Right triangle and a square
△
A
C
B
is right angled at
C
. Square
A
B
E
D
is drawn. If
A
C
=
2
0
and
C
B
=
2
1
, find the perimeter of
△
D
C
E
. If your answer can be expressed as
a
+
b
+
c
where
a
,
b
and
c
are positive co-prime integers and
a
and
c
are square free, find
a
+
b
+
c
.
The answer is 4232.
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2 solutions
Draw
C
G
perpendicular to
D
E
. Let point
F
be the intersection of
A
B
and
C
G
. By pythagorean theorem,
A
B
=
2
0
2
+
2
1
2
=
2
9
.
Since △ A F C ∼ △ A C B , we have : 2 0 C F = 2 9 2 1 or C F = 2 9 4 2 0 . Apply pythagorean theorem twice: A F = 2 0 2 − ( 2 9 4 2 0 ) 2 = 2 9 4 0 0 and B F = 2 1 2 − ( 2 9 4 2 0 ) 2 = 2 9 4 4 1 . Now, C G = C F + 2 9 = 2 9 1 2 6 1 .
The perimeter of △ D C E is
( 2 9 1 2 6 1 ) 2 + ( 2 9 4 0 0 ) 2 + 2 9 + ( 2 9 1 2 6 1 ) 2 + ( 2 9 4 4 1 ) 2 = 2 0 8 1 + 2 9 + 2 1 2 2 .
The desired answer is 2 0 8 1 + 2 9 + 2 1 2 2 = 4 2 3 2 .
If you take C as the origin then D is (20,41) while E is (41,21). Then the required perimeter = √(20²+41²)+√(41²+21²)+√(20²+21²) = √2081 + √2122 + 29. Hence, a + b + c = 4232