Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

The lengths of the tangents from $A,B$ and $C$ to the incircle are in A.P., then?

• A. $r_1,r_2,r_3$ are in H . P.
• B. $r_1,r_2,r_3$ are in A . P.
• C. $a,b,c$ are in A . P.
• D. $\cos A=\frac{4c-3b}{2c}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $r_1,r_2,r_3$ are in H . P.
C $a,b,c$ are in A . P.
D $\cos A=\frac{4c-3b}{2c}$

such that $Z-y=(y-x)or(y-z)=(x-y)\to \left(1\right)s=\frac{a+b+c}{2}=\frac{2(x+y+z)}{2}=s=y+2y=3y\to \left(ii\right)a=y+z,b=x+z,c=x+y\to \left(iii\right)b-a(x-y),c-b=(y-z)$

From (1) $b-a=c-b2b=a+c$

a,b,c, are in A.P.

$r_1,r_2,r_3$ are the ex centre of triangle

we know $r_1=\frac{s(s-c)(s-b)}{△},r_2=\frac{S(s-a)(s-c)}{△},r_3=\frac{s(s-a)(s-b)}{△}$

from (1),(11),(111)

$r_1=\frac{\left(3y\right).[3y-(y+X\left)\right][3y-(x+y\left)\right]}{△}=\frac{\left(3y\right)[2y-X].\left[y\right]}{△}{r}_2=\frac{\left(3y\right)[3y-(y+z\left)\right][3y-(x+y\left)\right]}{△}=\frac{\left(3y\right)(2y-z)(2y-X)}{△}{r}_3=\frac{\left(3y\right)[3y-(y+z\left)\right][3y-(x+y\left)\right]}{△}=\frac{\left(3y\right)(2y-z)\left(y\right)}{△}\frac{1}{r_1}+\frac{1}{r_3}=\frac{△}{3y\left(y\right)}[\frac{1}{(24-X)}+\frac{1}{(24-Z)}]=\frac{△}{\left(3y^2\right)}\left[\frac{4y-(x+x)}{(24-X)(24-z)}\right]=△\frac{△(4y-2y)}{\left(3y\right)\left(y\right).(24-X)(24-z)}=\frac{2△y}{\left(3y\right)\left(y\right)(24-X)(24-z)}=\frac{z}{r_3}{r}_1,r_2,r_3\cos A=\frac{b^2+c^2-a^2}{2bc}=(X+Z{)}^2+(x+4{)}^2-(y+z{)}^{2}2(x+z\left)\right(c)=(x+z{)}^2+(x-z)(x+z)+2y\left)2.\right(x+z)C=(x+z{)}^2+(x-z)(x+z)+(X+Z)2.(x+z)C=(X+Z)\frac{(x+z)+(X-Z)}{2.(X+Z)C}=\frac{X-Z}{2C}==4x+4x+2Z-3\frac{(X+Z)}{2C}=\frac{4X+2\left(2y\right)-3b}{2C}=\frac{4(X+4)-3b}{2C}=\frac{4C-3b}{2C}=\frac{4(X+4)-3b}{2C}=\frac{4C-3b}{2C}$