Italy TST 2000/2
Let $ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$ . Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$ . Show that $FC$ bisects the $\angle DFE$ .
Re: Italy TST 2000/2
The key move is to prove that $AE\perp DM$. Then $DCEF$ is cyclic with $CD = CE$ and the result follows.
- What is the value of the contour integral around Western Europe?
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
- Zero.
- Why?
- Because all the poles are in Eastern Europe.
Revive the IMO marathon.
Re: Italy TST 2000/2
An easy solution:
$\frac{CD} {DA}=\frac{CE} {EB}\Leftrightarrow DE\parallel AB$.$\frac{CD}{DE}=\frac{1}{\sqrt{2}}\Leftrightarrow DE=\sqrt{2}CD.AB=\sqrt{2}AC=3\sqrt{2}DC$..$\therefore \frac{DE}{AB}=\frac{1}{3}$.
$\Delta DFE \sim \Delta AFM$.$\therefore \frac{DE}{AM}=\frac{DF}{FM}=\frac{EF}{FA}=\frac{2} {3}$.Let $CF\cap AB=N$.By $\text{Menelaus's Theorem}$,we get that
$\frac{AC}{CD}\times \frac{DF}{FM}\times \frac{MN}{NA}=1\Leftrightarrow \frac{MN}{NA}$=$\frac{1}{2}$.Now,applying stewart's theorem,we get that
$AF^{2}=\frac{18CE^{2}}{5}$ and $FM^{2}=\frac{90CE^{2}}{100}$.$\therefore (\frac{FM}{AF})^{2}=\frac{1}{4}=(\frac{MN}{NA})^{2}\Leftrightarrow \frac{FM}{AF}=\frac{MN}{NA}$.$\therefore$ $CF$ is the bisector of $\angle DFE$.
$\frac{CD} {DA}=\frac{CE} {EB}\Leftrightarrow DE\parallel AB$.$\frac{CD}{DE}=\frac{1}{\sqrt{2}}\Leftrightarrow DE=\sqrt{2}CD.AB=\sqrt{2}AC=3\sqrt{2}DC$..$\therefore \frac{DE}{AB}=\frac{1}{3}$.
$\Delta DFE \sim \Delta AFM$.$\therefore \frac{DE}{AM}=\frac{DF}{FM}=\frac{EF}{FA}=\frac{2} {3}$.Let $CF\cap AB=N$.By $\text{Menelaus's Theorem}$,we get that
$\frac{AC}{CD}\times \frac{DF}{FM}\times \frac{MN}{NA}=1\Leftrightarrow \frac{MN}{NA}$=$\frac{1}{2}$.Now,applying stewart's theorem,we get that
$AF^{2}=\frac{18CE^{2}}{5}$ and $FM^{2}=\frac{90CE^{2}}{100}$.$\therefore (\frac{FM}{AF})^{2}=\frac{1}{4}=(\frac{MN}{NA})^{2}\Leftrightarrow \frac{FM}{AF}=\frac{MN}{NA}$.$\therefore$ $CF$ is the bisector of $\angle DFE$.
"Questions we can't answer are far better than answers we can't question"
Re: Italy TST 2000/2
i think my solution is too much boring let's apply Menelaus's and Menelaus's ;
let , $CF\cap AB=N ; BF\cap AC=P ; BF\cap CM=R ; CF\cap BD=S ; CM\cap BD=T$
Now
$\frac{BD}{DT}\frac{TF}{FA}\frac{AM}{MB}=1\Leftrightarrow \frac{AF}{FT}=\frac{BD}{DT}$
$\frac{AC}{CD}\frac{DS}{ST}\frac{TF}{FA}=1 \Leftrightarrow \frac{DS}{ST}=\frac{1}{3}\frac{AF}{FT}=\frac{1}{3}\frac{BD}{DT}$
$\frac{DB}{BT}\frac{TE}{EA}\frac{AC}{CD}=1\Leftrightarrow \frac{DB}{BT}=\frac{1}{3}\frac{EA}{ET}$
so, we have $\frac{DS}{ST}=\frac{DB}{BT}$
$\therefore (D,S,T,B) harmonic$
now, $\frac{AC}{CD}\frac{DF}{FM}\frac{MN}{NA}=1 \Leftrightarrow \frac{MN}{NA}=\frac{CD}{AC}\frac{FM}{DF}=
\frac{1}{3}\frac{3}{2}=\frac{1}{2}$
and, ceva for first time $\frac{BN}{NA}\frac{AP}{PC}\frac{CE}{EB}=1\Leftrightarrow \frac{AP}{PC}=1$
lastly, $\frac{AB}{BM}\frac{MR}{RC}\frac{CP}{PA}=1\Leftrightarrow \frac{MR}{RC}=\frac{BM}{AB}=\frac{1}{2}$
so, $\frac{MN}{NA}=\frac{MR}{RC}=\frac{CE}{EB}=\frac{1}{2}$
$\Delta NCM\sim \Delta RBM\sim \Delta EAC \Leftrightarrow \angle NCM=\angle RBM=\angle EAC$
$\therefore MFCB $ concyclic.
$\therefore \angle SFB=\angle CMB=90$
finally $FS$ bisects $\angle DFT$
so, $FC$ bisects $\angle DFE$
let , $CF\cap AB=N ; BF\cap AC=P ; BF\cap CM=R ; CF\cap BD=S ; CM\cap BD=T$
Now
$\frac{BD}{DT}\frac{TF}{FA}\frac{AM}{MB}=1\Leftrightarrow \frac{AF}{FT}=\frac{BD}{DT}$
$\frac{AC}{CD}\frac{DS}{ST}\frac{TF}{FA}=1 \Leftrightarrow \frac{DS}{ST}=\frac{1}{3}\frac{AF}{FT}=\frac{1}{3}\frac{BD}{DT}$
$\frac{DB}{BT}\frac{TE}{EA}\frac{AC}{CD}=1\Leftrightarrow \frac{DB}{BT}=\frac{1}{3}\frac{EA}{ET}$
so, we have $\frac{DS}{ST}=\frac{DB}{BT}$
$\therefore (D,S,T,B) harmonic$
now, $\frac{AC}{CD}\frac{DF}{FM}\frac{MN}{NA}=1 \Leftrightarrow \frac{MN}{NA}=\frac{CD}{AC}\frac{FM}{DF}=
\frac{1}{3}\frac{3}{2}=\frac{1}{2}$
and, ceva for first time $\frac{BN}{NA}\frac{AP}{PC}\frac{CE}{EB}=1\Leftrightarrow \frac{AP}{PC}=1$
lastly, $\frac{AB}{BM}\frac{MR}{RC}\frac{CP}{PA}=1\Leftrightarrow \frac{MR}{RC}=\frac{BM}{AB}=\frac{1}{2}$
so, $\frac{MN}{NA}=\frac{MR}{RC}=\frac{CE}{EB}=\frac{1}{2}$
$\Delta NCM\sim \Delta RBM\sim \Delta EAC \Leftrightarrow \angle NCM=\angle RBM=\angle EAC$
$\therefore MFCB $ concyclic.
$\therefore \angle SFB=\angle CMB=90$
finally $FS$ bisects $\angle DFT$
so, $FC$ bisects $\angle DFE$