Very Nice Problem. with simple angle chasing, we find that triangles $CEG$ and $MDB$ are similar. so, $GE/EC = MD/MB$. (*) again with simple angle chasing, we find that triangles $CEF$ and $AMD$ are similar. so, $EF/DM = CE/AM$. (**) so, by (*) and (**), we have $GE/EF = MA/MB = t/(t-1)$. E.L

here is my 'quick & dirty' solution: Out of projective considerations the ratio in question could be one of only 6 Salmon ratios: t , t -1 , 1 / t , 1/ (t - 1) , t / (t - 1) , (t - 1) / t A simple checking of limiting positions leads to the ratio t / (t - 1) Cute problem! Thank you. M.T.

I got $\frac{GE}{EF}=\frac{t}{1-t}$. I think this is right since $\frac{BM}{AB}=1-t$ and using the ratios from the similar triangles previously mentioned one obtains:\[ (1) \frac{BM}{MD}=\frac{CE}{EG} \]\[ (2) \frac{AM}{MD}=\frac{CE}{EF} \]Dividing $(1)$ by $(2)$ we obtain\[ \frac{EF}{EG}=\frac{BM}{AM} \]\[ \Leftrightarrow \frac{EG}{EF}=\frac{AM}{BM} \]\[ \Leftrightarrow \frac{EG}{EF}=\frac{AM}{AB} \cdot \frac{AB}{BM} \]Substituting from **\[ \Leftrightarrow \frac{EG}{EF}=\frac{t}{1-t} \]

Assume that $ M\in AE$ since the other case is analogous. Let $ N$ and $ N'$ be on $ EC$ and $ AC$, repectively, such that $ BNN'//GF$. Then $ \angle MDE=\angle GEA=\angle BEF=\angle NBE$ so $ MDBN$ is cyclic. Hence $ \angle MNE=\angle EBD=\angle ACD$, so $ MN//AC$. Using the parallel lines, $ GE: EF=N'N: NB=AM: MB=t: 1-t$.

I will present a proof using trigonometry. Notations:$\angle{DCB}=\theta,\angle{DME}=x$ Proof:Applying sine rule in $\triangle{AMD}$ we get $\frac{AM}{sin(x+\theta)}=\frac{AD}{sinx}$ We also have $\frac{AB}{sinC}=\frac{AD}{sin(C-\theta)}$ Dividing the two relations we get $t=\frac{AM}{AB}=\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC}$ Applying sine rule in $\triangle{EFC}$ we get $\frac{EF}{sin\theta}=\frac{EC}{sin(x+\theta)}$ Applying sine rule in $\triangle{EGC}$ we get $\frac{EG}{sin(C-\theta)}=\frac{EC}{sin(x-C+\theta)}$ Dividing the two relations we get $\frac{EG}{EF}=\frac{sin(C-\theta)sin(x+\theta)}{sin(x-C+\theta)sin\theta}$. Again $\frac{t}{1-t}=\frac{\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC}}{1-\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC}} =\frac{sin(x+\theta)sin(C-\theta)}{sinxsinC-sin(x+\theta)sin(C-\theta)} =\frac{2sin(x+\theta)sin(C-\theta)}{cos(C-x)-cos(C+x)-cos(C-x+2\theta)+cos(C+x)} =\frac{sin(x+\theta)sin(C-\theta)}{sin\theta sin(x-C+\theta)}$ From these relations we get $\frac{EG}{EF}=\frac{t}{1-t}$.We are done!!!

Dear Mathlinkers, this problem appears again at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=498513 Sincerely Jean-Louis

Just Notice, $\Delta EGC \sim \Delta MDB$ and, $\Delta ECF \sim \Delta MAD \implies EC \cdot MD= EF \cdot AM=EG \cdot MB \implies \frac{EG}{EF}=\frac{AM}{MB}=\boxed{\frac{t}{1-t}}$

A sweet problem. My solution is as follows: Let's first recognize some similarities among triangles. Since $ABCD$ is cyclic, $\angle{ECF}=\angle{DCB}=\angle{DAB}$. Furthermore, due to the tangent, $\angle{CEF}=\angle{GED}=\angle{EMD}$. Thus, $\triangle{CEF} \sim \triangle{AMD}$, leading to the following equality of ratios: $\frac{EF}{CE} = \frac{MD}{AM} (1)$ Once again since $ABCD$ is cyclic, $\angle{GCE}=\angle{ACD}=\angle{ABD}=\angle{MBD}$. In addition, because of the supplementary angles, $\angle{CEG}=\angle{BMD}$. Therefore $\triangle{CGE} \sim \triangle{BDM}$, leading to the following equality of ratios: $\frac{EG}{CE} = \frac{DM}{BM} (2)$ Dividing $(1)$ by $(2)$ gives us the following equality: $\frac{EF}{EG}=\frac{BM}{AM}=\frac{AB-AM}{AM}=\frac{AB}{MA}-\frac{AM}{AM}=\frac{1}{t}-1=\frac{1-t}{t}$ Therefore, $\frac{EG}{EF}=\frac{t}{1-t}$ $\blacksquare$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/A%20variant%20of%20IMO%201990.pdf Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, see here Sincerely Jean-Louis

Claim 1: $\triangle ECG\sim \triangle MBD$. Proof: Angle Chasing. $\angle ECG = \angle DCA = \angle DBA = \angle DBM$ and $\angle GEC = \angle GED = \angle EMD = \angle BMD$. $\square$. Claim 2: $\triangle CEF\sim \triangle AMD$ Proof: $\angle CEF = \angle GED = \angle DMA$ and $\angle ECF = \angle DCB = \angle DAB = \angle DAM$. $\square$. Thus, \[\frac{EG}{EF} = \frac{\frac{EG}{EC}}{\frac{EF}{EC}} = \frac{\frac{MD}{MB}}{\frac{MD}{AM}}=\frac{AM}{MB} = \frac{t}{1-t} \]$\blacksquare$.

The answer is $\frac{EG}{EF}=\frac{t}{t-1}$. From two triangle similarities $\triangle{CEG} \sim \triangle{BMD}$ and $\triangle{CEF}\sim \triangle{AMD}$ we get $$\frac{EG}{MD}=\frac{CE}{BM}, \frac{EF}{MD}=\frac{CE}{AM}\Rightarrow \frac{\frac{EG}{MD}}{\frac{EF}{MD}}=\frac{\frac{CE}{BM}}{\frac{CE}{AM}}\Rightarrow \frac{EG}{EF}=\frac{AM}{BM}.$$which is easy to derive once we know $\frac{AM}{AB}=t$.