BdMO National Higher Secondary 2007/8

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BdMO
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BdMO National Higher Secondary 2007/8

Unread post by BdMO » Sun Feb 06, 2011 10:19 pm

Problem 8:
Two parallel chords of a circle have length $10$ and $14$. The distance between them is $6$. The chord parallel to these chords and half way between them has length $\sqrt a$. Find $a$.

photon
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Re: BdMO National Higher Secondary 2007/8

Unread post by photon » Sun Feb 06, 2011 11:00 pm

In common sense,a=144.
How much a chord goes near to the center parallel to its previous position,it is increased in both sides equally.
I'm trying to get a perfect geometric solution... :)
Last edited by photon on Mon Feb 07, 2011 9:00 am, edited 1 time in total.
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Tahmid Hasan
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Re: BdMO National Higher Secondary 2007/8

Unread post by Tahmid Hasan » Sun Feb 06, 2011 11:10 pm

my ans is 184 and it has a very tough geometric logic and many cases :ugeek:
বড় ভালবাসি তোমায়,মা

photon
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Re: BdMO National Higher Secondary 2007/8

Unread post by photon » Mon Feb 07, 2011 8:55 am

'Very tough geometric logic'-i don't understand that word..... :?
Try not to become a man of success but rather to become a man of value.-Albert Einstein

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Zzzz
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Re: BdMO National Higher Secondary 2007/8

Unread post by Zzzz » Mon Feb 07, 2011 10:40 am

photon wrote:...
How much a chord goes near to the center parallel to its previous position,it is increased in both sides equally.
...
Sorry, its wrong :| Its better if you try to find some logical solution. Sometimes some properties seem obvious to us, but don't believe those until you get a proof.
Every logical solution to a problem has its own beauty.
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Cryptic.shohag
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Re: BdMO National Higher Secondary 2007/8

Unread post by Cryptic.shohag » Thu Feb 10, 2011 2:29 am

Let O be the center of the circle and AL, BM, CN be the 3 chords such that AL=14, BM=\[\sqrt{a}\] and CN=10. Let OD, OE, OF be perpendiculars on AL, BM, CN respectively. Let OD=P, then OE=P+3 and OF=P+6[considering that they are all on the same side of the diameter].
As OD, OE, OF are perpendiculars on AL, BM, CN respectively from center O, so AD=7, BE=\[\frac{\sqrt{a}}{2}\] and CF=5. If we connect O with the points A, B, C we get OA=OB=OC=R(Radius of the circle).
Now using Pythagoras's Theorem we get,
\[OA^2=OD^2+AD^2\: \Rightarrow R^2=P^2+49\]....... (i)
\[OB^2=OE^2+BE^2\: \Rightarrow R^2=(P+3)^2+(\frac{\sqrt{a}}{2})^2\: \Rightarrow R^2=P^2+6P+9+\frac{a}{4}\]...... (ii)
\[OC^2=OF^2+CF^2\: \Rightarrow R^2=(P+6)^2+5^2 \: \Rightarrow R^2=P^2+12P+61\]....... (iii)
From (i) and (iii) we get,
\[P^2+49=P^2+12P+61\: \Rightarrow 12P=-12\: \Rightarrow P=-1\]
Here negative value of P means that AL is on the other side of the diameter than BM and CN.
From (i) and (ii) we get,
\[P^2+49=P^2+6P+9+\frac{a}{4}\: \Rightarrow \frac{a}{4}=40-6P\: \Rightarrow \frac{a}{4}=40-6\times ({-1})\: \Rightarrow \frac{a}{4}=46\: \Rightarrow a=184.......\]
And it's the answer......... :)
God does not care about our mathematical difficulties; He integrates empirically. ~Albert Einstein

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