Circles $ C_1$ and $ C_2$ with centers at $ O_1$ and $ O_2$ respectively meet at points $ A$ and $ B$. The radii $ O_1B$ and $ O_2B$ meet $ C_1$ and $ C_2$ at $ F$ and$ E$. The line through $ B$ parallel to $ EF$ intersects $ C_1$ again at $ M$ and $ C_2$ again at $ N$. Prove that $ MN = AE + AF$.
Problem
Source: 17-th Iranian Mathematical Olympiad 1999/2000
Tags: geometry, rectangle, trapezoid, geometry proposed
Andreas
14.12.2005 17:12
$\angle BAE = 90^o = \angle BAF$. $AB$ is in common between $\Delta BAF$ and $\Delta BAE$, so $A \in FE$. Now $\angle AEB = \angle EBN$ and $\angle BNE = \angle BAE = 90^o$. For ASA $\Delta BAE \equiv \Delta BNE$ $\Longrightarrow$ $AE = BN$. Also $\Delta BAF \equiv \Delta BMF$ $\Longrightarrow$ $AF = BM$. Adding this $MN = AF + AE$.
Mashimaru
11.02.2009 19:07
Since $ \widehat{BAF}=\widehat{BAE}=\widehat{ABM}=\widehat{ABN}$ we obtain that $ ABMF$ and $ ABNE$ are rectangles, i.e., $ AF=MB, AE=NB$. Thus: $ MN=MB+BN=AE+AF$ and we are done.
Mashimaru
11.02.2009 20:15
I think the problem has been stated wrong. It should have been $ O_1B \cap (O_2) = \{B,E\}$ and $ O_2B \cap (O_1) = \{B,F\}$
Amir Hossein
28.01.2012 22:11
Please someone post a solution to this problem (to the correct version). Thanks.
Luis González
28.01.2012 23:30
Isosceles triangles $\triangle BFO_1$ and $\triangle BEO_2$ with apices $O_1,O_2$ are clearly similar $\Longrightarrow$ $\angle FO_1E=\angle EO_2F$ $\Longrightarrow$ $EFO_1O_2$ is cyclic $\Longrightarrow$ $\angle BFE= \angle BO_1O_2.$ But $\angle BO_1O_2=\frac{_1}{^2}\angle BO_1A=\angle BMA,$ thus $\angle BMA=\angle BFE$ $\Longrightarrow$ $AM \parallel BF,$ i.e. $ABFM$ is an isosceles trapezoid with congruent diagonals $AF=BM.$ Similarly, $AE=BN$ and the conclusion follows.