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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.
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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