164. Some metrical problems (5) in a circle.

by Virgil Nicula, Oct 21, 2010, 3:13 PM

PP1. Prove that in any $\triangle ABC$ with circumcircle $w=C(O,R)$ exists the equivalence $A=60^{\circ}\iff OI^2=2R\cdot\sin\frac {|B-C|}{4}$ (M. Chirita). Generalization.

Proof. Since $\sum\cos A=1+\frac rR$ obtain $A=60^{\circ}\iff$ $\cos B+\cos C=\frac 12+\frac rR\iff$ $\cos\frac {B-C}{2}=\frac 12+\frac rR\iff$ $1-\cos\frac {B-C}{2}=\frac 12-\frac rR\iff$

$2\sin^2\frac {B-C}{4}=\frac {R-2r}{2R}\iff$ $4R^2\sin^2\frac {B-C}{4}=R^2-2Rr\iff$ $4R^2\sin^2\frac {B-C}{4}=OI^2\iff$ $OI=2R\sin \frac{|B-C|}{4}$ .

Particular cases for $A=60^{\circ}$ . Denote $D\in BC$ for which $AD\perp BC$ and $F\in AB$ for which $CF\perp AB$ .

$\begin{array}{ccccccc}
OI=b & \iff & 2R\sin B=2R\sin\frac {|B-C|}{4} & \iff & \sin B=\sin\frac {|B-C|}{4} & \iff & B=20^{\circ}\ \wedge\ C=100^{\circ}\\\\
OI=BH & \iff & 2R|\cos B|= 2R\sin\frac{|B-C|}{4} & \iff & |\cos B|= \sin\frac{|B-C|}{4} & \iff & B=80^{\circ}\ \wedge\ C=40^{\circ}\\\\
OI=2\cdot DF & \iff & 2R|\sin 2B|=2R\sin\frac{|B-C|}{4} & \iff & |\sin 2B|=\sin\frac{|B-C|}{4} & \iff & B=12^{\circ}\ \wedge\ C=108^{\circ}\end{array}$ .

Quote:
Generalization. Prove that in any $\triangle ABC$ with circumcircle $w=C(O,R)$ and incircle $C(I,r)$

exists the equivalence $\boxed{\sin \frac A2=\lambda\iff OI^2=(2\lambda -1)^2\cdot R^2+8\lambda R^2\cdot\sin^2\frac {B-C}{4}}$ .

Remarks.

$\blacktriangleright\ OI^2\ge (2\lambda -1)^2\cdot R^2\ \implies\ \frac {2r}{R}$ $\le 4\sin\frac A2\left(1-\sin\frac A2\right)\le 1$ with left equality $\iff$ $B=C$ .

$\blacktriangleright\ OI^2\ge 8\lambda R^2\cdot\sin^2\frac {B-C}{4}\ \implies\ \frac {2r}{R}\le$ $ 1-8\sin\frac A2\sin ^2\frac {B-C}{4}\le 1$ with left equality $\iff$ $A=60^{\circ}$ .

Proof. $\sin\frac A2=\lambda\iff \cos A=1-2\lambda^2$ . From $\sum\cos A=1+\frac rR$ obtain $A=2\arcsin\lambda\iff$ $-2\lambda^2+\cos B+\cos C=\frac rR\iff$

$2\lambda\cos\frac {B-C}{2}=2\lambda^2+\frac rR\iff$ $2\lambda\left(1-2\sin^2\frac {B-C}{4}\right)=$ $2\lambda^2+\frac rR\iff$ $4\lambda R^2\left(1-2\sin^2\frac {B-C}{4}\right)=4\lambda^2R^2+2Rr$ $\iff$

$4\lambda R^2\left(1-2\sin^2\frac {B-C}{4}\right)=4\lambda^2R^2+R^2-OI^2\iff$ $OI^2=(2\lambda -1)^2R^2+8\lambda R^2\sin^2\frac {B-C}{4}$ .



PP2. Let $ABC$ be a triangle with circum radius $R$ , inradius $r$ and semiperimeter $s$ and let $x$ be a real number.

$\blacktriangleright$ Prove that exists the relation $\boxed{4R\cdot\sin \frac {A+x}{2}\sin\frac {B+x}{2}\sin\frac {C+x}{2}=2R\cdot\sin x\sin\frac x2+r\cdot\cos\frac x2+s\cdot \sin \frac x2}$ .

$\blacktriangleright$ Solve the equation $s=2R\cdot\sin x+r\cdot\cot\frac x2$ , where $x\in\left(0,\pi\right)$ (Virgil Nicula).


Proof. Denote $\boxed{E=4R\cdot\sin \frac {A+x}{2}\sin\frac {B+x}{2}\sin\frac {C+x}{2}}$ . Observe that $E=2R\sin\frac {C+x}{2}\left(\cos\frac {A-B}{2}-\cos\frac {A+B+2x}{2}\right)=$

$R\left(\sin\frac {A+C-B+x}{2}+\sin\frac {B+C-A+x}{2}-\sin\frac {A+B+C+3x}{2}+\sin \frac {A+B-C+x}{2}\right)=$

$R\left[\cos\left(A-\frac x2\right)+\cos\left(B-\frac x2\right)+\cos\left(C-\frac x2\right)-\cos\frac {3x}{2}\right]=$ $R\left[(\cos A+\cos B+\cos C)\cdot \cos\frac x2+(\sin A+\sin B+\sin C)\cdot \sin \frac x2-\cos\frac {3x}{2}\right]=$ $(R+r)\cdot\cos\frac x2+s\cdot\sin\frac x2-R\cos\frac {3x}{2}=$ $s\cdot\sin\frac x2+r\cdot\cos\frac x2+R\left(\cos\frac x2-\cos\frac {3x}{2}\right)\implies$

$\boxed{E=2R\cdot \sin x\sin\frac x2+r\cdot \cos\frac x2+s\cdot \sin\frac x2}$ because $\sum\cos A=1+\frac rR$ and $\sum\sin A=\frac sR$ . For the second point of the conclusion observe that for

$x:=-x$ the above proved identity becomes $4R\cdot\sin \frac {A-x}{2}\sin\frac {B-x}{2}\sin\frac {C-x}{2}=$ $2R\cdot \sin x\sin\frac x2+r\cdot \cos\frac x2-s\cdot \sin\frac x2=$

$\sin\frac x2\left(2R\cdot \sin x+r\cdot\cot\frac x2-s\right)$ . Thus, our equation $s=2R\cdot \sin x+r\cdot\cot\frac x2$ , where $x\in (0,\pi )$

is equivalently with the equation $\sin \frac {A-x}{2}\sin\frac {B-x}{2}\sin\frac {C-x}{2}=0$ , i.e. $x\in\{A,B,C\}$ .



PP3. For the cyclical convex quadrilateral with the circumcircle $w=C(O,R)$ denote $\begin{array}{c}
E\in AC\cap BD\\\
F\in AD\cap BC\end{array}$ and $\begin{array}{c}
AB=a\ ,\ BC=b\ ,\ CD=c\\\
DA=d\ ,\ AC=e\ ,\ BD=f\end{array}$ . Prove that :

$\blacktriangleright\ \ \frac {EA}{ad}=\frac {EB}{ba}=\frac {EC}{cb}=\frac {ED}{dc}=$ $\frac {e}{ad+bc}=\frac {f}{ab+cd}=\frac {e+f}{(a+c)(b+d)}=$ $\frac {f-e}{(a-c)(b-d)}=\sqrt{-\frac {p_w(E)}{abcd}}$ .

$\blacktriangleright\ \ \frac {FA}{ae}=\frac {FB}{af}=\frac {FC}{ce}=\frac {FD}{cf}=$ $\frac {d}{cf-ae}=\frac {b}{ce-af}=\frac {b+d}{(c-a)(f+e)}=$ $\frac {d-b}{(c+a)(f-e)}=\sqrt{\frac {p_w(F)}{acef}}$ .

$\blacktriangleright\ \ p_w(E)=-\frac {abcdef}{(ab+cd)(ad+bc)}\ \ ;\  p_w(F)=\frac {abcdef}{(cf-ae)(ce-af)}$ , where the powers of $E$ , $F$ w.r.t. $w$ is $p_w(E)$ , $p_w(F)$ respectively.


Proof. Prove easily all !


PP4. Let $ABC$ be a triangle with the incircle $w=C(I,r)$ . Denote $\begin{array}{c}
D\in BC\cap w\\\
E\in CA\cap w\\\
F\in AB\cap w\end{array}$ . Consider two points $M\in (AF)$ , $N\in (AE)$ . Prove that :

$\blacktriangleright\ \ MN$ is tangent to $w$ $\iff$ $(s-b)\cdot\frac {MA}{MB}+(s-c)\cdot\frac {NA}{NC}=(s-a)$ .

$\blacktriangleright\ \ MN$ is tangent to $w$ $\implies$ $4R\cdot [MDN]=(s-a)\cdot BM\cdot CN$ , where $2s=a+b+c$ and $R$ is the length of the circumradius (Toshio Seimyia).


Proof. Denote $\left\|\begin{array}{c}
AM=x\ ,\ AN=y\\\
T\in MN\cap w\end{array}\right\|$ . From $\left\|\begin{array}{c}
AE=AF=s-a\\\
IE=IF=IT=r\end{array}\right\|$ and $[MAN]+[AEIF]=$ $2\left([AIM]+[AIN]\right)$ obtain

$xy\cdot\sin A+2r(s-a)=2r(x+y)$ $\iff$ $xy\cdot\frac {2sr}{bc}+2r(s-a)=2r(x+y)$ $\iff$ $\boxed{sxy+bc(s-a)=bc(x+y)}\ (1)$ .

On other hand $(s-b)\cdot\frac {MA}{MB}+(s-c)\cdot\frac {NA}{NC}=(s-a)$ $\iff$ $(s-b)\cdot\frac {x}{c-x}+$ $(s-c)\cdot\frac {y}{b-y}=(s-a)$ $\iff$

$x(b-y)(s-b)+y(c-x)(s-c)=$ $(s-a)(c-x)(b-y)$ $\iff$ $xb(s-b)-xy(s-b)+yc(s-c)-xy(s-c)=$

$bc(s-a)-(s-a)(cy+bx)+xy(s-a)$ $\iff$ $bc(x+y)=sxy+bc(s-a)$ , i.e. the relation $(1)$ .

Otherwise. From $\left\|\begin{array}{c}
MN=2(s-a)-(x+y)\\\
MN^2=x^2+y^2-2xy\cdot\cos A\end{array}\right\|$ obtain $4(s-a)^2+(x+y)^2-4(s-a)(x+y)=x^2+y^2-2xy\cdot\cos A$ $\iff$

$4(s-a)^2+2xy(1+\cos A )=4(s-a)(x+y)$ $\iff$ $(s-a)^2+xy\cdot\cos^2\frac A2=(s-a)(x+y)$ $\iff$

$bc(s-a)^2+xys(s-a)=bc(s-a)(x+y)$ $\iff$ $bc(s-a)+sxy=bc(x+y)$ , i.e. the relation $(1)$ .

$\blacktriangleright\ \ [MDN]=S-[MAN]-[BDM]-CDN]$ $\implies$ $2\cdot [MDN]=(bc-xy)\cdot\sin A-(s-b)(c-x)\cdot\sin B-(s-c)(b-y)\cdot\sin C=$

$bc(\sin A+\sin B+\sin C)-$ $s(c\cdot\sin B+b\cdot\sin C)-$ $xy\cdot\sin A+x(s-b)\sin B+y(s-c)\cdot\sin C=$

$bc\cdot\frac sR-2s\cdot\frac {bc}{2R}-xy\cdot \frac {a}{2R}+$ $x(s-b)\cdot \frac {b}{2R}+y(s-c)\cdot \frac {c}{2R}$ . I used the relation $h_a=c\cdot\sin B=b\cdot\sin C=\frac {bc}{2R}$ . Therefore,

$4R\cdot[MDN]=xb(s-b)+yc(s-c)-xya=$ $xb(s-b)+yc(s-c)-xy[(s-b)+(s-c)]=$ $x(s-b)(b-y)+y(s-c)(c-x)=$

$(c-x)(b-y)\left[(s-b)\cdot\frac {x}{c-x}+(s-c)\cdot\frac {y}{b-y}\right]=$ $(c-x)(b-y)(s-a)=$ $(s-a)\cdot BM\cdot CN$ .

In conclusion, $4R\cdot [MDN]=BM\cdot (s-a)$ , i.e. the ratio $\frac {[MDN]}{BM\cdot CN}=\frac {s-a}{4R}$ is constant.

Remarks.

$\blacktriangleright$ The relation $(1)$ can write equivalently $(bc-sx)(bc-sy)=$ $bc(s-b)(s-c)\ (2)$ . Observe that $[MAN]$ is maximum $\iff$ $\frac 12\cdot xy\sin A$

is maximum $\iff$ $xy$ is maximum $\iff$ the sum $(x+y)$ is maximum from the relation $(1)$ $\iff$ he sum $(bc-sx)+(bc-sy)=$ $2bc-s(x+y)$ is

minimum. From the relation $(2)$ results that the product $(bc-sx)(bc-sy)$ is constant. Thus, the area $[MDN]$ is maximum $\iff$ $x=y$ , i.e. $AM=AN$ .

$\blacktriangleright$ If the triangle $ABC$ is equilateral, then the relation $(1)$ becomes $\frac {MA}{MB}+\frac {NA}{NC}=1$ . For $P\in CM\cap BN$ , $D_1\in AP\cap BC$ obtain

(from the Van Aubel's relation) $\frac {PA}{PD_1}=\frac {MA}{MB}+\frac {NA}{NC}=1$ , i.e. $PA=PD_1$ what means that the point $P$ belongs to the $A$-middle line.



PP5. Let $ABC$ be a triangle $ABC$ with circumcircle $C(O,R)$ and incircle $C(I,r)$ . Consider an inner point $T$ of given triangle and three circles with same radius $\rho$

which pass through $T$ and each from them is tangent only to two sides of $\triangle ABC$ . Prove that $T\in OI$ and $\frac 1R+\frac 1r=\frac {1}{\rho}$ . (Aram Yagubyants, Rostov, Rusia).


Proof. Denote : the circle $w_1=C(I_1,\rho )$ which is tangent to $AB$ , $AC$ , the circle $w_2=C(I_2,\rho )$ which is tangent to $BA$ , $BC$ and the circle $w_3=C(I_3,\rho )$ which is

tangent to $CA$ , $CB$ . Since $ABC\sim I_1I_2I_3$ obtain that $\frac {\rho}{R}=\frac {I_2I_3}{a}$ . Observe that $I_2I_3=a-\rho \left(\cot \frac B2+\cot\frac C2\right)=$ $a-\frac {\rho\cos\frac A2}{\sin\frac B2\sin\frac C2}=$ $a-\frac {as\rho}{S}=$

$a-\frac {a\rho}{r}=$ $\frac {a(r-\rho )}{r}$ . Thus, $\frac{\rho}{R}=\frac {r-\rho}{r}=\frac {r}{R+r}$ , i.e. $\rho (R+r)=Rr$ $\iff$ $\frac {1}{\rho}=\frac 1R+\frac 1r$ . Since $I$ is incenter of $\triangle I_1I_2I_3$ obtain $IT\parallel IO$ , i.e. $T\in OI$ .
This post has been edited 82 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:38 PM

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The relation $s=2R\cdot\sin\phi+r\cdot\cot\frac {\phi}2$ is equivalent to a

interesting identity i.e. $\boxed{\ \prod\left(\tan\frac A2-\tan\frac {\phi}2\right)=0\ }$ .

Therefore, $\boxed{\ \phi\in\{A,B,C\}\ }$ .

See here where it appears as an inequality ... :) (the upper product is $\le$ or $\ge 0$ )
This post has been edited 1 time. Last edited by Mateescu Constantin, Oct 21, 2010, 3:54 PM

by Mateescu Constantin, Oct 21, 2010, 3:44 PM

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Nice, very nice ! It is an interesting coincidence ! Thank you, M.C.

by Virgil Nicula, Oct 21, 2010, 4:06 PM

Own problems or extensions/generalizations of some problems which was posted here.

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