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Q.

Circles C1 and C2 are externally tangent and they both are internally tangent to the circle C3. The radii of C1 and C2 are 4 and 10 respectively, and the centres of the three circles are collinear. A chord of C3 is also a common internal tangent of C1 and C2. Given that the length of the chord is mnp, where, m,n and p are positive integers, m and p are relatively prime and n is not divisible by the square of any prime, find the value of m+n+p.

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a

405

b

504

c

205

d

502 

answer is A.

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Detailed Solution

Question ImageLet the centres’ of the three triangles be O, O1 and O2 respectively and there radius be
OP(r1)=4
O1P(r2)=10
O2P(r3)=r1+r2
r3=4+10
r3=14
 Construction: AB is the tangent touching both the circles C1 and C2 and the line joining all the centres of the circles meet at an extension out of the circle C3 at X.
 Now, we can see that as all the three triangles are on the same plane, therefore,
 According to the similar right triangles theorem, which says, “if an altitude is drawn from the hypotenuse to the right-angled triangle, then the two triangles formed will be similar to each other.
 In the diagram, OT, O1T1 and O2T2 are the altitudes drawn to the triangles ΔXOT, ΔXO1T1 and ΔXO2T2 respectively.
 Hence,
ΔXOTΔXO1T1ΔXO2T2
 Now, equating the lines of the similar triangles,
XOOT=XO1O1T1=XO2O2T2
XP+OP4=XP+O1PO1T1=XP+O2P10
XP+44=XP+14O1T1=XP+1810
4XP+14XP+4O1T1=XP+1810
4XP+56XP+4O1T1=XP+1810
After solving, we get,
XPO1T1=163587
Therefore,
XP=163
O1T1=587
 Now, in ΔBO1T1,
By Pythagoras Theorem,
AB=2AT1
AB=2r32-O1T12
AB=2142-5872
AB=2196-336449
AB=29604-336449
AB=2624049
AB=24×4×3907×7
AB=2×47390
AB=87390
 Now, comparing with mnp, we get,
m=8
n=390
p=7
 Hence,
m+n+p=8+390+7=405
 
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