Question

A chord of a circle has length 3n, where n is a positive integer. The segment cut off by the chord has height n, as shown. What is the smallest value of n for which the radius of the circle ia also a positive integer?

Answer 1

Let radius of the circle is a positive integer e.g. r
Given, chord length ,PQ = 3n
so, PB = BQ = PQ/2 = 3n/2 = 1.5n
OB = OA - AB = r - n
OQ = radius of circle = r

now from Pythagoras theorem,
in ∆OBQ , OB² + BQ² = OQ²
now, (r - n)² + (1.5n)² = r²
r² + n² - 2rn + 2.25n² = r²
3.25n² - 2rn = 0
3.25n - 2r = 0
r = 3.25n/2 = 1.625n

because r is positive integer so, we have to choose a smallest positive integer of n.
let's choose n = 4
then, r = 1.625 × 4 = 6.5 ≠ integer
take n = 8 , r = 1.625 × 8 = 13 = a positive integer .

hence, smallest intger value of $\bf{n = 8}$ by which we get positive integer of r e.g., 13 unit.

Answer 2

Let the centre and radius of the circle be O and r

Then OA = OB = r and OM = r −n.

Since the radius perpendicular to a chord bisects the chord,

AM =  3n /2

Now triangle OAM is right-angled, so we can use Pythagoras’ Theorem to give :

r^2 = (3n/2)^2 + (r-n)^2

= 9n^2/4 + r^2 - 2rn + n^2

And Hence

8rn = 13n^2

Since n is positive and so non-zero, we can divide both sides by n to get

8r = 13n

so that

r = 13n/8

For both r and n to be integers, n needs to be divisible by 8. But n is positive, so the smallest  possible value of n is 8 (giving r = 13).