Page **3** of **4**

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Wed Aug 28, 2013 10:13 pm**

by **brintodibyendu**

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-(4pi-4p)=>4q=4p>=p=q

so the p-q>==p-p=0

so there wiil be no difference!!!

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Wed Aug 28, 2013 10:50 pm**

by **Samiun Fateeha Ira**

Mursalin wrote:You can click on the original-problem-link 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.

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Wed Aug 28, 2013 10:52 pm**

by **Samiun Fateeha Ira**

photon wrote:@ Samiun Fateeha Ira , apply power of a point any other way.

By the way , did it occur to you that $AB=48$ ?

well, yes!

AB= AD+DE+EB= 3+39+6= 48.

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Wed Aug 28, 2013 11:07 pm**

by **Mursalin**

@ Ira:

Here is a hint.

You arrived at these equations:

$x(x+y)=126$,

$z(y+z)=702$

Put $48-z=x+y$ and

$48-x=y+z$.

Now subtract the former from the latter.

Hope this helps!

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Wed Aug 28, 2013 11:32 pm**

by **Tahmid**

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

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Wed Aug 28, 2013 11:40 pm**

by **Fatin Farhan**

Solution to problem 1

BE.BD= BF.BG

=> 6*(39+6)=BF(BF+21)

=>270=BF^2+21BF

=> BF^2+21BF-270=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)}(48-x)=702

=> 48^2-48(x+y+x)+x(x+y)=702

=> 48^2-48(x+y+x)+126=702

=> 48(x+y+x)=1728

=> $x+y+x=36$

Again

y=x+y-x

=> y^2=(x+y-x)^2

=> y^2=(x+y+x)-4x(x+y)=36^2-4*126=792.

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Sat Jan 03, 2015 7:18 pm**

by **tanmoy**

$Solution$ $of$ $4$:

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 $GD=72$.$\therefore CD=120$.

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Mon Jan 05, 2015 4:11 pm**

by **tanmoy**

$\text{Solution of problem 6}$:

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.

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Mon Jan 05, 2015 11:22 pm**

by ***Mahi***

Please use \text{} in LaTeX to write something within it.

### Re: [OGC1] Online Geometry Camp: Day 4

Posted: **Tue Jan 06, 2015 12:35 pm**

by **tanmoy**

*Mahi* wrote:Please use \text{} in LaTeX to write something within it.

ভাইয়া,বুঝলাম না।