[OGC1] Online Geometry Camp: Day 4

Discussion on Bangladesh National Math Camp
photon
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by photon » Wed Aug 28, 2013 5:22 pm

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 .
Try not to become a man of success but rather to become a man of value.-Albert Einstein

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

Unread post by nowshin » Wed Aug 28, 2013 5:42 pm

via,ami latex use korte partesina. tai details e kivabe likhbo bujhtesina.

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

Unread post by nowshin » Wed Aug 28, 2013 5:43 pm

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.

Tusher Chakraborty
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Tusher Chakraborty » Wed Aug 28, 2013 6:01 pm

nowshin wrote:via,ami latex use korte partesina. tai details e kivabe likhbo bujhtesina.
খাতায় লিখে তারপর ছবি তুলে attach করে দাও।
"Your present circumstances don't determine where you can go; they merely determine where you start." -Nido Qubein

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

Unread post by Tahmid » Wed Aug 28, 2013 7:04 pm

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, $BC=101$

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

Unread post by Mursalin » Wed Aug 28, 2013 7:11 pm

@ Tahmid: perfect!
"Be boring. It's the only way get work done."

- Austin Kleon

Ayantika rinti bose
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Ayantika rinti bose » Wed Aug 28, 2013 9:31 pm

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 4pi-4p.and let the upper shadow be R.then it is 4 pi-(4 pi-4p)=4r.so, it is p=r.so,a=0

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Samiun Fateeha Ira
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Re: [OGC1] Online Geometry Camp: Day 4

Unread post by Samiun Fateeha Ira » Wed Aug 28, 2013 9:37 pm

problem 01 :

BE.BD= BF.BG
or, 6*(39+6)=BF(BF+21)
or, BF^2+21BF-270=0
so, BF=9
so, CG=48-21-9=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?

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

Unread post by Mursalin » Wed Aug 28, 2013 9:54 pm

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?
"Be boring. It's the only way get work done."

- Austin Kleon

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

Unread post by photon » Wed Aug 28, 2013 9:57 pm

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

By the way , did it occur to you that $AB=48$ ? :idea:
Try not to become a man of success but rather to become a man of value.-Albert Einstein

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