Negotiating a Corridor

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
Hello Rich,

Ask and you shall receive.... :D

You are walking down a long straight corridor that is 2 meters wide. You reach another corridor that branches off at a right angle. You enter this corridor, and find it is 4 metres wide. What is the longest straight rod that can be carried horizontally around the corner from the 2 meter corridor into the 4 metre corridor?

This question of course is open to all forum members who enjoy a good puzzle... :D

Ron.
 
Hi Ron,

Thanks i was getting withdrawal symptoms... and this will keep me occupied this afternoon while i'm being good and not doing anything like the doctor said ( had some moles taken off this morning ;-))
 
Hi Rich,

Glad to oblige.... :D

I am sure it will keep you entertained this afternoon, and it is always wise to do what the Doctor said.. :wink:

Hope the results all come back clear too!!

Ron.
 
pretty sure it'll be fine... I have hundreds of moles and this is the first time theyve taken any off and they said it was a precaution rather than anything serious looking...
 
Hi,

I think it's 6.927 metres - not sure on my rationale that the angle should be 60 degrees - chose taht because of the ratio betwenn the widths of the 2 corridors. If they were identical the angle of the rod would be 45 degrees- so at a ratio of 2:1 i think it should be 60 ( but here's where you tell me i'm wrong ;-)) and then a bit of trig gets you to 2.3094 - similar triangles takes that to 6.927.

Rich.
 
Bearing in mind the rod can only be carried in the horizontal, I suspect the longest rod would be that which would sit at 45 degrees between corridors. Assuming this (which is based on nothing more than gut feeling) and applying some simple Pythagoras-

sq root (4+4) + sq root (16 +16) = ( sq root 8 ) + ( sq root 32 ) = 8.485m


Go on now............... tell me why I am wrong.
 
rockdemon wrote,...
I think it's 6.927 metres - not sure on my rationale that the angle should be 60 degrees - chose taht because of the ratio betwenn the widths of the 2 corridors. If they were identical the angle of the rod would be 45 degrees- so at a ratio of 2:1 i think it should be 60 ( but here's where you tell me i'm wrong ) and then a bit of trig gets you to 2.3094 - similar triangles takes that to 6.927.
Click to expand...

Hello Rich,

Hmmm,...the angle isn't 60 degrees... :( and the length that you have of 6.927m is actually on the short side, so you have room to spare. So...thinking cap back on.. :wink:

JVY wrote,...
Bearing in mind the rod can only be carried in the horizontal, I suspect the longest rod would be that which would sit at 45 degrees between corridors. Assuming this (which is based on nothing more than gut feeling) and applying some simple Pythagoras-

sq root (4+4) + sq root (16 +16) = ( sq root 8 ) + ( sq root 32 ) = 8.485m
Click to expand...

Hello JVY,

Your suspicion that the longest rod will sit at an angle of 45 degrees between the corridors,...hmmm,..sorry that isn't right. The length you have at 8.485m is actually too long, so unfortunately it won't fit. I'll let you think a bit longer, so thinking cap back on... :wink:

Working the mind is a good thing,
Ron.
 
i'm fast coming to the conclusion that the answer to this lies in a differential equation with the length of rod for a given angle. I've not forgotten... Just still thinking:)


rich
 
Just some of my workings/trials

Triangles ACB, DCF, and EFB are all similar

AC = AD + AC

AB = AE + EB

BC2 = AC2 +AB2

If DC = 4, EB=2 (similar triangles)

If DC = 3, EB=2/3*4=2.67

So if DC = x, EB = 2/x*4

So if AC = 4 + x, AB = 2 + 2/x*4

My answer does NOT allow the rod to have any thickness however. It's simply a point to point thing.
 

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I'm not sure at all!

Usually Quattro and Rockdemon hit not that far off target, so I'm more inclined to go with them.

It's got to be in the 8m region, if "6.927m is actually on the short side" and "8.485m is actually too long"

As a shot in the dark, I will say 8.2m thereabouts! :p
 
:)

Come lunch time i'll see if i can try the differentiation method i talked about above. At the point the rate of change in stick length is 0, the stick should be at it's shortest and from this angle work out the maximum length....
 
When side AC of the triangle is 6.51 metres long, BC is 8.323921726
When side AC of the triangle is 6.52 metres long, BC is 8.323876382
When side AC of the triangle is 6.53 metres long, BC is 8.323924272

I just put the equations from above into a spreadsheet and got it to add .01 to it and find the length of BC. At 6.52, BC stops getting shorter, and at 5.63 it starts to get longer again.

If you plot all of the figures on a graph, it shows a definte curve peaking out at 6.52 metres (AC) and 8.323876382 for BC.

I suppose it could be construed as 3.32388

Richard
 
Yep you're right :O) (told you i was probably wrong.... ;o))

The length of the line is given by (from your diagram)

2/sin x + 4/cos x where x is the angle C on your diagram. (SOH CAH TOA)

I had to look up my cots and secs and cosecs - but this works out as 2 cosec x+4sec x.

Again i had to look it up to differentiate it but,
this gave me -2cosec x cot x + 4sec x tan x.

This is zero at 0.671 rads - from graphing(theres some great online graphical calculators now!), i didnt solve it . (38.44 degrees)

At .671 rads
2/sin .671 + 4/ cos .671 we get
3.2166 + 5.1072 = 8.324 :O)

Rich.
 
rockdemon said:
Yep you're right :O) (told you i was probably wrong.... ;o))

The length of the line is given by (from your diagram)

2/sin x + 4/cos x where x is the angle C on your diagram. (SOH CAH TOA)

I had to look up my cots and secs and cosecs - but this works out as 2 cosec x+4sec x.

Again i had to look it up to differentiate it but,
this gave me -2cosec x cot x + 4sec x tan x.

This is zero at 0.671 rads - from graphing(theres some great online graphical calculators now!), i didnt solve it . (38.44 degrees)

At .671 rads
2/sin .671 + 4/ cos .671 we get
3.2166 + 5.1072 = 8.324 :O)

Rich.
Click to expand...

yes, I get that also! :)
 
Well the answer is indeed 8.32m to two decimal places. Well done Richard (quattro) for obtaining the correct answer, Rich, JVY and Darth all for giving it a go and for confirmation.. :D

appexp.gif


From the diagram, let p = 2 and q = 4, BR = L1, RA = L2, Angle BA0 = x.

The desired length is actually the minimum length L = L1 + L2, expressing L as a function of x
L(x) = 4 cosec(x) + 2 sec(x) = 4/sin(x) + 2/cos(x)

The domain of L is the open interval 0 < x < PI/2

Differentiate,...dL/dx = - 4cos(x)/sin^2(x) + 2sin(x)/cos^2(x)

For the critical point, let dL/dx = 0 and solve for x

This gives tan(x) = 2^1/3 thus x = arctan 2^1/3....ie angle x is the inverse tan of the cube root of 2. Use the first derivative test and also test the end points of the domain in L'(x)

This shows that the angle x is 0.899908348 radians = 51.56 degrees. Let x = approx 0.9,...substituting into L(x) gives the length of the rod which is 4/sin(0.9) + 2/ cos(0.9) = 8.32m to two decimal places. This is the minimum length of the rod L which is also the maximum length it can be in order to satisfy the question.

Well done everyone... :D

Ron.
 
One thing that did confuse me until just now was how come the two Rich's got 2/sin x + 4/cos x , whereas I got 4/sin x + 2/cos x.

I just realised that they were using the 'other' angle (They took the 4m corridor to be the adjacent, and the 2m to be the opposite), whereas Ron and I (independently) worked the with the 4m being opposite and 2m as being adjacent.

Both yield the same result though!

Using Ron's method (and mine!) of getting 51.56 degrees

sin (x) = cos (x - 90) and cos (x) = sin (x + 90)

4/sin (51.56) = 4/cos (-38.44) = 4/cos (38.44) cos (-x) = cos (x)
2/cos (51.56) = 2/sin (141.56) = 2 /sin (180 - 141.56) = 2/sin (38.44) sin (x) = sin (180 - x)

51.56 deg = 0.9 rads

38.44 deg = 0.671 rads which is of course what the two Rich's got!
 
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