428. Some metrical problems from the contests II.

by Virgil Nicula, Sep 25, 2015, 7:54 PM

PP10. For a convex $ABCD$ which is inscribed in $w$ denote $E\in AB\cap CD$ and $F\in AD\cap BC$ . Prove that $p_w(E)+p_w(F)=EF^2$ , where $p_w(X)$ is the power of $X$ w.r.t. $w$ .

Proof. Let the circumcircles $e$ , $f$ of $\triangle BCE$ , $\triangle BAF$ respectively and $\{B,L\}=e\cap f$ . Thus, $\left\{\begin{array}{ccc} \widehat{BLF} & \equiv & \widehat{BAD}\\\\ \widehat{BLE} & \equiv & \widehat{BCD}\end{array}\right\|$ $\bigoplus\implies$ $\left(\widehat{BLF}\right)+\left(\widehat{BLE}\right)=180^{\circ}\implies$

$\boxed{L\in EF}\implies$ $\boxed{LE+LF=EF}\ (*)$ . In conclusion, $\left\{\begin{array}{ccc} p_w(E) & = & EA\cdot EB=EL\cdot EF\\\\ p_w(F) & = & FB\cdot FC=FL\cdot FE\end{array}\right\|$ $\bigoplus\ \stackrel{(*)}{\implies}\ p_w(E)+p_w(F)=EF^2\ .$

PP11. Let a semicircle $w=\mathcal S(O,R)$ with the diameter $[AB]$ . Denote $\{C,D\}\subset w$ so that $CO\perp AB$ , $D\in\overarc{BC}$ and $E\in AD\cap OE\ .$ Prove that $\frac {EC\cdot DB}{OE\cdot CD}=\sqrt 2\ .$

Proof 1. Suppose w.l.o.g. $CA=CB=1$ , i.e. $AB=\sqrt 2$ . Observe that $\frac {EC}{EO}=\frac {AC}{AO}\cdot\frac {\sin \widehat{EAC}}{\sin\widehat{EAO}}=\sqrt 2\cdot\frac {\sin \widehat{DAC}}{\sin\widehat{DAB}}=\sqrt 2\cdot\frac {DC}{DB}\implies$ $\frac {EC\cdot DB}{OE\cdot CD}=\sqrt 2\ .$

Proof 2. Suppose w.l.o.g. $CA=CB=1$ , i.e. $AB=\sqrt 2$ . Observe that $\frac {EC}{EO}=\frac {DC}{DO}\cdot\frac {\sin \widehat{EDC}}{\sin\widehat{EDO}}=$ $\frac {DC}{AO}\cdot\frac {\sin 45^{\circ}}{\sin\widehat{DAO}}=$ $\frac {DC}{AO}\cdot \frac {\sqrt 2}2\cdot\frac {AB}{DB}=$ $\frac {DC}{DB}\cdot\sqrt 2\implies$ $\frac {EC\cdot DB}{OE\cdot CD}=\sqrt 2\ .$

Proof 3. Suppose w.l.o.g. $CA=CB=1$ , i.e. $AB=\sqrt 2$ . Apply the Ptolemy's theorem to the cyclic $ACDB\ :\ AC\cdot BD+AB\cdot CD=AD\cdot BC$ $\iff$

$\boxed{BD+CD\sqrt 2=AD}\ (*)\ .$ Therefore, $\triangle DAB\sim\triangle OAE\iff$ $\frac {DA}{OA}=\frac {DB}{OE}\iff$ $DB\cdot OC=DA\cdot OE\iff$ $DB\cdot (OE+EC)=DA\cdot OE\iff$

$OE\cdot (DA-DB)=DB\cdot EC\ \stackrel{(*)}{\iff}$ $\ OE\cdot CD\cdot\sqrt 2=$ $DB\cdot EC\iff$ $\frac {EC\cdot DB}{OE\cdot CD}=\sqrt 2\ .$

Proof 4. Suppose w.l.o.g. $CA=CB=1$ , i.e. $AB=\sqrt 2$ . Apply the Menelaus' theorem to the transversal $\overline {AEI}$ and $\triangle OCB\ :\ \frac {AO}{AB}\cdot\frac {IB}{IC}\cdot\frac {EC}{EO}=1\iff$ $\boxed{\frac {IB}{IC}\cdot\frac {EC}{EO}=2}\ (1)\ .$

Therefore, $\frac {IC}{IB}=\frac {DC}{DB}\cdot\frac {\sin\widehat{IDC}}{\sin\widehat{IDB}}\iff$ $\boxed{\frac {IC}{IB}=\frac {DC}{DB}\cdot\frac {\sqrt 2}2}\ (2)\ .$ From the product of the relations $(1)$ and $(2)$ obtain that $\frac {EC}{EO}=\frac {DC}{DB}\cdot\sqrt 2$ , i.e. $\frac {EC\cdot DB}{OE\cdot CD}=\sqrt 2\ .$

An easy extension. Let $ACDB$ be an convex cyclical quadrilateral. For a point $F\in (AB)$ denote $E\in AD\cap CF$ . Provethat $\frac {EC\cdot DB}{EF\cdot DC}=\frac {AC}{AF}$ .

Proof. $\frac {EC}{EF}=\frac {AC}{AF}\cdot\frac {\sin\widehat{EAC}}{\sin\widehat{EAF}}=\frac {AC}{AF}\cdot\frac {DC}{DB}\implies$ $\frac {EC\cdot DB}{EF\cdot DC}=\frac {AC}{AF}$ . In the particular case when $F:=O$ - circumcenter of $ACDB$ and $CF\perp AB$ get the proposed problem.

PP12. Let $\triangle ABC$ with incircle $w=C(I,r)$ . Let $\left\{\begin{array}{c} E\in AC\cap w\\\ F\in AB\cap w\end{array}\right|$ . Circle with diameter $[BC]$ cut $EF$ in $\{G,H\}$ so that $G\in (FH)$ . Prove that $\frac {[AFG]}{[AEH]}=\left|\frac {a-b}{a-c}\right|$ .

Proof. Is well-known that $I\in BH\cap CG$ and prove easily that $m\left(\widehat {HBG}\right)=\frac A2$ and $GH=BC\cdot\sin\widehat HBG$ , i.e. $GH=a\cdot\sin\frac A2$ . Apply the theorem of Sinus :

$\left\{\begin{array}{cccccc} \triangle BFH\ : & \frac {FH}{\sin\frac B2}=\frac {BF}{\sin \frac C2} & \implies & FG=FH-GH=\frac {(s-b)\sin \frac B2}{\sin\frac C2}-a\cdot\sin \frac A2 & \implies & FG=|a-b|\cdot\sin\frac A2\\\\ \triangle CGE\ : & \frac {EG}{\sin\frac C2}=\frac {CE}{\sin \frac B2} & \implies & EH=EG-HG=\frac {(s-c)\sin \frac C2}{\sin\frac B2}-a\cdot\sin \frac A2 & \implies & EH=|a-c|\cdot\sin\frac A2\end{array}\right|$ .

In conclusion, $\frac {[AFG]}{[AEH]}=\frac {AF\cdot FG\cdot\sin\widehat{AFG}}{AE\cdot EH\cdot\sin\widehat{AEH}}=\frac {FG}{CE}\implies$ $\frac {[AFG]}{[AEH]}=\left|\frac {a-b}{a-c}\right|\ .$

PP13. Let $ABC$ be a triangle so that $b\ne c$ . The medians $AD$ , $BE$ , $CF$ ( where $D\in BC$ , $E\in CA$ , $F\in AB$ )

intersect again the circumcircle $w$ of $ABC$ at $L$ , $M$ , $N$ respectively. Prove that $LM = LN\ \Longleftrightarrow\ b^2 + c^2 = 2a^2$ .

Proof. Apply the power of the points $D$ , $E$ , $F$ w.r.t. circle $w$ :

$\left\{\begin{array}{ccccc} DL\cdot DA = DB\cdot DC & \implies & LD = \frac {a^2}{4m_a} & \implies & LA = LD + m_a = \frac {b^2 + c^2}{2m_a} \\ \\ EM\cdot EB = EC\cdot EA & \implies & ME = \frac {b^2}{4m_b} & \implies & MB = ME + m_b = \frac {c^2 + a^2}{2m_b} \\ \\ FN\cdot FC = FA\cdot FB & \implies & NF = \frac {c^2}{4m_c} & \implies & NC = NF + m_c = \frac {a^2 + b^2}{2m_c}\end{array}\right\}$

Apply the equivalence of areas and Ptolemeu's therem to the quadrilaterals $ABLC$ , $BCMA$ , $CANB$ :

$\left\{\begin{array}{cccccc} \left\|\begin{array}{c} c\cdot LB = b\cdot LC \\ \\ b\cdot LB + c\cdot LC = a\cdot LA\end{array}\right\| & \implies & \left\|\begin{array}{c} LB = \frac {ab}{b^2 + c^2}\cdot LA \\ \\ LC = \frac {ac}{b^2 + c^2}\cdot LA\end{array}\right\| & \implies & \left\|\begin{array}{c} LB = \frac {ab}{2m_a} \\ \\ LC = \frac {ac}{2m_a}\end{array}\right\| \\ \\ \left\|\begin{array}{c} a\cdot MC = c\cdot MA \\ \\ c\cdot MC + a\cdot MA = b\cdot MB\end{array}\right\| & \implies & \left\|\begin{array}{c} MC = \frac {bc}{c^2 + a^2}\cdot MB \\ \\ MA = \frac {ba}{c^2 + a^2}\cdot MB\end{array}\right\| & \implies & \left\|\begin{array}{c} MC = \frac {bc}{2m_b} \\ \\ MA = \frac {ba}{2m_b}\end{array}\right\| \\ \\ \left\|\begin{array}{c} b\cdot NA = a\cdot NB \\ \\ a\cdot NA + b\cdot NB = c\cdot NC\end{array}\right\| & \implies & \left\|\begin{array}{c} NA = \frac {ca}{a^2 + b^2}\cdot NC \\ \\ NB = \frac {cb}{a^2 + b^2}\cdot NC\end{array}\right\| & \implies & \left\|\begin{array}{c} NA = \frac {ca}{2m_c} \\ \\ NB = \frac {cb}{2m_c}\end{array}\right\|\end{array}\right\}$ Apply Ptolemeu's theorem to $LBNC$ , $LCMB$ :

$\left\{\begin {array}{c} a\cdot LN = LB\cdot NC + LC\cdot NB = \frac {ab(a^2 + b^2 + c^2)}{4m_am_c} \\ \\ a\cdot LM = LB\cdot MC + LC\cdot MB = \frac {ac(a^2 + b^2 + c^2)}{4m_am_b}\end{array}\right\}$ . Therefore, $LN = LM$ $\Longleftrightarrow$ $bm_b = cm_c$ $\Longleftrightarrow$ $b^2 + c^2 = 2a^2$ .

Virgil Nicula wrote:

Generalization. Let $ABC$ be a triangle, $b\ne c$ with the circumcircle $w$ . For an interior point $P(x,y,z)$ denote the second intersections $L$ , $M$ , $N$ of the

circle $w$ with $AP$ , $BP$ , $CP$ respectively. Prove that $LM = LN\ \Longleftrightarrow\ x\left[\left(zb^4 - yc^4\right) + (y - z)b^2c^2\right] = a^2\left[z(x + z)b^2 - y(x + y)c^2\right]$ .

Remark. I denoted by $P(x,y,z)$ the point $P$ with the the barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$ . Here are two particular cases.

$1.\ \blacktriangleright\ \ P = G(1,1,1)$ - centroid : $LM = LN\ \Longleftrightarrow\ b^2 + c^2 = 2a^2$ .

$2.\ \blacktriangleright\ \ P = I(a,b,c)$ - incenter : $LM = LN\ \Longleftrightarrow\ \emptyset$ a.s.o.

PP14. Given an acute triangle $ABC$ with the orthocenter $H$ and the midpoint $M$ of $[BC]$ . Denote the projection $P$ of $H$ on $AM$ . Show that $MA\cdot MP=MB^2$ .

Proof 1 (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 2. 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 the relations

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

PP15. Let $\triangle ABC$ and $D\in (BC)$ so that $AD\perp BC$ . Let perimeters $P$ , $X$ , $Y$ of $\triangle ABC$ , $\triangle ABD$ , $\triangle ACD$ respectively. Prove that $AB\perp AC\Longleftrightarrow P^2=X^2+Y^2$ .

Proof 1. Denote $2s=a+b+c$ , $AD=h$ , $DB=x$ and $DC=y$ , i.e. $x+y=a$ . Therefore, $P^2=X^2+Y^2\iff$ $(c+x+h)^2+(b+y+h)^2=(a+b+c)^2\iff$

$[(c+x+h)+(b+y+h)]^2=(a+b+c)^2+2(c+x+h)(b+y+h)\iff$ $2(s+h)^2=2s^2+(c+x)(b+y)+2hs+h^2\iff$

$\ (1)\boxed{h(a+b+c)+h^2=bc(1+\cos B)(1+\cos C)}\iff$ $2s\cdot \frac {2S}{a}+\left(\frac {2S}{a}\right)^2=4bc\cdot\frac {s(s-b)}{ac}\cdot\frac {s(s-c)}{ab}\iff$ $4asS+4S^2=4s^2(s-b)(s-c)\iff$

$ar+r^2=(s-b)(s-c)\iff$ $r(s-a)(a+r)=sr^2\iff$ $(s-a)(a+r)=sr\iff$ $r=s-a\iff$ $\frac r{s-a}=1\iff$ $\tan\frac A2=1\iff$ $A=90^{\circ}\iff AB\perp AC$ .

Proof 2. Denote $2s=a+b+c$ , $AD=h$ , $DB=x$ and $DC=y$ , i.e. $x+y=a$ . Therefore, $P^2=X^2+Y^2\iff$ $(c+x+h)^2+(b+y+h)^2=(a+b+c)^2=$

$[(c+x)+(b+y)]^2\iff$ $(c+x)^2+(b+y)^2+2h[(c+x)+(b+y)]+2h^2=$ $(c+x)^2+(b+y)^2+2(c+x)(b+y)\iff$

$(2)\ \left|\boxed{h^2+h(a+b+c)=(c+x)(b+y)}\right|\ :\ (bc)\iff$ $\frac hb\cdot\frac hc+\frac {2S}{bc}+\frac hc+\frac hb=$ $\left(1+\frac xc\right)\left(1+\frac yb\right)\iff$ $\sin C\sin B+\sin A+\sin B+\sin C=$

$1+\cos B+\cos C+\cos B\cos C\iff$ $\sum \sin A=1-\cos A+\cos B+\cos C\iff$ $4\prod\cos\frac A2=2\sin^2\frac A2+2\cos\frac {B+C}{2}\cos\frac {B-C}{2}\iff$

$4\prod\cos\frac A2=2\sin\frac A2\left(\cos\frac {B+C}{2}+\cos\frac {B-C}{2}\right)\iff$ $4\prod\cos\frac A2=4\sin\frac A2\cos\frac B2\cos\frac C2\iff$ $\tan\frac A2=1\iff A=90^{\circ}\iff$ $AB\perp AC$ .

Remark. Since $\left\{\begin{array}{c} x=c\cdot\cos B\\\ y=b\cdot\cos C\end{array}\right|$ means that the relations $(1)$ and $(2)$ are equivalently (the first and second proofs have a common point).

Denote the incenters $U$ , $V$ of the triangles $ABD$ , $ACD$ respectively and $\left\{\begin{array}{c} X\in BU\cap AD\\\ Y\in CV\cap AD\end{array}\right\|$ . Since $a=(c+x)+(b+y)$ the relation $(2)$ is equivalently with

$[h+(c+x)][h+(b+y)]=2(c+x)(b+y)\iff$ $\left(1+\frac{h}{c+x}\right)\left(1+\frac {h}{b+y}\right)=2\iff$ $\left(1+\frac{UX}{UB}\right)\left(1+\frac {VY}{VC}\right)=2\iff$ $\frac {BX}{BU}\cdot\frac {CY}{CV}=2$ .

Proof 3 (trigonometric). Suppose w.l.o.g. that $2R=1$ , where $R$ is the length of the circumradius. Thus, $\left\{\begin{array}{ccc} a=\sin A & ; & h=\sin B\sin C\\\\ b=\sin B & ; & c=\sin C\\\\ DB=\sin C\cos B & ; & DC=\sin B\cos C\end{array}\right\|$ and $a=DB+DC=$

$\sin C\cos B+\sin B\cos C=\sin A$ . Therefore, $P^2=X^2+Y^2\iff$ $\sin^2C(1+\sin B+\cos B)^2+$ $\sin^2B(1+\sin C+\cos C)^2=$ $(\sin A+\sin B+\sin C)^2\iff$

$\sin^2C\left(2\cos^2\frac B2+2\sin\frac B2\cos\frac B2\right)^2+$ $\sin^2B\left(2\cos^2\frac C2+2\sin\frac C2\cos\frac C2\right)^2=$ $\left(\sin A+\sin B+\sin C\right)^2\iff$ $4\sin^2C\cos^2\frac B2\left(\sin\frac B2+\cos\frac B2\right)^2+$

$4\sin^2B\cos^2\frac C2\left(\sin\frac C2+\cos\frac C2\right)^2=\left(4\cos\frac A2\cos\frac B2\cos\frac C2\right)^2\iff$ $\sin^2\frac C2\cos^2\frac C2\cos^2\frac B2(1+\sin B)+$ $\sin^2\frac B2\cos^2\frac B2\cos^2\frac C2(1+\sin C)=$ $\cos^2\frac A2\cos^2\frac B2\cos^2\frac C2$

$\iff$ $\sin^2\frac C2(1+\sin B)+$ $\sin^2\frac B2(1+\sin C)=$ $\cos^2\frac A2\iff$ $(1-\cos C)(1+\sin B)+$ $(1-\cos B)(1+\sin C)=1+\cos A\iff$ $\sin B+\sin C-\sin A=$

$\sum\cos A-1$ . Using the identity $\sum \cos A=1+\frac rR$ obtain that $P^2=X^2+Y^2\iff$ $\frac {b+c-a}{2R}=\frac rR\iff$ $\frac {r}{s-a}=1\iff$ $\tan\frac A2=1\iff$ $A=90^{\circ}\iff$ $AB\perp AC$ .

PP16. $ABCD$ - parallelogram $\implies\ (\forall )\ P\ ,\ PA^2+PC^2=$ $PB^2+PD^2+2\cdot AB\cdot AD\cdot\cos\alpha$ , where $m\left(\widehat{BAD}\right)=\alpha$ .

Proof 1 (metric). Let $O\in AC\cap BD$ . So $2\left[\left(PA^2+PC^2\right)-\left(PB^2+PD^2\right)\right]=$ $\left(4\cdot PO^2+AC^2\right)-\left(4\cdot PO^2+BD^2\right)=$ $AC^2-BD^2=$ $AC^2+BD^2-2\cdot BD^2=$

$2\cdot \left(AB^2+AD^2\right)-2\cdot BD^2=2\cdot\left(AB^2+AD^2-BD^2\right)\implies$ $\left(PA^2+PC^2\right)-\left(PB^2+PD^2\right)=AB^2+AD^2-BD^2=$ $2\cdot AB\cdot AD\cdot\cos\alpha\implies$

$\boxed{PA^2+PC^2=PB^2+PD^2+2\cdot AB\cdot AD\cdot\cos\alpha}$ .

Proof 2 (vectorial). $\left\{\begin{array}{c} \overrightarrow{AB}=\overrightarrow{DC}=\vec a\\\\ \overrightarrow{AD}=\overrightarrow{BC}=\vec b\end{array}\right\|\implies$ $\left\{\begin{array}{ccc} \overrightarrow{AP}=\vec a+\overrightarrow{BP} & \implies & AP^2=a^2+2\vec a\cdot\overrightarrow{BP}+BP^2\\\\ \overrightarrow{CP}=-\vec a+\overrightarrow{DP} & \implies & CP^2=a^2-2\vec a\cdot\overrightarrow{DP}+DP^2\end{array}\right\|\bigoplus$ $\implies$ $AP^2+CP^2=$

$2a^2+2\vec a\cdot(\overrightarrow{BP}-\overrightarrow{DP})+BP^2+DP^2=$ $BP^2+DP^2+2a^2+2\vec a\cdot\overrightarrow{BD}=$ $BP^2+DP^2+2\vec a(\vec a+\overrightarrow{BD})\implies$ $\boxed{AP^2+CP^2=BP^2+DP^2+2\vec a\cdot\vec b}$ .

PP17. Circle $w(O,r)$ meet at $X$ and $Y$ the circle $C$ and $O\in C$ . Point $Z\in C$ and belongs to the outside of $w$ so that $XZ=13$ , $OZ=11$ and $YZ=7$ . Find the radius of circle $w$ .

Lemma. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ . The line $AI$ cut again the circumcircle of $\triangle ABC$ in the point $S$ . Prove that $IS^2+AB\cdot AC=AS^2$ .

Proof 1 (synthetical). Prove easily that $\triangle ABI_a\sim\triangle AIC$ . Thus, $\frac {AB}{AI}=\frac {AI_a}{AC}\iff$ $AI\cdot (AI+2\cdot IS)=bc\iff$ $(AS-IS)(AS+IS)=bc\iff$

$AS^2=IS^2+bc$ . I used the well-known property $SB=SI=SC=SI_a$ . Observe that the power of the point $A$ w.r.t. the circle $C(S,SI)$ is equally to $bc$ .

Proof 2 (synthetical). Denote $D\in AS\cap BC$ . Prove easily that $SI=SB$ and $\triangle ABS\sim\triangle ADC\iff$ $\frac {AB}{AD}=\frac {AS}{AC}\iff$ $\boxed{AD\cdot AS=AB\cdot AC}\ (*)$ .

Therefore, $\triangle SBD\sim \triangle SAB\iff$ $\frac {SB}{SA}=\frac {SD}{SB}\iff$ $SA\cdot SD=SB^2\iff$ $SA(SA-AD)=SI^2\iff$ $IS^2+SA\cdot AD=SA^2\stackrel{(*)}{\iff}$ $IS^2+AB\cdot AC=AS^2$ .

Proof 3 (trigonometrical). $IS^2+AB\cdot AC=AS^2\iff$ $AB\cdot AC=AS^2-SB^2\iff$ $\sin C\sin B=$ $\sin ^2\left(B+\frac A2\right)-\sin^2\frac A2\iff$

$\sin B\sin C=\sin \left(B+\frac A2+\frac A2\right)\sin\left(B+\frac A2-\frac A2\right)\iff$ $\sin B\sin C=\sin (B+A)\sin B$ , what is truly.

Proof 4 (metrical). Denote the projections $U$ , $V$ of $S$ on the sidelines $AB$ , $AC$ respectively. Prove easily that $\triangle SBU\equiv\triangle SCV\implies$

$BU=CV$ and $SA^2-SI^2=$ $SA^2-SB^2=UA^2-UB^2=$ $(UA-UB)(UA+UB)=AB\cdot (VA+CV)=$ $AB\cdot AC\implies$ $IS^2+AB\cdot AC=AS^2$ .

Remark. The proposed problem is equivalently with upper lemma. Indeed, if $I$ is the incenter of $\triangle XYZ$ , then

$I\in OZ\cap w$ and $OI=OX=OY=r$ . In conclusion, $r^2+ZX\cdot ZY=ZO^2\iff$ $r^2+13\cdot 7=11^2\iff$ $\boxed{r=\sqrt {30}}$ .

PP18. The incircle $I$ of the triangle $ABC$ touches $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ respectively. Denote $L\in BC\cap EF$ .

The line $EF$ intersects $BI$ , $CI$ , $DI$ at $M$ , $N$ , $K$ respectively. Prove that $\frac {LM}{LN}=\frac {KM}{KN}=\frac {KF}{KE}=$ $\frac {DM}{DN}=\frac {b}{c}$ .

Proof. Prove easily that or it's well-known that the pentagons $BDINF$ , $CDIME$ are cyclically. Thus, $KM\cdot KE=KI\cdot KD=KN\cdot KF\implies$ $\boxed{\frac {KM}{KN}=\frac {KF}{KE}}\ (1)$ .

Also, $\widehat{MDK}\equiv\widehat{MDI}\equiv\widehat{KEI}$ $\equiv\widehat{FEI}\equiv\widehat{KFI}\equiv$ $\widehat{NFI}\equiv\widehat{NDI}\equiv\widehat{NDK}\implies$ $\widehat{MDK}\equiv\widehat{NDK}\implies$ $\boxed{\frac {DM}{DN}=\frac {KM}{KN}}\ (2)$ .

Otherwise, $\frac {KF}{KE}=\frac {IF}{IE}\cdot\frac {\sin\widehat {KIF}}{\sin\widehat{KIE}}=$ $\frac {\sin B}{\sin C}=\frac bc$ . Other hands, $\frac {DM}{DN}=\frac {IC\cdot \sin \widehat{DCM}}{IB\cdot\sin\widehat{DBN}}=$ $\frac {\sin\frac B2\cos\frac B2}{\sin\frac C2\cos\frac C2}=\frac {\sin B}{\sin C}=\frac {AC}{AB}$ . In conclusion, $\boxed{\frac {KF}{KE}=\frac {DM}{DN}=\frac {b}{c}}\ (3)$ .

Remark. Denote $L\in BC\cap EF$ . Apply the Menelaus' relation to the transversal $\overline{LBC}/\triangle AFE$ and obtain that $\frac {LF}{LE}\cdot \frac {CE}{CA}\cdot\frac {BA}{BF}=1\implies$ $\frac {LF}{LE}=\frac {b(s-b)}{c(s-c)}$ . Thus, $\frac {LB}{LC}=\frac {s-b}{s-c}$

and $\left\{\begin{array}{c} LF\cdot LN=LB\cdot LD\\\ LE\cdot LM=LC\cdot LD\end{array}\right\|\implies$ $\frac {LF}{LE}\cdot\frac {LN}{LM}=\frac {LB}{LC}\implies$ $\frac{b(s-b)}{c(s-c)}\cdot \frac {LN}{LM}=\frac {s-b}{s-c}\implies$ $\boxed{\ \frac {LM}{LN}=\frac bc\ }$ .

PP19. Suppose that exists a point $P$ which belongs to the inside of the square $ABCD$ so that $PA=1$ , $PB=2$ and $PC=3$ . Ascertain the measure of the angle $\widehat{APB}$ .

Proof 1 (metric). Denote $AB=x$ and $m\left(\widehat{ABP}\right)=\phi$ . Apply the generalized Pythagoras' teorem in the triangles $\left\{\begin{array}{ccc} \triangle ABP & \implies & 4x\cos\phi =x^2+3\\\\ \triangle CBP & \implies & 4x\sin \phi =x^2-5\end{array}\right\|\implies$

$\boxed{x^2>5}$ and $16x^2=\left(x^2+3\right)^2+\left(x^2-5\right)^2\iff$ $16t^2=2t^2-4t+34$ , where $x^2=t>0$ $\iff$ $t^2-10t+17=0\ ,\ t>5\iff$ $t=\boxed{x^2=5+2\sqrt 2}$ .

Proof 2 (synthetic). See upper figure. Rotate $\triangle APB$ to get $\triangle CQB$ . Then $\triangle PBQ$ is isosceles right, hence $PQ=2\sqrt{2}$ . Obtain that

$PQ^2+QC^2=8+1=9=PC^2$ . Therefore $\angle PQC=90^\circ$ , so finally $\angle APB=\angle CQB=90^\circ+45^\circ=135^\circ$ .

PP20. Let the square $ABCD$ and the points $\left\{\begin{array}{cc} E\in (CD)\ : & DE=3\cdot EC\\\ F\in (BC)\ : & BF=2\cdot FC\end{array}\right\|$ . Denote $M\in AE\cap DF$ . Prove that $MA=AB$ and ascertain the measure of the angle $\widehat{CMF}$ .

Proof (hyperspace.rulz). Suppose w.l.o.g. $AB=12$ . Obtain that $CE=3$ , $DE=9$ , $AE=15$ , $CF=4$ . Denote $K\in (DE)$ such that $DK=4$ , $KE=5$ .

Since $\frac{KD}{KE}=\frac{4}{5}=\frac{12}{15}=\frac{AD}{AE}$ obtain that $\widehat{KAD}\equiv\widehat{KAE}$ . Therefore, $m(\angle FDC)=m(\angle KAD)=m(\angle KAE)=\frac 12\cdot m(\angle DAE)$ . From here, a simple angle chase

shows that $AM=AD=AB=12$ (thus showing that $MA=AB$ ). Note that, since $DE=9$ , we have that $CE=3=$ $15-12=AE-AM=ME$ . Using this,

a simple angle chase again yields that $\boxed{m(\angle CMF)=45^0}$ . Indeed, $KD=KM=4\implies$ $m\left(\widehat{KDM}\right)=m\left(\widehat{KMD}\right)=u\implies$ $m\left(\widehat{EKM}\right)=2u$ ;

$MK\perp ME\ (EK=5\ ,\ KM=4\ ,\ ME=3)\implies$ $m\left(\widehat{KEM}\right)=90-2u$ ; $EC=EM=3$ $\implies$ $m\left(\widehat{EMC}\right)=m\left(\widehat{ECM}\right)=45-u$ $\implies$

$\boxed{m\left(\widehat{MCF}\right)=45+u}$ ; $m\left(\widehat{CFD}\right)=\boxed{m\left(\widehat{CFM}\right)=90-u}$ . In conclusion, $m\left(\widehat{CMF}\right)=45$ . Remark. Prove easily that the midpoint $N$ of $[AD]$

belongs to $CM$ . Thus, $\tan\widehat{FCM}=2$ , $\tan\widehat{CFM}=3$ and $3+2+1=3\cdot 2\cdot 1$ , i.e. $\tan\widehat{CMF}=1\iff m(\angle CMF)=45^{\circ}$ .

PP21. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and the midpoint $M$ of $[AC]$ . Let $N$ so that $NB\perp BI$ and $NM\perp AC$ . Prove that $2\cdot\cot\widehat{BNI}=\left|\cot\frac C2-\cot\frac A2\right|=\frac {|a-c|}r$ .

Proof 1. Suppose w.l.o.g. $A>C$ . Denote the circumcircle $\alpha =\mathbb C(O,R)$ of $\triangle ABC$ and let $[NS]$ be its diameter. Prove easily that $M\in NS$ . Therefore,

$\cot\widehat{BNI}=\frac {NB}{BI}=$ $\frac {2R\sin\frac {A-C}2}{\frac r{\sin\frac B2}}=$ $\frac {2R\sin\frac {A-C}2\cos\frac {A+C}2}r=$ $\frac {R(\sin A-\sin C)}r=$ $\frac {a-c}{2r}$ , i.e. $\boxed{\cot\widehat{BNI}=\frac {a-c}{2r}}\ (1)\ .$ Denote $D\in BC\cap w\ .$ Thus,

$\cot\frac C2-\cot\frac A2=\frac {CD}{DI}-\frac {AD}{DI}=$ $\frac {s-c}r-\frac {s-a}r=\frac {a-c}r$ , i.e. $\boxed{\cot\frac C2-\cot\frac A2=\frac {a-c}r}\ (2)\ .$ From the relations $(1)$ and $(2)$ obtain the required relation.

Proof 2. Let $P\in BC$ so that $BP\perp BC$ and $U\in BP$ , $V\in NS$ so that $I\in UV\parallel AC$ . Thus $BIVN$ is cyclic, i.e. $\widehat{BNI}\equiv \widehat{BVI}\ \stackrel{IM\parallel BV}{\equiv}\ \widehat{IMD}$ . Thus, $\cot\widehat{BNI}=$

$\cot\widehat{BVI}=$ $\frac {UV}{UB}=\frac {PM}{h_b-r}=\frac {\left|a^2-c^2\right|}{2b(h_b-r)}=\frac {(a+c)|a-c|}{2r(2s-b)}=\frac {|a-c|}{2r}\implies$ $\cot\widehat{BNI}=\frac {|a-c|}{2r}\ .$ Otherwise. $\cot\widehat{BNI}=\cot\widehat{IMD}=$ $\frac {MD}{DI}=\frac {|a-c|}{2r}$ a.s.o.

PP22. Let an $B$ -isosceles $\triangle ABC$ and $D\in (AC)$ so that $A=C=\alpha$ and $\left\{\begin{array}{ccc} DB=DA\sqrt 2 & ; &m\left(\widehat{DBA}\right)=\theta\\\ DC=DA\sqrt 3 & ; & m\left(\widehat{DBC}\right)=\omega\end{array}\right\|$ . Prove that $\omega =2\theta$ .

Proof 1. Suppose w.l.o.g. $DA=1$ , i.e. $DB=\sqrt 2$ and $DC=\sqrt 3$ . From the well-known relation $BA^2=BD^2+DA\cdot DC$ get $BA^2=2+\sqrt 3$ , i.e. $BA=BC=\frac {1+\sqrt 3}{\sqrt 2}$ .

Let the midpoint $M$ of $[AC]$ and $m\left(\widehat{DBM}\right)=\delta\ .$ Thus, $D\in (AM)$ , $MA=MC=\frac {\sqrt 3+1}2$ and $MD=\frac {\sqrt 3-1}2$ . Thus, $\cos\alpha =\frac {AM}{AB}=\frac {\frac {1+\sqrt 3}2}{\frac {1+\sqrt 3}{\sqrt 2}}=\frac {\sqrt 2}2\implies$ $\boxed{\alpha =\frac {\pi}4}$ , i.e.

$MB=MA=\frac {1+\sqrt 3}2$ and $\tan\delta =\frac {MD}{MB}=\frac {\frac {\sqrt 3-1}2}{\frac {\sqrt 3+1}2}=\frac {\sqrt 3-1}{\sqrt 3+1}=2-\sqrt 3$ , i.e. $\boxed{\delta =\frac {\pi}{12}}$ . In conclusion, $\left\{\begin{array}{ccc} \theta =\frac {\pi}2-\alpha -\delta =\frac {\pi}4-\frac {\pi}{12} & \implies & \theta =\frac {\pi}6\\\\ \omega =\frac {\pi}2-\alpha +\delta =\frac {\pi}4+\frac {\pi}{12} & \implies & \omega=\frac {\pi}3\end{array}\right\|$ $\implies \omega =2\theta$ .

Proof 2. Suppose w.l.o.g. $DA=1$ , i.e. $DB=\sqrt 2$ and $DC=\sqrt 3$ . Observe that $\theta +\omega +2\alpha =\pi$ , i.e. $\frac {\theta +\omega}2=\frac {\pi}2-\alpha$ and $\frac {\omega -\theta}2=\delta\implies$ $\left\{\begin{array}{ccc} \theta & = & \frac {\pi}2-\alpha -\delta\\\ \omega & = & \frac {\pi}2-\alpha +\delta\end{array}\right\|$ .

Apply the theorem of Sines $:\ \frac {DA}{\sin\widehat{DBA}}=\frac {DB}{\sin\widehat{DAB}}=\frac {DB}{\sin\widehat{DCB}}=\frac {DC}{\sin\widehat{DBC}}\implies$ $\frac {DA}{\sin\theta}=\frac {DB}{\sin\alpha}=\frac {DC}{\sin\omega}\implies$ $\frac {1}{\cos(\alpha +\delta)}=\frac{\sqrt 2}{\sin\alpha}=\frac {\sqrt 3}{\cos (\alpha -\delta)}\implies$

$\frac 1{\cos\delta -\sin\delta\tan\alpha}=\frac {\sqrt 2}{\tan\alpha}=\frac {\sqrt 3}{\cos\delta +\sin\delta \tan\alpha}\implies$ $\tan\alpha =\frac {\sqrt 2\cos\delta}{1+\sqrt 2\sin\delta}=\frac {\sqrt 2\cos\delta}{\sqrt 3-\sqrt 2\sin\delta}\implies$ $\delta =\frac {\pi}{12}\ \wedge\ \alpha =\frac {\pi}4$ $\implies$ $\theta=\frac {\pi}6\ \wedge\ \omega =\frac {\pi}3\implies w=2\theta$ .

PP23. Let an $A$ -isosceles $\triangle ABC$ with $A=120^{\circ}$ and $a=\sqrt 5$ . There is $P\in\mathrm{int(ABC)}$ so that $PB=1\ \mathrm{and}\ PC=\sqrt 2$ . Prove that $m\left(\widehat{APC}\right)=60^{\circ}$ .

Proof. Denote $D\in BP\cap AC$ and the midpoint $M$ of $[BC]$ , i.e. $MB=MC=\frac {\sqrt 5}2$ and $2\cdot AM=AB=AC=\sqrt{\frac 53}$ . Apply the generalized Pythagoras' theorem $:$

$BC^2=PB^2+PC^2-2\cdot PB\cdot PC\cdot\cos \widehat{BPC}\implies$ $5=1+2-2\cdot\sqrt 2\cos\widehat{BPC}\implies$ $\cos\widehat{BPC}=-\frac {\sqrt 2}2\implies$ $\boxed{m\left(\widehat{BPC}\right)=135^{\circ}}$ . Apply the theorem of the median

$[PM]$ in $\triangle BPC\ :\ 4\cdot PM^2=2\left(PB^2+PC^2\right)-BC^2=2(1+2)-5=1\implies$ $PM=\frac 12$ . Since $PB^2+PM^2=1+\frac 14=\frac 54=BM^2\implies PM\perp PB\implies$

$m\left(\widehat{CPM}\right)=m\left(\widehat{CPD}\right)=45^{\circ}$ . Let $\left\{\begin{array}{c} m\left(\widehat{PBC}\right)=x\\\ m\left(\widehat{PAM}\right)=z\\\ m\left(\widehat{APD}\right)=y\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc} m\left(\widehat{PAB}\right)=60^{\circ}-z\\\ m\left(\widehat{PBA}\right)=30^{\circ}-x\\\ m\left(\widehat{APC}\right)=45^{\circ}+y\end{array}\right\|$ , where $\boxed{y=90^{\circ}-(x+z)}$ and $\tan x=\frac {PM}{PB}=\frac 12\implies$ $\boxed{\tan x=\frac 12}\ (1)$

Apply theorem of Sines in $\triangle APM\ :\ \frac {AM}{\sin\widehat{APM}}=\frac {PM}{\sin\widehat{PAM}}$ $\iff$ $\frac {\frac 12\sqrt{\frac 53}}{\sin (90^{\circ}+y)}=$ $\frac 1{2\sin z}\iff$ $\sqrt 5\sin z=$ $\sqrt 3\sin (x+z)\iff$ $\sqrt 5\tan z=\sqrt 3(\sin x+\cos x\tan z)$

$\iff$ $\tan z=$ $\frac {\sqrt 3\sin x}{\sqrt 5-\sqrt 3\cos x}=$ $\frac {\sqrt 3\cdot\frac 1{\sqrt 5}}{\sqrt 5-\sqrt 3\cdot\frac 2{\sqrt 5}}=$ $\frac {\sqrt 3}{5-2\sqrt 3}$ $\implies$ $\boxed{\tan z=\frac {\sqrt 3}{5-2\sqrt 3}}\ (2)$ . From the relation $(1)$ and $(2)$ obtain that $\tan (x+z)=\frac {\tan x+\tan z}{1-\tan x\tan z}=$

$\frac {\frac 12+\frac {\sqrt 3}{5-2\sqrt 3}}{1-\frac 12\cdot \frac {\sqrt 3}{5-2\sqrt 3}}=$ $\frac 5{10-5\sqrt 3}=$ $\frac 1{2-\sqrt 3}=$ $2+\sqrt 3$ $\implies$ $\tan (x+z)=2+\sqrt 3$ $\iff$ $x+z=$ $75^{\circ}\iff$ $y=$ $90^{\circ}-(x+z)=$ $90^{\circ}-75^{\circ}=15^{\circ}\iff$ $m\left(\widehat{APC}\right)=60^{\circ}$ .

PP24. Let $ABCD$ be a square with $AB=1$ . Denote its center $O$ , the point $Q\in (BC)$ so that $\frac {QB}{QC}=k$ , the symmetrical point $P$ of $Q$ w.r.t. $O$

and the point $M\in (CD)$ so that $MD=p<1$ . Construct the square $PQRS$ so that $M\in (AR)$ . Prove that $\left\{\begin{array}{ccc} \tan\widehat{ADS} & = & 1-k\\\\ p & = & \frac 2{2k+1}\end{array}\right\|$ .

Particular case. $k=\frac 32$ and $p=\frac 12$ .

Proof. Let $:$ the projection $Y$ of $Q$ on $AD\ ;$ the projection $X$ of $S$ on $AD\ ;$ the projections $U$ , $V$ of $R$ on $BC$ , $AD$ respectively $;\ \left\{\begin{array}{c} PA=QC=n\\\\ PD=BQ=m\end{array}\right\|$ , where $\left\{\begin{array}{ccc} m+n & = & 1\\\\ \frac mn & = & k\end{array}\right\|$ .

Thus, $DY=n$ , $PY=m-n$ . Prove easily that $\triangle PXS\equiv\triangle QUR\implies$ $AV=2m+n$ and $\left\{\begin{array}{ccc} DX=n & ; & CU=m\\\\ DV=m & ; & AV=2m+n\end{array}\right\|$ . Thus, $\tan\widehat {MAD}=MD/DA=p\implies$

$\frac {RV}{AV}=p\implies$ $RV=p(2m+n)\implies$ $SX=UR=UV-RV=1-p(2m+n)\implies$ $\tan\widehat {XDS}=\frac {SX}{DX}=\frac {1-p(2m+n)}n$ . Hence $\tan\widehat{ADS}=-\tan\widehat{XDS}=$

$-\frac {1-p(2m+n)}n\implies$ $\tan\widehat{ADS}=\frac {p(2m+n)-(m+n)}n=p(2k+1)-(k+1)\implies$ $(k+1)+\tan\widehat{ADS}=p(2k+1)\ (*)$ . Observe that $\triangle PQY\equiv\triangle SPX\implies$

$PY=SX\implies$ $m-n=1-p(2m+n)\implies$ $m-n=m+n-p(2m+n)\implies$ $p(2m+n)=2n\implies$ $\boxed{p=\frac {2}{2k+1}}\ \stackrel{(*)}{\implies}\ \boxed{\tan\widehat{ADS}=1-k}$ .

PP25. Let $\triangle ABC$ with circumcircle $\alpha =\mathbb C(O,R)$ , incicle $w=\mathbb C(I,r)$ , $A$ -excircle $w_a=\mathbb C\left(I_a,r_a\right)$ , $\left\{\begin{array}{ccc} D\in BC & ; & AD\perp AI\\\\ P\in DI_a & ; & IP\perp DI_a\end{array}\right\|$ and $K\in (IP)\cap \alpha$ . Prove that $KI=KP$

Proof. Let $\{A,N\}=\{A,D\}\cap\alpha$ , the diameter $[NS]$ of $\alpha\ (N-$ north $,\ S-$ south $)$ , the midpoint $M$ of $[BC]$ , $R\in NI\cap I_aD$ and $E\in AI\cap BC$ where $\{S,I_a\}\subset AI$ . Thus,

$\left\{\begin{array}{cccccc} ABI_a\sim AIC & \implies & \frac {AB}{AI}=\frac {AI_a}{AC} & \implies & AI\cdot AI_a=bc & (1)\\\\ ABN\sim ADC & \implies & \frac {AB}{AD}=\frac {AN}{AC} & \implies & AN\cdot AD=bc & (2)\end{array}\right\|\implies$ $AN\cdot AD=AI\cdot AI_a\implies$ $\frac {AI}{AN}=\frac {AD}{AI_a}\implies$ $\triangle ANI\sim \triangle AI_aD$ $\implies$ $\widehat{ANI}\equiv\widehat{AI_aD}\implies$

$\widehat{ANR}\equiv\widehat{AI_R}\implies$ $ANI_aR$ is cyclically $\implies NI\perp I_aD\implies R\equiv P$ $\implies P\in NI$ . Therefore, $\left\{\begin{array}{c} KS\perp KN\\\\ KN\perp I_aD\end{array}\right\|$ $\implies KS\parallel I_aP\ (SI=SI_a)\implies KI=KP$ .

Remark. $\left\{\begin{array}{ccccc} ABS\sim AEC & \implies & \frac {AB}{AE}=\frac {AS}{AC} & \implies & AE\cdot AS=bc\ (*)\\\\ ANS\sim AED & \implies & \frac {AN}{AE}=\frac {AS}{AD} & \stackrel{(*)}{\implies} & AN\cdot AD=bc\end{array}\right\|\ .$

PP26. The midpoints of $[AC]$ , $[AB]$ are $E$ and $F$ respectively. A straight line is drawn through $A$ cutting $(CF)$ , $(BE)$ and $(BC)$ at $P$ , $R$ and $D$ respectively.

Prove that $\frac {1}{AP} + \frac {1}{AR} = \frac {3}{AD}$ . An easy extension. $\left\|\begin{array}{c} D\in BC\ \ ,\ \ E\in CA\ \ ,\ \ F\in AB \\ \\ P\in AD\cap CF\ \ ,\ \ R\in AD\cap BE\end{array}\right\|\ \implies\ \frac {\overline {AE}}{\overline {EC}}\cdot\frac {1}{\overline {AR}} + \frac {\overline {AF}}{\overline {FB}}\cdot\frac {1}{\overline {AP}} =$ $\left(1 + \frac {\overline {AE}}{\overline {EC}} + \frac {\overline {AF}}{\overline {FB}}\right)\cdot\frac {1}{\overline {AD}}$ .

Proof. Apply the Menelaus' theorem to the mentioned transversals and the adequate triangles :

$\left\|\begin{array}{ccccc} \overline {BRE}/ \triangle ACD & : & \frac {\overline {BD}}{\overline {BC}}\cdot\frac {\overline {EC}}{\overline {EA}}\cdot\frac {\overline {RA}}{\overline {RD}} = 1 & \implies & \frac {\overline {EA}}{\overline {EC}}\cdot\frac {\overline {RD}}{\overline {RA}} = \frac {\overline {BD}}{\overline {BC}} \\ \\ \overline {CPF}/\triangle ABD & : & \frac {\overline {CD}}{\overline {CB}}\cdot\frac {\overline {FB}}{FA}\cdot\frac {\overline {PA}}{\overline {PD}} = 1 & \implies & \frac {\overline {FA}}{\overline {FB}}\cdot\frac {\overline {PD}}{\overline {PA}} = \frac {\overline {DC}}{\overline {BC}} \end{array}\right\|\ \implies$ $\boxed {\ \frac {\overline {EA}}{\overline {EC}}\cdot\frac {\overline {RD}}{\overline {RA}} + \frac {\overline {FA}}{\overline {FB}}\cdot\frac {\overline {PD}}{\overline {PA}} = 1\ }\ (*)$ $\implies$

$\frac {\overline {EA}}{\overline {EC}}\cdot\left(1 + \frac {\overline {AD}}{\overline {RA}}\right) + \frac {\overline {FA}}{\overline {FB}}\cdot\left(1 + \frac {\overline {AD}}{\overline {PA}}\right) = 1$ $\implies$ $\frac {\overline {EA}}{\overline {EC}}\cdot\frac {1}{\overline {AR}} + \frac {\overline {FA}}{\overline {FB}}\cdot\frac {1}{\overline {AP}} = \left(\frac {\overline {EA}}{\overline {EC}} + \frac {\overline {FA}}{\overline {FB}} - 1\right)\cdot\frac {1}{\overline {AD}}$ $\implies$ $\boxed {\ \frac {\overline {AE}}{\overline {EC}}\cdot\frac {1}{\overline {AR}} + \frac {\overline {AF}}{\overline {FB}}\cdot\frac {1}{\overline {AP}} = \left(1 + \frac {\overline {AE}}{\overline {EC}} + \frac {\overline {AF}}{\overline {FB}}\right)\cdot\frac {1}{\overline {AD}}\ }$ .

Particular cases.

$1\ \blacktriangleright\ \frac {\overline {EA}}{\overline {EC}} =\frac {\overline {FA}}{\overline {FB}} =m\ \implies\ \frac {1}{\overline {AP}} + \frac {1}{\overline {AR}} =\frac {2m-1}{m}\cdot \frac {1}{\overline {AD}}$ .

$2\ \blacktriangleright\ \left\|\begin{array}{c} \frac {\overline {EA}}{\overline {EC}} = - 1 \\ \\ \frac {\overline {FA}}{\overline {FB}} = - 1\end{array}\right\|\ \implies\ \frac {1}{AP} + \frac {1}{AR} = \frac {3}{AD}$ (the proposed problem - the previous case for $m=-1$ ).

$3\ \blacktriangleright\ P\equiv R\ \stackrel {(*)}{\implies}\ \frac {\overline {PA}}{\overline {PD}} = \frac {\overline {EA}}{\overline {EC}} + \frac {\overline {FA}}{\overline {FB}}$ (the remarkable Van Aubel's relation).

PP27 (IMO 2010). Let an interior $P$ of $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ , where $a\ne b$ . Denote the second intersections

$(K,L,M)$ of $w$ with $AP$ , $BP$ , $CP$ respectively . The tangent line $CC$ at $C$ to $w$ meets $AB$ at $S$ . Prove that $SC = SP\implies MK = ML$ .

Proof. Consider the circle which is tangent to $w$ in $C$ and to $SP$ in $P$ . Denote $\{X,Y\}=SP\cap w$ , where $X\in \overarc{AL}$ . From the well-known property

obtain $\widehat {MCX}\equiv\widehat {MCY}$ , i.e. $\overarc{MX}=\overarc{MY}$ . Since $SP^2=SC^2=SA\cdot SB$ obtain $SP^2=SA\cdot SB$ $\implies$ $\widehat {SPA}\equiv\widehat{SBP}$ , i.e. $SP$ is tangent to

the circumcircle of $\triangle APB\iff$ $\overarc{XL}=\overarc{YK}$ . In conclusion, $\overarc{ML}=\overarc{MX}+\overarc{XL}=$ $\overarc{MY}+\overarc{YK}=\overarc{MK}$ $\implies$ $\overarc{ML}=\overarc{MK}\iff$ $ML=MK$ .

PP28. Let $\triangle ABC$ and $D\in (BC)\ .$ The bisectors of $\widehat{ADB}\ ,$ $\widehat{ADC}$ cut $AB\ ,$ $AC$ at $E\ ,$ $F$ respectively. Prove that $\boxed{AB\cdot AE+AC\cdot AF=AD\cdot BC <=> AB\perp AC}\ (*)\ .$

Proof 1. Apply the Stewart's relation $:\ \boxed{a\cdot AD^2+a\cdot DB\cdot DC=b^2\cdot DB+c^2\cdot DC}\ (1)\ .$ Since $\left\{\begin{array}{ccc} AE & = & \frac {c\cdot AD}{DA+DB}\\\\ AF & = & \frac {b\cdot AD}{DA+DC}\end{array}\right|\ ,$ the relation $(*)$ becomes

$\frac {c^2\cdot AD}{DA+DB}+\frac {b^2\cdot AD}{DA+DC}=a\cdot AD$ $\iff$ $\frac {c^2}{DA+DB}+\frac {b^2}{DA+DC}=a$ $\iff$ $c^2(DA+DC)+b^2(DA+DB)=$ $a(DA+DB)(DA+DC)$ $\iff$

$\left(b^2+c^2\right)\cdot AD+\underline{c^2\cdot DC+b^2\cdot DB}=$ $a^2\cdot AD+\underline{a\cdot AD^2+a\cdot DB\cdot DC}\ \stackrel{(1)}{\iff}\ \left(b^2+c^2\right)\cdot AD=$ $a^2\cdot AD$ $\iff$ $b^2+c^2=a^2\iff AB\perp AC\ .$

Proof 2. Let $\left\{\begin{array}{ccc} m\left(\widehat{EDA}\right)=m\left(\widehat{EDB}\right) & = & x\\\\ m\left(\widehat{FDA}\right)=m\left(\widehat{FDC}\right) & = & y\end{array}\right\|\ ,$ where $x+y=\frac {\pi}2\ .$ Thus, the relation $(*)$ $\iff$ $c\cdot \frac {AE}{AD}+b\cdot \frac {AF}{AD}=a\iff$ $c\cdot \frac {\sin x}{\sin (B+x)}+b\cdot \frac {\sin y}{\sin (C+y)}=$

$c\cdot\cos B+b\cdot\cos C$ $\iff$ $c\cdot\left(\frac {\tan x}{\sin B+\cos B\tan x}-\cos B\right)+$ $b\cdot \left(\frac {\tan y}{\sin C+\cos C\tan y}-\cos C\right)=0\iff$ $c\cdot\frac {\tan x\left(1-\cos^2B\right)-\sin B\cos B}{\sin B+\cos B\tan x}+$

$b\cdot\frac {\tan y\left(1-\cos^2C\right)-\sin C\cos C}{\sin C+\cos C\tan y}=0\iff$ $\frac {c\cdot\sin B\left(\tan x\sin B-\cos B\right)}{\sin B+\cos B\tan x}+$ $\frac {b\cdot\sin C\left(\tan y\sin C-\cos C\right)}{\sin C+\cos C\tan y}=0\ \stackrel{b\sin C=c\sin B}{\iff}\ \frac {\tan x\sin B-\cos B}{\sin B+\cos B\tan x}+$

$\frac {\tan y\sin C-\cos C}{\sin C+\cos C\tan y}=0$ $\iff$ $\frac {\tan x\tan B-1}{\tan B+\tan x}+$ $\frac {\tan y\tan C-1}{\tan C+\tan y}=0$ $\iff$ $\tan (B+x)=-\tan (C+y)\iff$ $B+C=\frac{\pi}2\iff$ $A=90^{\circ}\iff AB\perp AC\ .$

Proof 3. Let $\left\{\begin{array}{ccc} m\left(\widehat{EDA}\right)=m\left(\widehat{EDB}\right)=x & ; & \{x,y\}\subset\left(0,\frac {\pi}2\right)\\\\ m\left(\widehat{FDA}\right)=m\left(\widehat{FDC}\right)=y & ; & x+y=\frac {\pi}2\end{array}\right\|\ .$ Thus, $(*)$ $\iff$ $c\cdot \frac {AE}{AD}+b\cdot \frac {AF}{AD}=a\iff$ $c\cdot \frac {\sin x}{\sin (B+x)}+b\cdot \frac {\sin y}{\sin (C+y)}=a$ $\iff$

$\sin C\cdot \frac {\tan x}{\sin B+\cos B\tan x}+\sin B\cdot \frac {\tan y}{\sin C+\cos C\tan y}=\sin A\iff$ $\tan x\cdot \frac {\sin (A+B)}{\sin B+\cos B\tan x}+\tan y\cdot \frac {\sin (A+C)}{\sin C+\cos C\tan y}=\sin A\iff$

$\tan x\cdot \frac {\sin A+\cos A\tan B}{\tan B+\tan x}+\tan y\cdot \frac {\sin A+\cos A\tan C}{\tan C+\tan y}=\sin A\iff$ $\boxed{\tan x\cdot \frac {\tan A+\tan B}{\tan B+\tan x}+\tan y\cdot \frac {\tan A+\tan C}{\tan C+\tan y}=\tan A}\ (1)\ .$ Therefore, denote

$\left\{\begin{array}{cccc} \tan x=u\ ,\ \tan y=v & ; & \{u,v\}\subset\mathbb R^*_+\ ,\ uv=1 & (2)\\\\ \tan A=\alpha\ ,\ \tan B=\beta\ ,\ \tan C=\gamma & ; & \alpha (\beta\gamma -1 )=\beta +\gamma & (3)\end{array}\right\|\ .$ The relation $(1)$ becomes $\frac {u(\alpha +\beta )}{\beta +u}+\frac {v(\alpha +\gamma)}{\gamma+v}=\alpha\iff$ $u\beta (\gamma +v)+v\gamma (\beta +u)=$

$\alpha [(\beta +u)(\gamma +v)-u(\gamma +v)-v(\beta +u)]\iff$ $\beta\gamma (u+v)+uv(\beta +\gamma )=\alpha (\beta \gamma -uv)\ \stackrel{(2)}{\iff}$ $\boxed{\beta\gamma (u+v)+uv(\alpha +\beta +\gamma )=\alpha\beta\gamma}\ (4)\ .$ Since $\alpha +\beta +\gamma =\alpha\beta\gamma$

the relation $(4)$ becomes $\beta\gamma (u+v)+uv\alpha\beta\gamma =\alpha\beta\gamma\iff$ $u+v=\alpha (1-uv)\iff$ $\alpha =\frac {u+v}{1-uv}\iff$ $\tan A=\tan (x+y)\iff$ $A=x+y\iff$ $A=90^{\circ}\ .$

$\mathrm{END}$

This post has been edited 283 times. Last edited by Virgil Nicula, Jun 20, 2016, 8:37 PM

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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

428. Some metrical problems from the contests II.

by Virgil Nicula, Sep 25, 2015, 7:54 PM

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