Incircle Riddle
A triangle A B C has an incircle O , and D , E , F are the touching points on the triangle tangents, as shown above.
The 3 side lengths of A B C are pairwise coprime integers, where points D , E , F each divide those sides into 2 smaller integer lengths as well. Moreover, the triangle's perimeter can be written as both the sum and the difference of 2 perfect squares.
If the incircle's radius is 3, what is the least possible area of the triangle A B C ?
The answer is 60.
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1 solution
I had posted an alternating solution . But it is edited out !!!!!
So I am posting it here.
Let us see if ABC is a right triangle at B, with inradius r = 3.
At B, the two tangents and corresponding radii at points of tangency
form a 3x3 square.
So the side have to be >3.
Let us see a few pythagorean triples, r=3 and sides >3.
r
=
1
/
2
(
a
+
b
+
c
)
1
/
2
a
b
=
a
+
b
+
c
a
b
.
(
5
,
1
2
,
1
3
)
.
.
.
.
.
.
r
=
5
+
1
2
+
1
3
5
∗
1
2
=
2
.
(
7
,
2
4
,
2
5
)
.
.
.
.
.
.
r
=
7
+
2
4
+
2
5
7
∗
2
4
=
3
.
.
.
.
.
.
.
.
B
u
t
7
+
2
4
+
2
5
=
5
6
i
s
n
o
t
a
s
u
m
a
n
d
a
d
i
f
f
e
r
e
n
c
e
o
f
t
w
o
s
q
u
a
r
e
s
.
(
8
,
1
5
,
1
7
)
.
.
.
.
.
.
r
=
8
+
1
5
+
1
7
4
∗
1
5
=
3
.
.
.
.
.
.
.
.
.
8
+
1
5
+
1
7
=
4
0
=
6
2
+
2
2
=
1
1
2
−
9
2
.
A
r
e
a
Δ
A
B
C
=
2
1
∗
8
∗
1
5
=
6
0
.
The angles angle bisectors make with the sides are A/2, B/2, C/2.
Also A/2+B/2+C/2=90.
So CotA/2=3/k, CotB/2=3/m, CotC/2=3/n, where k, m, n are tangent lengths from respective vertices.
But CotP+CotQ+CotR=CotP * CotQ * CotR when P+Q+R=90.
So 3/k+3/m+3/n=3/k * 3/m * 3/n.
⟹ 9 = k + m + n k ∗ m ∗ n .
Solving (k,m,n)=(3, 5,12).
So the sides are (8, 15, 17), a pythagorean triple.
So area =1/2 * 8 * 15= 6 0 .
8+15+17=40=6 * 6 + 2 * 2=11 * 11 - 9 * 9.