[OGC1] Online Geometry Camp: Day 4

 Posts: 5
 Joined: Wed Aug 28, 2013 12:19 pm
Re: [OGC1] Online Geometry Camp: Day 4
here main circle area is 4pi.suppose one of the common circle place is p.then the four side is 4p.then the side place is q.so 4q=4pi(4pi4p)=>4q=4p>=p=q
so the pq>==pp=0
so there wiil be no difference!!!
so the pq>==pp=0
so there wiil be no difference!!!
 Samiun Fateeha Ira
 Posts: 23
 Joined: Sat Aug 24, 2013 7:08 pm
 Location: Dhaka, Bangladesh
Re: [OGC1] Online Geometry Camp: Day 4
Mursalin wrote:You can click on the originalproblemlink and see the solution but I don't recommend that. It's not good to give up on a problem that easily!
Everything you did is correct. And you shouldn't get multiple solutions. Would you like a hint?
i don't want to see the solution. just give a hint.
 Samiun Fateeha Ira
 Posts: 23
 Joined: Sat Aug 24, 2013 7:08 pm
 Location: Dhaka, Bangladesh
Re: [OGC1] Online Geometry Camp: Day 4
well, yes!photon wrote:@ Samiun Fateeha Ira , apply power of a point any other way.
By the way , did it occur to you that $AB=48$ ?
AB= AD+DE+EB= 3+39+6= 48.
Re: [OGC1] Online Geometry Camp: Day 4
@ Ira:
Here is a hint.
You arrived at these equations:
$x(x+y)=126$,
$z(y+z)=702$
Put $48z=x+y$ and
$48x=y+z$.
Now subtract the former from the latter.
Hope this helps!
Here is a hint.
You arrived at these equations:
$x(x+y)=126$,
$z(y+z)=702$
Put $48z=x+y$ and
$48x=y+z$.
Now subtract the former from the latter.
Hope this helps!
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Re: [OGC1] Online Geometry Camp: Day 4
solution of problem 4:
first let the centre of the circle is O.
now, $\angle EBF+\angle GCF=180$
$\therefore \frac{1}{2}\angle EBF+\frac{1}{2}\angle GCF=90$
or,$\angle OBF+\angle OCF=90$
or,$\angle BOC=90$
samely we get $\angle BOC=\angle AOD=90$
now let GC=FC=x and GD=HD=y.
use pythagorus's theorem in $\Delta BOC$
so,$OB^{2}+OC^{2}=BC^{2}$
or,$OF^{2}+FB^{2}+OF^{2}+FC^{2}=(BF+FC)^{2}$
or,$12^{2}+3^{2}+12^{2}+x^{2}=(3+x)^{2}$
or,$144+9+144+x^{2}=x^{2}+6x+9$
or,$288=6x$
or,$x=48$
samely by using pythagorus's theorem in $\Delta AOD$, we get y=72
so x+y=48+72=120
or,CD=120
first let the centre of the circle is O.
now, $\angle EBF+\angle GCF=180$
$\therefore \frac{1}{2}\angle EBF+\frac{1}{2}\angle GCF=90$
or,$\angle OBF+\angle OCF=90$
or,$\angle BOC=90$
samely we get $\angle BOC=\angle AOD=90$
now let GC=FC=x and GD=HD=y.
use pythagorus's theorem in $\Delta BOC$
so,$OB^{2}+OC^{2}=BC^{2}$
or,$OF^{2}+FB^{2}+OF^{2}+FC^{2}=(BF+FC)^{2}$
or,$12^{2}+3^{2}+12^{2}+x^{2}=(3+x)^{2}$
or,$144+9+144+x^{2}=x^{2}+6x+9$
or,$288=6x$
or,$x=48$
samely by using pythagorus's theorem in $\Delta AOD$, we get y=72
so x+y=48+72=120
or,CD=120
 Fatin Farhan
 Posts: 75
 Joined: Sun Mar 17, 2013 5:19 pm
 Location: Kushtia,Bangladesh.
 Contact:
Re: [OGC1] Online Geometry Camp: Day 4
Solution to problem 1
BE.BD= BF.BG
=> 6*(39+6)=BF(BF+21)
=>270=BF^2+21BF
=> BF^2+21BF270=0
BF=9;CG=18
Let AI=x, HI=y, CH=z.
x(x+y)=126; z(y+z)=702; x+y+z=48.
Now
z(y+z)=702
=> {(x+y+z)(x+y)}{(x+y+z)x}=702
=> {48(x+y)}(48x)=702
=> 48^248(x+y+x)+x(x+y)=702
=> 48^248(x+y+x)+126=702
=> 48(x+y+x)=1728
=> $x+y+x=36$
Again
y=x+yx
=> y^2=(x+yx)^2
=> y^2=(x+y+x)4x(x+y)=36^24*126=792.
BE.BD= BF.BG
=> 6*(39+6)=BF(BF+21)
=>270=BF^2+21BF
=> BF^2+21BF270=0
BF=9;CG=18
Let AI=x, HI=y, CH=z.
x(x+y)=126; z(y+z)=702; x+y+z=48.
Now
z(y+z)=702
=> {(x+y+z)(x+y)}{(x+y+z)x}=702
=> {48(x+y)}(48x)=702
=> 48^248(x+y+x)+x(x+y)=702
=> 48^248(x+y+x)+126=702
=> 48(x+y+x)=1728
=> $x+y+x=36$
Again
y=x+yx
=> y^2=(x+yx)^2
=> y^2=(x+y+x)4x(x+y)=36^24*126=792.
"The box said 'Requires Windows XP or better'. So I installed L$$i$$nux...:p"
Re: [OGC1] Online Geometry Camp: Day 4
$Solution$ $of$ $4$:
The solution given by Tahmid vaia is easy and nice.Here is another path in which I have solved it:
Join $G,E$.$GE$ is a diameter of the inscribed circle.Suppose,$O$ is the center of the circle.Again,join $E,F;F,G;O,F;O,B$ and $O,C$.Applying $Pythagorus's$ theorem,we get,$OB=3\sqrt{17}$.Applying $Ptolemy's$ theorem in cyclic quadrilateral $OFBE$,we get,$EF=\frac{24}{\sqrt{17}}$.In,right angled triangle $EFG$,$EG=24,EF=\frac{24}{\sqrt{17}}.\therefore GF=\frac{96}{\sqrt{17}}$.Suppose,$GF\cap OC=T$.Using Power of point in cyclic quadrilateral $OGCF$,we get,$GT \times TF=OT \times TC.\therefore OT \times TC=\frac{2304}{17}$.Now,$OT=\frac{12}{\sqrt{17}}.\therefore TC=\frac{192}{\sqrt{17}}.\therefore GC=48$.In the same method,we get $GD=72$.$\therefore CD=120$.
The solution given by Tahmid vaia is easy and nice.Here is another path in which I have solved it:
Join $G,E$.$GE$ is a diameter of the inscribed circle.Suppose,$O$ is the center of the circle.Again,join $E,F;F,G;O,F;O,B$ and $O,C$.Applying $Pythagorus's$ theorem,we get,$OB=3\sqrt{17}$.Applying $Ptolemy's$ theorem in cyclic quadrilateral $OFBE$,we get,$EF=\frac{24}{\sqrt{17}}$.In,right angled triangle $EFG$,$EG=24,EF=\frac{24}{\sqrt{17}}.\therefore GF=\frac{96}{\sqrt{17}}$.Suppose,$GF\cap OC=T$.Using Power of point in cyclic quadrilateral $OGCF$,we get,$GT \times TF=OT \times TC.\therefore OT \times TC=\frac{2304}{17}$.Now,$OT=\frac{12}{\sqrt{17}}.\therefore TC=\frac{192}{\sqrt{17}}.\therefore GC=48$.In the same method,we get $GD=72$.$\therefore CD=120$.
"Questions we can't answer are far better than answers we can't question"
Re: [OGC1] Online Geometry Camp: Day 4
$\text{Solution of problem 6}$:
Suppose,$AM\cap CD=P$ and the line parallel to $CM$ meets $DM$ at $L$.
Now,in quadrilateral $ADMC$,$\angle MCA+\angle MDA$=$\angle MCD+\angle DCA+\angle MDC+\angle ADC=\angle CAB+\angle DCA+\angle BAD$+$\angle ADC$=$180^{\circ}$.$\therefore$ quadrilateral $ADMC$ is cyclic.Now,$\angle CMD=\angle KLD.\therefore \angle KAD+\angle KLD=180^{\circ}$.$\therefore$ $AKLD$ is cyclic.Now,$\angle KAD+\angle KLD=\angle KAB+\angle BAD+\angle KLD=\angle KAB+\angle PDL+\angle KLD$.Again $\angle DPL+\angle PDL+\angle KLD=180^{\circ}$.$\therefore \angle KAB=\angle DPL.\therefore \angle DPL+\angle DPK=\angle KAB+ \angle DPK=180^{\circ}$.$\therefore$ $KABP$ is cyclic.
$\therefore \angle KBA=\angle KPA=\angle MPL=\angle CMP=\angle CMA=\angle CDA$.Then, by the $\text{converse of alternate segment theorem}$,$BK$ is tangent to the $2nd$ circle.
Suppose,$AM\cap CD=P$ and the line parallel to $CM$ meets $DM$ at $L$.
Now,in quadrilateral $ADMC$,$\angle MCA+\angle MDA$=$\angle MCD+\angle DCA+\angle MDC+\angle ADC=\angle CAB+\angle DCA+\angle BAD$+$\angle ADC$=$180^{\circ}$.$\therefore$ quadrilateral $ADMC$ is cyclic.Now,$\angle CMD=\angle KLD.\therefore \angle KAD+\angle KLD=180^{\circ}$.$\therefore$ $AKLD$ is cyclic.Now,$\angle KAD+\angle KLD=\angle KAB+\angle BAD+\angle KLD=\angle KAB+\angle PDL+\angle KLD$.Again $\angle DPL+\angle PDL+\angle KLD=180^{\circ}$.$\therefore \angle KAB=\angle DPL.\therefore \angle DPL+\angle DPK=\angle KAB+ \angle DPK=180^{\circ}$.$\therefore$ $KABP$ is cyclic.
$\therefore \angle KBA=\angle KPA=\angle MPL=\angle CMP=\angle CMA=\angle CDA$.Then, by the $\text{converse of alternate segment theorem}$,$BK$ is tangent to the $2nd$ circle.
Last edited by tanmoy on Tue Jan 06, 2015 9:32 pm, edited 2 times in total.
"Questions we can't answer are far better than answers we can't question"
Re: [OGC1] Online Geometry Camp: Day 4
Please use \text{} in LaTeX to write something within it.
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Re: [OGC1] Online Geometry Camp: Day 4
ভাইয়া,বুঝলাম না।*Mahi* wrote:Please use \text{} in LaTeX to write something within it.
"Questions we can't answer are far better than answers we can't question"