TS EAMCET 03 May 2019 Shift 2 Solved Paper

Given that p1,p2 and p3 are the altitudes of the triangles from the edges A,B and C respectively.
Therefore, the area of the triangle is given by
∆=‌

1

2

ap1=‌

1

2

bp2=‌

1

2

cp3
and
r1=‌

∆

s−a

,r2=‌

∆

s−b

,r3=‌

∆

s−c

and
r=‌

∆

s

The required expression is given by.
‌

1

r12

+‌

1

r22

+‌

1

r32

+‌

1

r2

=‌

(s−a)2

∆2

+‌

(s−b)2

∆2

+‌

(s−c)2

∆2

+‌

(s)2

∆2

=‌

1

∆2

(4s2−2s(a+b+c)+(a2+b2+c2))
=‌

1

∆2

(4s2−4s2+a2+b2+c2)
=‌

a2+b2+c2

∆2

=‌

4∆2

∆2

(‌

1

p12

+‌

1

p22

+‌

1

p32

)
=4(‌

1

p12

+‌

1

p22

+‌

1

p32

)