Awesome geometry -5
In
△
A
B
C
,
A
C
=
1
3
,
B
C
=
1
4
,
A
B
=
1
5
.
If points J A , J B , J C , are the ex-centers and point I is the in-center of △ A B C . Then find area of △ J A J B J C
Details and Assumptions :-
∙ The asked area is of triangle having vertices J A , J B , J C .
The answer is 341.25.
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2 solutions
Exactly the same!
L
e
t
M
b
e
t
h
e
m
i
d
−
p
o
i
n
t
o
f
B
C
.
T
r
i
a
n
g
l
e
A
B
C
i
s
m
a
d
e
u
p
o
f
t
w
o
t
r
i
a
n
g
l
e
s
C
A
M
=
1
3
.
1
2
.
5
a
n
d
B
A
M
=
1
5
.
1
2
.
9
.
∴
a
r
e
a
A
B
C
,
Δ
=
1
/
2
∗
1
4
∗
1
2
=
8
4
.
2
s
=
a
+
b
+
c
=
4
2
.
A
r
e
a
t
r
i
a
n
g
l
e
J
A
J
B
J
C
,
Δ
J
=
2
∗
r
2
∗
s
a
b
c
∗
Δ
=
(
2
∗
r
)
∗
(
r
∗
s
)
a
b
c
∗
Δ
=
2
∗
r
a
b
c
=
2
∗
2
∗
2
∗
s
Δ
a
b
c
=
4
∗
8
4
1
4
∗
1
3
∗
1
5
∗
4
2
=
3
4
1
.
2
5
.
Sorry I missed the link where I got the formula. I will post it as soon as I find it.
If S is the area of △ABC and S(J) is the area of △J(A)J(B)J(C), the nice equality holds: r·S(J) =2R·S . Here r,R represent the radius of the in-circle and the circumcircle of △ABC respectively. And then we can write S(J) as an algebraic expression of a,b,c (which are the sides of △ABC) easily.
In the given image I have plotted the triangle on the Cartesian plane and found out the coordinates of the ex-centers.
Here the formula for ex-centers in a △ A B C with A ( x 1 , y 1 ) , B ( x 2 , y 2 ) and C ( x 3 , y 3 ) & a = B C , b = C A and c = A B ,
J A = ( − a + b + c − a x 1 + b x 2 + c x 3 , − a + b + c − a y 1 + b y 2 + c y 3 )
Now we could have figured out the are by Area formula.