## BdMO National Higher Secondary 2007/8

BdMO
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### BdMO National Higher Secondary 2007/8

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

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

Tahmid Hasan
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### Re: BdMO National Higher Secondary 2007/8

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

photon
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### Re: BdMO National Higher Secondary 2007/8

'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

Zzzz
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### Re: BdMO National Higher Secondary 2007/8

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.

Cryptic.shohag
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### Re: BdMO National Higher Secondary 2007/8

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