242. A general geometrical problems.

by Virgil Nicula, Mar 4, 2011, 5:23 PM

PP1. Let $ ABC$ be a triangle. Consider the points $ D\in (AC)$ , $ M\in (BD)$ . Denote the properties :

$ \left|\ \begin{array}{cc} \mathrm{p\ : } & m(\angle {MCB})\ = \ m(\angle{MBA}) \\ \\ \mathrm{q\ : } & AM\perp MC \\ \\ \mathrm{r\ : } & \frac {DA}{DC} = \frac {c^2\cdot \left(a^2 + c^2 - b^2\right)}{a^2\cdot \left(b^2 + c^2\right) - \left(b^2 - c^2\right)^2}\end{array}\right|\ .$ Prove that $ (\ p\ \wedge\ q\ \rightarrow\ r\ )\ \ \wedge\ \ (\ p\ \wedge\ r\ \rightarrow\ q\ )\ \ \wedge\ \ (\ q\ \wedge\ r\ \rightarrow\ p\ )\ .$

Some interesting particular cases : $ \left|\ \begin{array}{ccc} b = c & \Longrightarrow & \frac {DA}{DC} = \frac 12 \\ \\ A = 90^{\circ} & \Longrightarrow & \frac {DA}{DC} = \frac {c^2}{2b^2}\end{array}\ \right|$ AND $ \left|\ \begin{array}{ccc} C = 90^{\circ} & \Longrightarrow & \frac {DA}{DC} = \left(\frac cb\right)^2 \\ \\ B = 60^{\circ} & \Longrightarrow & \frac {DA}{DC} = \frac {c^2}{a(a + c)}\end{array}\ \right|\ .$


PP2 (Toshio Seimiya). 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 11 times. Last edited by Virgil Nicula, Nov 22, 2015, 12:46 PM

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