A Parallelogram and 2 Circles
In Parallelogram ABCD,
A
B
=
1
3
,
B
C
=
1
1
, and
A
C
=
2
0
. Two congruent circles are inscribed in
△
A
C
D
and
△
A
C
B
as shown in the figure.
The shortest distance between the centers of the two inscribed circles can be expressed in the form a b . Where b is a square-free integer. Find the value of a + b .
- Found this somewhere.
The answer is 12.
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2 solutions
would you please explain the part 1 3 − ( 2 0 / 2 + x ) = 1 1 − ( 2 0 / 2 − x ) .
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I think it is a typo. 1 3 2 − ( 2 0 / 2 + x ) 2 = 1 1 2 − ( 2 0 / 2 − x ) 2
C o s 2 ϕ = 2 ∗ 1 1 ∗ 2 0 1 1 2 + 2 0 2 − 1 3 2 = 5 4 . C o s ϕ = 2 1 ∗ ( C o s 2 ϕ + 1 ) = 1 0 9 , S i n ϕ = 1 0 1 r = P e r i m e t e r 2 ∗ A r e a = 1 1 + 1 3 + 2 0 2 ∗ 4 1 ∗ 4 4 ∗ 2 2 ∗ 1 8 ∗ 4 = 3 . I F / A F = 3 / A F = C o s ϕ S i n ϕ , ∴ A F = 3 ∗ 3 = 9 .
p = ( A B + B C + A C ) / 2 = ( 1 3 + 1 1 + 2 0 ) / 2 = 2 2 .
r =congruent circle radius
r = ( ( p − A B ) ( p − B C ) ( p − A C ) / p ) 1 / 2 = ( ( 2 2 − 1 3 ) ( 2 2 − 1 1 ) ( 2 2 − 2 0 ) / 2 2 ) 1 / 2 = 3 .
d =The shortest distance between the centeres
x=distance between center of AC and the intersection point of perpendicular line from center of circle to AC which is also the radius of circle
( d / 2 ) 2 = x 2 + r 2 .
( d / 2 ) 2 = x 2 + 9 .
On the other hand :
1 3 − ( 2 0 / 2 + x ) = 1 1 − ( 2 0 / 2 − x ) .
x = 1 .
Therefore :
( d / 2 ) 2 = 1 + 9 = 1 0 .
d = 2 X ( 1 0 ) 1 / 2 .
a = 2 , b = 1 0 .
a + b = 2 + 1 0 = 1 2 .