Author |
Topic: Really Hard Precalculus Project (Read 865 times) |
|
warriors837
Guest
|
As the diagram attached suggests, a rectangular beam of length L and width W has become tightly wedged because it would not fit when someone tried to move it around a corner formed by two intersecting corridors, one of width, a, and the other of width b. The beam makes an angle of theta, with one wall, as shown, Express the length of the beam L in terms of the other quantities a, b, W, and theta.
The diagram is an attached file.
|
|
 IP Logged |
 |
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender: 
Posts: 13730
|
 |
Re: Really Hard Precalculus Project
« Reply #1 on: May 19th, 2009, 8:40am » |
 Quote  Modify
|
You can solve it by drawing a few triangle in the diagram.
And if I managed not to screw it up, you can get:
a/y =sin(t)
b/x =cos(t)
v/w=tan(t)
w/u=tan(t)
L=x+y-u-v
L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t)
I extended the line of lower side of the beam, u and v are the extra pieces on both ends, and x and y are the length from the corner to the ends of the beam.
|
| « Last Edit: May 19th, 2009, 8:42am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Immanuel_Bonfils
Junior Member
Posts: 114
|
|
Re: Really Hard Precalculus Project
« Reply #2 on: May 19th, 2009, 1:29pm » |
Quote Modify
|
If X is the distance from the beams lowest corner to the convex edge of the corridors intersect, we have :
blocking the vertical corridor, X cos  +W sin = b , and
blocking the horizontal corridor, (L-X) sin +W cos = a .
Eliminating X we get L = b sec + a csc - 2 W csc(2) .
|
| « Last Edit: May 19th, 2009, 1:30pm by Immanuel_Bonfils » |
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Really Hard Precalculus Project
« Reply #3 on: May 20th, 2009, 1:31pm » |
Quote Modify
|
on May 19th, 2009, 8:40am, towr wrote:| L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t) |
|
... = a/sin(t) + b/cos(t) - w/(sin(t)cos(t))
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Really Hard Precalculus Project
« Reply #4 on: May 21st, 2009, 12:47am » |
Quote Modify
|
on May 20th, 2009, 1:31pm, Grimbal wrote:
... = a/sin(t) + b/cos(t) - w/(sin(t)cos(t)) |
|
And that equal's Immanuel's result, yeah.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Really Hard Precalculus Project
« Reply #5 on: May 21st, 2009, 12:59am » |
Quote Modify
|
Uh, right. I am not familiar with sec and csc. I never use them.
|
| « Last Edit: May 21st, 2009, 1:00am by Grimbal » |
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Really Hard Precalculus Project
« Reply #6 on: May 21st, 2009, 1:53am » |
Quote Modify
|
on May 21st, 2009, 12:59am, Grimbal wrote:| Uh, right. I am not familiar with sec and csc. I never use them. |
|
Yeah, me neither. But you could have simplified sin(t)cos(t) to 2sin(2t) [edit]1/2 sin(2x)[/edit]; that way you use one less trigonometric function.
|
| « Last Edit: May 21st, 2009, 8:22am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
rmsgrey
Uberpuzzler
    
Gender:
Posts: 2873
|
|
Re: Really Hard Precalculus Project
« Reply #7 on: May 21st, 2009, 5:58am » |
Quote Modify
|
on May 21st, 2009, 1:53am, towr wrote:
Yeah, me neither. But you could have simplified sin(t)cos(t) to 2sin(2t); that way you use one less trigonometric function.
|
|
... or you could have simplified it to (1/2)sin(2t) and had the added bonus of being correct 
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Really Hard Precalculus Project
« Reply #8 on: May 21st, 2009, 8:25am » |
Quote Modify
|
on May 21st, 2009, 5:58am, rmsgrey wrote:
... or you could have simplified it to (1/2)sin(2t) and had the added bonus of being correct |
|
Meh. I was perfectly content to leave it as L=a/sin(t) + b/cos(t) - w tan(t) - w/tan(t)
Needless manipulation just breeds errors.
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Really Hard Precalculus Project
« Reply #9 on: May 21st, 2009, 9:47am » |
Quote Modify
|
I personally dislike the introduction of sin(2t) which gives the impression there is an angle of 2·theta somewhere in the diagram.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Really Hard Precalculus Project
« Reply #10 on: May 21st, 2009, 10:23am » |
Quote Modify
|
on May 21st, 2009, 9:47am, Grimbal wrote:| I personally dislike the introduction of sin(2t) which gives the impression there is an angle of 2·theta somewhere in the diagram. |
|
In that case, isn't it better to just keep the terms separate? Because what does sin(t)cos(t) correspond to in the diagram?
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Really Hard Precalculus Project
« Reply #11 on: May 21st, 2009, 2:22pm » |
Quote Modify
|
Actually, it's the third height of a right triangle with hypotenuse 1 and angle t (the other 2 being sin(t) and cos(t). It even gives some nice symmetry to my formula. 
But my point was that sin(2t) looks ugly to me. It doesn't recombine in an obvious way with sin(t) and cos(t). By keeping sin(t) and cos(t), you can factor them out, and get for instance:
L = (a·b - (a-w/cos(t))·(b-w/sin(t)))/w
which you might or might not find useful.
|
| « Last Edit: May 21st, 2009, 2:23pm by Grimbal » |
IP Logged |
|
|
|
warriors837
Guest
|
Here is the complete work, just look at the attachment...
|
| « Last Edit: May 22nd, 2009, 9:04am by warriors837 » |
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Really Hard Precalculus Project
« Reply #13 on: May 22nd, 2009, 2:56pm » |
Quote Modify
|
on May 22nd, 2009, 9:03am, warriors837 wrote:| Here is the complete work, just look at the attachment... |
|
You can simplify things a bit more quickly by using the fact that tan(90 - t) = 1/tan(t).
And it isn't entirely clear from the diagram that x and y include the segments u and v respectively; it now seems they end where the beam ends rather than go up to the wall. (Which admittedly wasn't clear from my description either )
|
|
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print
Remove
Proof.doc