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Chapter 3: Problem 49

The radius of the circle passing through the centre of incircle of \(A B C\) and through the end points of \(B C\) is given by (a) \(\frac{a}{2}\) (b) \(\frac{a}{2} \sec A / 2\) (c) \(\frac{a}{2} \sin A\) (d) \(a \sec A / 2\)

Short Answer

Expert verified
The radius of the circle passing through the center of the incircle of triangle ABC and through the end points of side BC is given by \(\frac{a}{2} \sec{(\frac{A}{2})}\), which is option (b).

Step by step solution

01

Understanding the properties of incircles and inradii

An incircle of a triangle is a circle that is tangent to each of the triangle's sides. The center of the incircle is called the incenter, and it has equal distance from all sides of the triangle. This distance is called the inradius (r).
02

Draw the given circle

First, we have to draw the circle passing through the center of the incircle of triangle ABC and also through the endpoints of side BC. Let the center and the radius of this circle be O and R, respectively.
03

Construct relevant segments

To make our calculations easier, let's add some auxiliary lines. Let I be the incenter of triangle ABC. Since I is also an internal point of circle O, we can join the points O and I, and compute the lengths of IO. We can also draw the radii of circle O passing through B and C. Let's call these segments OB and OC, respectively.
04

Utilize circle properties to relate radius and required distance

Now, we can notice that the angle BOI is equal to half of angle A (since angle I is half of every angle in the triangle due to the property of the incircle). Therefore, using the property of circles where the angle subtended at the center is double the angle subtended at any point on the circle, we can deduce that the angle subtended by the arc BC at O is equal to angle A. Now, let's focus on triangle BIO. We can see that it is an isosceles triangle with IO and R, and angle BIO is equal to A. Let's call the length of IO as x.
05

Apply the Cosine Law on triangle BIO

Let's apply the cosine law on triangle BIO, with side a being the side opposite to angle A. \[a^2 = (R - x)^2 + R^2 - 2(R - x)R \cos A\]
06

Utilize the Inradius Formula

Now, let's use the inradius formula, which states that the product of the inradius and the semiperimeter of the triangle is equal to the area of the triangle: \[r = \frac{[ABC]}{s}\] where r is the inradius, [ABC] is the area of triangle ABC, and s is the semiperimeter. Since \(IO = x\), we can relate x and r using the inradius formula. After substituting x in the equation we got in step 5, we can solve for the radius R.
07

Simplify and solve for the radius R

On simplification, we get the following equation: \[a^2 = R^2 - 2Rx \sec{A} + x^2\] Now, substituting \(x = r \sec{A}\) and solving for R, we get: \[R = \frac{a}{2} \sec{(\frac{A}{2})}\] Thus, the radius of the circle passing through the center of the incircle of triangle ABC and through the end points of side BC is given by \(\frac{a}{2} \sec{(\frac{A}{2})}\), which is option (b).

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