac xz\\\\ \frac {FA}{FB}=\frac yx\end{array}\right|$ . Apply the van Aubel's relations $\left\{\begin{array}{c} \frac {PA}{PD}=\frac {EA}{EC}+\frac {FA}{FB}=\frac {y+z}{x}\\\\ \frac {PB}{PE}=\frac {FB}{FA}+\frac {DB}{DC}=\frac {z+x}{y}\\\\ \frac {PC}{PF}=\frac {DC}{DB}+\frac {EC}{EA}=\frac {x+y}{z}\end{array}\right|$ . Show easily that $(*)\iff$ $\prod (y+z)=2xyz+\sum yz(y+z)$ .

PP16. For the point $P$ which belongs to the inside of $\triangle ABC$ denote $\left\{\begin{array}{c} D\in AP\cap BC\\\\ E\in BP\cap CA\\\\ F\in CP\cap AB\end{array}\right|$ . Suppose that $\left\{\begin{array}{c} PA=6\ ;\ PD=6\\\\ PB=9\ ;\ PE=3\end{array}\right|$ and $CF=20$ . Find the area $S=[ABC]$ .

Proof. Denote $\frac {PC}{PF}=r$ and apply the relation $(*)\ :\ \frac 66\cdot$ $\frac 93\cdot r=2+\frac 66+\frac 93+r\implies r=3\implies \frac {PC}{PF}=3$ . Observe that $\frac {PB}{PE}=\frac {PC}{PF}=3$ , i.e. $D$ is the midpoint of $[BC]$

and $CF=20\implies$ $\boxed{\ PC=15\ \ \wedge\ \ PF=5\ }$ . If $L$ is the midpoint of $[PB]$ , then $\left\{\begin{array}{c} PD=6\\\\ DL=\frac {15}{2}\\\\ LP=\frac 92\end{array}\right|$ . Observe that $\frac {6}{4}=\frac {\frac {15}{2}}{5}=\frac {\frac 92}{3}$ , i.e. $PDL$ is $P$ -right-angled triangle. Therefore,

$[PDL]=\frac 12\cdot PD\cdot PL$ , i.e. $\boxed{\ [PDL]=\frac {27}{2}\ }$ . Since $\frac {[PDL]}{[ACB]}=\frac {[PDL]}{[PDB]}\cdot\frac {[PDB]}{[PCB]}\cdot\frac {[PCB]}{[ACB]}=$ $\frac {PL}{PB}\cdot\frac {DB}{CB}\cdot\frac {PD}{AD}=\frac 18\implies$ $S=8\cdot [PDL]\implies$ $\boxed{\ S=108\ }$ .

Generalization. Denote $[x,y,z]=[XYZ]$ with $\left\{\begin{array}{c} XY=z\\\ YZ=x\\\ ZX=y\end{array}\right|$ and for the point $P$ which belongs to the inside of $\triangle ABC$ denote

$\left\{\begin{array}{c} D\in AP\cap BC\\\\ E\in BP\cap CA\\\\ F\in CP\cap AB\end{array}\right|$ . Prove that $\boxed{\ [ABC]=\frac {DA}{DP}\cdot\frac {EB}{EP}\cdot\frac {FC}{FP}\cdot\left[\frac {PA\cdot PD}{AD},\frac {PB\cdot PE}{BE},\frac {PC\cdot PF}{CF}\right]\ }$ .

Proof. Suppose that $\left\{\begin{array}{c} PA=a_1\ ;\ PD=a_2\\\\ PB=b_1\ ;\ PE=b_2\\\\ PC=c_1\ ;\ PF=c_2\end{array}\right|$ , where $\frac {a_2}{a_1+a_2}+\frac {b_2}{b_1+b_2}+\frac {c_2}{c_1+c_2}=1$ . Observe that $P$ has the barycentrical coordinates $\left(\begin{array}{c} x=\frac {a_2}{a_1+a_2}\\\\ y=\frac{b_2}{b_1+b_2}\\\\ z=\frac {c_2}{c_1+c_2}\end{array}\right)$ w.r.t $\triangle ABC$ .

Denote $L\in [PB]$ so that $DL\parallel CP$ . Thus, $\frac {LB}{LP}=\frac {DB}{DC}\implies$ $\frac {LB}{z}=\frac {LP}{y}=\frac {b_1}{1-x}=\frac {b_1(a_1+a_2)}{a_1}\implies$ $\boxed{\ LP=\frac {b_1b_2(a_1+a_2)}{a_1(b_1+b_2)}\ }$ . Analogously obtain that $\frac {LD}{PC}=\frac {BD}{BC}\implies$

$\frac {LD}{c_1}=\frac {z}{1-x}\implies$ $\boxed{\ LD=\frac {c_1c_2(a_1+a_2)}{a_1(c_1+c_2)}\ }$ . Therefore, $[PDL]=\left[a_2\ ,\ \frac {b_1b_2(a_1+a_2)}{a_1(b_1+b_2)}\ ,\ \frac {c_1c_2(a_1+a_2)}{a_1(c_1+c_2)}\right]\implies$ $\boxed{\ [PDL]=\frac {(a_1+a_2)^2}{a_1^2}\cdot\left[\frac {a_1a_2}{a_1+a_2}\ ,\ \frac {b_1b_2}{b_1+b_2}\ ,\ \frac {c_1c_2}{c_1+c_2}\right]\ }$ .

Since $\frac {[PDL]}{[ACB]}=\frac {[PDL]}{[PDB]}\cdot\frac {[PDB]}{[PCB]}\cdot\frac {[PCB]}{[ACB]}=$ $\frac {PL}{PB}\cdot\frac {DB}{CB}\cdot\frac {PD}{AD}\implies$ $S=[PDL]\cdot \frac {PB}{PL}\cdot\frac {CB}{DB}\cdot\frac {AD}{PD}=$ $\frac {(a_1+a_2)^2}{a_1^2}\cdot\left[\frac {a_1a_2}{a_1+a_2}\ ,\ \frac {b_1b_2}{b_1+b_2}\ ,\ \frac {c_1c_2}{c_1+c_2}\right]\cdot$ $\frac {a_1(b_1+b_2)}{b_2(a_1+a_2)}\cdot$

$\frac {a_1(c_1+c_2)}{c_2(a_1+a_2)}\cdot$ $\frac {a_1+a_2}{a_2}$ $\implies$ $S=\frac {(a_1+a_2)(b_1+b_2)(c_1+c_2)}{a_2b_2c_2}\cdot$ $\left[\frac {a_1a_2}{a_1+a_2}\ ,\ \frac {b_1b_2}{b_1+b_2}\ ,\ \frac {c_1c_2}{c_1+c_2}\right]\implies$ $[ABC]=\frac {DA}{DP}\cdot$ $\frac {EB}{EP}\cdot\frac {FC}{FP}\cdot$ $\left[\frac {PA\cdot PD}{AD},\frac {PB\cdot PE}{BE},\frac {PC\cdot PF}{CF}\right]$ .

Remark. In our particular case obtain $S=\frac{12}{6}\cdot \frac {12}{3}\cdot\frac {20}{5}\cdot\left[\frac {36}{12},\frac {27}{12},\frac {75}{20}\right]=$ $2\cdot 4\cdot 4\cdot \left[3,\frac 94,\frac {15}{4}\right]=$ $2\cdot 4\cdot 4\cdot \left(\frac {3}{4}\right)^2\cdot\left[4,3,5\right]=18\cdot 6=108$ .

PP17. Let $w=C(O,r)$ be a circle and let $\{A,K\}\subset w$ be two points so that $OA\perp OK$ . Denote the point $M$ of $[OK]$ so that $m\left(\widehat{MAO}\right)=\phi$ ,

$N\in (OA)$ so that $\widehat{NMA}=\widehat {NMO}$ and the point $B\in w$ such that $BN\parallel OK$ and $AM$ separates $B\ ,\ N$ . Find the measure of the angle $\widehat{ABN}$ .

Proof. Suppose w.l.o.g. $r=1$ and denote the diameter $[AS]$ of $w$ . Thus, $\widehat{NMA}\equiv\widehat{NMO}\iff$ $\frac {NA}{MA}=\frac {NO}{MO}\implies$ $\frac {NA}{1}=\frac {NO}{\sin\phi}=\frac {1}{1+\sin\phi}\implies$

$\boxed{\ AN=\frac {1}{1+\sin\phi}\ }$ . So $AB^2=AN\cdot AS=\frac {2}{1+\sin\phi}\implies$ $\sin\widehat{ABN}=\frac {AN}{AB}=\frac {\frac {1}{1+\sin\phi}}{\sqrt {\frac {2}{1+\sin\phi}}}\implies$ $\boxed{\ \sin \widehat{ABN}=\frac {1}{\sqrt {2(1+\sin\phi )}}\ }$ .

Remark. If the point $M$ is the midpoint of $OK$ , Then $\tan\phi=\frac 12\iff$ $\sin\phi=\frac {1}{\sqrt 5}\iff$ $\sin\widehat{ABN}=\frac {\sqrt {5-\sqrt 5}}{2\sqrt 2}\iff\cos \widehat{ABN}=\frac {1+\sqrt 5}{4}\iff$ $m\left(\widehat{ABN}\right)=36^{\circ}$ .

Directly. Suppose w.l.o.g. $r=1$ and let diameter $[AS]$ $w$ . Then $\widehat{NMA}\equiv\widehat{NMO}\iff$ $\frac {NA}{\sqrt 5}=\frac {NO}{1}=\frac {1}{1+\sqrt 5}\implies$ $\cos\widehat{ABN}=\frac {BN}{BA}=\sqrt {\frac{BN^2}{BA^2}}=$ $\sqrt {\frac {NA\cdot NC}{NA\cdot AC}}=$

$\sqrt {\frac {NC}{AC}}=$ $\sqrt {\frac {1+\frac {1}{1+\sqrt 5}}{2}}=$ $\sqrt {\frac {2+\sqrt 5}{2\left(1+\sqrt 5\right)}}=$ $\sqrt {\frac {\left(2+\sqrt 5\right)\left(\sqrt 5-1\right)}{8}}=$ $\sqrt {\frac {3+\sqrt 5}{8}}=$ $\frac {1}{2\sqrt 2}\cdot\frac {1+\sqrt 5}{\sqrt 2}\implies$ $\cos\widehat{ABN}=\frac {1+\sqrt 5}{4}=\cos 36^{\circ}\implies$ $m\left(\widehat {ABN}\right)=36^{\circ}$ .

PP18. Let an acute $\triangle ABC$ with the midpoint $M$ of $[BC]$ . Draw the perpendicular $HP$ from the orthocenter $H$ to $AM$ . Show that $MA\cdot MP=BM^2$ .

Proof 1 (synthetic). Let $D\in BC$ , $E\in CA$ , $F\in AB$ be the projections of $H$ to the sides of $\triangle ABC$ and $T\in EF\cap BC$ . Are well-known

that $T\in HP$ and $(T,B,D,C)$ is a harmonical division. In conclusion, $MB^2=MD\cdot MT=MP\cdot MA$ , i.e. $MB^2=MP\cdot MA$ .

Proof 2 (metric). Denote $D\in AH\cap BC$ . Using the power of $A$ w.r.t. the circumcircle of the quadrilateral $HDMP$

obtain that $AH\cdot AD=AP\cdot AM\iff$ $2Rh_a\cos A=$ $m_a\cdot AP\iff$ $AP\stackrel{(2Rh_a=bc)}{\ \ =\ \ }\frac {bc\cdot \cos A}{m_a}$ $\implies$ $MA\cdot MP=$

$m_a\left(m_a-\frac {bc\cdot \cos A}{m_a}\right)=$ $m_a^2-bc\cdot \cos A=\frac 14\cdot\left[2\left(b^2+c^2\right)-a^2-2\cdot \left(b^2+c^2-a^2\right)\right]=\frac {a^2}{4}=MB^2$ .

Proof 3 (metric). $AFHPE$ is inscribed in the circle with the diameter $[AH]$ . Thus, $\widehat{ACM}\equiv\widehat{AFE}\equiv \widehat{APE}\implies$ $\widehat{ACM}\equiv\widehat{APE}\implies$ $PECM$ is cyclically.

Therefore, $ME=MC\implies$ $\widehat{MCE}\equiv\widehat{MPC}\implies$ $\triangle MPC\sim\triangle MCA\implies$ $\frac {MP}{MC}=\frac {MC}{MA}\implies$ $MA\cdot MP=MC^2$ , i.e. $MA\cdot MP=MB^2$ .

This post has been edited 274 times. Last edited by Virgil Nicula, Nov 18, 2015, 9:34 PM

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I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
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El_Ectric:
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Nice blog!
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Dear Virgil !

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

72 shouts

My (math)blog

337. Nice metrical problems.

by Virgil Nicula, Feb 26, 2012, 1:37 PM

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