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Triangle $\Delta ABC$ is inscribed in a circle of radius one unit. If the internal angle bisectors of angles $\angle A, \angle B,\angle C$ meets the circle at the points $A_1,B_1,C_1$ respectively. Find value of $$S=\frac{\sum AA_1 \cos\left(\frac{A}{2}\right)}{\sum \sin A}$$ enter image description here

My try:

Letting $BC=a, AB=c, AC=b$

We have $$AD=\frac{2bc}{b+c}\cos\left(\frac{A}{2}\right)$$

Hence $$AA_1=AD+DA_1=\frac{2bc}{b+c}\cos\left(\frac{A}{2}\right)+DA_1$$

But how to find $DA_1$?

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1 Answer 1

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Note that $A_1,$ $B_1,$ $C_1$ are the midpoints of $\overset{\huge\frown}{BC},$ $\overset{\huge\frown}{CA},$ $\overset{\huge\frown}{AB}$ respectively.

Therefore, considering $\triangle A_1BC,$ $A_1B=A_1C$. Now we can use law of cosines in $\triangle AA_1B$ and $\triangle AA_1C$ to get,

$$(A_1B)^2=(AA_1)^2+c^2-2\cdot AA_1\cdot c\cos\frac A2$$ $$(A_1C)^2=(AA_1)^2+b^2-2\cdot AA_1\cdot b\cos\frac A2$$

Using these two, we can easily get $$AA_1\cos\left(\frac A2\right)=\frac{(b+c)}2$$

Also from law of sines we have $$\frac{\sin A}a=\frac{\sin B}b=\frac{\sin C}c=2R$$ where $R$ is the radius of the circumcircle.

Hence, $$\frac{\sum AA_1\cos\left(\frac A2\right)}{\sum\sin A}\equiv R\cdot\frac{\sum(b+c)}{\sum a}$$

Can you take it from here?

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