Solved Examples on Circles Connected with Triangle

Illustration 1: A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circum circle of the triangle PQR at the point T. If S is not the center of the circum circle then which of the following relation is true? (2008)

1. 1/PS + 1/ST < 2/√QS.SR                              2. 1/PS + 1/ST > 2/√QS.SR

3. 1/PS + 1/ST < 4/QR                                       4. 1/PS + 1/ST > 4/QR

Solution: Let us suppose that a straight line through the vertex P of a given ?PQR intersects the side QR at the point S and the circum circle of the triangle PQR at the point T.

Now, the points P, Q, R and T are concyclic and hence PS.ST = QS.SR

Now we know that A.M > G.M

So, we have, (PS + ST)/ 2 > √PS.ST

1/ PS + 1/ ST > 2/ √PS.ST = 2/ √QS.SR

Also, (SQ + QR)/ 2 > √QS.SR

So, QR/2 > √QS.SR

1/√QS.SR > 2/QR

Hence, 2/ √QS.SR > 4/ QR

So, we obtain 1/ PS + 1/ ST > 2/ √QS.SR > 4/ QR.

Illustration 2: If in a triangle ABC,

(2 cos A)/a + (cos B)/b + (2 cos C)/c = a/bc + b/ca.

Find the value of angle A. (1993)

Solution: The given condition is

(2 cos A)/a + (cos B)/b + (2 cos C)/c = a/bc + b/ca.     ….. (1)

We know that in ?ABC, by cosine rule we have

cos A = b² + c² - a² /2bc 

and so, cos B = c² + a² - b² /2ac

and cos C = a² + b² - c² /2ab 

We now substitute these values in equation (1) and hence obtain,

[2 (b² + c² - a²)/ 2abc] + [(c² + a² - b²) /2abc] + [2(a² + b² - c²)/ 2abc] = a/bc + b/ca

This gives [2(b² + c² - a²) + (c² + a² - b²) + 2(a² + b² - c²)] / abc

= (a² + b²)/abc 

This means 3b² + c² + a² = 2a² + 2b²

Hence, b² + c² = a².

Hence we get the measure of angle A is 90°.

Illustration 3: I_n is the area of n-sided regular polygon inscribed in a circle of unit radius and O_n be the area of the polygon circumscribing the given circle, then prove that

I_n = O_n/2 (1 + √{1 – (2I_n/n)²}) (2003)

Solution: The given circle is of unit radius. An n-sided polygon of area I_n is inscribed inside a circle and another polygon of area O_n is circumscribed around the given circle.

We know that I_n = nr²/2 .sin 2π/n  

Hence, since r = 1, so 2I_n / n = sin 2π/n    ….. (1) 

Also, O_n = nr² tan π/2  

Hence, O_n/ n = tan π/n  

Therefore 2I_n / O_n =   (sin 2π/n) / (tan π/n)

Hence, I_n / O_n = cos²π/n  

                       = (1 + cos 2π/n)/2

Therefore, from equation (1) we obtain,

I_n / O_n = {1 + √[1 – (2I_n/ n)²]}/ 2

Hence, I_n = (O_n/ 2). (1 + √[1 – (2I_n/ n)²]

Hence proved.