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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
In this puzzle we have to find radius of blue circle which is drawn exterior of 2 circles with radius 24 and 8 and inside semicircle. Centers of two circles are end points of diameter of semicircle.
Lets make this puzzle little bit easier by removing unnecessary parts from this figure.
We don't need full circles having radius 24 and 8.
Take a look at following figure:
⇒ Diameter of quarter circle is 28+8 = 32
⇒ Radius = 16
Now draw lines joining centers of two circles having radius 24 and 8 and small circle.
Let radius of small circle is r.
Now draw line passing through center of semicircle and center of small circle.
Circle with center A is touching to circle with center C externally
⇒ AC = 24+r
Circle with center B is touching to circle with center C externally
⇒ BC = 8+r
Circle with center O is touching to circle with center C internally
⇒ OC = 16-r
Now we can observe, angle forming on either side of OC at point O are complementary
Let Angle BOC = x
⇒ Angle AOC = 180-x
Now consider Triangle BOC and apply Cosine rule at angle x,
\[cos(x)=\frac{16^2+(16-r)^2-(8+r)^2}{2\times16\times(16-r)}\]
Now consider Triangle AOC and apply Cosine rule at angle (180-x),
\[cos(180-x)=\frac{16^2+(16-r)^2-(24+r)^2}{2\times16\times(16-r)}\]
\[-cos(x)=\frac{16^2+(16-r)^2-(24+r)^2}{2\times16\times(16-r)}\]
\[cos(x)=\frac{(24+r)^2-16^2-(16-r)^2}{2\times16\times(16-r)}\]
Now compare both equation,
cos(x) is same on left side of both equations,
⇒ Right hand sides are equal,
\[\frac{16^2+(16-r)^2-(8+r)^2}{2\times16\times(16-r)}=\frac{(24+r)^2-16^2-(16-r)^2}{2\times16\times(16-r)}\]
\[16^2+(16-r)^2-(8+r)^2=(24+r)^2-16^2-(16-r)^2\]
After some careful algebraic calculations, you will get:
128r = 284 -------(r^2 will be cancelled completely during first step)
⇒ r = 3
