[OGC1] Online Geometry Camp: Day 4
Re: [OGC1] Online Geometry Camp: Day 4
Problem 6
Two circles intersect at $AB$ . A line through $B$ intersect the first circle at $C$ and the second circle at $D$. The tangents to the first circle at $C$ and the second at $D$ intersect at $M$ . Through the intersection point of $AM$ and $CD$ , there passes a line parallel to $CM$ and intersecting $AC$ at $K$ . Prove that $BK$ is tangent to the second circle .
Two circles intersect at $AB$ . A line through $B$ intersect the first circle at $C$ and the second circle at $D$. The tangents to the first circle at $C$ and the second at $D$ intersect at $M$ . Through the intersection point of $AM$ and $CD$ , there passes a line parallel to $CM$ and intersecting $AC$ at $K$ . Prove that $BK$ is tangent to the second circle .
Try not to become a man of success but rather to become a man of value.Albert Einstein
Re: [OGC1] Online Geometry Camp: Day 4
via,ami latex use korte partesina. tai details e kivabe likhbo bujhtesina.
Re: [OGC1] Online Geometry Camp: Day 4
3rd problem tar sltn
Let AD and BE intersect at G, which is the centroid of the triangle. We know that AG=2GD and BG=2GE.
Applying the Pythagorean theorem on triangles BGD and AGE, we get that
4GD2+GE2=(AC2)2, 4GE2+GD2=(BC2)2.
Hence, AB2=4GD2+4GE2=45[(AC2)2+(BC2)2]=10201. Thus, AB=101.
Let AD and BE intersect at G, which is the centroid of the triangle. We know that AG=2GD and BG=2GE.
Applying the Pythagorean theorem on triangles BGD and AGE, we get that
4GD2+GE2=(AC2)2, 4GE2+GD2=(BC2)2.
Hence, AB2=4GD2+4GE2=45[(AC2)2+(BC2)2]=10201. Thus, AB=101.

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Re: [OGC1] Online Geometry Camp: Day 4
খাতায় লিখে তারপর ছবি তুলে attach করে দাও।nowshin wrote:via,ami latex use korte partesina. tai details e kivabe likhbo bujhtesina.
"Your present circumstances don't determine where you can go; they merely determine where you start." Nido Qubein
Re: [OGC1] Online Geometry Camp: Day 4
solution of problem 3:
let the intersecting point of AD and BE is G. that means G is the centroid of $\Delta ABC$.
so,AG=2GD and BG=2GE;
now , $AG^{2}+GE^{2}=AE^{2}$
or, $AG^{2}+GE^{2}=(\frac{AC}{2})^{2}$
or, $(2GD)^{2}+GE^{2}=(69.5)^{2}$
or, $4GD^{2}+GE^{2}=4830.25$.....1st result
and $BG^{2}+GD^{2}=(\frac{BC}{2})^{2}$
or, $BG^{2}+GD^{2}=89^{2}$
or, $(2GE)^{2}+GD^{2}=89^{2}$
or, $4GE^{2}+GD^{2}=7921$.....2nd result
1st result+2nd result=$4GD^{2}+GE^{2}+4GE^{2}+GD^{2}=4830.25+7921$
or, $5GD^{2}+5GE^{2}=12751.25$
or, $GD^{2}+GE^{2}=2550.25$
or, $DE^{2}=2550.25$
or, $DE=50.5$
or, $2*DE=2*50.5$
or, $BC=101$
let the intersecting point of AD and BE is G. that means G is the centroid of $\Delta ABC$.
so,AG=2GD and BG=2GE;
now , $AG^{2}+GE^{2}=AE^{2}$
or, $AG^{2}+GE^{2}=(\frac{AC}{2})^{2}$
or, $(2GD)^{2}+GE^{2}=(69.5)^{2}$
or, $4GD^{2}+GE^{2}=4830.25$.....1st result
and $BG^{2}+GD^{2}=(\frac{BC}{2})^{2}$
or, $BG^{2}+GD^{2}=89^{2}$
or, $(2GE)^{2}+GD^{2}=89^{2}$
or, $4GE^{2}+GD^{2}=7921$.....2nd result
1st result+2nd result=$4GD^{2}+GE^{2}+4GE^{2}+GD^{2}=4830.25+7921$
or, $5GD^{2}+5GE^{2}=12751.25$
or, $GD^{2}+GE^{2}=2550.25$
or, $DE^{2}=2550.25$
or, $DE=50.5$
or, $2*DE=2*50.5$
or, $BC=101$
Re: [OGC1] Online Geometry Camp: Day 4
@ Tahmid: perfect!
"Be boring. It's the only way get work done."
 Austin Kleon
 Austin Kleon

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 Joined: Sun Aug 25, 2013 10:52 pm
 Location: dinajpur,dhaka,bangladesh
Re: [OGC1] Online Geometry Camp: Day 4
the area of large circle is 4 pi.and let the intersectors of the 4 small circles be P.then the area of 4 small circles area 4pi4p.and let the upper shadow be R.then it is 4 pi(4 pi4p)=4r.so, it is p=r.so,a=0
 Samiun Fateeha Ira
 Posts: 23
 Joined: Sat Aug 24, 2013 7:08 pm
 Location: Dhaka, Bangladesh
Re: [OGC1] Online Geometry Camp: Day 4
problem 01 :
BE.BD= BF.BG
or, 6*(39+6)=BF(BF+21)
or, BF^2+21BF270=0
so, BF=9
so, CG=48219=18
suppose, AI=x, HI=y, CH=z
Now, i have 3 equations:
x(x+y)=126, z(y+z)=702, x+y+z=48
but, i am stucked here. i am getting a lot of solution sets. what should i do & where is my mistake?
BE.BD= BF.BG
or, 6*(39+6)=BF(BF+21)
or, BF^2+21BF270=0
so, BF=9
so, CG=48219=18
suppose, AI=x, HI=y, CH=z
Now, i have 3 equations:
x(x+y)=126, z(y+z)=702, x+y+z=48
but, i am stucked here. i am getting a lot of solution sets. what should i do & where is my mistake?
Re: [OGC1] Online Geometry Camp: Day 4
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?
Everything you did is correct. And you shouldn't get multiple solutions. Would you like a hint?
"Be boring. It's the only way get work done."
 Austin Kleon
 Austin Kleon
Re: [OGC1] Online Geometry Camp: Day 4
@ Samiun Fateeha Ira , apply power of a point any other way.
By the way , did it occur to you that $AB=48$ ?
By the way , did it occur to you that $AB=48$ ?
Try not to become a man of success but rather to become a man of value.Albert Einstein