Lightoj 1072 solution in English

Problem Link: Lightoj 1072

Click here to see the discussion in Bangla

In this problem, we are given:

• The radius of a large outer circle
  $R$

• The number of smaller circles
  $n$
   that fit inside the larger circle, touching it

We need to find the radius of each smaller circle 

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Step 1: Angle subdivision

We know a full circle is 

$360^\circ$

.

If there are 

$n$

small circles inside, they divide the large circle into 

$n$

equal sectors.

So, the angle at the center of the large circle subtended by one small circle is:

$$\frac{360^\circ}{n} = \frac{180^\circ}{n} = \frac{\pi}{n}$$

Step 2: Using trigonometry

Consider the triangle formed by:

• The center of the large circle

• The center of one small circle

• The point of tangency on the large circle

In this triangle:

• The opposite side = 
  $r$
  (radius of the small circle)

• The hypotenuse = 
  $R - r$

So:

$$\sin\theta = \frac{r}{R - r}, \quad \theta = \frac{\pi}{n}$$

                                   

Step 3: Solve for 

$r$

$$\sin\left(\frac{\pi}{n}\right) = \frac{r}{R - r}$$ $$r = \sin\left(\frac{\pi}{n}\right)(R - r)$$ $$r = R\sin\left(\frac{\pi}{n}\right) - r\sin\left(\frac{\pi}{n}\right)$$ $$r + r\sin\left(\frac{\pi}{n}\right) = R\sin\left(\frac{\pi}{n}\right)$$ $$r(1 + \sin(\tfrac{\pi}{n})) = R\sin\left(\tfrac{\pi}{n}\right)$$ $$\boxed{r = \frac{R \sin(\pi/n)}{1 + \sin(\pi/n)}}$$

Thus, the radius of each inner small circle is:

$$r = \frac{R \sin(\pi/n)}{1 + \sin(\pi/n)}$$

Would you like me to also provide a short Python function that computes 

$r$

for given 

$R$

and 

$n$

?

Source Code:

Another problem like this, 

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Source Code : SourceCodeLink