anonymous
  • anonymous
***
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
|dw:1439559062484:dw|
anonymous
  • anonymous
not all of them will pass through the center though...
ganeshie8
  • ganeshie8
Notice that the center of circle is the "circumcenter" of the triangle we're working on

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
so basically the problem is equivalent to finding the triangles such that the circumcenter lies interior to the triangle
anonymous
  • anonymous
oh, yes
ganeshie8
  • ganeshie8
for what triangles do we have circumcenter interior to the triangle ?
anonymous
  • anonymous
acute triangles
ganeshie8
  • ganeshie8
yes, they would be all "acute" triangles so our job is to find the number of acute triangles and divide them by 84
ganeshie8
  • ganeshie8
we may use the relation between inscribed angle and central angle : \(\theta = \dfrac{\alpha}{2}\) |dw:1439560115696:dw|
ganeshie8
  • ganeshie8
not really sure how to approach this, im still thinking..
anonymous
  • anonymous
Thanks so much for helping me, sir. But I must go to sleep now, its getting late for me. I will think about this tomorrow.
ganeshie8
  • ganeshie8
Okay, have good sleep :) I'll try to post the solution over the night, pretty sure this is not that hard...
anonymous
  • anonymous
thanks so much! i really appreciate it!
Loser66
  • Loser66
|dw:1439562522979:dw|
Loser66
  • Loser66
Consider vertex A. From AB, we have 7 triangles, and among them just ABF has center inside of it. That is the probability to get the center inside of the triangle is 1/7 for vertex A Same for others Hence the total is 1/7^9 but we have to subtract the overlap parts. I meant \(\triangle ABC\) when consider node A will be overlap with \(\triangle BCA\) for node B.
Loser66
  • Loser66
Hence for node B, we have 1 triangle overlaps with node A for node C, we have 1triangles overlaps with node A \(\triangle ACD\), and 1 triangle overlaps with nod B \(\triangle CBA\) Same for other nodes and same argument, we have the logic 1st node --0 overlap 2nd node--1 overlap 3rd node---2 overlap ::::::::::::::::::::::::::::::: 9th node---8 overlap ------------------------ total 36 cases.
Loser66
  • Loser66
|dw:1439564084039:dw|
Loser66
  • Loser66
|dw:1439564259300:dw|
Loser66
  • Loser66
Ok, part 1) is done, now just calculate the possibility of it.
ganeshie8
  • ganeshie8
for part a, im getting \(\dfrac{30}{84}\) for partb, the general formula for probability is \(\dfrac{n+1}{2(2n-1)}\)
anonymous
  • anonymous
can you tell me how, i need to write an explanation with my answer...

Looking for something else?

Not the answer you are looking for? Search for more explanations.