klimenkov
  • klimenkov
Two points on the plane \(A\) and \(B\) are given. \(|AB| = 2\). \(C\) is a randomly picked point in the circle of the radius \(R\) with the center in the midpoint of \(AB\). What is the probability that the \(\triangle ABC\) has an obtuse angle?
Mathematics
katieb
  • katieb
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nipunmalhotra93
  • nipunmalhotra93
The angle will be obtuse if the point lies inside the smaller circle. So the ratio of the areas should be the answer. Sounds right?
nipunmalhotra93
  • nipunmalhotra93
no wait
klimenkov
  • klimenkov
|dw:1401618750847:dw|

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Miracrown
  • Miracrown
points A and B are fixed and they are 2 units apart |dw:1401618811902:dw| the center of our circle is the midpoint of segment AB
Miracrown
  • Miracrown
|dw:1401618885857:dw|
nipunmalhotra93
  • nipunmalhotra93
is the answer|dw:1401618840569:dw|
nipunmalhotra93
  • nipunmalhotra93
the answer is ratio of the area of the shaded region to total area.
Miracrown
  • Miracrown
we don't know the actual numerical radius of this circle , so we use R it could be R < 1 , in which case A , B would be outside the circle it could be R > 1 , in which case A , B are inside the circle or with R = 1 , A and Bb are on the circle
Miracrown
  • Miracrown
not exactly sure yet Let's try the case with R = 1 , so that A and B are on the circle . |dw:1401619069277:dw| C must be a point "in" the circle, so I interpret that to be in the interior of the circle.
Miracrown
  • Miracrown
|dw:1401619212117:dw|
nipunmalhotra93
  • nipunmalhotra93
@bswan this is obtuse |dw:1401619197813:dw|
Miracrown
  • Miracrown
There certainly "appears" to be an obtuse angle at C .
Miracrown
  • Miracrown
|dw:1401619275225:dw|
nipunmalhotra93
  • nipunmalhotra93
@bswan |dw:1401619255446:dw| and this is obtuse. And hence the answer I previously proposed.
nipunmalhotra93
  • nipunmalhotra93
It should be noted that the question is asking for an obtuse angle in the triangle. So the angles at A and B could be obtuse too. Which is possible only if C lies in the following shaded region: |dw:1401619378172:dw|
anonymous
  • anonymous
|dw:1401619217208:dw| so I would say the probability is the area of double shaded area vs total area of bigger circle
nipunmalhotra93
  • nipunmalhotra93
@myko we also need to include the area of the smaller circle.
anonymous
  • anonymous
it is included
anonymous
  • anonymous
bigger circle includes the small one
nipunmalhotra93
  • nipunmalhotra93
I meant that the double shaded area must cover the smaller circle too.
anonymous
  • anonymous
no
Miracrown
  • Miracrown
so, it looks like it need to be qualified with the condition that R > 1 .
anonymous
  • anonymous
yes
nipunmalhotra93
  • nipunmalhotra93
@myko and @Miracrown |dw:1401619632107:dw| is angle C not obtuse here?
anonymous
  • anonymous
nowyou need to calculate what is the angle in black and later integrate from negative to positive of this angle to find the area|dw:1401619688850:dw|
anonymous
  • anonymous
△ABC means the angle at vertex B @nipunmalhotra93
Miracrown
  • Miracrown
C "is" obtuse for that diagram, yes
nipunmalhotra93
  • nipunmalhotra93
@myko dude... that means Triangle ABC.
anonymous
  • anonymous
oh, never mind. You right
nipunmalhotra93
  • nipunmalhotra93
|dw:1401619854190:dw|
anonymous
  • anonymous
read it bad
nipunmalhotra93
  • nipunmalhotra93
np... that happens... :)
Miracrown
  • Miracrown
If you choose C to be in the black-shaded region , is where it appears that we will "not" get an obtuse angle for C . |dw:1401619882929:dw|
anonymous
  • anonymous
so then R must include the condition\(AC^2+CB^2>4\) from one side and what I said befor from the other.
anonymous
  • anonymous
<4 correction
Miracrown
  • Miracrown
I think you can just say that R > 1. If R <= 1 , the prob (obtuse angle) = 1 .
Miracrown
  • Miracrown
Very interesting problem, that was.

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