Faceman's silly math trivias

[Bozos of Bones]

SPOILER (maybe)

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¸Once. The blackhawk comes to the train, unloads a man, the guy goes to the connector between the locomotive and the cart sections and disconnects them. The locomotive thus moves faster because there is a sever weight decrease, and the carts slow because there is nothing to pull them. If there is an upslope, even better, cause the UH60 can use the low speed there to completely stop the carts.

[Bozos of Bones]

I've got one:

you have the equasion a=b+c

Presuming a, b and c are not 0, and using only mathematical functions(all levels) what do you have to do to get a=b?

Bonus points for answers under five lines of equasion. These bonus points are mental cookies that the solver should give him/herself.

[ 08-21-2003, 01:55 PM: Message edited by: Bozos of Bones ]

[Faceman]

Originally posted by Bozos of Bones: I've got one: you have the equasion a=b+c Presuming a, b and c are not 0, and using only mathematical functions(all levels) what do you have to do to get a=b?

erm...substract c ?

I don't quite get your question.
Assuming I'd go through with this it would look something like this

a=b+c
=> ...
.
.
. (less than five line, yeah )
=> a=b
=> a=b+c=a+c => a=a+c => c=0 -> contradiction

I assume you want to use some exploit in squaring, integrating, ...
to get to that solution but that wouldn't be proper math, would it.

Of course there's still
a=infinite b=infinite -> a=b+c infinite=infinite+c which would be correct

[Faceman]

Originally posted by Bozos of Bones: Once. The blackhawk comes to the train, unloads a man, the guy goes to the connector between the locomotive and the cart sections and disconnects them. The locomotive thus moves faster because there is a sever weight decrease, and the carts slow because there is nothing to pull them. If there is an upslope, even better, cause the UH60 can use the low speed there to completely stop the carts.

Well that wouldn't be fun anymore and it would rob us of such great films as Con Train or Under Siege 2.

So I'm going to solve it now:

answer: It probably can't be done
assume time to load the people into the blackhawk 3.33 min
assume they can be unloaded instantaneously
assume Chicago-Milwaukee is 90 miles (actual value) and straight (false assumption)
Blackhawk traveling speed 180mph (actual value)
Blackhawk fits 20 people (can actually take 11 fully equipped soldiers)
assuming you set your base at the best possible point which is obviously halfway between Chicago and Milwaukee.
If you calculate it through you get to pick up the last 20 people right at Milwaukee.

If you want to calculate it however with a base more far away and a higher speed (Blackhawk goes up to 225mph top speed)
you have to use a bit of trigonometry
you get 5 triangles
for every going back and forth you get one with the sides v*a (the way you go back from the train and v being the correlation between your speed and the trains speed: for 180 it would be 2) and v*b (for going to the train) with a and b being the distances the train covers while you fly
if you now draw a line x from your base to the point between a and b you can work with the sinuses as the angle #1 between 2a and x is 1/v the angle between a and x (same for b calling the top angle #2)
With that you can easily calculate the top angle between v*a and v*b taken that the neighbouring bottom angles are 180 degrees combined.
all five triangles are seperated by the distance 5 which is what you and the train cover while loading in the passengers. All you have to do now is to (via construction or calculation) show if and where it is possible to fit these 5 triangles together without exceeding the base (90miles Chicago-Milwaukee)

[Deejax]

Aren't you forgetting the distance you travel while picking up the passengers? The triangles you describe assume an instantaneous pickup of passengers.

In your "easy" calculation you take 3.33 minutes to pickup 20 passengers. In that time the train travels 5 miles.

The helicopter travels five times from the base to the train and four times back in the hour it takes the train to reach milwaukee. (you don't need to drop of the last group of passengers in the hour)

The distance traveled by the helicopter falls into two categories: Top speed and Train speed, for the distance travelled to and from the base and the distance traveled accompanying the train, 225 and 90 mph respectively.

The helicopter travels 9 different path from base to train and back. And I don't think they will make 'nice' triangles to calculate.

[Faceman]

Originally posted by Deejax: Aren't you forgetting the distance you travel while picking up the passengers? The triangles you describe assume an instantaneous pickup of passengers.

No I'm not. If you read my post carefully: all five triangles are seperated by the distance 5 which is what you and the train cover while loading in the passengers

The helicopter travels 9 different path from base to train and back. And I don't think they will make 'nice' triangles to calculate.

That depends on what "nice" triangles are but:
the first path TO the train is not part of one of my triangles
the first part BACK builds a triangle with the second part TO the train and so do the other pairs.
I'm trying to do an ASCII sketch

-------------------------B(ase)-----------------------------


Chicago______A___________S______________C_________ _Milwaukee

S is the point the train reaches when the helicopter reaches base (and I assume instantaneous unloading)
Now assuming a Blackhawk speed of 225mph (=2.5*90)
we get AB = 2.5*AS
and BC = 2.5*CS

AB/sinASB = AS/sinABS => 2.5*AS/sinASB = AS/sinABS => sinABS/2.5 = sinASB
same for the triangle BSC
BAS=180-ABS-arcsin(sinABS/2.5)
same for the triangle BSC
if you now introduce a height (orthogonal distance between B and AC) it
takes a little bit more trigonometry and a system of equations to figure it out.
It's not that difficult BUT it's certainly a LOT of work

[ 08-22-2003, 12:07 PM: Message edited by: Faceman ]

[Deejax]

Originally posted by Faceman: No I'm not. If you read my post carefully: all five triangles are seperated by the distance 5 which is what you and the train cover while loading in the passengers

Ooops, you're absolutely right, sorry. [img]graemlins/blush.gif[/img]

AB/sinASB = AS/sinABS => 2.5*AS/sinASB = AS/sinABS => sinABS/2.5 = sinASB

This I don't follow. Maybe my trigonometry is a bit rusty, but your first line is totally unclear to me.
AB/sinASB = AS/sinABS is the same as AB/AS = SinASB/SinABS
i.e. the ratio of two sides of a triangle is equal to the ratio of the sinuses of its opposite angles. That can't be right. Please explain, i'm lost.

[Faceman]

Originally posted by Deejax: quote: Originally posted by Faceman: No I'm not. If you read my post carefully: all five triangles are seperated by the distance 5 which is what you and the train cover while loading in the passengers

Ooops, you're absolutely right, sorry. [img]graemlins/blush.gif[/img]

AB/sinASB = AS/sinABS => 2.5*AS/sinASB = AS/sinABS => sinABS/2.5 = sinASB

This I don't follow. Maybe my trigonometry is a bit rusty, but your first line is totally unclear to me.
AB/sinASB = AS/sinABS is the same as AB/AS = SinASB/SinABS
i.e. the ratio of two sides of a triangle is equal to the ratio of the sinuses of its opposite angles. That can't be right. Please explain, i'm lost. [/QUOTE]It IS right and a basic trigonometric formula. I don't know the Englisch term but in German it's the "Sinussatz" opposed to the
"Cosinussatz" : c² = a² + b² + 2a*b*cosACB
which is the global sentence of the commonly known Pythagoras (if you put ACB = 90 you get it)


EDIT: Okay I searched a bit and found the English expressions: "law of sines" and "law of cosines" and also a page with the proof
http://mcraefamily.com/MathHelp/Geom...SinesProof.htm
this is the first page I found by luck so you may want to search further for better ones.

[Deejax]

Originally posted by Faceman: It IS right and a basic trigonometric formula. I don't know the Englisch term but in German it's the "Sinussatz" opposed to the "Cosinussatz" : c² = a² + b² + 2a*b*cosACB which is the global sentence of the commonly known Pythagoras (if you put ACB = 90 you get it)

Cool. I knew the "Cosinussatz" (which is "cosinusregel" in dutch i.e. cosinusrule) but I can't remember ever knowing the simple ratios equation. Memory is a strange thing. [img]smile.gif[/img]

Something to remember!
It does help the blackhawk problem, as you described. I should find some time and try to solve it. Don't know when though.

Have you thought up any new problems? I think they are pretty good, especially because they are not standard, dry, take-this-formula-and-these-constants-and-start-up-your-calculator type problems. My favourite is the exploding, rain-giving balloon.


[Edit] Something went wrong with the quote.

[ 08-26-2003, 03:01 AM: Message edited by: Deejax ]