268. Saraghin contest, 2011.

by Virgil Nicula, Apr 23, 2011, 3:00 PM

Proposed problem 20. Let $ABCD$ be a cyclical and tangential convex quadrilateral with the incircle $C(I,r)$ and the

circumcircle $C(O,R)$ . Denote the midpoints $M$ , $N$ of its diagonals $AC$ , $BD$ . Prove that $I\in MN$ and $\frac {IM}{IN}=\frac {AC}{BD}$ .


Proof (metric). $I\in MN$ because $[AMB]+[CMD]=[ANB]+[CND]=[AIB]+[CID]=\frac S2=\frac{rs}{2}$ , where $2s=a+b+c+d$

and the geometrical locus of the mobile point $L$ for which $[ALB]+[CLD]=k$ (constant) is a line. Denote $AC=e$ , $BD=f$ . Apply the theorem of median in :

$\left\|\begin{array}{ccc}
\triangle AIC\ : & 4\cdot IM^2=2\left(IA^2+IC^2\right)-e^2\\\
\triangle BID\ : & 4\cdot IN^2=2\left(IB^2+ID^2\right)-f^2\end{array}\right\|$ . Therefore, $\frac {IM}{IN}=\frac {e}{f}\iff$ $\frac {2\left(IA^2+IC^2\right)-e^2}{2\left(IB^2+ID^2\right)-f^2}=\frac {e^2}{f^2}\iff$

$\frac {IA^2+IC^2}{IB^2+ID^2}=\frac {e^2}{f^2}\iff$ $\frac {IA\cdot IC\cdot\cos\widehat{AIC}}{IB\cdot ID\cdot \cos\widehat{BID}}=\frac {e^2}{f^2}\iff$ $\frac {\frac {r}{\sin\frac A2}\cdot\frac {r}{\sin\frac C2}\cdot\cos\left(D+\frac {A+C}{2}\right)}{\frac {r}{\sin\frac B2}\cdot\frac {r}{\sin\frac D2}\cdot\cos\left(C+\frac {B+D}{2}\right)}=\frac {\sin^2B}{\sin^2A}\iff$

$\frac {\sin\frac B2\cdot\sin\frac D2\cdot\sin D}{\sin\frac A2\cdot\sin\frac C2\cdot\sin C}=\frac {\sin^2B}{\sin^2A}\iff$ $\frac {\sin\frac B2\cdot\sin\frac D2}{\sin\frac A2\cdot\sin\frac C2}=\frac {\sin B}{\sin A}\iff$ $\frac {\sin\frac D2}{\sin\frac C2}=\frac {\cos\frac B2}{\cos\frac A2}$ , what is truly because $\frac {B+D}{2}=\frac {A+C}{2}=90^{\circ}$ .

Remark. Are well-known and the following relations : $\frac {1}{e}+\frac {1}{f}=\frac {s}{4Rr}$ and $OI^2=R^2+r^2-r\cdot\sqrt {4R^2+r^2}$ (the Durrande's relation).

The power of $T\in AC\cap BD$ w.r.t. $w=C(O,R)$ is $p_w(T)=-\left(\frac {ef}{4R}\right)^2$ and $p_w(I)=-\frac {8R^2r^2}{ef}$ . See and
here


Proposed problem 21. Let $ABC$ be an $A$-right triangle with incenter $I$ and circumcircle $w$ . Denote: midpoints $M$ and $N$ of the sides $[AB]$ and $[AC]$ respectively;

intersections $L\in AB\cap CI$ and $K\in AC\cap BI$ ; second intersections $P$ , $Q$ of the circle $w$ with $CI$ , $BI$ respectively. Prove that $MQ\cap NP\cap PQ\ne\emptyset$ .


Proof (metric). Observe that $\boxed{S=[ABC]=s(s-a)=(s-b)(s-c)=\frac 12\cdot bc}$ and $\boxed{IA^2=2(s-a)^2=(a-b)(a-c)}$ , where $a+b+c=2s$ .

Denote $\left\|\begin{array}{c}
\frac B2=m\left(\widehat {KBA}\right) =m\left(\widehat {KBC}\right)=x\\\\
\frac C2=m\left(\widehat {LCA}\right)=m\left(\widehat {LCB}\right)=y\end{array}\right\|$ , where $x+y=\frac {\pi}{4}$ and $\boxed{\tan x\cdot\tan y=\frac {s-a}{s}}$ . Prove easily that $\left\|\begin{array}{c}
\frac {LA}{b}=\frac {LB}{a}=\frac {c}{a+b}\\\\
\frac {KA}{c}=\frac {KC}{a}=\frac {b}{a+c}\\\\
MB=\frac c2\ ;\ NC=\frac b2\end{array}\right\|$ $\implies$

$\left\|\begin{array}{ccc}
KA=\frac {bc}{a+c} & ; & LA=\frac {bc}{a+b}\\\\
ML=\frac {c(a-b)}{2(a+b)} & ; & NK=\frac {b(a-c)}{2(a+c)}\end{array}\right\|$ $\implies$ $\left\|\begin{array}{ccc}
\frac {PL}{PC}=\frac {AL}{AC}\cdot\frac {\sin\widehat {PAL}}{\sin\widehat{PAC}}=\frac {c}{a+b}\cdot \frac {\sin y}{\sin \left(90^{\circ}+y\right)} & \implies & \frac {PL}{PC}=\frac {c\cdot\tan y}{a+b}\\\\
\frac {QK}{QB}=\frac {AK}{AB}\cdot\frac {\sin\widehat {QAK}}{\sin\widehat{QAB}}=\frac {b}{a+c}\cdot \frac {\sin x}{\sin \left(90^{\circ}+x\right)} & \implies & \frac {QK}{QB}=\frac {b\cdot\tan x}{a+c}\end{array}\right\|$ . For $X\in MQ\cap LK$ ,

$Y\in NP\cap LK$ apply the Menelaus' theorem to the transversals $\left\|\begin{array}{cccc}
\overline{MXK}/\triangle BLK\ : & \frac {QK}{QB}\cdot\frac {MB}{ML}\cdot\frac {XL}{XK}=1 & \implies & \frac {XL}{XK}=\frac {(a+c)(a-b)}{b(a+b)\tan x}\\\\
\overline{NYP}/\triangle CKL\ : & \frac {PL}{PC}\cdot\frac {NC}{NK}\cdot\frac {YK}{YL}=1 & \implies & \frac {YL}{YK}=\frac {c(a+c)\tan y}{(a+b)(a-c)}\end{array}\right\|$ .

In conclusion, $MQ\cap NP\cap PQ\ne\emptyset\iff$ $X\equiv Y\iff$ $\frac {(a+c)(a-b)}{b(a+b)\tan x}=\frac {c(a+c)\tan y}{(a+b)(a-c)}\iff$ $(a-b)(a-c)=bc\cdot\tan x\tan y\iff$

$2(s-a)^2=\frac {bc(s-a)}{s}\iff$ $2s(s-a)=bc\iff$ $(b+c)^2-a^2=2bc$ , what is truly.
This post has been edited 47 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:35 AM

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