in a triangle $ABC$, $I$ is the incenter. $BI$ and $CI$ cut the circumcircle of $ABC$ at $E$ and $F$ respectively. $M$ is the midpoint of $EF$. $C$ is a circle with diameter $EF$. $IM$ cuts $C$ at two points $L$ and $K$ and the arc $BC$ of circumcircle of $ABC$ (not containing $A$) at $D$. prove that $\frac{DL}{IL}=\frac{DK}{IK}$.(25 points)
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Tags: geometry, circumcircle, LaTeX, geometric transformation, reflection, incenter, parallelogram
mihai miculita
07.08.2010 15:17
HINT: $ \mbox{The circle with diameter }[EF]\mbox{-is Apollonius circle of }\\ \triangle{EID}\left(\mbox{and }\triangle{FID}\right) \Rightarrow \frac{|DL|}{|IL|}=\frac{|ED|}{|EI|}=\frac{|DK|}{|IK|}. $
vladimir92
07.08.2010 21:06
Good problem! this is my solution which is different than mihai's one.
Let $D_1$ be a point on the arc $\overarc{BC}$ not containing $A$ such that $\angle{ID_1F}=\frac{1}{2}\angle{ABC}$. let $D_1F$ , $D_1I$ and $D_1E$ meet $IB$, $EF$ and $IC$ at $I_1$, $M_1$ and $I_2$ respectively. From the fact that $\angle{ID_1F}=\frac{1}{2}\angle{ABC}$ follow directly that $\angle{ID_1E}=\frac{1}{2}\angle{ACB}$, then $\angle{I_2D_1I_1}=180-\angle{I_2II_1}$ therfore, the quadrilateral $II_1D_1I_2$ is cyclic, that gives $\angle{EI_1I_2}=\angle{ID_1E}=\frac{1}{2}\angle{ACB}=\angle{FEI_1}$ then $(I_1I_2) \parallel (EF)$ then from Ceva on $\triangle{EFD}$ follow that $M_1F=M_1E$ then $M_1=M$ and $D_1=D$, thus $\angle{IDF}=\angle{EFC}$, We have $\angle{IKF}=90-\frac{1}{2}\angle{DMF} \Rightarrow \angle{IFK}=90-\angle{FID}+\frac{1}{2}\angle{DMF}$ and we have $\angle{IFD}=180-\angle{FID}-\angle{IDF}$ then $\angle{IFD}-2\angle{IFK}=\angle{EFC}-\angle{IDF}=0$ that's suffice to say that $KF$ is the internal angle bissector of $\angle{IFD}$ and we have $(LF) \perp (FK)$ then the pencil $F(L,I,K,D)$ is harmonic, from which follow that: $ \frac{DL}{IL}=\frac{DK}{IK}$.
cnyd
12.08.2010 17:38
@vladimir 92 I think your solution is wrong because; $\angle FEI_{1}=\angle FI_{2}I_{1},not \angle E_{1}I_{2}$
vladimir92
12.08.2010 22:54
cnyd wrote: @vladimir 92 I think your solution is wrong because; $\angle FEI_{1}=\angle FI_{2}I_{1},not \angle E_{1}I_{2}$ Why? if you draw a diagram, you get that $\angle FEI_{1}=\angle FI_{2}I_{1}$ if and only if $FEI_2I_1$ is concyclic, which isn't always true. I read the part you mentioned on my solution and it isn't wrong.
sunken rock
13.11.2010 03:28
I am on board my ship now, so I have little time to write down my beautiful solution; it is no latex-ed, so I am sorry, but I was quite excited to communicate it! Well, it took me a lot of time to find a different solution, and here it is! In my drawing L is between I and D and I noted: N the second intersection of MI with the circle (ABC) , D’ the midpoint of the arc BC containing D in (ABC) and B’=BE x FD’. There are well known facts: a) I is the orthocenter of $\triangle$D’EF – hence B’ lies on the circle C(M,ME) and, also, B’ is the middle of BI, so BI = 2B’I (1) b) the reflection of the orthocenter in the midpoint of a side lies on the circumcircle of the subject triangle, hence IN=2IM ( 2 ). Do the above facts need proof? From the power of I wrt (ABC): BI.IE=DI.IN ( 3 ); with (1) and (2) we get B’I.IE=DI.IM ( 4 ); this shows that EMB’D is cyclic and, since ME=MB’=ML as rays of the circle C(M,ME), it follows that MD is the angle bisector of <EDB’ and L the incenter of EDB’, hence EL and EK are the bisectors of the <DEB. I think it is not bad at all! Best regards, sunken rock
ORamirez
23.08.2011 19:30
In the figure below, area of triangles and is given. Find the area of triangle in terms of area of these three triangles.
goodar2006
23.08.2011 19:44
ORamirez wrote: In the figure below, area of triangles and is given. Find the area of triangle in terms of area of these three triangles. I can't undrestad anything of your words, and BTW, where is the figure?? are you posting a new problem?? here is not the place to do so....
TelvCohl
22.10.2014 09:34
My solution: Let $ X $ be the intersection of $ MI $ and $ \odot (ABC) $ (different from $ D $). Let $ E'= \odot (EF) \cap CI, $ $ F'= \odot (EF) \cap BI, $ $ Y=EE' \cap FF' $ . Since $ XFIE $ is a parallelogram , so we get $ XE=FI=FA $ and $ XF=EI=EA $ . i.e. $ AX \parallel EF $ or $ X $ is the midpoint of arc $ BAC $ in $ \odot (ABC) $ Since $ EE', FF' $ is the perpendicular bisector of $ CI, BI $ , so we get $ Y $ is the midpoint of arc $ BC $ in $ \odot (ABC) $ and $ YD \perp MD $ , Since $ Y $ lie on the polar of $ I $ with respect to $ \odot (EF) $ , so $ YD $ is the polar of $ I $ with respect to $ \odot (EF) $ , hence we get $ (I,D;K,L)=-1 $ . i.e. $ \frac {DL}{IL}=\frac {DK}{IK} $ Q.E.D Remark: From a property of Mixtilinear circle we can get $ D $ is the tangent point of $ A $ -mixtilinear circle with $ \odot (ABC) $
steppewolf
13.09.2015 10:55
Let $S$ be the midpoint of the arc $BAC$. We have that $$\angle EFS = \frac{1}{2}(180^{\circ} - \angle BAC - \angle CBA) = \frac{1}{2} \angle ACB = \angle BEF$$so $FS \parallel IE$. Similarly $ES \parallel IF$ so $I$, $M$ and $S$ are collinear and $MS$ = $MI$. Now $ML^{2} = ME \cdot MF = MS \cdot MD = MI \cdot MD$ and because $M$ is the midpoint of $KL$ we have that $(K,L;I,D) = -1$ or $\frac{DL}{IL} = \frac{DK}{IK}$.