Dhaka Higher Secondary 2010/2 (Secondary 2010/9)

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Moon
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Re: Find the area??!!

Unread post by Moon » Mon Jan 03, 2011 9:24 am

OK...let me clarify...I did not say that you should not use ceva. I tried to say that after you solve a problem with a cool theorem, make sure you understand WHY the whole thing is true.
It is not bad to solve problems with those theorems, but if you don't understand why you work with $\phi(n)$ while trying to find the last digit, you'll face huge problem in future.

In fact this is true for me. I knew that euler totient function stuff when I was in class 8. However, I did not understand it fully, therefore I could not use it with full power. :)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

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Labib
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Re: Find the area??!!

Unread post by Labib » Mon Jan 03, 2011 10:27 pm

@ Moon vai, Why do we use Euler's Totient Theorem?? :)
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BdMO
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Dhaka Higher Secondary 2010/2

Unread post by BdMO » Tue Jan 18, 2011 2:03 pm

2010_2.png
2010_2.png (9.01 KiB) Viewed 2263 times
As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$.

the arrivals
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Re: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)

Unread post by the arrivals » Tue Feb 01, 2011 1:43 pm

yes moon perhaps stated the just statement
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN

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Re: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)

Unread post by Shifat » Tue Aug 16, 2011 4:15 am

Moon vaia I think I have a solution but not sure about it.. here it is shown. But first, let the point is O and the lines are AD, BE and CF.

From the figure we can see Triangle AOB and BOD have the same height. So AOB:BOD= AO:OD=2:1(same height means the area is propotionate to their bases), which indicates the point O is the centre of mass(vorkendro) , So we can find BO:OE=2:1 which means tringle AEO is 35 as well. And again triangle AOC and DOC have the same height and AO:OD=2:1... so triangle DOC should be (84+35)/2= 59.5..
Now we add them and the result is 283.5........(I am horribly confused about the solution but seems logical to me..) :| :?:

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Re: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)

Unread post by photon » Tue Aug 16, 2011 3:34 pm

Shifat wrote: which indicates the point O is the centre of mass(vorkendro) ,:
it is incorrect.these cevians aren't median,as area (AOF) doesn't equal to AREA (BOF)
Try not to become a man of success but rather to become a man of value.-Albert Einstein

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Re: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)

Unread post by Shifat » Sat Aug 20, 2011 1:10 am

Cevian mane ki?? yes I get my mistakes, 2:1 holei je median hote pare na eta mathai ashe nai... sorry

Shifat
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Re: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)

Unread post by Shifat » Sat Aug 20, 2011 4:33 am

By the way I am posting another solution, this time the result is different though... but more convincing to me.please check it if it is right or not:- The point and the lines are mentioned before..but we let the area of AOE is x and the area of COD is y.
Now we saw.. AO:OD=2:1, and also AF:FB=4:3...
Comparng AOC and COD we find:- Triangle AOC=2 Triangle COD So:- 2y=84+x..........(1)
again Comparing AFC and BFC we find:- 3(triangle AFC)=4(triangle BFC) So:- 3(84+40+x)=4(y+35+30)...(2)
Solving (1) and (2) we can find x=56 and y=70...
Now the result is 315. :?: :?:

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