BdMO National Higher Secondary 2011/8

[rakeen]

8.I’ve figured out couple of info from this problem. But unable to connect them to find the result. We can see these info(from the figure) :

BD=CD ,AO=FO , AD||GB

ACD হল সমবাহু ত্রিভুজ।
(ADG) =(ADB) and also CGB সমবাহু ত্রিভুজ।

AB=GD ACDE ব্রিত্তস্থ চতুরভুজ। <ACF=<ADF and <CDA=<CFA

CF intersects GB equally.

[Mehfuj Zahir]

Ans has been already posted.Please Check.

[rakeen]

he..he.. ami to post gula na dekhei disilam

[Tahmid Hasan]

this was a really tough problem ,no contestant in bdmo could fully solve it and lived to tell the tale 3:)

[Tahmid Hasan]

Solution: $BD=CD$ and $GB \parallel AD$.
So $AG=AC$.
So $\triangle ABG \cong \triangle ABC$.
Since $\angle A$ = 90°.$AB$ is a median of $\triangle GBC $ and $D$ is the midpoint of $BC$.
So $F$ is the centroid of $\triangle GBC$.
Join $B$ , $O$ and extend such it intersects $AC$ at $L$.
now ceva! .$\frac{AL.CD.BF}{CL.BD.AF}=1$
.so $\frac{AL}{CL}=\frac{1}{2}$.
So, $\frac{GK}{CL}=2$
Extend $CF$ to $CK$ such hat it intersects $BG$ at $K$.
Again ceva.$\frac {GL.CR.BK}{CL.BR.GK}=1$.
So $\frac{BR}{CR}=2$.
So$\frac{BF}{AF}=\frac{BR}{CR}$.
So $FR||AC$.
So $\triangle FGR=\triangle FCR$.
So $\triangle GDR= \triangle FDC$.
$\triangle FBD$ and $\triangle FDC$ have equal base and same opposite vertex .so they have equal area.
$\triangle FBD=\frac{1}{2}.\frac{2}{3}AB.\frac{1}{2}BC.sin \angle ABC$=$\frac {1}{3} \triangle ABC$.
So $AB.AC=\frac{12}{\sqrt 15}$.........(yahoo!)
$\triangle AFO$ and $\triangle ABD$ are similar.
so $\frac{AO}{AB}=\frac{AF}{AD}$[$AF=\frac {1}{3}AB$ and $AD=\frac{1}{2}BC$]
So $\frac{1}{2}.BC.AO=\frac{1}{3}AB^2$
Hence $AO=\frac{2AB^2}{3BC}$
$OD=\frac{1}{2}BC-\frac{2AB^2}{3BC}=\frac{3AC^2-AB^2}{6BC}$
so $\frac{AO}{OD}=\frac{4AB^2}{3AC^2-AB^2}=\frac{AC.sin\angle ACF}{\frac{1}{2}BC.sin\angle BCF}$..........(hurrah!)
$ \frac {AF}{BF} = \frac {AC.sin \angle ACF}{BC.sin \angle BCF}$.
so $\frac{sin \angle ACF}{sin. \angle BCF}=\frac{1}{2}$.
Input that in (hurrah!) and get $\frac{AB}{AC}=\frac{\sqrt 3}{\sqrt 5}$
Mix (yahoo!) and (hurrah!) and get $AC=2$,$AB=\frac{2.\sqrt 3}{\sqrt 5}$,$BC=\frac {4.\sqrt 2}{\sqrt 5}$.
[P.S. 2:30 hours to solve,1 hour to LaTeX]

[Shifat]

Ceva's theorem chara ki eta prove kora jabe na!!!!

[Tahmid Hasan]

Ceva's theorem chara ki eta prove kora jabe na!!!!

চেষ্টা করে দেখুন,হলেও ত হইতে পারে

[Shifat]

Rakeen vai ADC and BGC konotai shomobahu triangle hobe na......., hoile AF=FO impossible..

[Nadim Ul Abrar]

1.Menelaus dia AB=OC
2.cava dia DR=RC/2=BC/6
3.GD^2=(3/2AC)^2+(1/2AB)^2
4.GR^2=(AB^2+25AC^2)^(1/2)/3
5.AC^2=15/9 AB^2

1 theke 5 porjonto use kore ;

DR=(6)^(1/2)/9 AB
GD=2AB
GR=(128/3)^(1/2)/3 AB .

Now (s(s-a)(s-b)(s-c))^(1/2) e DR,GD,GR er value gulo boshie .
AB=6/(15)^(1/2)
so AC=2
and BC=4(6/15)^(1/2)