Dhaka Higher Secondary 2011/6

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2011/6

Unread post by BdMO » Fri Jan 28, 2011 10:29 pm

One circle is touching another circle internally. The inner circle is also tangent to a diameter of the outer circle which makes an angle of $60 {\circ}$ with the common tangent of the circles. Radius of the outer circle is 6, what is the radius of the inner circle?

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TIUrmi
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Re: Dhaka Higher Secondary 2011/6

Unread post by TIUrmi » Sun Jan 30, 2011 1:06 pm

Let A be the intersecting point of the common tangent and the radius of large circle that is touching the small circle. B be the internal touching point of the two circles and E be the tangency point of the smaller circle with the larger radius. C is the centre of large circle and D is the centre of small circle.

$DE = DB$, $AE = AB$, So $\angle EAD = \angle DAB$
$\frac {AC}{AB} = \frac {CD}{DB}$ as AD is the angle bisector.
$\frac {AC}{AB} = sec 60 = \frac {CD}{DB} = 2$
So, $CD = 2. DB$
$CB = CD + DB$ = $3. DB$
$DB = 2$
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter

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