320. Some problems with metrical relations.

by Virgil Nicula, Oct 1, 2011, 12:44 PM

PP1. Let $ABC$ be an $A$ -isoceles triangle. For a mobile point $P\in BC$ so that $B\in (PC)$ , denote the length $r_1$

of the inradius of $\triangle APB$ and the length $r_2$ of the $P$ -exradius of $\triangle APC$ . Prove that the sum $r_1+r_2$ is constant.

Proof 1 (metric). $AD\perp PB\iff$ $AP^2-AB^2=DP^2-DB^2\iff$ $AP^2-AB^2=(PD-DB)\cdot (PD+DC)\iff$ $\boxed{AP^2=PB\cdot PC+AB^2}\ (*)$ .

Denote $\delta_{BC}(A)=h$ . I"ll show that $r_1+r_2=h$ . Therefore, $\frac {r_1}{h}+\frac {r_2}{h}=1\iff$ $\frac {PB}{PA+PB+AB}+\frac {PC}{PA+PC-AC}=1\iff$ $PB\cdot (PA+PC-AC)+$

$PC\cdot (PA+PB+AB)=$ $(PA+PB+AB)\cdot (PA+PC-AC)\iff$ $PB\cdot (PA+PC-AC)=(PA+PB+AB)\cdot (PA-AC)\iff$

$PB\cdot PC=(PA+AB)\cdot (PA-AC)\iff$ $AP^2=PB\cdot PC+AB^2$ , i.e. the relation $(*)$ .

Remark 1 . If $r_1'$ is the length of the inradius of $\triangle PAC$ and $r_2'$ is the length of the $P$ -exinradius of $\triangle PAB$ , then also

$r_1'+r_2'=h$ because the relation $(*)$ and the proof of the equivalence is symmetrically in the vertices $B$ and $C$ .

Remark 2 . Can prove otherwise the relation $(*)$ . Denote the second intersection $R$ of $AP$ with the circumcircle of $\triangle ABC$ and $X\in AB$ so that $B\in (AX)$ .

Observe that $\widehat {PRB}\equiv$ $\widehat{ACB}\equiv$ $\widehat{ABC}\equiv$ $\widehat {PBX}$ , i.e. $\widehat {PRB}\equiv\widehat {PBX}$ , what means that the line $AB$ is tangent to the circumcircle of $\triangle PRB\implies$

$AB^2=AP\cdot AR=$ $AP\cdot (AP-PR)=$ $AP^2-PA\cdot PR=$ $AP^2-PB\cdot PC\implies$ $AP^2=PB\cdot PC+AB^2$ , i.e. the relation $(*)$ .

Proof 2. Let $I_{1}$ be the incircle and $I_{2}$ the excircle and let $CI_{2}$ intersect the line parallel to $BC$ which passes through $A$ in $B'$ . It can be easily seen that the angle formed

by the lines $AI_{1}$ and $AI_{2}$ is constant and is equal to the angle $BAB'$ . Consider the rotation with center $A$ and angle $BAB'$ so that $B$ goes to $B'$ and it is possible since

$B'AC$ is isosceles with vertex $A$ . Now, triangle $AI_{1}B$ will go to triangle $AI_{2} B'$ and so the inradius of $AI_{1}B$ is equal to the inradius of $AI_{2}B'$ and so the length of $r_{1}$

can be added to $r_{2}$ to get the length between the line parallel to $BC$ and passing through $A$ and $BC$ which is the result claimed.

PP2. Let $w_k=C(O_k)\ ,\ k\in \overline {1,3}$ be two by two secant circles. Denote $\left\{\begin{array}{c} \left\{A,D\right\}=w_2\cap w_3\\\\ \left\{B,E\right\}=w_3\cap w_1\\\\ \left\{C,F\right\}=w_1\cap w_2\end{array}\right|$ . Prove that $AF\cdot BD\cdot CE=AE\cdot BF\cdot CD\ \ (*)$ .

Proof. Denote the radical center $R\in AD\cap BE\cap CF$ of the given circles. Thus, $\left\{\begin{array}{ccc} \triangle ARF\sim\triangle CRD & \implies & \frac {AF}{CD}=\frac {RF}{RD}\\\\ \triangle BRD\sim\triangle ARE & \implies & \frac {BD}{AE}=\frac {RD}{RE}\\\\ \triangle CRE\sim\triangle BRF & \implies & \frac {CE}{BF}=\frac {RE}{RF}\end{array}\right\|\bigodot\implies\ (*)$ .

PP3. Let $ABCD$ be a convex quadrilateral. Prove that the set of the points $X$ so that

$XA^2+XB^2+CD^2=XB^2+XC^2+AD^2=$ $XC^2+XD^2+AB^2=XD^2+XA^2+BC^2$ .

Proof $\ .\ (\exists )\ X$ so that $\left\{\begin{array}{c} XA^2+XB^2+CD^2=XB^2+XC^2+AD^2\\\\ XC^2+XD^2+AB^2=XD^2+XA^2+BC^2\end{array}\right|\ \bigoplus\ \iff\ AB^2$ $+CD^2=AD^2+BC^2\iff AC\perp BD$ .

If $AB\not\perp BD$ , then $X\in\emptyset$ . Suppose that $AB\perp BD$ and denote $\left\{\begin{array}{ccc} AB=a & ; & BC=b\\\\ CD=c & ; & DA=d\end{array}\right|$ . In this case

the given relations becomes $\left\{\begin{array}{ccc} XA^2-XC^2=DA^2-CD^2 & \iff & X\in BD\\\\ XB^2-XD^2=AB^2-DA^2 & \iff & X\in AC\end{array}\right|\ \iff\ X$ $\in AC\cap BD\ .$

PP4. Let $ABC$ be a triangle with inradius $r$ and let $w_1=C(I_1,\rho)$ and $w_2=C(I_2,\rho)$ be two circles so

that $[BC]$ , $[BA]$ are tangent to $w_1$ and $[CB]$ , $[CA]$ are tangent to $w_2$ . Prove that $\frac{2}{BC}=\frac{1}{\rho}-\frac{1}{r}$ .

Proof. Denote $\left\{P_1,P_2\right\}\subset [BC]$ so that $I_1P_1\perp BC$ and $I_2P_2\perp BC$ and $BP_1=x$ , $CP_2=y$ . Observe that $a=BC=$

$BP_1+P_1P_2+P_2C=$ $\rho\cdot \cot\frac B2+2\rho+\rho\cdot \cot\frac C2\implies$ $\rho=\frac {a}{2+\frac {s-b}{r}+\frac {s-c}{r}}=$ $\frac {a}{2+\frac ar}=$ $\frac {ar}{2r+a}\implies$ $\boxed{\frac 1{\rho}=\frac 1r+\frac 2a}$ .

PP5. Find all points $M$ in a plane such that $2\cdot MA^2-3\cdot MB^2=AB^2$ , where $A$ and $B$ are two given points.

Proof. Define the point $C\in AB$ so that $B\in (AC)$ and $BC=2\cdot AB$ . Denote $AB=l$ and apply the Stewart's relation to the cevian $AC$ and

$\triangle MAB\ : MA^2\cdot 2l+$ $MC^2\cdot l=MB^2\cdot 3l+l\cdot 2l\cdot 3l\iff$ $2\cdot MA^2+MC^2=3\cdot MB^2+6l^2$ . Therefore, $2\cdot MA^2-3\cdot MB^2=AB^2\iff$

$6l^2-MC^2=l^2\iff$ $MC=l\sqrt 5$ (constant). In conclusion, since the point $C$ is fixed and $MC$ is constant the required points $M$ belong to the circle $(C,l\sqrt 5)$ .

Generalization. Find all points $M$ in a plane such that $\alpha\cdot MA^2+\beta\cdot MB^2=k$ (constant), where $A$ and $B$ are two given points and $\{\alpha , \beta\}\subset\mathbb R^*$ , $\alpha +\beta > 0$ .

Proof. Denote $AB=a$ . Suppose w.l.o.g. that $\alpha +\beta =1$ and exist only two cases :

$\blacktriangleright$ The first case : $\alpha >0\ \wedge\ \beta >0$ . Denote $C\in (AB)$ so that $\frac {CA}{\beta}=\frac {CB}{\alpha}=a$ and apply the Stewart's relation to the cevian $MC$ and $\triangle MAB\ :$

$\alpha\cdot MA^2+\beta\cdot MB^2=MC^2+\alpha\beta a^2$ . Therefore, $\alpha\cdot MA^2+\beta\cdot MB^2=k\iff$ $MC^2=k-\alpha\beta a^2$ . Since the point $C$ is fixed and

$MC$ is constant obtain in this case that the point $M$ belongs to the circle $C(C,\rho )$ , where $\rho =\sqrt {k-\alpha\beta a^2}$ $\iff$ $k\ge \alpha\beta a^2$ , i.e. $\frac {1}{\alpha}+\frac {1}{\beta}\ge \frac {a^2}{k}$ .

$\blacktriangleright$ The second case : $\alpha <0<\beta$ . Denote $C\in AB$ so that $B\in (AC)\ ,\ \frac {CA}{\beta}=\frac {CB}{-\alpha}=a$ and apply the Stewart's relation to

the cevian $MC$ and $\triangle MAB\ :\ -\alpha\cdot MA^2+MC^2=\beta\cdot MB^2-\alpha\beta a^2$ . Therefore, $\alpha\cdot MA^2+\beta\cdot MB^2=k\iff$

$MC^2=k-\alpha\beta a^2$ . Since the point $C$ is fixed and $MC$ is constant obtain in this case that the point $M$ belongs to the circle $C(C,\rho )$ ,

where $\rho =\sqrt {k-\alpha\beta a^2}$ $\iff$ $k\ge \alpha\beta a^2$ , i.e. $\frac {1}{\alpha}+\frac {1}{\beta}\le \frac {a^2}{k}$ .

PP6. Let $w_1=C(O_1,r_1)$ and $w_2=C(O_2,r_2)$ be the circles through $A$ and which touch $BC$ at $B\ ,\ C$ respectively. Prove that $r_1r_2= R^2$ .

Proof. Denote $\{A,P\}=w_1\cap w_2$ and $S\in AP\cap BC$ . Observe that $m(\angle SPB)=B$ and $m(\angle SPC)=C$ . Therefore,

$\left\{\begin{array}{c} 2R\sin C= c=2r_1\sin B\\\\ 2R\sin B=b=2r_2\sin C\end{array}\right|\implies$ $\left\{\begin{array}{c} R\sin C=r_1\sin B\\\\ R\sin B=r_2\sin C\end{array}\right|\ \bigodot\implies$ $R^2=r_1r_2$ . Remark that $\frac {r_1}{r_2}=\left(\frac cb\right)^2$ and

$\frac {r_1}{c^2}=\frac {r_2}{b^2}=$ $\frac {\sqrt{r_1r_2}}{bc}=\frac {R}{bc}=\frac {1}{2h_a}=\frac {a}{4S}$ . Thus, $r_1+r_2=\frac {a\left(b^2+c^2\right)}{4S}\ge \frac {abc}{2S}=2R\implies$ $r_1+r_2\ge 2R$ .

Remark. Let $O\ ,\ O_1\ ,\ O_2$ be the centers of the circumcircle and the circles $w_1$ m $w_2$ . Prove easily that $\triangle AO_1O\sim\triangle ABC\sim\triangle AOO_2\implies r_1r_2=R^2$ .

PP7. Let $BE$ and $CF$ be angle bisectors in $\triangle ABC$ . It's also given an arbitrary point $M\in EF$

and the distancies $x,y,z$ between $M$ and $BC$ , $CA$ , $AB$ respectively. Prove that $x+y=z$ .

Proof. Prove easily that $(a+c)\cdot AE=(a+b)\cdot AF=bc$ . Thus, $[EAF]=[MAE]+[MAF]\iff$

$AE\cdot AF\cdot \sin A=y\cdot AE+z\cdot AF\iff$ $y(a+b)+z(a+c)=2S$ , where $S=[ABC]=$

$\frac 12\cdot bc\sin A$ is the area of $\triangle ABC$ . In conclusion, $\left\{\begin{array}{c} (y+z)a+(yb+zc)=2S\\\\ xa+(yb+zc)=2S\end{array}\right\|\ \implies\ y+z=x$ .

PP8. Let $ABC$ be a triangle with $A\le 90^{\circ}$ . Denote the orthocenter $H$ of $\triangle ABC$ , the midpoint $M$ of the side $[BC]$ and the

projection $P$ of $H$ on $AM$ . Prove that the line $BC$ is a common tangent to the circumcircles of $APB$ , $APC$ and $\frac {PB}{PC}=\frac {AB}{AC}$ .

Proof. Denote $D\in BC$ so that $AD\perp BC$ . Observe that $HDMP$ is cyclical quadrilateral, i.e. $AH\cdot AD=AP\cdot AM\iff$ $AP\cdot AM=bc\cdot\cos A$ . Thus,

$MP\cdot MA=(MA-PA)\cdot MA=$ $MA^2-AP\cdot AM=$ $m_a^2-bc\cdot\cos A=$ $\frac {2\left(b^2+c^2\right)-a^2}{4}-\frac {b^2+c^2-a^2}{2}\implies$ $MP\cdot MA=\frac {a^2}{4}\implies$

$\boxed {MP\cdot MA=MB^2}\ (*)$ . Thus, $\left\{\begin{array}{c} \widehat{PCM}\equiv\widehat{CAM}\\\\ \widehat{PBM}\equiv\widehat{BAM}\end{array}\right\|$ $\implies$ $BC$ is a common tangent to circumcircles of $APB$ , $APC$ and $\frac {PB}{c}=\frac {MP}{MB}=\frac {MP}{MC}=\frac {PC}{b}$ .

Generalization (Petrisor Neagoe). Let $ABC$ be a triangle. A circle $\left(M\right)$ through the points $B$ , $C$ intersects $AC$ , $AB$ at $E$ , $F$ respectively.

Denote $H\in BE\cap CF$ and $P\in AM$ , $HP\bot AM$ . Prove that $BC$ is tangent to the circumcircles of $\Delta APB\ ,\ APC$ and $\frac{PB}{PC}=\frac{AB}{AC}$ .

Proof. Denote $V\in EF\cap BC$ . Observe that $AV$ is the polar of $H$ w.r.t. $(M)\Longrightarrow MH\bot AV$ and $AH$ is the polar of $V$ w.r.t. $(M) \Longrightarrow MV\bot AH$ .

Therefore, $H$ is the orthocenter of $\Delta AVM$ $\Longrightarrow VH\bot AM$ . So, $V\in HP$ . Denote $K\in MH\cap AV$ . Thus, $MH\cdot MK=MP\cdot MA$

and $MH\cdot MK=$ $MB^{2}=$ $MC^{2}\Longrightarrow MP\cdot MA=MB^{2}=MC^{2}\Longrightarrow$ $MB$ is tangent to the circumcircle of $\Delta APB$ and $MC$ is

tangent to the circumcircle of $\Delta APC$ . On other hand, $\frac{PB}{AB}=\frac{sin\left(\angle PAB\right)}{sin\left(\angle APB\right)}=\frac{sin\left(\angle PBM\right)}{sin\left(\angle MPB\right)}=\frac{PM}{MB}\Longrightarrow\frac{PB}{AB}=\frac{PM}{MB}\ (*)$ .

Similarly $\frac{PC}{AC}=\frac{PM}{MC}\ (**)$ . The relations $(*),(**)$ and $MB=MC\Longrightarrow\frac{PB}{AB}=\frac{PC}{AC}\Rightarrow\frac{PB}{PC}=\frac{AB}{AC}$ .

PP9. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ . Denote the inradii $r_1,r_2,r_3$

of $\triangle BIC,\triangle CIA,\triangle AIB$ respectively. Prove that $AI+BI+CI = \sum_{sym} \frac{a(r-r_1)}{2r_1}$ .

Proof. $2\sum AI+\sum a=\sum (BI+CI+BC)=$ $\sum\frac {2\cdot [BIC]}{r_1}=\sum\frac {ar}{r_1}\implies$ $AI+BI+CI = \sum_{sym} \frac{a(r-r_1)}{2r_1}$ .

PP10. Let $ABC$ be a triangle with $A-B= \frac{2\pi}{3}$ and $R=8r$ . Find $\cos C$ .

Proof. I"ll use the remarkable identity $\sin\frac A2\sin\frac B2\sin\frac C2=\frac {r}{4R}$ , i.e. $R=8r\iff$ $\prod\sin\frac A2=\frac {1}{32}\iff$

$32\cdot\prod\sin\frac A2=1\iff$ $16\cdot\sin\frac C2\left(\cos\frac {A-B}{2}-\cos\frac {A+B}{2}\right)=1\iff$ $16\cdot\sin\frac C2\left(\frac 12-\sin\frac C2\right)=1\iff$

$16\sin^2\frac C2 -8\sin\frac C2+1=0\iff$ $\left(4\sin\frac C2-1\right)^2=0$ . Answer : $\left\{\begin{array}{c} \sin \frac C2=\frac 14\\\\ A-B=120^{\circ}\end{array}\right\|\implies$ $\cos C=1-2\sin^2\frac C2=\frac 78$ .

Remark. $\prod\sin\frac A2=\prod\sqrt {\frac {(s-b)(s-c)}{bc}}=$ $\frac {(s-a)(s-b)(s-c)}{abc}=\frac {sr^2}{4Rrs}=\frac {r}{4R}$ . Analogously $\boxed{\frac 1r\cdot\prod\sin\frac A2=\frac 1s\cdot\prod\cos\frac A2=\frac {1}{4R}}$ .

PP11. Two circles with radii $r_1$ and $r_2$ are placed so that they are tangent to each other and a straight line. A third circle

is nestled between them so that it is tangent to the first two circles and the line. Find the radius $r$ of the third circle.

Proof. $\left[2\left(\sqrt {rr_1}+\sqrt{rr_2}\right)\right]^2+\left(r_1-r_2\right)^2=$ $\left(r_1+r_2\right)^2\implies$ $r=\frac {r_1r_2}{\left(\sqrt {r_1}+\sqrt{r_2}\right)^2}\implies$ $\boxed{\frac {1}{\sqrt r}=\frac {1}{\sqrt{r_1}}+\frac {1}{\sqrt{r_2}}}\ (*)$

Remark. Generally, if three circles $w_k=C(I_k,r_k)\ ,\ k\in\overline{1,3}$ are exterior tangent to each other, then the radius $r$ of

the circle $w=C(I,r)$ which is nestled between them so that is tangent to the first three circles is given by the relation

$r=\frac {r_1r_2r_3}{r_1r_2+r_2r_3+r_3r_1+2\sqrt{r_1r_2r_3(r_1+r_2+r_3)}}$ . For $r_3\rightarrow\infty$ obtain upper relation $(*)$ , i.e.

$r=\frac {r_1r_2}{r_1+r_2+\sqrt {r_1r_2}}=$ $\left(\frac {\sqrt {r_1r_2}}{\sqrt {r_1}+\sqrt{r_2}}\right)^2\implies$ $\sqrt r=\frac {\sqrt {r_1r_2}}{\sqrt {r_1}+\sqrt{r_2}}\implies$ $\frac {1}{\sqrt r}=\frac {1}{\sqrt{r_1}}+\frac {1}{\sqrt{r_2}}$ .

PP12. Let $\triangle ABC$ be an equilateral with sidelength $l$ . We draw a semicircle with the diameter $BC$ outside of triangle. A line $d$ is tangent

to the semicircle such that the distance from $A$ to $d$ is $a$ . Compute the distancies $b\ ,\ c$ from the remaining vertices $B\ ,\ C$ to the line $d$ .

Proof. Denote : the midpoint $D$ of $[BC]$ ; the point $X\in BC\cap d$ and suppose w.l.o.g. that $C\in (BX)$ ; $T\in d$ so that $DT\perp d$ ; $P\in d$ so that $AP\perp d$ ; $\phi =m\left(\widehat{DAP}\right)$ . Observe that

$AD=\frac {l\sqrt 3}{2}$ and $a=\frac l2+\frac {l\sqrt 3}{2}\cdot\cos\phi$ , i.e. $\cos\phi =\frac {2a-l}{l\sqrt 3}\implies$ $\sin\phi =\sqrt{\frac {2\left(l^2+2al-2a^2\right)}{3l^2}}$ . Thus, $\{b,c\}=\left\{\frac l2\cdot\left(1\pm\sin\phi\right)\right\}\iff$ $\left\{b,c\right\}=\frac l2\pm\sqrt {\frac {l^2+2al-2a^2}{6}}$ .

Compatibility test : $l\ \le\ 2a\ \le\ (1+\sqrt 3) l$ .

PP13. Let $ABCD$ be a rectangle with $AB=a$ and $AD=b$ . Consider the points $X\in [AB]$ , $Y\in [BC]$ , $Z\in [CD]$ ,

$T\in [DA]$ so that $XYZT$ is a rectangle and $BX=x\ ,\ BY=y\ ,\ XY=l\ ,\ XT=L$ . Calculate $L=f(a,b,l)$ .

Proof. $\triangle AXT\equiv\triangle CZY\sim\triangle BYX\equiv\triangle DTZ\implies$ $\left\{\begin{array}{ccc} \frac {AX}{BY}=\frac {XT}{YX}=\frac {TA}{XB} & \iff & \frac {a-x}{y}=\frac Ll=\frac {b-y}x\\\\ BX^2+BY^2=XY^2 & \iff & x^2+y^2=l^2\end{array}\right|$ .

Thus, $\left\{\begin{array}{c} L\cdot x+l\cdot y=bl\\\\ l\cdot x+L\cdot y=al\end{array}\right|\ (*)\ \implies$ $\left\{\begin{array}{c} x=\frac {l(bL-al)}{L^2-l^2}\\\\ y=\frac {l(aL-bl)}{L^2-l^2}\end{array}\right|$ and $x^2+y^2=l^2\iff$ $\boxed{(bL-al)^2+(aL-bl)^2=\left(L^2-l^2\right)^2}$ .

Remark. $(*)\implies\left\{\begin{array}{c} y+x=l\cdot \frac {a+b}{L+l}\\\\ y-x=l\cdot\frac {a-b}{L-l}\end{array}\right|\implies$ $\left\{\begin{array}{c} 2x=l\cdot\left(\frac {a+b}{L+l}-\frac {a-b}{L-l}\right)\\\\ 2y=l\cdot\left(\frac {a+b}{L+l}+\frac {a-b}{L-l}\right)\end{array}\right|\implies$ $\boxed{\left(\frac {a+b}{L+l}\right)^2+\left(\frac {a-b}{L-l}\right)^2=2}$ .

PP14. Let $\triangle ABC$ so that $AB\perp AC$ . Consider $\left\{\begin{array}{ccc} D\in (BC) & ; & \frac {DB}{DC}=k>\frac {c^2}{b^2}\\\\ E\in (AD) & ; & BE\perp AD\end{array}\right|$ . Prove that $\cot\widehat{CED}=k\cdot\frac bc-\frac cb$ .

Proof. Denote $P\in (BC)$ so that $AP\perp BC$ . Hence $\frac {PB}{PC}=\frac {c^2}{b^2}$ and is evidently $k>\frac {c^2}{b^2}\iff$ $D\in (PC)$ and $m\left(\widehat {BDA}\right)<90^{\circ}\iff$

$E\in (AD)$ . Denote $\left\{\begin{array}{c} m\left(\widehat{DAB}\right)=\alpha\\\\ m\left(\widehat{CED}\right)=x\end{array}\right|$ . Thus, on the one hand $k=\frac {DB}{DC}=\frac {AB}{AC}\cdot\frac {\sin \widehat{DAB}}{\sin\widehat{DAC}}=\frac cb\cdot\tan\alpha\iff$ $\boxed{\tan\alpha=\frac {bk}{c}}\ (1)$

and on the other hand $k=\frac {DB}{DC}=\frac {\sin \widehat{ECB}}{\sin\widehat{EBC}}\cdot\frac {\sin \widehat{DEB}}{\sin\widehat{DEC}}=$ $\frac {\sin \left(90^{\circ}-\alpha +C-x\right)}{\sin (\alpha -C)}\cdot\frac {1}{\sin x}=$ $\frac {\cos\left[\left(\alpha -C\right)+x\right]}{\sin (\alpha -C)}\cdot\frac {1}{\sin x}\implies$

$\sin x=\frac 1k\cdot \frac {\cos (\alpha -C)\cos x-\sin (\alpha -C)\sin x}{\sin (\alpha -C)}\implies$ $\tan x=\frac {1-\tan (\alpha -C)\tan x}{k\tan (\alpha -C)}$ , i.e. $\boxed{\tan x=\frac {1}{(k+1)\tan (\alpha -C)}}\ (2)$ .

Observe that $\tan (\alpha -C)=\frac {\tan \alpha -\tan C}{1+\tan\alpha\tan C}\stackrel{(1)}{\ =\ }$ $\frac {\frac {bk}{c}-\frac cb}{1+\frac {bk}{c}\cdot\frac cb}\implies$ $\boxed{\tan (\alpha -C)=\frac {b^2k-c^2}{bc(1+k)}}\ (3)$ . Therefore, the relation $(2)$ becomes

$\tan x=\frac {bc}{b^2k-c^2}\iff$ $\cot x=\frac {b^2k-c^2}{bc}=$ $\frac {kb}{c}-\frac cb$ $\implies$ $\cot x=k\cdot\frac bc-\frac cb\iff$ $\cot\widehat{CED}=k\cdot\frac bc-\frac cb$ .

PP15. Let $\triangle ABC$ and the points $M\in (AB)$ , $N\in (BC)$ , $P\in (CA)$ , $O\in AN\cap MP$ . Denote $x=\frac{MA}{MB}$ , $y=\frac{NB}{NC}$ , $z=\frac{PC}{PA}$ , $t=\frac{OA}{ON}$ . Prove that $\frac{1+xyz}{1+y}=\frac{x}{t}$ .

Proof 1. I"ll use an well-known property $\boxed{\frac {MB}{MA}\cdot NC+\frac {PC}{PA}\cdot NB=\frac {ON}{OA}\cdot BC}\ (*)$ . Therefore, $\frac{1+xyz}{1+y}=\frac{x}{t}\iff$ $1+\frac {MA}{MB}\cdot \frac {NB}{NC}\cdot\frac {PC}{PA}=$

$\frac {\boxed{BC}}{NC}\cdot\frac {MA}{MB}\cdot\boxed{\frac {ON}{OA}}\ \ (1)\ \ \stackrel{(*)}{\iff}$ $1+\frac {MA}{MB}\cdot \frac {NB}{NC}\cdot\frac {PC}{PA}=$ $\frac {1}{NC}\cdot\frac {MA}{MB}\cdot\left(\frac {MB}{MA}\cdot NC+\frac {PC}{PA}\cdot NB\right)$ , what is truly.

Remark. Denote $T\in BP\cap CM$ and $S\in AB\cap CO$ . Apply the Menelaus' theorem to the transversal $\overline{COS}$ and $\triangle ABN$ : $\frac {CB}{CN}\cdot \frac {ON}{OA}\cdot\frac {SA}{SB}=1\iff \frac {CB}{CN}\cdot\frac {ON}{OA}=\frac {SB}{SA}$ .

Thus, the relation $(1)$ becomes $\boxed{1+\frac {MA}{MB}\cdot\frac {NB}{NC}\cdot\frac {PC}{PA}=\frac{MA}{MB}\cdot\frac {SB}{SA}}\ (2)$ . In the particular case $T\in AN$ when $\frac {MA}{MB}\cdot\frac {NB}{NC}\cdot\frac {PC}{PA}=1$ , the previous relation becomes $\frac {MA}{MB}=2\cdot\frac {SA}{SB}$ .

In this case it is well-known the relation $\boxed{\frac {TA}{TN}=2\cdot\frac {OA}{ON}}\ (2)$ - a characterization of the harmonical division $(A,O,T,N)$ . Thus, the proposed problem is a generalization of the well-known relation $(2)$ .

$(*)\blacktriangleright$ Remark. If $MP\parallel BC$ , then the relation $(*)$ is truly. If $MP\not\parallel BC$ , then suppose w.l.o.g. that $S\in BC\cap MP$ so that $B\in (SC)$ . For $d\parallel BC\ ,\ A\in d$ denote $R\in d\cap MP$ . Observe that

$\frac {MB}{MA}\cdot NC+\frac {PC}{PA}\cdot NB=\frac {ON}{OA}\cdot BC\iff$ $\frac {MB}{MA}\cdot NC+\frac {PC}{PA}\cdot NB=\frac {ON}{OA}\cdot BC\iff$ $\frac {SB}{AR}\cdot NC+\frac {SC}{AR}\cdot NB=\frac {SN}{AR}\cdot BC\iff$ $SB\cdot NC+SC\cdot NB=SN\cdot BC$

$\iff$ $(SN-BN)\cdot NC+(SN+NC)\cdot NB=SN\cdot BC\iff$ $SN\cdot NC+SN\cdot NB=SN\cdot BC\iff$ $NC+NB=BC$ , what is truly.

Proof 2 (with areas). Denote $S=[ABC]$ . Therefore, $\frac {[AMP]}{S}=\frac {[AMP]}{[ABC]}=$ $\frac {AM}{AB}\cdot\frac {AP}{AC}=$ $\frac {x}{x+1}\cdot\frac {1}{z+1}$ $\implies$ $[AMP]=\frac {x\cdot S}{(x+1)(z+1)}$ .

$\blacktriangleright\ \frac {[AOM]}{S}=$ $\frac {[AOM}{[ABN]}\cdot \frac {[ABN]}{[ABC]}=$ $\frac {AM}{AB}\cdot \frac {AO}{AN}\cdot\frac {BN}{BC}=$ $\frac {x}{x+1}\cdot\frac {t}{t+1}\cdot\frac {y}{y+1}$ $\implies$ $[AOM]=\frac {xyt\cdot S}{(x+1)(y+1)(t+1)}$ .

$\blacktriangleright\ \frac {[AOP]}{S}=$ $\frac {[AOP]}{[ANC]}\cdot\frac {[ANC]}{[ABC]}=$ $\frac {AO}{AN}\cdot \frac{AP}{AC}\cdot \frac {CN}{CB}=$ $\frac {t}{t+1}\cdot\frac {1}{z+1}\cdot\frac {1}{y+1}$ $\implies$ $[AOP]=\frac {t\cdot S}{(y+1)(z+1)(t+1)}$ .

Hence $[AOM]+[AOP]=[AMP]\iff$ $\frac {xyt}{(x+1)(y+1)(t+1)}+\frac {t}{(y+1)(z+1)(t+1)}=\frac {x}{(x+1)(z+1)}\iff$ $xyt(1+z)+t(1+x)=$

$x(1+y)(1+t)\iff$ $xyt+xyzt+t+xt=x+xy+xt+xyt\iff$ $xyzt+t=x+xy\iff$ $t(1+xyz)=x(1+y)\iff$ ${1+xyz\over 1+y}={x\over t}$ .

PP16. Let $w=C(O,r)$ be a circle with the diameter $[AB]$ . Let $K\in (OA)$ be a fixed point Denote the tangent line $t$ at $A$ to the circle $w$ . For a mobile chord

$CD\ne AB$ so that $K\in (CD)$ denote $P\in BC\cap t$ and $Q\in BD\cap t$ . Prove that the product $AP \cdot AQ$ is constant, regardless of the choice of chord $CD$ .

Proof. Denote $AK=d$ (constant). It's well-known or can show easily the relation $\frac {d}{2r-d}=\frac {KA}{KB}=\frac {CA}{CB}\cdot\frac {\sin\widehat{KCA}}{\sin\widehat{KCB}}=\frac {CA}{CB}\cdot\frac {DA}{DB}\implies$ $\boxed{\frac {CA}{CB}\cdot\frac {DA}{DB}=\frac {d}{2r-d}}\ (*)$ .

Method 1 (synthetic). $\left\{\begin{array}{ccc} \triangle APB\sim\triangle CAB & \implies & \frac {AP}{AB}=\frac {CA}{CB}\\\\ \triangle AQB\sim\triangle DAB & \implies & \frac {AQ}{AB}=\frac {DA}{DB}\end{array}\right\|\ \stackrel{(*)}{\implies}\ AP\cdot AQ=\frac {4r^2d}{2r-d}$ (constant).

Method 2 (trigonometric). $\left\{\begin{array}{ccc} m\left(\widehat{ABP}\right)=x & \implies & AP=2r\tan x\\\\ m\left(\widehat{ABQ}\right)=y & \implies & AQ=2r\tan y\\\\ \tan x\tan y=\frac {CA}{CB}\cdot\frac {DA}{DB} & \stackrel{(*)}{\implies} & \tan x\tan y=\frac {d}{2r-d}\end{array}\right\|\implies$ $AP\cdot AQ=4r^2\cdot \frac {d}{2r-d}$ .

Remark. Since $BC\cdot BP=BD\cdot BQ=BA^2$ obtain that the quadrilateral $PQDC$ is cyclically. Here is an inversion with pole $B$ and modulus $BA^2$ , i.e. $w \rightarrow PQ$ .

Since $AP\cdot AQ$ is constant, then $PQ=AP+AQ$ is minimum $\iff AP=AQ=2r\sqrt{\frac {d}{2r-d}}\iff$ $CD\perp AB$ , i.e. $CD\parallel t$ .

PP17. Let $A$ , $B$ be two fixed points on a given circle $w$ . Denote $C\in AA\cap BB$ , where $XX$ is the tangent at $X\in w$ to the circle $w$ .

For a mobile point $P\in w$ denote its projections $L$ , $M$ , $N$ on $AB$ , $BC$ , $CA$ respectively. Prove that $PL^2=PM\cdot PN$ .

Proof. The quadrilateral $PLAN$ is inscribed in the circle with the diameter $[PA]$ and the quadrilateral $PLBM$ is inscribed in the circle with the diameter $[PB]$ . Therefore,

$\left\{\begin{array}{ccc} \widehat{PLN}\equiv\widehat{PAN}\equiv\widehat{PBL}\equiv\widehat{PML}\equiv x &\implies & \widehat{PLN}\equiv\widehat{PML}\\\\ \widehat{PNL}\equiv\widehat{PAL}\equiv\widehat{PBM}\equiv\widehat{PLM}\equiv y & \implies & \widehat{PNL}\equiv\widehat{PLM}\end{array}\right\|$ $\implies$ $\triangle PLN\sim\triangle PML$ $\implies$ $\frac {PL}{PM}=\frac {PN}{PL}$ $\implies$ $\boxed{PL^2=PM\cdot PN}$ .

Denote $S\in PL\cap MN$ . Observe that $\widehat{LPN}\equiv$ $\widehat{LAC}\equiv$ $\widehat{BAC}\equiv$ $\widehat{ABC}\equiv$ $\widehat{LPM}\implies$ the ray $[PL$ is the $P$ -angled bisector of $\triangle MPN$ .

From the relations $\frac {SM}{SN}=\frac {LM}{LN}\cdot\frac {\sin y}{\sin x}=\left(\frac {LM}{LN}\right)^2$ obtain that $\frac {SM}{SN}=\left(\frac {LM}{LN}\right)^2$ , i.e. the ray $[LS$ is the $L$ -symmedian of the triangle $MLN$ .

PP18. Let $A$ -right $\triangle ABC$ . Denote the foot $D\in (AC)$ of the $B$ -angled bisector. Suppose that $BD=7$ and $CD=15$ . Find $AD$ .

Proof 1 (trigonometric - similar with DoctorB's). Denote the projection $E$ of $D$ on $BC$ , $m\left(\widehat{DBC}\right)=\phi$ and $DA=DE=x$ . Observe that $m\left(\widehat{CDE}\right)=2\phi$ . Therefore, $\sin\phi=\sin\widehat{DBC}=$

$\sin\widehat{DBE}=$ $\frac {DE}{DB}=\frac x7$ and $\cos 2\phi=\cos\widehat{CDE}=$ $\frac {DE}{DC}=\frac x{15}$ . Thus, $\cos 2\phi =1-2\sin^2\phi\implies$ $\frac x{15}=1-\frac {2x^2}{49}\implies$ $30x^2+49x-735=0\implies$ $\boxed{x=\frac {21}5}$ .

Proof 2 (metric). Let $E\in BC\ ,\ DE\perp BC$ . Thus $\left\{\begin{array}{ccc} AB\perp AC & \implies & x^2+c^2=49\\\\ \triangle CDE\sim\triangle CBA & \implies & \frac {15}{a}=\frac xc=\frac {a-c}{x+15}\end{array}\right|$ $\implies$

$c^2=\frac {x^2(x+15)}{15-x}$ $\implies$ $x^2(x+15)=(15-x)\left(49-x^2\right)$ $\implies$ $30x^2+49x-735=0$ $\implies$ $\boxed{x=\frac {21}{5}}$ .

Proof 3 (metric). Denote $DA=x$ . Prove easily that $\left\{\begin{array}{c} \frac x{15}=\frac ca\iff ax=15c\\\\ x^2+c^2=49\\\\ 49+15x=ac\end{array}\right|\implies$ $49x+15x^2=15\left(49-x^2\right)\implies$ $30x^2+49x-735=0$ $\implies$ $\boxed{x=\frac {21}{5}}$ .

PP19. Let $\triangle ABC$ with $AB\perp AC$ , altitude $AH$ , incenter $I$ , the projection $K$ of $I$ on $BC$ and $L\in IC$ such that $IK=LK$ . Show that $L$ is the incenter of $\triangle AHC$ .

Proof (Sunken Rock). Let incircle touch $AC$ at $M$ . As constructed, $IKLM$ is a rhombus. Hence $ML\bot BC$ and $L$ is the orthocenter of $\triangle CMK$ . Let $MN, KP$ be altitudes of $C$ -isosceles $\Delta CMK$ . To prove

that $L$ is the incenter of $\Delta AHC$ we need to prove that $P$ and $K$ are homologue points into triangles $\Delta ACH$ , $\Delta BCA$ or $\frac{CP}{CK}=\frac{AC}{BC}$ . But this is obvious, since $KP\parallel AB$ . Nice problem and its proof !

PP20. Let a rectangle $ABCD$ and the midpoints $M$ , $N$ of $[AB]$ , $[BC]$ respectively. Prove $\boxed{CM\perp DN\iff AB=AD}$ .

Proof 1. $CM\perp DN$ $\iff$ $\widehat {BCM}\equiv\widehat{CDN}$ $\iff$ $\triangle BCM\sim\triangle CDN$ $\iff$ $\frac {MB}{NC}=\frac {BC}{CD}$ $\iff$ $\frac {AB}{AD}=\frac {AD}{AB}$ $\iff$ $AB=AD$ .

Proof 2. Let $X\in CM\cap DN$ and $P\in CM\cap BC$ . Thus, $\frac 14=\frac {NC}{PD}=$ $\frac {XN}{XD}=$ $\left(\frac {CN}{CD}\right)^2\implies$ $\left(\frac {\frac 12\cdot AD}{AB}\right)^2\implies$ $AB=AD$ .

Proof 3. Denote $X\in CM\cap DN$ and $Z\in CM$ so that $NZ\parallel AB$ . Thus, $\frac {XN}{XD}=\frac {NZ}{CD}$ $\iff$ $\left(\frac {CN}{CD}\right)^2=\frac 14$ $\iff$ $\left(\frac {\frac 12\cdot AD}{AB}\right)^2=\frac 14$ $\iff$ $AB=AD$ .

Proof 4. Let $X\in CM\cap DN$ and $R\in CD$ so that $MR\parallel AD$ . Thus, the quadrilaterals $BNXM$ , $AMXRD$ are cyclically and $MX$ is the radical axis of their

circumcircles. Therefore, $C\in MX\iff$ $CN\cdot CB=CR\cdot CD\iff$ $\frac 12\cdot CB\cdot CB=\frac 12\cdot CD\cdot CD\iff$ $CB=CD\iff$ $AB=AD$ .

Proof 5. Let $A(0,0)$ , $B(0,2b)$ , $C(2a,2b)$ and $D(2a,0)$ , where $a>0$ , $b>0$ . Thus, $M(0,b)$ , $N(a,2b)$ and the slopes $\left\{\begin{array}{c} s(CM)=\frac {b}{2a}\\\\ s(DN)=\frac {-2b}{a}\end{array}\right|$ .

In conclusion, $MC\perp ND\iff$ $s(CM)\cdot s(DN)=-1\iff$ $\frac {b}{2a}\cdot\frac {-2b}{a} =-1\iff$ $a^2=b^2\iff$ $a=b\iff$ $AB=AD$ .

An easy extension. Let $ABCD$ be an isosceles trapezoid so that $AD\parallel BC$ . Let $M\in (AB)$ , $N\in (BC)$ so that $\frac {BM}{BA}=\frac {CN}{CB}\ (*)$ and $X\in CM\cap DN$ . Prove $\boxed{\widehat{MXN}\equiv \widehat{BAD}\iff AB=BC}$ .

Proof. $\widehat{MXN}\equiv \widehat{BAD}$ $\iff$ $\widehat {BCM}\equiv\widehat{CDN}$ $\iff$ $\triangle BCM\sim\triangle CDN$ $\iff$ $\frac {MB}{NC}=\frac {BC}{CD}\stackrel{(*)}{\iff}$ $\frac {BA}{CB}=\frac {BC}{CD}\iff$ $BC^2=BA\cdot CD$ $\iff$ $AB=BC$ .

PP21 (worthhawholebean). Let $\triangle ABC$ . Suppose that exist $\{D,E\}\subset (BC)$ so that $AD\perp BC$ and $E\in (DC)$ , $DE=BD$ so that $\tan \widehat {EAC}$ ,

$\tan \widehat{DAC}$ , $\tan \widehat{BAC}$ form a geometric progression and $\cot \widehat{DAC}$ , $\cot\widehat{EAC}$ , $\cot \widehat{DAE}$ form an arithmetic progression. Prove that $E$ is the midpoint of $[BC]$ .

Proof. Denote $\left\{\begin{array}{c} m\left(\widehat{DAB}\right)=x\\\\ m\left(\widehat{DAC}\right)=y\end{array}\right|$ . Thus, $\left\{\begin{array}{cc} \tan^2y=\tan (y-x)\tan (y+x) & (1)\\\\ 2\cot (y-x)=\cot x+\cot y & (2)\end{array}\right|$ . Observe that $(1)\iff$ $\tan^2y=\frac {\tan^2y-\tan^2x}{1-\tan^2x\tan^2y}\iff$

$\tan^4y\tan^2x=\tan^2x\iff$ $\tan y=1\iff$ $\boxed{y=45^{o}}$ . Therefore, $(2)\iff$ $2\cot \left(45^{o}-x\right)=\cot x+\cot 45^{o}\iff$ $\frac {2(1+\tan x)}{1-\tan x}=$

$\frac {1}{\tan x}+1\iff$ $\boxed{\tan x=\frac 13}$ . In conclusion, $y=45^{o}\implies$ $AD=DC$ and $\tan x=\frac 13\implies$ $\frac {DB}{DA}=\frac 13\implies\boxed{\frac {DB}{DC}=\frac 13}\iff EB=EC$ .

PP22 (NMO - Bulgaria, 1997). Let $\triangle ABC$ with the circumcircle $w$ and $M\in (AC)$ , $N\in (AB)$ , $D\in MN\cap w$ . Prove that $\boxed{\ \left|\frac b{DB}\cdot\frac {NB}{NA}-\frac c{DC}\cdot\frac {MC}{MA}\right|=\frac a{DA}\ }$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{ADN}\right)=x$ . Apply an well-known relations $\left\{\begin{array}{ccc} \frac {NB}{NA}=\frac {DB}{DA}\cdot\frac {\sin\widehat{NDB}}{\sin\widehat{NDA}} & \implies & \frac {DA}{DB}\cdot\frac {NB}{NA}=\frac {\sin (C-x)}{\sin x}\\\\ \frac {MC}{MA}=\frac {DC}{DA}\cdot\frac {\sin\widehat{MDC}}{\sin\widehat{MDA}} & \implies & \frac {DA}{DC}\cdot\frac {MC}{MA}=\frac {\sin (B+x)}{\sin x}\end{array}\right|$ . Therefore, $\left|\frac b{DB}\cdot\frac {NB}{NA}-\frac c{DC}\cdot\frac {MC}{MA}\right|=\frac a{DA}\iff$

$\left|b\cdot \frac {\sin (C-x)}{\sin x}-c\cdot\frac {\sin (B+x)}{\sin x} \right|=a\iff$ $\left|\sin B\cdot \sin (C-x)-\sin C\cdot\sin (B+x)\right|=|\sin x\cdot \sin A|\iff$ $\left|\sin x(\sin B\cos C+\sin C\cos B\right|=|\sin x\cdot \sin A|$ , what is truly.

Proof 2 (synthetic). Apply the Ptolemy's relation in the cyclical quadrilateral $ABCE$ , where $\{D,E\}=MN\cap w\ :\ \boxed{a\cdot AE+b\cdot BE=c\cdot CE}\ (*)$ and the well-known relations $\left\{\begin{array}{c} \frac {NB}{NA}=\frac {BE\cdot BD}{AE\cdot AD}\\\\ \frac {MC}{MA}=\frac {CE\cdot CD}{AE\cdot AD}\end{array}\right|$ . Thus, $\left|\frac b{DB}\cdot\frac {NB}{NA}-\frac c{DC}\cdot\frac {MC}{MA}\right|=\frac a{DA}\iff$ $\left|b\cdot \frac {DA}{DB}\cdot\frac {BE\cdot BD}{AE\cdot AD}-c\cdot \frac {DA}{DC}\cdot\frac {CE\cdot CD}{AE\cdot AD}\right|=a\iff$ $\left|b\cdot\frac {BE}{AE}-c\cdot \frac {CE}{AE}\right|=a\iff$ $|b\cdot BE-c\cdot CE|=a\cdot AE$ what is truly from $(*)$ .

PP23. Let $\triangle ABC$ with the centroid $G$ . Let midpoint $M$ of $[BC]$ , $E\in (AB)$ , $F\in (AC)$ so that $G\in EF$ and $H\in BC$ , $K\in AC$ so that $E\in HK\parallel AM$ . Prove that $\left\{\begin{array}{cc} 1\blacktriangleright & HK+HE=2AM\\\\ 2\blacktriangleright & \frac{AE}{AB}+\frac{AF}{AC} \ge \frac 43\end{array}\right\|$ .

Proof.

$\blacktriangleright\ \left\{\begin{array}{ccccc} EH\parallel AM & \implies & \frac {EH}{AM}=\frac {BH}{BM}\\\\ KH\parallel AM & \implies & \frac {KH}{AM}=\frac {HC}{MC} \end{array}\right|$ $\implies$ $\frac {EH}{AM}+\frac {KH}{AM}=\frac {BH+HC}{\frac 12\cdot BC}\implies$ $\boxed{HK+HE=2\cdot AM}$ .

$\blacktriangleright$ I"ll use an well-known property $\frac {AB}{AE}+\frac {AC}{AF}=3$ . Therefore, $\left(\frac {AE}{AB}+\frac {AF}{AC}\right)\cdot \left(\frac {AB}{AE}+\frac {AC}{AF}\right)\ge 4\implies$ $\boxed{\frac {AE}{AB}+\frac {AF}{AC}\ge \frac 43}$ .

An easy extension. For an interior point $P$ of $\triangle ABC$ denote $M\in AP\cap BC\ ,\ \frac {MB}{MC}=\frac xy$ , two points $E\in (AB)\ ,\ F\in (AC)$ so that

$P\in EF$ and $H\in BC$ , $K\in AC$ so that $E\in HK\parallel AM$ . Prove that $\left\{\begin{array}{cc} 1\blacktriangleright & x\cdot HE+y\cdot HK=(x+y)\cdot AM\\\\ 2\blacktriangleright & x\cdot \frac{AE}{AB}+y\cdot \frac{AF}{AC} \ge \frac {4xy}{x+y}\cdot\frac {AP}{AM}\end{array}\right\|$ .

Proof.

$\blacktriangleright\ \frac {MB}{x}=\frac {MC}{y}=\frac a{x+y}$ . Thus, $\left\{\begin{array}{ccccc} EH\parallel AM & \implies & \frac {EH}{AM}=\frac {BH}{BM} & \implies & x\cdot \frac {HE}{AM}=\frac {x+y}{a}\cdot BH\\\\ KH\parallel AM & \implies & \frac {KH}{AM}=\frac {HC}{MC} & \implies & y\cdot \frac {HK}{AM}=\frac {x+y}{a}\cdot HC\end{array}\right|$ $\implies$

$x\cdot \frac {EH}{AM}+y\cdot \frac {KH}{AM}=x+y\implies$ $\boxed{x\cdot HE+y\cdot HK=(x+y)\cdot AM}$ .

$\blacktriangleright$ I"ll use an well-known property $\boxed{y\cdot \frac {AB}{AE}+x\cdot \frac {AC}{AF}=(x+y)\cdot\frac {AM}{AP}}\ (*)$ . Therefore, $\left(x\cdot \frac {AE}{AB}+y\cdot \frac {AF}{AC}\right)\cdot \left(y\cdot \frac {AB}{AE}+x\cdot \frac {AC}{AF}\right)\ge 4xy\implies$

$\boxed{x\cdot \frac {AE}{AB}+y\cdot \frac {AF}{AC}\ge \frac {4xy}{x+y}\cdot\frac {AP}{AM}}$ . Particular case. If $P:=I$ (incentre), then $\left\{\begin{array}{c} x=c\\\ y=b\end{array}\right\|$ and there are the relations $\left\{\begin{array}{cc} 1\blacktriangleright & \frac {HE}{b}+\frac {HK}{c}=2\cos\frac A2\\\\ 2\blacktriangleright & AE+AF\ge \frac {2}{\cos\frac A2}\cdot AI\end{array}\right\|$ .

===================================================================================================================================================

$(*)$ I"ll prove the remarkable relation $(*)$ . Denote $X\in EF$ so that $AX\parallel BC$ and $L\in BC\cap EF$ . Therefore, $\left\{\begin{array}{c} \frac{EB}{EA}=\frac {LB}{AX}\\\\ \frac {FC}{FA}=\frac {LC}{AX}\end{array}\right|\left|\begin{array}{c} \odot\ y\\\\ \odot\ x\end{array}\right\|\bigoplus\implies$

$y\cdot \frac {EB}{EA}+x\cdot \frac {FC}{FA}=\frac {y\cdot LB+x\cdot LC}{AX}=$ $\frac {y\cdot (LM-MB)+x\cdot (LM+MC)}{AX}=$ $(x+y)\cdot\frac {LM}{AX}$ $\implies$ $\boxed{y\cdot \frac {EB}{EA}+x\cdot \frac {FC}{FA}=(x+y)\cdot\frac {PM}{PA}}$ .

Therefore, $y\cdot\left(1+\frac {EB}{EA}\right)+x\cdot\left(1+\frac {FC}{FA}\right)=$ $(x+y)\cdot\left(1+\frac {PM}{PA}\right)$ $\implies$ $\boxed{y\cdot\frac {AB}{AE}+x\cdot\frac {AC}{AF}=(x+y)\cdot \frac {AM}{AP}}$ .

PP24. $\ell_1\parallel\ell_2\ ;\ \odot w_1=$ $C\left(I_1,r_1\right)$ tangent to $\ell_1$ , $\ell_2\ ;\ \odot w_2=$ $C\left(I_2,r_2\right)$ tangent to $\ell_1$ , $w_1\ ;\ \odot w_3$ $=C\left(I_3,r_3\right)$ tangent to $w_1$ , $w_2$ , $\ell_2$ . Prove that there is the relation $r_1^2=4r_2r_3$ .

Proof. Denote the points $\{A,B,C\}\subset \ell_1$ , $\{D,E,F\}\subset\ell_2$ , $G\in CF$ so that $I_1\in AD\perp \ell_1$ , $I_2\in BE\perp\ell_1$ , $I_3\in CF\perp \ell_1$ and $I_2G\parallel \ell_1$ . Prove easily that $AB=DE=2\sqrt {r_1r_2}$ ,

$DF=AC=2\sqrt {r_3r_1}$ . Thus, $\boxed{I_3I_2=r_2+r_3}$ , $GI_2=AB-DF=$ $2\sqrt{r_1r_2}-2\sqrt {r_3r_1}\implies$ $\boxed{GI_2=2\sqrt {r_1}\left(\sqrt {r_2}-\sqrt r_3\right)}$ and $GI_3=CF-GC-I_3F\implies$

$\boxed{GI_3=2r_1-r_2-r_3}$ . In conclusion, $GI_2^2=I_3I_2^2-GI_3^2\iff$ $4r_1\left(\sqrt {r_2}-\sqrt r_3\right)^2=\left(r_2+r_3\right)^2-\left(r_2+r_3-2r_1\right)^2\iff$ $\boxed{r_1^2=4r_2r_3}$ .

This post has been edited 216 times. Last edited by Virgil Nicula, Nov 19, 2015, 9:44 PM

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320. Some problems with metrical relations.

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