Spin the Segment
Circles Γ 1 and Γ 2 have radii of 106 and 200, respectively, and they intersect at two distinct points, A and B . A line passing through A intersects Γ 1 at P and Γ 2 at Q . If A B = 1 1 2 , find the maximum possible length of P Q .
The answer is 564.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Why will need to prove that max length of PQ is when PQ is perpendicular to AB.
Log in to reply
Let P be the point so that PQ is perpendicular to AB. By taking two points
P
1
and
P
2
, above and below P, we can prove that
P
1
Q
1
or
P
2
Q
2
is less than PQ.
Shortest distance would be when Q is on A and PQ is the diameter of the smaller circle. Hence the feeling is that PQ perpendicular to AB is the longest!!. Waiting for further explanations.
x
=
2
2
0
0
2
+
(
2
1
1
2
)
2
+
2
1
0
6
2
+
(
2
1
1
2
)
2
x
=
5
6
4
.
5
6
4
Just the same....... I have used ...\dfrac... and... \sqrt{...}... in Latex.
Max distance will be perpendicular to A B
x = 2 2 0 0 2 + ( 2 1 1 2 ) 2 + 2 1 0 6 2 + ( 2 1 1 2 ) 2
x = 5 6 4