Stopping on a dime ...
Let R be the fraction of the area of a regular octagon of side length 2 that lies within a distance of 1 of at least one of its vertices.
If R can be expressed as b a π ( c − d ) , where a , b , c , d are positive integers with a and b being coprime and c square-free, then find a + b + c + d .
The answer is 14.
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
Oops! I didn't consider the phrase 'at least' and found fraction for just 1 vertice .
So , my fraction came = 2 2 4 3 3 × ( 2 − 1 )
Sorry I forgot about the factor π when I first posted the problem. I hope that didn't cause anyone to waste one (or more) of their attempts. :(
Log in to reply
So , it becomes = 6 4 3 × π × ( 2 − 1 )
By the way nice problem!
Log in to reply
O.k., which multiplied by 8 gives the answer I have. So I guess that you used the approximation π = 7 2 2 before I added the factor π into the expression. Sorry about the mistake; I wish I could give you credit for answering the question since you clearly knew what you were doing.
Log in to reply
@Brian Charlesworth – Yeah! Its Ok , no problem. Its good that you realized it early .
Hi sir, I have a bit of problem in understanding Double Summations ? Can you please help me out in understanding them, It would be great if you could cite some sources from where I could learn them .
I have my JEE Exam coming up and I have seen some questions containing these . Wherever I search for, I only get double integrals, so I am at a loss .
Thanks for the same !!
Log in to reply
I'm at a bit of a loss, too. There don't seem to be any good sites out there dealing expansively with double summations/series. This , however, gives a good general description, and it might be all you need. Good luck on the JEE Exam. :)
Log in to reply
@Brian Charlesworth – Thanks a lot sir !!!
The octagon is composed of 8 identical isosceles triangles with a base length 2 and an apex angle of 4 5 ∘ . The area of the octagon is then
8 ∗ cot ( 2 2 . 5 ∘ ) = 8 ∗ ( 2 + 1 ) .
Now the region of the octagon that lies within a distance of 1 of at least one vertex of the octagon is composed of 8 non-overlapping, radius 1 circular sectors with central angles of 1 3 5 ∘ each. Now since 8 ∗ 1 3 5 = 3 ∗ 3 6 0 , the combined areas of these sectors is the same as that of 3 unit circles.
Thus R = 8 ( 2 + 1 ) 3 π = 8 3 π ( 2 − 1 ) , and so
a + b + c + d = 3 + 8 + 2 + 1 = 1 4 .