Circles

Geometry Level 3

Two circles with radius 64cm and 16cm touch each other externally. A circle with radius $r$ touches these two circles externally and also the common tangent. If $r$ can be expressed as $\frac{a}{b}$ for coprime positive integers, what is $a+b$ ?

The answer is 73.

1 solution

Shivam K
Mar 16, 2016

basic geometry

$\frac { 1 }{ \sqrt { r } } =\frac { 1 }{ \sqrt { 64 } } +\frac { 1 }{ \sqrt { 16 } } \\ \\ \frac { 1 }{ \sqrt { r } } =\frac { 3 }{ 8 } \\ \\ r=\frac { 64 }{ 9 } \\ \\ a+b=73\\ \\$

How do you get the first line?

Anik Mandal - 4 years, 9 months ago

@anik mandal Lets generalize it Two circles with radii a and b touch each other externally Another circle with radii c touches both these circles as well as the common tangent It is only possible when the circles lies between the gap of common tangent and circles on we can say between the intersection of the two circles and the intersection of these two circles with common tangent Now Let us begin C(a),C(b),C(c) are the three centres of circles Join C(a),C(c) Its length is a+c Draw radii of the three circles such that they are perpendicular with c.t. Draw a line parallel to common tangent passing through C(c) intersecting radii of circle(a) at d.and circle(b) at e Now C(a),d,C(c) is a right angle triangle Therefore Lengh of dC(c) is = 2√ac. Similarly eC(c) is= 2√ab Therefore de is (2√ac + 2√ab) Now draw a line parallel to de passing through C(b) intersecting circle (a) at f Therefore C(a)f = a-b fC(b)=de C(a)C(b) =a+b These 3 points make a right angle triangle de=2√ab Now 2√ab=2√ac +2√bc Dividing both the sides by 2√abc We get 1/√c = 1/√a + 1/√b

Shivam K - 4 years, 5 months ago

Circles

The answer is 73.

1 solution

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