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|dw:1439559062484:dw|
not all of them will pass through the center though...
Notice that the center of circle is the "circumcenter" of the triangle we're working on

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so basically the problem is equivalent to finding the triangles such that the circumcenter lies interior to the triangle
oh, yes
for what triangles do we have circumcenter interior to the triangle ?
acute triangles
yes, they would be all "acute" triangles so our job is to find the number of acute triangles and divide them by 84
we may use the relation between inscribed angle and central angle : \(\theta = \dfrac{\alpha}{2}\) |dw:1439560115696:dw|
not really sure how to approach this, im still thinking..
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.
Okay, have good sleep :) I'll try to post the solution over the night, pretty sure this is not that hard...
thanks so much! i really appreciate it!
|dw:1439562522979:dw|
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.
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.
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|dw:1439564259300:dw|
Ok, part 1) is done, now just calculate the possibility of it.
for part a, im getting \(\dfrac{30}{84}\) for partb, the general formula for probability is \(\dfrac{n+1}{2(2n-1)}\)
can you tell me how, i need to write an explanation with my answer...

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