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```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: In is the area of n-sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, then prove that

In = On/2 (1 + √{1 – (2In/n)2}) (2003)

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

We know that In = nr2/2 .sin 2π/n

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

Also, On = nr2 tan π/2

Hence, On/ n = tan π/n

Therefore 2In / On =   (sin 2π/n) / (tan π/n)

Hence, In / On = cos2π/n

= (1 + cos 2π/n)/2

Therefore, from equation (1) we obtain,

In / On = {1 + √[1 – (2In/ n)2]}/ 2

Hence, In = (On/ 2). (1 + √[1 – (2In/ n)2]

Hence proved.

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