427. Selected nice or well-known geometry problems.

by Virgil Nicula, Jul 25, 2015, 3:43 PM

Lemma 1. Let a circle $w=\mathcal C(O,r)$ and two points $A\in \mathrm{ext} w$ , $B\in \mathrm{int} w$ , where denote $\{C,D\}=AB\cap w$ and $p_w(X)$ - the power of $X$ w.r.t. $w$ . Prove

that $p_w(A)+p_w(B)=AB^2\iff$ $\frac {CA}{CB}=\frac {DA}{DB}$ , i.e. $C$ , $D$ are harmonic conjugate w.r.t. $A$ , $B$ and mark it that $(C,D;A,B)$ is a harmonic division.

Proof 1. Suppose w.l.o.g. that $C\in (AB)$ and denote $\left\{\begin{array}{ccc} AC & = & x\\\ CB & = & y\\\ BD & = & z\end{array}\right\|$ . Observe that $\left\{\begin{array}{ccc} DA & = & x+y+z\\\ AB & = & x+y\end{array}\right\|$ and $\left\{\begin{array}{ccccc} p_w(A) & = & \overline{AC}\cdot \overline{AD} & = & x(x+y+z)\\\\ p_w(B) & = & \overline{BC}\cdot \overline{BD} & = & -yz\end{array}\right\|$

In conclusion, $p_w(A)+p_w(B)=AB^2$ $\iff$ $x(x+y+z)-yz=(x+y)^2\ (*)$ $\iff$ $xz=y(x+y+z)\iff$ $\frac xy=\frac {x+y+z}z$ $\iff$ $\frac {CA}{CB}=\frac {DA}{DB}$ .

Proof 2. Denote $\{M,N\}\subset w$ so that $A\in MM\cap NN$ , where $XX$ is the tangent line in the point $X\in w$ to the circle $w$ . Thus, $p_w(A)+p_w(B)=AB^2\iff$

$AM^2+BO^2-r^2=AB^2\iff$ $BA^2-BO^2=MA^2-MO^2\iff$ $BM\perp AO\iff$ $B\in MN\iff$ $(A,B;C,D)$ is a harmonic division.

P1. Let $\triangle ABC$ with the orthocenter $H$ , the circumcircle $\alpha =\mathbb C(O,R)$ , the incircle $w=\mathbb C(I,r)$ and the $A$ -excircle $w_a$ . Denote $:$

$\left\{\begin{array}{ccc} P\in AH\cap BC & ; & D\in BC\cap w\\\\ L\in AI\cap BC & ; & D'\in BC\cap w_a\end{array}\right\|$ and $\left\{\begin{array}{c} \{A,S\}=AI\cap \alpha\\\\ \{A,A'\}=AO\cap\alpha\end{array}\right\|$ . Define $\left\{\begin{array}{c} X\in SP\cap LA'\\\\ Y\in SD\cap A'I\\\\ Z\in SD'\cap A'I_a\end{array}\right\|$ . Prove that $\{X,Y,Z\}\subset \alpha$ .

Proof. Prove easily or is well-known $:\ SB=SI=SC\ ;\ \boxed{IA\cdot IS=2Rr\ ;\ I_aA\cdot I_aS=2Rr_a}\ ($ power of $I$ , $I_a$ w.r.t. $\alpha )\ ;\ \boxed{2Rh_a=bc=AL\cdot AS=AI\cdot AI_a}\ (*)$ . Thus $:$

$\blacktriangleright\ \triangle APS\sim ALA'\iff\frac {AP}{AS}=\frac {AL}{AA'}\iff\frac {h_a}{AS}=$ $\frac {AL}{2R}\iff AL\cdot AS=2Rh_a\ \mathrm{(\underline{truly})}\implies$ $\widehat{ASP}\equiv\widehat {AA'L}\implies$ $\widehat{ASX}\equiv\widehat {AA'X}\implies X\in\odot(ASA')\equiv\alpha\ .$

$\blacktriangleright\ \triangle IDS\sim AIA'\iff\frac {ID}{AI}=\frac {IS}{AA'}\iff\frac r{AI}=$ $\frac {IS}{2R}\iff IA\cdot IS=2Rr\ \mathrm{(\underline{truly})}\implies$ $\widehat{ISD}\equiv\widehat {AA'I}\implies$ $\widehat{ASY}\equiv\widehat {AA'Y}\implies Y\in \odot (ASA')\equiv\alpha\ .$

$\blacktriangleright\ \triangle I_aD'S\sim AI_aA'\iff\frac {I_aD'}{AI_a}=\frac {IS}{AA'}\iff\frac {r_a}{AI_a}=$ $\frac {IS}{2R}\iff I_aA\cdot IS=2Rr_a\ \mathrm{(\underline{truly})}\implies$ $\widehat{I_aSD'}\equiv\widehat {AA'I_a}\implies$ $\widehat{ASZ}\equiv\widehat {AA'Z}\implies Z\in \odot (ASA')\equiv\alpha\ .$

Remark. $\left\{\begin{array}{ccccccccccc} \frac r{IA} & = & \sin\frac A2 & = & \frac {SC}{2R} & = & \frac {IS}{2R} & \implies & IA\cdot IS & = & 2Rr\\\\ \frac {r_a}{I_aA} & = & \sin\frac A2 & = & \frac {SC}{2R} & = & \frac {I_aS}{2R} & \implies & I_aA\cdot I_aS & = & 2Rr_a\end{array}\right\|$ . Now I"ll prove the relations $(*)\ :$

$\left\{\begin{array}{ccccccc} \triangle ABP\sim\triangle AA'C & \iff & \frac {AB}{AA'}=\frac {AP}{AC} & \iff & \frac c{2R}=\frac {h_a}b & \iff & 2R\cdot h_a=bc\\\\ \triangle ABS\sim\triangle ALC & \iff & \frac {AB}{AL}=\frac {AS}{AC} & \iff & \frac c{AL}=\frac {AS}b & \iff & AL\cdot AS=bc\\\\ \triangle ABI_a\sim\triangle AIC & \iff & \frac {AB}{AI}=\frac {AI_a}{AC} & \iff & \frac c{AI}=\frac {AI_a}b & \iff & AI\cdot AI_a=bc\end{array}\right\|$ $\implies 2R\cdot h_a=bc=AL\cdot AS=AI\cdot AI_a$ .

P2. Let $\triangle ABC$ with orthocenter $H$ , circumcircle $\alpha =\mathbb C(O,R)$ , incircle $w=\mathbb C(I,r)$ and $A$ -excircle $w_a$ . Let $\{A,S\}=AI\cap \alpha$ , $D\in BC\cap w$

and $\odot\begin{array}{cccc} \nearrow & P\in AH\cap BC & ; & X\in SP\cap \alpha\\\\ \rightarrow & L\in AI\cap BC & ; & Y\in SD\cap \alpha\\\\ \searrow & D'\in BC\cap w_a & ; & Z\in SD'\cap\alpha\end{array}$ . Prove that $LX\cap IY\cap I_aZ\ne\emptyset$ and $[AT]$ is a diameter of $\alpha$ , where $T\in LX\cap IY\cap I_aZ$ .

P3. Let $\triangle ABC$ with the circumcircle $\alpha$ and a mobile point $M\in\alpha$ . Denote $\left\{\begin{array}{ccc} N & \in & MB\cap AC\\\\ P & \in & MC\cap AB\end{array}\right\|$ . Prove that the line $NP$ passes by a fixed point.

Proof. Denote $F\in BB\cap CC$ , where $XX$ means the line what is tangent to the circle $\alpha$ at the point $X\in\alpha$ . I"ll apply the

Pascal's theorem to the degenerated cyclical hexagon $ABBMCC\ :\ \odot\begin{array}{ccc} \nearrow & P\in AB\cap MC & \searrow\\\\ \rightarrow & F\in BB\cap CC & \rightarrow\\\\ \searrow & N\in BM\cap CA & \nearrow\end{array}\odot \implies F\in NP$ .

P4 (European Mathematical Cup - 2012). Let an acute $\triangle ABC$ with $B>C$ , the orthocentre $H$ and the orthic triangle $H_aH_bH_c$ . Denote the midpoint $M$ of $[AH]$ ,

the intersection $D\in AH\cap H_bH_c$ and the symmetrical point $E$ of $D$ w.r.t. $BC$ . Prove that $D$ is the orthocenter of $BMC$ an $MBEC$ is a cyclical quadrilateral.

Lemma. Let $\triangle ABC$ with $A$ -altitude $AD$ , where $D\in BC$ and $X\in AD$ . Then $X$ is the orthocenter of $\triangle ABC\iff$ $\frac {AD}{XD}=\tan B\tan C\iff$ $DB\cdot DC=AD\cdot XD\ .$

Proof of the lemma. Denote $E\in BX\cap AC$ and apply the Menelaus' theorem to the transversal $\overline{BXE}/\triangle ADC\ :\ \frac{BD}{BC}\cdot\frac {EC}{EA}\cdot\frac {XA}{XD}=1\ (1)\ .$ Therefore, $X$ is the

orthocenter of $\triangle ABC\iff$ $BE\perp AC\ \stackrel{(1)}{\iff}\ \frac{c\cos B}{a}\cdot\frac {a\cos C}{c\cos A}\cdot\frac {XA}{XD}=1\iff$ $\frac {XA}{XD}=\frac {\cos A}{\cos B\cos C}\iff$ $\frac {AD}{XD}=\tan B\tan C\iff$ $DB\cdot DC=AD\cdot XD\ .$

Proof 1. Let the midpoint $L$ of $[BC]$ . The Euler's circle of $\triangle ABC$ is the circumcircle of $MH_aH_bH_cL$ with the diameter $[ML]$ , the ray $[H_aM$ is the bisector of the angle $\widehat{H_bH_aH_c}$

and $MH_b=MH=MH_c\implies$ $H$ is the incenter of $\triangle H_aH_bH_c$ . Observe that $H_aBH_c\sim H_aH_bC$ , i.e. $\frac {H_aB}{H_aH_b}=\frac {H_aH_c}{H_aC}\iff$ $\boxed{H_aH_b\cdot H_aH_c=H_aB\cdot H_aC}\ (*)\ .$ Apply

an well known identity $\boxed{H_aD\cdot H_aM=H_aH_b\cdot H_aH_c}\ (1)$ , i.e. $H_aE\cdot H_aM=H_aH_b\cdot H_aH_c\ \stackrel{(*)}{=}\ H_aB\cdot H_aC$ , i.e. $H_aE\cdot H_aM=H_aB\cdot H_aC\iff MBEC$ is cyclic.

$\blacktriangleright$ Apply upper lemma to $\triangle BMC\ :\ D$ is orthocenter of $\triangle BMC\ \iff\ \frac {AH_a}{DH_a}=\tan \widehat{MBC}\tan \widehat{MCB}\iff$ $H_aB\cdot H_aC=MH_a\cdot DH_a$ - true, see relations $(*\wedge 1)\ .$

Remark. The relations $(*)$ and $(1)$ appear as application of the property PP10 from here.

Proof 2. I"ll use an well known implication $\ :\ \boxed{D\in H_bH_c\implies \frac {H_cB}{H_cA}\cdot H_aC+\frac {H_bC}{H_bA}\cdot H_aB=\frac {DH_a}{DA}\cdot BC}\ .$ So $\frac {a\cos B}{b\cos A}\cdot b\cos C+\frac {a\cos C}{a\cos A}\cdot c\cos B=\frac {DH_a}{DA}\cdot a\iff$

$\frac {DH_a}{DA}=\frac {2\cos B\cos C}{\cos A}\iff$ $\frac {DA}{\cos A}=\frac {DH_a}{2\cos B\cos C}=\frac {h_a}{\cos (B-C)}\implies$ $\boxed{DH_a=\frac {2h_a\cos B\cos C}{\cos (B-C)}}\ (1)$ . Let the midpoint $L$ of $[BC]$ and the diameter $[AA']$ of the

circumcircle $\mathbb C(O,R)$ . Is well known or prove easily that $[ML]$ is the diameter of the Euler's circle w.r.t. $\triangle ABC$ , $ML=R$ and $ML\parallel AO$ with $m\left(\widehat{HAO}\right)=B-C$ $\implies$

$MH_a=ML\cos (B-C)$ i.e. $\boxed{MH_a=R\cos (B-C)}\ (2)$ . Thus $H_aM\cdot H_aE=H_aM\cdot H_aD\ \stackrel{(1\wedge 2)}{=}\ R\cos (B-C)\cdot \frac {2h_a\cos B\cos C}{\cos (B-C)}=2Rh_a\cos B\cos C\implies$

$H_aM\cdot H_aE=bc\cos B\cos C=c\cos B\cdot b\cos C=H_aB\cdot H_aC\implies$ $H_aM\cdot H_aE=H_aB\cdot H_aC\implies\ MBEC$ is cyclic. I used the well known identity $\boxed{2Rh_a=bc}$ .

$\blacktriangleright$ (Stefan Tudose) Division $(A, D,H,H_a)$ is harmonically. Apply its characterization $MD\cdot MH_a=MH^2$ , i.e. $\boxed{4\cdot MD\cdot MH_a=AH^2}\ (3)$ . Let $X\in MC\ , BX\perp MC$ .

Then $MX\cdot MC=\frac{MB^2+MC^2-BC^2}2=$ $\frac{\frac{2\cdot \left(AB^2+HB^2\right)-AH^2}4+\frac{2\cdot \left(AC^2+HC^2\right)-AH^2}4-BC^2}2=$ $\frac{BA^2+CA^2-2\cdot BC^2+HB^2+HC^2-HA^2}4$ . Therefore,

$\left\{\begin{array}{c} CH\perp AB\iff HA^2-HB^2=CA^2-CB^2\\\\ BH\perp AC\iff HA^2-HC^2=BA^2-BC^2\end{array}\right\|$ $\implies$ $MX\cdot MC=\frac{AH^2}{4}\ \stackrel{(3)}{=}\ MD\cdot MH_a$ , i.e. $DXCH_a$ is cyclic. But $D\in MH_a\implies$ $D$ is orthocenter of $\triangle BMC$ .

P5 (Kvant M734). Let $\triangle ABC$ with the circumcircle $w$ and the $A$ -bisector $AD$ , where $D\in w$ . Denote $E\in AB$ so that $DE\perp AB$ . Prove that $2\cdot AE=AB+AC$ .

Proof 1. Denote $F\in AB$ so that $B\in (AF)$ and $BF=AC$ . Observe that $ABDC$ is cyclically $\implies \boxed{\widehat{DBF}\equiv\widehat{DCA}}\ (*)$ . Therefore, $\left\{\begin{array}{ccc} BD & = & CD\\\ BF & = & CA\end{array}\right\|\ \stackrel{(*)}{\implies}$

$\triangle BDF\ \stackrel{(s.a.s)}{\equiv}\ \triangle CDA$ $\implies$ $DF=DA$ , i.e. the triangle $ADF$ is $D$ -isosceles $\implies$ the point $E$ is the midpoint of $[AF]$ , i.e. $2\cdot AE=AB+AC\ .$

Proof 2. Suppose w.l.o.g. $B\ne C$ , i.e. $b\ne c$ . Prove easily that $\left\{\begin{array}{ccc} B>C & \iff & B\in (AE)\\\ B<C & \iff & E\in (AB)\end{array}\right\|$ . Suppose w.l.o.g. $B>C$ , i.e. $B\in (AE)$ . Let $F\in (AC)$

so that $DF\perp AC$ . Observe that $\left\{\begin{array}{ccc} DB & = & DC\\\ DE & = & DF\end{array}\right\|$ $\implies\triangle DBE\ \stackrel{(s.a.s)}{\equiv}\ \triangle DCF\implies$ $BE=CF=x$ . In conclusion, $AE=AF\iff$

$AB+BE=AC-CF\iff$ $c+x=b-x\iff$ $2x=b-c\iff$ $AE=c+x=c+\frac {b-c}2\iff$ $AE=\frac {b+c}2$ , i.e. $2\cdot AE=AB+AC\ .$

P6 (M. O. Sanchez). Let a trapezoid $ABCD$ with $AD\parallel BC\ ,\ BC<AD$ and the midpoints $M\ ,\ N$ of $[BC]\ ,\ [AD]$ respectively, where $\left\{\begin{array}{ccc} AC & = & 10\\\\ BD & = & 12\\\\ MN & = & 5\end{array}\right\|\ .$ Find $[ABCD]\ .$

Proof 1 (Baris Altay). Let $X\in AB\ ,\ Y\in CD$ so that $I\in XY$ and $XY\parallel AD$ . Hince $IX=IY$ obtain that $I\in MN$ . Denote $\left\{\begin{array}{ccc} E\in AD & ; & BE\parallel AC\\\\ F\in AD & ; & BF\parallel MN\end{array}\right\|$ . Observe that

$\left\{\begin{array}{ccc} :\ [ABE] & = & [ABC]\\\\ :\ [ABD] & = & [ACD]\end{array}\right\|$ $\bigoplus\implies$ $[EBD]=[ABE]+[ABD]=[ABC]+[ACD]=[ABCD]\implies$ $\boxed{[EBD]=[ABCD]}\ (*)\ .$ Prove easily that

$FE=FD=AD+BC\ ,\ BF=MN=5$ and $BE=AC=10$ . Apply the theorem of $B$ -median $BF$ in $\triangle EBD\ :\ 4\cdot BF^2=2\cdot \left(BE^2+BD^2\right)-ED^2\iff$

$4\cdot 5^2=2\cdot\left(10^2+12^2\right)-ED^2\iff$ $\boxed{ED=2\sqrt{97}}\ (1)\ .$ Apply theorem of Cosines $:\ 2\cdot BE\cdot BD\cdot\cos\widehat{EBD}=$ $BE^2+BD^2-ED^2\ \stackrel{(1)}{\iff}\ 2\cdot 10\cdot 12\cdot\cos\widehat{EBD}=$

$10^2+12^2-4\cdot 97\iff$ $\cos\widehat{EBD}=-\frac 35\iff$ $\sin\widehat{EBD}=\frac 45\implies$ $2\cdot[ABCD]\ \stackrel{(*)}{=}\ 2\cdot [EBD]=BE\cdot BD\cdot\sin\widehat{EBD}=10\cdot 12\cdot\frac 45\implies [ABCD]=48\ .$

Proof 2. Denote $\left\{\begin{array}{ccc} AD=2a & ; & BC=2b\\\ AB=c & ; & CD=d\\\ AC=e & ; & BD=f\end{array}\right\|$ and two points $\left\{\begin{array}{ccc} E\in AD & ; & ME\parallel AC\\\\ F\in AD & ; & MF\parallel BD\end{array}\right\|$ . Observe that $MCAE$ and $MBDF$ are parallelograms $\implies$

$\left\{\begin{array}{ccc} AE=MC=DF=MB=a & ; & ME=AC=e=10\\\ MF=BD=f=12 & ; & NE=NF=a+b\\\ EF=2(a+b) & = & AD+BC\end{array}\right\|$ , i.e. $MN$ is the $M$ -median in the $\triangle EMF$ . Apply the theorem of $M$ -median $MN$ in

$\triangle EMF\ :\ \boxed{4\cdot MN^2+EF^2=2\cdot \left(ME^2+MF^2\right)}\ (*)\iff$ $4\cdot 5^2+EF^2=2\cdot\left(10^2+12^2\right)\iff$ $\boxed{EF=2\sqrt{97}}\ (1)\ .$ Apply theorem of Cosines $:$

$2\cdot ME\cdot MF\cdot\cos\widehat{EMF}=$ $ME^2+MF^2-EF^2\ \stackrel{(1)}{\iff}\ 2\cdot 10\cdot 12\cdot\cos\widehat{EMF}=$ $10^2+12^2-4\cdot 97\iff$ $\cos\widehat{EBD}=-\frac 35\iff$

$\sin\widehat{EMF}=\frac 45\implies$ $[ABCD]=[EMF]=\frac 12\cdot ME\cdot MF\cdot\sin\widehat{EMF}=\frac 12\cdot 10\cdot 12\cdot\frac 45\implies [ABCD]=48\ .$

Remark 1. The relation $\boxed{4\cdot MN^2+(AD+BC)^2=2\cdot \left(AC^2+BD^2\right)}\ (*)$ is well known. I"ll present its another proof. Denoted the midpoints

$M$ , $N$ of $[BC]$ , $[AD]$ respectively $,\ \left\{\begin{array}{ccccc} AD=a & ; & AB=c & ; & AC=e\\\\ BC=b & ; & CD=d & ; & BD=f\end{array}\right\|$ and apply Euler's relations in the trapezoid $ABCD\ :$

$\left\{\begin{array}{ccc} \left(a^2+b^2\right)+\left(c^2+d^2\right) & = & \left(e^2+f^2\right)+(a-b)^2\\\\ \left(a^2+b^2\right)+4\cdot MN^2 & = & \left(c^2+d^2\right)+\left(e^2+f^2\right)\end{array}\right\|$ $\bigoplus\implies$ $\boxed{(AD+BC)^2+4\cdot MN^2=2\cdot \left(AC^2+BD^2\right)}\ (*)\ .$

Remark 2. I'll find the area $S=f\left(b,c,m_a\right)$ of $\triangle ABC$ , where $b\ ,\ c\ ,\ m_a$ are known. Denote the midpoint $M$ of $[BC]$ and the projection $D$ of $A$ on $BC$ . Prove easily

that $AD\perp BC\iff$ $b^2-c^2=DC^2-DB^2=2a\cdot DM\iff \boxed{DM=\frac {b^2-c^2}{2a}}$ . Thus, $AM^2=AD^2+DM^2\iff$ $m_a^2=h_a^2+\left(\frac {b^2-c^2}{2a}\right)^2\iff$

$4a^2m_a^2=4a^2h_a^2+\left(b^2-c^2\right)^2\iff$ $\boxed{4a^2m_a^2=16S^2+\left(b^2-c^2\right)^2}\iff$ $4m_a^2\left[2\left(b^2+c^2\right)-4m_a^2\right]=16S^2+\left(b^2-c^2\right)^2\iff$

$\boxed{16\left(S^2+m_a^4\right)+\left(b^2-c^2\right)^2=8m_a^2\left(b^2+c^2\right)}$ . Example. $b=12\ ,\ c=10$ and $m_a=5\implies$ $16S^2+16\cdot 5^4+\left(12^2-10^2\right)^2=$ $8\cdot 5^2\left(12^2+10^2\right)$ $\implies$

$S^2+625+(36-25)^2=50(36+25)$ $\iff$ $S^2=3050-746$ $\iff$ $S^2=2304=12\cdot 12\cdot 16=48^2\iff \boxed{\ S=48\ }$ .

P7. Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ and a mobile point $M\in \mathcal P$ (plane). The lines $MA$ , $MB$ ,

$MC$ meet again $w$ in $X$ , $Y$ , $Z$ respectively. Find the locus of $M$ for which $\frac {MA}{MX}+\frac {MB}{MY}+\frac {MC}{MZ}=3$ .

Proof. $\overline{MA}\cdot\overline{MX}=\overline{MB}\cdot \overline{MY}=\overline{MC}\cdot\overline{MZ}=p_w(M)=MO^2-R^2$ - the power of $M$ w.r.t. $w\implies$ $3=\frac {MA}{MX}+\frac {MB}{MY}+\frac {MC}{MZ}=$ $\frac {MA^2+MB^2+MC^2}{\left|p_w(M)\right|}=$

$\frac 1{\left|MO^2-R^2\right|}\cdot\left[3\cdot MG^2+\frac {a^2+b^2+c^2}3\right]$ , where $G$ is the centroid of $\triangle ABC\implies$ $\boxed{\left|MO^2-R^2\right|=MG^2+\frac {a^2+b^2+c^2}9}\ (*)$ . Appear two cases $:$

$\blacktriangleright\ M\in \mathrm{ext}(w)\iff MO>R\iff MO^2-MG^2=R^2+\frac {a^2+b^2+c^2}9$ (constant) $\iff$ the locus is a line $d\perp OG\ ;$

$\blacktriangleright\ M\in \mathrm{int}(w)\iff MO<R\iff MO^2+MG^2=R^2-\frac {a^2+b^2+c^2}9$ (constant) $\iff$ the locus is a circle with the center in the midpoint of $[OG]\ .$

P8. Let the cyclical convex quadrilateral $ABC$ with the circumcircle $w=\mathbb C(O,R)$ and a mobile point $M\in \mathcal P$ (plane). The lines $MA$ , $MB$ ,

$MC$ and $MD$ meet again $w$ in $X$ , $Y$ , $Z$ and $T$ respectively. Find the locus of $M$ for which $\frac {MA}{MX}+\frac {MB}{MY}+\frac {MC}{MZ}+\frac {MD}{MT}=4$ .

Proof. Denote the midpoints $E$ , $F$ and $G$ $($ centroid of $ABCD)$ of $[AC]$ , $[BD]$ and $[EF]$ respectively and $\left\{\begin{array}{ccccc} AB=a & ; & BC=b & ; & AC=e\\\\ CD=c & ; & DA=d & ; & BD=f\end{array}\right\|$ . Apply the theorem of median

in the triangles $:\ \left\{\begin{array}{cccc} \triangle AMC\ : & 2\cdot\left(MA^2+MC^2\right) & = & 4\cdot ME^2+e^2\\\\ \triangle BMD\ : & 2\cdot\left(MB^2+MD^2\right) & = & 4\cdot MF^2+f^2\end{array}\right\|$ $\bigoplus\implies$ $\sum MA^2=$ $2\left(ME^2+MF^2\right)+\frac {e^2+f^2}2=$ $4\cdot MG^2+EF^2+\frac {e^2+f^2}2=$

$4\cdot MG^2+\frac {\sum a^2-\left(e^2+f^2\right)}4+\frac {e^2+f^2}2\implies$ $\boxed{\sum MA^2=4\cdot MG^2+\frac 14\cdot\left[\sum a^2+\left(e^2+f^2\right)\right]}\ (*)$ . From the power $p_w(M)$ of the point $M$ w.r.t. the circle $w$ obtain $:$

$\overline{MA}\cdot\overline{MX}=\overline{MB}\cdot \overline{MY}=\overline{MC}\cdot\overline{MZ}=\overline {MD}\cdot\overline{MT}=p_w(M)=MO^2-R^2$ $\implies$ $4=\frac {MA}{MX}+\frac {MB}{MY}+\frac {MC}{MZ}+\frac {MD}{MT}=$ $\frac {MA^2+MB^2+MC^2+MD^2}{\left|p_w(M)\right|}=$

$\frac {4\cdot MG^2+4k}{\left|MO^2-R^2\right|}$ , where $G$ is the centroid of $\triangle ABC$ and $16k\ \stackrel{(*)}{=}\ \sum a^2+\left(e^2+f^2\right)$ $\implies$ $\boxed{\left|MO^2-R^2\right|=MG^2+k}$ . Appear two cases $:$

$\blacktriangleright\ M\in \mathrm{ext}(w)\iff MO>R\iff MO^2-MG^2=R^2+k$ (constant) $\iff$ the locus is a line $d\perp OG\ ;$

$\blacktriangleright\ M\in \mathrm{int}(w)\iff MO<R\iff MO^2+MG^2=R^2-k$ (constant) $\iff$ the locus is a circle with the center in the midpoint of $[OG]\ .$

Lemma. Let $\triangle ABC$ with the circumcircle $w=\mathcal C(O,R)$ and $U\in AB$ , $V\in AC$ . Then $AO\perp UV$ $\iff$ $BUVC$ is cyclically.

.Proof. Let $L\in AA$ so that $AC$ separates $B$ and $L$ . Thus, $\underline{AO\perp UV}\iff$ $\underline{UV\parallel AA}\iff$ $\widehat{UBC}\equiv\widehat{ABC}\equiv\underline{\widehat{LAC}\equiv \widehat{AVU}}\iff$ $\widehat{UBC}\equiv\widehat{AVU}\iff$ $\underline{BUVC}$ is cyclically.

P9. Consider $\triangle ABC$ and let $D\in (BC)$ be the foot of the altitude from the vertex $A$ . The circle with diameter $[AD]$ intersects the sides $AB$

and $AC$ at points $X$ and $Y$ respectively. Prove that the altitude from $A$ in triangle $AXY$ passes through the circumcenter $O$ of triangle $ABC$ .

.Method 1. Let $P\in XY\ ,\ AP\perp XY$ . Thus, $\widehat{AXY}\equiv\widehat{ADY}\equiv\widehat{ACD}\equiv\widehat{ACB}\implies$ $\widehat{AXY}\equiv\widehat{ACB}\implies$ $BXYC$ is cyclic $\implies\ XY$ is antiparallel to $BC\ \stackrel{\mathrm{lemma}}{\implies}\ O\in AP$ .

.Method 2. $\left\{\begin{array}{ccccc} DX\perp AB & \implies & AX\cdot AB=AD^2\\\\ DY\perp AC & \implies & AY\cdot AC=AD^2\end{array}\right\|\implies$ $AX\cdot AB=AY\cdot AC\implies$ $BXYC$ is cyclic $\implies XY$ is antiparallel to $BC\ \stackrel{\mathrm{lemma}}{\implies}\ O\in AP$ .

Lemma (Ruben Dario). Let $\triangle ABC$ with the circumcircle $\alpha =\mathcal (O,R)$ and the incircle $w=\mathcal C(I,r)$ . Denote $\left\{\begin{array}{ccc} E\in BI\cap AC\\\\ F\in CI\cap AB\end{array}\right\|$ .

Prove that $P\in (EF)\ \iff\ \delta_{CA}(P)+\delta_{AB}(P)=\delta_{BC}(P)$ , where $\delta_d(X)$ is the distance of the point $X$ to the line $d$ .

.Proof. Denote $\left\{\begin{array}{ccc} \delta_{BC}(P) & = & x\\\\ \delta_{CA}(P) & = & y\\\\ \delta_{AB}(P) & = & z\end{array}\right\|$ . Thus, $[BPC]+[CPA]+[APB]=2\cdot [ABC]\iff$ $\boxed{ax+by+cz=2S}\ (*)$ and $\left\{\begin{array}{cccc} \frac {FA}b=\frac {FB}a=\frac c{a+b} & \implies & AF=\frac {bc}{a+b} & (1)\\\\ \frac {EA}c=\frac {EC}a=\frac b{a+c} & \implies & AE=\frac {bc}{a+c} & (2)\end{array}\right\|$ .

Therefore, $P\in (EF)\iff$ $[AFP]+[AEP]=[AFE]\iff$ $z\cdot AF+y\cdot AE=AF\cdot AE\cdot\sin A\iff\ \stackrel{1\wedge 2}{\iff}\ z\cdot \frac {bc}{a+b}+y\cdot \frac {bc}{a+c}=\frac {bc}{a+b}\cdot \frac {bc}{a+c}\cdot\sin A\iff$

$\frac z{a+b}+\frac y{a+c}=\frac {bc}{(a+b)(a+c)}\cdot\sin A\iff$ $z(a+c)+y(a+b)=2S\iff$ $a(y+z)+(by+cz)=2S\stackrel{(*)}{\iff}\ a(y+z)+(2S-ax)=2S\iff$ $\boxed{y+z=x}\ .$

Remark. I"ll present some particular cases $:$

$\blacktriangleright\ P:=O$ (circumcenter). Hence $O\in (EF)\iff$ $2R\cos B+2R\cos C=2R\cos A\iff$ $\boxed{\cos B+\cos C=\cos A}\iff 1+\frac rR=2\cos A\iff$

$2R\cos A=R+r\iff$ $2\left(R-\frac {r_a-r}{2}\right)=R+r\iff$ $2R-r_a+r=R=r\iff$ $\boxed{\ R=r_a\ }\ .$ I used the well-known identity $\boxed{\cos A+\cos B+\cos C=1+\frac rR}\ .$

$\blacktriangleright\ P:=G$ (centroid). Hence $G\in (EF)\iff$ $\left\{\begin{array}{ccc} 3x & = & h_a\\\\ 3y & = & h_b\\\\ 3z & = & h_c\end{array}\right\|\iff$ $h_b+h_c=h_a\iff \boxed{\frac 1b+\frac 1c=\frac 1a}\ .$

$\blacktriangleright\ P:=H$ (orthocenter). Hence $H\in (EF)\iff$ $\left\{\begin{array}{ccc} x & = & 2R\cos B\cos C\\\\ y & = & 2R\cos C\cos A\\\\ z & = & 2R\cos A\cos B\end{array}\right\|\iff$ $\cos C\cos A+\cos A\cos B=\cos B\cos C\iff$ $\boxed{\frac 1{\cos B}+\frac 1{\cos C}=\frac 1{\cos A}}\ .$

P10. Let $\triangle ABC$ with circumcircle $\alpha =\mathcal (O,R)$ and incircle $w=\mathcal C(I,r)$ . Denote $\left\{\begin{array}{ccc} E\in BI\cap AC\\\\ F\in CI\cap AB\end{array}\right\|$ . Prove that $O\in EF\ \iff\ \cos B+\cos C=\cos A\ \iff\ R=r_a$ .

.Method 1. Let $\boxed{b\cos B+c\cos C=k}\ (*)$ and apply Cristea's relation $:\ O\in EF\iff \frac {FB}{FA}\cdot DC+\frac {EC}{EA}\cdot DB=\frac {OD}{OA}\cdot BC\iff$ $\frac ab\cdot DC+\frac ac\cdot DB=\frac {OD}{OA}\cdot a\iff$

$\boxed{\frac {DC}b+\frac {DB}c=\frac {OD}{OA}}\ (1)$ . Observe that $\frac {DB}{c\cos C}=\frac {DC}{b\cos B}\ \stackrel{*}{=}\ \frac ak\implies$ $\odot\begin{array}{cccccc} \nearrow & \frac {DB}c & = & \frac {a\cos C}k & (2) & \searrow\\\\ \searrow & \frac {DC}b & = & \frac {a\cos B}k & (3) & \nearrow\end{array}\odot$ and $\frac {OD}{OA}=\frac {BD}{BA}\cdot\frac {\sin\widehat{OBD}}{\\sin\widehat{OBA}}\ \stackrel{2}{=}\ \frac {a\cos C}k\cdot\frac {\cos A}{\cos C}$ $\iff$

$\boxed{\frac {OD}{OA}=\frac {a\cos A}k}\ (4)$ . Therefore, the relation $(1)$ becomes $\frac {a\cos B}k+\frac {a\cos C}k=\frac {a\cos A}k$ , i.e. $\boxed{\cos B+\cos C=\cos A}\ \iff$ $2\cos\frac {B-C}2\sin\frac A2=1-2\sin^2\frac A2\iff$

$2\sin\frac A2\left(\cos\frac {B-c}2+\cos\frac {B+C}2\right)=1\iff$ $4\sin\frac A2\cos\frac B2\cos\frac C2=1\iff$ $4\sqrt {\frac {(s-b)(s-c)}{bc}\cdot\frac{s(s-b)}{ac}\cdot\frac {s(s-c)}{ab}} =1\iff$ $4s(s-b)(s-c)=abc\iff$

$4s\cdot (s-a)(s-b)(s-c)=abc\cdot (s-a)\iff$ $4s\cdot sr^2=4Rsr\cdot (s-a)\iff$ $sr=R(s-a)\iff$ $\boxed{\ R=r_a\ }\ .$

.Method 2. I"ll use barycentrical coordinates $:\ O(\sin 2A,\sin 2B,\sin 2C)\ ;\ E(a.0.c)\ ;\ F(a,b,0)$ . Thus, $O\in EF\iff$ $\left|\begin{array}{ccc} \sin 2A & \sin 2B & \sin 2C\\\\ a & b & 0\\\\ a & 0 & c\end{array}\right|=0\iff$ $bc\sin 2A=$

$ac\sin2B+ab\sin 2C\iff$ $bc\sin A\cdot \cos A=ac\sin B\cdot\cos B+ab\sin C\cdot\cos C\iff$ $2S\cdot \cos A=2S\cdot \cos B+2S\cdot \cos C\iff$ $\cos A=\cos B+\cos C$ a.s.o.

Remark. If $P(\alpha ,\beta ,\gamma )$ , then $P\in (EF)\iff$ $\left|\begin{array}{ccc} \alpha & \beta & \gamma\\\\ a & b & 0\\\\ a & 0 & c\end{array}\right|=0\iff$ $\gamma ab+\beta ac=\alpha bc\iff$ $\boxed{\frac {\beta}b+\frac {\gamma}c=\frac {\alpha}a}\ .$

P11. Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ . The $A$ -bisector cut the sideline $BC$ at $D$ and meet again

the circle $w$ at $M\ .$ Denote the projection $P$ of the midpoint $S$ of $[BC]$ on $AM$ . Prove that $4\cdot AP\cdot DM=BC^2\ .$

Proof 1. Suppose $b>c\ ,$ denote the diameter $[MN]$ of $w\ ,$ the midpoint $S$ of $[BC]\ ,$ $m(\angle ADB)=\phi$ and $L\in AN\cap BC\ .$ Since $\triangle SDM\sim\triangle SNL$ obtain that $\frac {SD}{SM}=\frac {SN}{SL}$

and $BS^2=SM\cdot SN\implies$ $SD\cdot SL=BS^2\ .$ Observe that $AP=SL\cos\phi$ and $DM\cos\phi =SD\ .$ In conclusion, $AP\cdot DM=SL\cdot SD=SB^2\implies AP\cdot DM=\frac {a^2}{4}\ .$

Proof 2. $\frac {DB}c=\frac {DC}b=\frac a{b+c}\implies$ $\left\{\begin{array}{ccc} DB & = & \frac {ac}{b+c}\\\\ DC & = & \frac {ab}{b+c}\end{array}\right\|$ $\bigodot\implies$ $DB\cdot DC=\frac {a^2bc}{(b+c)^2}\ .$ Is well known that $AD=\frac {2bc\cos\frac A2}{b+c}\ .$ From the power of the point $D$

w.r.t. $w$ obtain that $DB\cdot DC=DA\cdot DM\implies$ $\frac {a^2bc}{(b+c)^2}=DM\cdot\frac {2bc\cos\frac A2}{b+c}\implies$ $\boxed{DM=\frac{a^2}{2(b+c)\cdot\cos\frac{A}{2}}}\ (1)\ .$ Denote $\left\{\begin{array}{ccc} R\in AD & ; & BR\perp AD\\\\ Q\in AD & ; & CQ\perp AD\end{array}\right\|\ .$

Prove easily that the point $P$ is the midpoint of $[RQ]\implies$ $\boxed{AP=\frac{AR+AQ}{2}=\frac {b+c}2\cdot\cos\frac{A}{2}}\ (2)\ .$ From the relations $(1)$ and $(2)$ obtain that the required relation.

P12. Let a convex $ABCD$ with $I\in AC\cap BD$ so that $AB=BC=CD$ and $IA=ID$ . Prove that $IB\ne IC\implies$ $m\left(\widehat{AIB}\right)=60^{\circ}$ .

Proof. Denote $\left\{\begin{array}{c} AB=BC=CD=x\\\\ IA=ID=y\\\\ IB=u\ ;\ IC=v\end{array}\right\|$ . Apply the Stewart's theorem to the isosceles $:\ \left\{\begin{array}{cccc} \triangle ABC\ : & BI^2+IA\cdot IC=BA^2 & \implies & u^2+yv=x^2\\\\ \triangle DCB\ : & CI^2+IB\cdot ID=CD^2 & \implies & v^2+yu=x^2\end{array}\right\|$ $\implies$

$u^2+yv=x^2=v^2+yu\iff$ $\left(u^2-v^2\right)=y(u-v)\ \stackrel{u\ne v}{\iff}\ y=u+v\iff$ $x^2=u^2+v(u+v)\iff$ $x^2=u^2+uv+v^2\iff$ $m\left(\widehat{AIB}\right)=60^{\circ}$ .

P13. Let an acute $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)\ ;\ [AT]\ -$ the diameter of $w\ ;\ TT\ -$ the tangent

line to $w\ ;$ the points $X\in AB\ ,\ Y\in AC$ so that $O\in XY$ and $OX=OY$ . Prove that $XY\cap TT\cap BC\ne\emptyset$ .

Proof 1. Denote $\left\{\begin{array}{c} P\in XY\cap BC\\\\ R\in TT\cap BC\end{array}\right\|$ . So $\left\{\begin{array}{ccccc} OX=OY & \iff & \frac {YA}{XA}=\frac {\sin\widehat{TAB}}{\sin\widehat{TAC}} & \iff & \boxed{\frac {YA}{XA}=\frac {\cos C}{\cos B}}\\\\ OX=OY & \iff & p_w(X)=p_w(Y) & \iff & \boxed{\frac {XB}{YC}=\frac {YA}{XA}}\end{array}\right\|$ $\implies$ $\boxed{\frac {YA}{XA}=\frac {\cos C}{\cos B}=\frac {XB}{YC}}\ (*)$ . Apply Menelaus' theorem

to $\overline{XYP}/\triangle ABC\ :\ \frac {PC}{PB}\cdot\frac {XB}{XA}\cdot\frac {YA}{YC}=1\iff$ $\frac {PB}{PC}=\frac {XB}{XA}\cdot \frac {YA}{YC}$ $\implies$ $\frac {PB}{PC}=\frac {XB}{YC}\cdot \frac {YA}{XA}\ \stackrel{(*)}{\iff}$ $\boxed{\frac {PB}{PC}=\left(\frac {\cos C}{\cos B}\right)^2}\ (1)$ . Apply an well known relation to the ray $[TR$

and $\triangle BTC\ :\ \frac {RB}{RC}=$ $\frac {TB}{TC}\cdot\frac {\sin\widehat{RTB}}{\sin\widehat{RTC}}=$ $\left[\frac {\sin\left(90^{\circ}-C\right)}{\sin\left(90^{\circ}-C\right)}\right]^2$ $\implies$ $\boxed{\frac {RB}{RC}=\left(\frac {\cos C}{\cos B}\right)^2}\ (2)$ . From the relations $(2)$ and $(3)$ obtain that $P\equiv R$ , i.e. $XY\cap TT\cap BC\ne\emptyset\ .$

Proof 2 (E.V. Diaz). I"ll use same notations from previous proof. Denote $\{U,V\}\subset w$ so that $X\in TU\cap AB$ and $Y\in TV\cap AC$ .

Apply the Pascal's theorem to $ABCUTT\ :\ \odot\begin{array}{cccc} \nearrow & X\in AB\cap UT & \searrow\\\\ \rightarrow & R\in BC\cap TT & \rightarrow\\\\ \searrow & O\in CU\cap TA & \nearrow\end{array}\odot\implies$ $R\in OX\implies XY\cap TT\cap BC\ne\emptyset\ .$

Proof 3. I"ll use same notations from previous proofs. Denote $S\in XY$ so that $TS\perp XY$ and $Z\in TS\cap BC$ , i.e. $\boxed{ZS\perp XY}\ (1)$ . Observe that $BXST$ is inscribed in the circle

with the diameter $[TX]$ and $CYST$ is inscribed in the circle with the diameter $[TY]$ . Thus, $\widehat{ZSC}\equiv \widehat {TSC}\equiv\widehat{TYC}\equiv\widehat{BAC}\equiv$ $\widehat{TXB}\equiv\widehat{TSB}\equiv\widehat{ZSB}$ $\implies$ $\boxed{\widehat{ZSC}\equiv\widehat{ZSB}}\ (2)$ .

Therefore, the relations $(1)$ and $(2)$ together means that the division $(B,C;Z,P)$ is harmonically, i.e. $TS$ is the polar of the point $P$ . In conclusion, $P\in TT$ .

Remark. Prove easily that $:\ O\in CU\cap BV\cap AT\ ;\ \left\{\begin{array}{ccc} XU=YC & ; & XB=YV\\\\ TB=AV & ; & TC=AU\end{array}\right\|\ ;\ \frac {PB}{PC}=$ $\left(\frac {TB}{TC}\right)^2=\left(\frac {\cos C}{\cos B}\right)^2$ .

P14 (Kadir Altintas - Turkey). Let $\triangle ABC$ with $\frac A4=\frac B2=C=\frac {\pi}7$ and incircle $w=\mathbb C(I,r)$ what touches $(BC,CA,AB)$ at $(D,E,F)$ respectively. Prove $DE=DF\ .$

Proof. Let $G\in AB\ ,$ $IG\parallel BC\ .$ By a simple angle chase obtain that $BG=GI=IE=IA=AF$ and $BF=AG=AE$ $\implies$ $\triangle DBF\ \stackrel{(s.a.s)}{\equiv}\ \triangle DAE$ $\implies DE=DF\ .$

P15 (M. O. Sanchez). Chord $CD\ ,$ $QT$ of a given circle $w$ meet at $P\ .$ The tangents at $Q\ ,$ $T$ cross $CD$ extended at $A\ ,$ $B$ respectively. Prove that $\frac 1{PA}-\frac 1{PB}=\frac 1{PC}-\frac 1{PD}\ .$

Proof. Denote $\left\{\begin{array}{ccc} PC=m & ; & PD=n\\\\ PA=a & ; & PB=b\end{array}\right\|$ and $S\in QT$ so that $BS=BT\ ,$ $T\in (PS)\ .$ Thus, $\left\{\begin{array}{ccc} AD=a+n & ; & AC=a-m\\\\ BD=b-n & ; & BC=b+m\end{array}\right\|$ and $\frac ab=$ $\frac {PA}{PB}=$ $\frac {AQ}{BS}=$ $\frac {AQ}{BT}$ $\implies$

$\left(\frac ab\right)^2=\left(\frac {AQ}{BT}\right)^2=$ $\frac {AD\cdot AC}{BD\cdot BC}=$ $\frac {(a+n)(a-m)}{(b-n)(b+m)}$ $\implies$ $\frac {(a+n)(a-m)}{a^2}=$ $\frac {(b+m)(b-n)}{b^2}\implies$ $\frac {a^2-a(m-n)-mn}{a^2}=$ $\frac {b^2+b(m-n)-mn}{b^2}$ $\implies$

$1-\frac {m-n}a-\frac {mn}{a^2}=$ $1+\frac {m-n}b-\frac {mn}{b^2}$ $\implies$ $(m-n)\left(\frac 1b+\frac 1a\right)=$ $mn\left(\frac 1{b^2}-\frac 1{a^2}\right)$ $\implies$ $\frac 1n-\frac 1m=\frac 1b-\frac 1a\implies$ $\frac 1a-\frac 1b=\frac 1m-\frac 1n\implies$ $\frac 1{PA}-\frac 1{PB}=\frac 1{PC}-\frac 1{PD}\ .$

P16 (Czech-Polish-Slovak, Match 2016). Let $ABC$ be an acute-angled triangle with $AB < AC$ . Tangent to its circumcircle $\Omega$ at $A$ intersects the line $BC$ at $D$ . Let $G$

be the centroid of $\triangle ABC$ and let $AG$ meet $\Omega$ again at $H \neq A\ .$ Suppose the line $DG$ intersects the lines $AB\ ,$ $AC$ at $E\ ,\ F$ respectively. Prove that $\widehat{EHG}\equiv\widehat{GHF}\ .$

P17. Let $\triangle ABC$ with $AB\ne AC$ , the circumcircle $w$ and the midpoint $S$ of the arc $\overarc{BC}\subset w$ for which $BC$ separates $A$ and $S$ . Prove that the relation

$SA^2=AB\cdot AC+SB\cdot SC$ and $(\forall )\ M\in\overarc{BC}\ ,\ MA^2=AB\cdot AC+MB\cdot MC\iff MB=MC$ , i.e. $AM$ is the $A$ -bisector of $\triangle ABC$ .

Proof.

$\blacktriangleleft 1\blacktriangleright\ SB=SI=SC\implies$ $SA^2=bc+SB\cdot SC=bc+SI^2\iff$ $SA^2-SI^2=bc\iff$ $(SA-SI)(SA+SI)=bc\iff$ $IA\cdot (SA+SI)=bc\iff$

$IA\cdot (2\cdot IS+IA)=bc\iff$ $2\cdot IA\cdot IS+IA^2=bc\iff$ $2\cdot p_w(I)+IA^2=bc\iff$ $4Rr+IA^2=bc\iff$ $IA^2=bc-4Rr\ ,$ what is true. Indeed, $s-a=IA\cdot\cos\frac A2$

$\iff$ $(s-a)^{\cancel 2}=$ $IA^2\cdot\frac {s\cancel{(s-a)}}{bc}\iff$ $\boxed{IA^2=\frac {bc(s-a)}s}\ (*)\ .$ Otherwise 1. $\underline{\underline{s\cdot IA}}^2=$ $s\left[(s-a)^2+r^2\right]=$ $s(s-a)^2+(s-a)(s-b)(s-c)=$

$(s-a)\left[s(s-a)+(s-b)(s-c)\right]=\underline{\underline{(s-a)bc}}\iff$ $\boxed{IA^2=\frac {bc(s-a)}s}\ .$ Otherwise 2. Let $D\in AS\cap BC$ and $\boxed{SB=SI=SC}\ (*)\ .$ Therefore,

$\left\{\begin{array}{cccc} AD\cdot AS & = & AB\cdot AC & (1)\\\\ \frac {DA}{DS} & = & \frac {AB\cdot AC}{SB\cdot SC} & (2)\end{array}\right\|$ $\implies$ $\underline{\underline{\frac 1{SA}}}\ \stackrel{(1)}{=}\ \frac {DA}{bc}\ \stackrel{(2)}{=}$ $\frac {DS}{SB\cdot SC}\ \stackrel{(*)}{=}\ \frac {DA+DS}{bc+SI^2}=$ $\underline{\underline{\frac {SA}{AB\cdot AC+SB\cdot SC}}}\implies$ $SA^2=AB\cdot AC+SB\cdot SC\ .$

$\blacktriangleleft 2\blacktriangleright$ I"ll use the relation $4RS=abc$ for $\triangle ABC$ with the area $S=[ABC]$ . Let $AM=m$ and $\left\{\begin{array}{ccc} MB & = & u\\\\ MC & = & v\end{array}\right\|$ . Thus, $[MAB]+[MAC]=[BAC]+[BMC]\iff$

$mcu+mbv=abc+auv\iff$ $\boxed{m(cu+bv)=a(bc+uv)}\ (*)$ . Apply Ptolemy's theorem $\boxed{bu+cv=am}\ (1)\ .$ Therefore, $MA^2=AB\cdot AC+MB\cdot MC\iff$

$m^2=bc+uv\ \stackrel{(*)}{\iff}\ am^2=m(cu+bv)\iff$ $cu+bv=am\ \stackrel{(1)}{\iff}\ cu+bv=bu+cv\iff$ $u(b-c)=v(b-c)\stackrel{(b\ne c)}{\iff}\ u=v\iff$ $MB=MC$ .

P18. Let $P$ be an interior point of $\triangle ABC$ for which denote $\left\{\begin{array}{ccc} D & \in & BC\cap AP\\\\ E & \in & CA\cap BP\\\\ F & \in & AB\cap CP\end{array}\right\|\ .$ Prove that $\frac {PA}{PD}\cdot\frac {PB}{PE}\cdot \frac {PC}{PF} =2+\frac {PA}{PD}+\frac {PB}{PE}+\frac {PC}{PF}\ .$

Proof. Since $\frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1$ then there are $(x,y,z)\subset\mathbb R^*_+$ so that $\left\{\begin{array}{ccc} \frac {DB}{DC} & = & \frac yz\\\\ \frac {EC}{EA} & = & \frac zx\\\\ \frac {FA}{FB} & = & \frac xy\end{array}\right\|\ .$ I''ll use Van Aubel's relation $:\ \frac {PA}{PD}=\frac {EA}{EC}+\frac {FA}{FB}\iff$

$\frac {PA}{PD}=\frac xz+\frac xy\ \mathrm{a.s.o.}$ Thus, $\prod\frac {PA}{PD}=2+\sum\frac {PA}{PE}\iff$ $\prod\frac{x(y+z)}{yz} =2+\sum\frac{x(y+z)}{yz}\iff$ $\prod (y+z)=2xyz+\sum x^2(y+z)\ ,$ what is true.

P19 (M.O. Sanchez). Let an acute $\triangle ABC$ with $A=60^{\circ}$ and the orthocenter $H.$ Prove that $\frac {AB+BC+CA}{HA+HB+HC}=\sqrt 3.$

Proof 1 (Baris Altay}. Denote the orthic $\triangle DEF,$ where $D\in (BC),$ $E\in (CA),$ $F\in (AB)$ and $HA=x,$ $HB=y,$ $HC=z.$ Observe that $\triangle BCE\sim\triangle AHE\iff$

$\frac {BC}{AH}=\frac {CE}{HE}\implies$ $\frac ax=\frac {\frac {z\sqrt 3}2}{\frac z2}$ $\implies$ $\boxed{a=x\sqrt 3}\ (1)$ and $\left\{\begin{array}{cccccc} b=AC=AE+EC & \implies & b=\frac c2+\frac {z\sqrt 3}2 & \implies & \boxed{2b=c+z\sqrt 3} & (2)\\\\ c=AB=AF+FB & \implies & c=\frac b2+\frac {y\sqrt 3}2 & \implies & \boxed{2c=b+y\sqrt 3} & (3)\end{array}\right\|\ \stackrel{\oplus 1\wedge 2\wedge 3}{\implies}\ \frac {a+b+c}{x+y+z}=\sqrt 3.$

Proof 2 (own). Observe that $\left(\frac B2+\frac C2\right)+\frac A2=\frac {\pi}2$ $\implies$ $\tan\frac A2\cdot \tan\left(\frac B2+\frac C2\right)=1$ $\implies$ $\tan\frac A2\cdot \frac {\tan\frac B2+\tan\frac C2}{1-\tan \frac B2\tan\frac C2}=1$ $\implies$ $\boxed{\sum\tan\frac B2\tan\frac C2=1}\ (1)$ and

$\left\{\begin{array}{cccccc} \sum \tan\frac A2 & = & \sum\frac {r_a}s=\frac 1s\cdot\sum r_a=\frac {4R+r}{s} & \implies & \boxed{\sum\tan\frac A2=\frac {4R+r}s} & (2)\\\\ \prod \tan\frac A2 & = & \prod\frac r{s-a}=\frac {r^3}{\prod (s-a)}=\frac {r^3}{sr^2} & \implies & \boxed{\prod\tan\frac A2=\frac rs} & (3)\end{array}\right\|\ \stackrel{1\wedge 2\wedge 3}{\implies}\ \boxed{f(t)\equiv st^3-(4R+r)t^2+st-r=0}\ (*)$ $\odot\begin{array}{ccc} \nearrow & \tan\frac A2 & \searrow\\\\ \rightarrow & \tan\frac B2 & \rightarrow\\\\ \searrow & \tan\frac C2 & \nearrow\end{array}\odot$

Thus, $\frac {a+b+c}{x+y+z}=\frac {2s}{2R(\cos A+\cos B+\cos C)}=\frac {s}{R\left(1+\frac rR\right)}=\frac s{R+r}.$ Thus, $\boxed{\frac {a+b+c}{x+y+z}=\sqrt 3\iff s=(R+r)\sqrt 3}\ .$ On other hand $60^{\circ}\in \{A,B,C\}\iff$

$f\left(\frac {\sqrt 3}{3}\right)=0\iff$ $\frac {s\sqrt 3}9-\frac {4R+r}3+\frac {s\sqrt 3}3-r=0\iff$ $s\sqrt 3-3(4R+r)+3s\sqrt 3-9r=0\iff$ $4s\sqrt 3-12(R+r)=0\iff$ $\boxed{s=(R+r)\sqrt 3}.$

P20. Let $\triangle ABC$ with the orthocenter $H.$ Prove that the chain of the equivalencies $:\ s=2R+r\iff$ $ABC$ is $A$ -right triangle $\iff$ $HA+r_a=s$ (standard notations).

Proof. $\boxed{HA+r_a=2R+r}\iff$ $2R(1-\cos A)=r_a-r\iff$ $4R\cdot\frac {(s-b)(s-c)}{bc}=\frac S{s-a}-\frac Ss\iff$ $4Rs(s-a)(s-b)(s-c)=abcS\iff$

$4RS^2=4RS^2$ what is true. Hence $HA+r_a=s\iff 2R+r=s$ a.s.o. Generally, in any triangle $ABC$ there is the identity $\boxed{\frac {AB+BC+CA}{HA+HB+HC}=\frac {2R+r}{R+r}}\ .$

P21. Let $\triangle ABC$ with $\left\{\begin{array}{c} AB=c=6\\\ BC=a=7\\\ AC=b=8\end{array}\right\|$ and $w=\mathbb C(X,x)$ with the center $X\in (BC)$ , $B\in w$ and which is tangent to $AC$ . Find the length $x$ of its radius.

Proof. Show easily that $\triangle ABC$ is acute. Let the projection $D$ of $A$ on $BC$ and $AD=h_a$ . Since $BX=x=\delta_{AC}(X)$ , i.e. $DC=a-x$ obtain that $h_a(a-x)=2[ADC]=xb,$ i.e.

$\boxed{x=\frac {ah_a}{b+h_a}}\ (*)$ . Thus, $DB=u\ ,\ DC=v$ and $AD\perp BC\iff$ $DC^2-DB^2=AC^2-AB^2\iff$ $v^2-u^2=28$ and $u+v=7$ . Hence $\left\{\begin{array}{ccc} v+u & = & 7\\\ v-u & = & 4\end{array}\right\|$ , i.e.

$\frac u3=\frac v{11}=\frac 12$ . Obtain easily that $h_a^2+u^2=c^2\iff$ $h_a^2=6^2-\frac 94=$ $\frac {12^2-3^2}{4}=\frac {15\cdot 9}{4}$ $\implies \boxed{h_a=\frac {3\sqrt{15}}2}$ . Using the relation $(*)$ obtain that $\boxed{x=\frac {21\left(16\sqrt{15}-45\right)}{121}}$ .

This post has been edited 486 times. Last edited by Virgil Nicula, Nov 16, 2016, 6:57 AM

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by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
by CSPAL, May 27, 2020, 4:17 PM
impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
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!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

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

72 shouts

My (math)blog

427. Selected nice or well-known geometry problems.

by Virgil Nicula, Jul 25, 2015, 3:43 PM

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