18. O problema a lui Toshio Seimiya.

by Virgil Nicula, Apr 21, 2010, 1:29 AM

Quote:
An equivalent enunciation. Let $ ABCD$ be a convex quadrilateral. Suppose $ OA < OC$ and $ OD < OB$ , where $ O\in AC\cap BD$. Construct the midpoints $ M$ , $ N$

of $ [BD]$ , $ [AC]$ respectively, the intersections $ E\in MN\cap AB$ , $ F\in DC\cap MN$ , $ P\in CE\cap BF$ and the midpoint $ L$ of $ EF$ . Prove that $ L\in OP$ .

Proof. $ \left\|\begin{array}{c} OA=a\ ,\ OB=b\\\\ OC=c\ ,\ OD=d\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} c>a\ ;\ \frac {OA}{2a}=\frac {OC}{2c}=\frac {CA}{2(c+a)}=\frac {NA}{c+a}=\frac {ON}{c-a}\\\\ b>d\ ;\ \frac {OD}{2d}=\frac {OB}{2b}=\frac {BD}{2(b+d)}=\frac {MD}{b+d}=\frac {OM}{b-d}\end{array}\right\|\ (1)$ . Apply Menelaos' theorem to :

$ \left\|\begin{array}{cccccc} \overline {NME}/AOB\ : & \frac {NO}{NA}\cdot\frac {EA}{EB}\cdot\frac {MB}{MO}=1 & \implies & \frac {EA}{EB}=\frac {c+a}{c-a}\cdot\frac {b-d}{b+d} & \implies & \frac {EA}{(c+a)(b-d)}=\frac {EB}{(b+d)(c-a)}=\frac {AB}{2(bc-ad)}\\\\ \overline {MNF}/DOC\ : & \frac {MO}{MD}\cdot\frac {FD}{FC}\cdot\frac {NC}{NO}=1 & \implies & \frac {FC}{FD}=\frac {b-d}{b+d}\cdot\frac {c+a}{c-a} & \implies & \frac {FC}{(c+a)(b-d)}=\frac {FD}{(b+d)(c-a)}=\frac {DC}{2(bc-ad)}\end{array}\right\|$
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 23, 2015, 6:59 AM

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