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Problem 363
Posted: Fri May 04, 2012 2:41 pm
by olopierpa
Hello,
when I try to access
Problem 363 (
View Problem), my browser (I.E. 9) says that there's a problem with the page (no pun intended!) and refuses to open it. Unfortunately, I cannot use another browser. What's that is special about this problem that may be causing this?
Thanks!
Re: Problem 363
Posted: Fri May 04, 2012 3:57 pm
by jaap
There is a java applet on that page. I think there may be some kind of server configuration problem because I cannot see the applet in Firefox on my current computer (it complains about a missing plugin, but cannot find what plugin it needs for mimetype=application/x-java-vm) even though I can normally see java applets just fine. If I try to download the jar file directly, it comes in as gobbledygook text, not as a binary file.
Anyway, here is the full problem text for you. Note that I haven't put in all the subscripts, so P0 means P
0 etc., but other than that it should be readable. The point O is the origin of course.
A cubic Bézier curve is defined by four points: P0, P1, P2 and P3.
The curve is constructed as follows:
On the segments P0P1, P1P2 and P2P3 the points Q0,Q1 and Q2 are drawn such that P0Q0/P0P1=P1Q1/P1P2=P2Q2/P2P3=t (t in [0,1]).
On the segments Q0Q1 and Q1Q2 the points R0 and R1 are drawn such that Q0R0/Q0Q1=Q1R1/Q1Q2=t for the same value of t.
On the segment R0R1 the point B is drawn such that R0B/R0R1=t for the same value of t.
The Bézier curve defined by the points P0, P1, P2, P3 is the locus of B as Q0 takes all possible positions on the segment P0P1. (Please note that for all points the value of t is the same.)
In the applet to the right you can drag the points P0, P1, P2 and P3 to see what the Bézier curve (green curve) defined by those points looks like. You can also drag the point Q0 along the segment P0P1.
From the construction it is clear that the Bézier curve will be tangent to the segments P0P1 in P0 and P2P3 in P3.
A cubic Bézier curve with P0=(1,0), P1=(1,v), P2=(v,1) and P3=(0,1) is used to approximate a quarter circle.
The value v>0 is chosen such that the area enclosed by the lines OP0, OP3 and the curve is equal to π/4 (the area of the quarter circle).
By how many percent does the length of the curve differ from the length of the quarter circle?
That is, if L is the length of the curve, calculate 100*(L-π/2)/(π/2).
Give your answer rounded to 10 digits behind the decimal point.
Re: Problem 363
Posted: Fri May 04, 2012 5:05 pm
by olopierpa
Thank you very much, jaap!
Re: Problem 363
Posted: Wed Nov 25, 2015 10:26 pm
by pj6444
Quick question about the question itself. What is meant by the lines OP0 and OP3? I don't quite understand where the O is coming from.
Thanks all
Re: Problem 363
Posted: Wed Nov 25, 2015 10:45 pm
by mpiotte
pj6444 wrote:Quick question about the question itself. What is meant by the lines OP0 and OP3? I don't quite understand where the O is coming from...
"O" is the center of the circle.
Re: Problem 363
Posted: Sun Aug 25, 2019 8:35 am
by RishadanPort
I attempted to click on the external website page listed on the page -- and no application appears.
I'm using Google Chrome -- It doesn't even prompt me for anything.
Re: Problem 363
Posted: Sun Aug 25, 2019 12:45 pm
by hk
Unfortunately the applet is in Java. Many browsers don't even tell you that they don't support Java any more.
At the time the problem was published in 2011every browser supported Java.
So alas, for you nothing to be viewed.
Re: Problem 363
Posted: Sun Aug 25, 2019 9:58 pm
by RishadanPort
Hm. I get 10 decimal places accurate...
my answer has L is less than Pi/2, which results in L - PI/2 as negative.
does the question want absolute value of that? or perhaps im just flat wrong.
**
EDIT **
I was flat out wrong and an idiot
Re: Problem 363
Posted: Sun Aug 25, 2019 10:05 pm
by RishadanPort
Can I get a confirmation that
v = 0. ***784**
where * is hidden number
Re: Problem 363
Posted: Sun Aug 25, 2019 11:16 pm
by RishadanPort
nm I figured it out and solved the problem.
v = 0.****784**** is correct for anyone that is wondering...