Considering the price of gas.

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TchrWill

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A commuter, who lives in Ardsville, drives to work in Campton every day at the same speed (the car has a governor on it). Campton is generally in a southeasterly direction from his home in Ardsville. He can take three routes to work, one through Bardsville, 10 miles due north of Campton, one through Davis, 10 miles due west of Campton, and one directly to Campton. He makes it to work from Ardsville direct to Campton in 30 minutes; from Ardsville through Bardsville to Campton in 35 minutes; and from Ardsville through Davis to Campton in 40 minutes. How fast does he drive?

Enjoy :)
 
Code: 
       A
                (b)
                           B



   (d)        (c)
                         (10)




D            (10)         C
I'm always nervous with "due" directions; does my diagram indicate what you mean?
Let b = AB, c = AC and d = AD; then:
b + 10 takes 35 minutes, c takes 30 minutes and d + 10 takes 40 minutes ; correct?
Angle BCD = 90 degrees; correct?
 


I used a system of three equations in three unknowns:

[spoiler:m4m48g4w].

Put C at the origin

Put D at (0, 10)

Put B at (10, 0)

Let r = the rate in miles per minutes

Put A at (x, y) in Quadrant I on a circle of radius 30r centered at C

Eqn1: x^2 + y^2 = (30r)^2

Eqn2: x^2 + (y - 10)^2 = (40r - 10)^2

Eqn3: (x - 10)^2 + y^2 = (35r - 10)^2

.[/spoiler:m4m48g4w]

My answer:

[spoiler:m4m48g4w].

r = 0.6474 (rounded) which is about 38.8 mph, governor 8-)

.[/spoiler:m4m48g4w]

 
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Re:

mmm4444bot said:
Denis said:
I'm always nervous with "due" directions
Why is that?
Skipping class?
Financial difficulties?
:twisted:
Click to expand...
Click to expand...
Well, in my school days in the 50's, "due" had nothing to do with maths!
 
Denis said:
Code: 
       A
                (b)
                           B



   (d)        (c)
                         (10)




D            (10)         C
I'm always nervous with "due" directions; does my diagram indicate what you mean?
Let b = AB, c = AC and d = AD; then:
b + 10 takes 35 minutes, c takes 30 minutes and d + 10 takes 40 minutes ; correct?
Angle BCD = 90 degrees; correct?
Click to expand...

Right on the money.
 
Re:

mmm4444bot said:
I used a system of three equations in three unknowns:

[spoiler:1detqdyc].

Put C at the origin

Put D at (0, 10)

Put B at (10, 0)

Let r = the rate in miles per minutes

Put A at (x, y) in Quadrant I on a circle of radius 30r centered at C

Eqn1: x^2 + y^2 = (30r)^2

Eqn2: x^2 + (y - 10)^2 = (40r - 10)^2

Eqn3: (x - 10)^2 + y^2 = (35r - 10)^2

.[/spoiler:1detqdyc]

My answer:

[spoiler:1detqdyc].

r = 0.6474 (rounded) which is about 38.8 mph, governor 8-)

.[/spoiler:1detqdyc]
Click to expand...

mmm4444bot gets the Gold Medal with the speed of 38.8 mph.

This problem first appeared in Popular Science in September 1939 and caused quite a stir among puzzle enthusiasts, then, and again in 1977 when it was republished by popular request. The mail room was deluged with responses producing two solutions, only one of which fits the scenario, using at least five distinctly different approaches.

Some might come up with, and try to justify, the answer of 119.81 MPH, as it is a legitimate root of the solved quadratic. It can be readily argued that the 119.81 MPH answer does not fit the figure as defined in the problem statement but another figure can be defined that satisfies the answer. The solution can be derived from analytical geometry, internal triangles, area formulas, the Law of Cosines, and constructing a rectangle around the extremeties.

If any respondents are interested in the whole interesting story of the problem, I have the 1977 article which I would be more than happy to copy and send to anyone interested. It shows all of the methods of solution.
 
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Denis said:
in my school days in the 50's, "due" had nothing to do with maths

What about all of those past-due tuition notices?
Click to expand...

Anyway, welcome to the 21st century. :wink:

 
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TchrWill said:
The solution can be derived from analytical geometry, internal triangles, area formulas, the Law of Cosines, and constructing a rectangle around the extremeties.

I have the 1977 article which I would be more than happy to copy and send to anyone interested.
Click to expand...

I started out using the Law of Cosines (to set up a function in r), but I abandoned that approach for what I thought was an easier way.

Are you offering to copy and send the article electronically? If so, I could PM an e-mail address to you. I'd like to see the area and rectangle approaches.

 
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Re:

mmm4444bot said:
TchrWill said:
The solution can be derived from analytical geometry, internal triangles, area formulas, the Law of Cosines, and constructing a rectangle around the extremeties.

I have the 1977 article which I would be more than happy to copy and send to anyone interested.
Click to expand...

I started out using the Law of Cosines (to set up a function in r), but I abandoned that approach for what I thought was an easier way.

Are you offering to copy and send the article electronically? If so, I could PM an e-mail address to you. I'd like to see the area and rectangle approaches.
Click to expand...

Sorry for the tardiness. Just got back home.

I would be happy to mail you a copy of the 4 pages (have a copy machine) or scan them into my PC and send them as attachements to an E-Mail.

Your call.
 
Last edited by a moderator:
Made up a "general case" solution :
minutes going via B = u
minutes going via D = v
minutes directly to C = w
distance from B and D to C = e
speed per minute = r

r = [-b - SQRT(b^2 - 4ac)] / (2a) : the "+" solution produces negative distance, so n/a
where:
a = u^4 + v^4 + 2w^2(w^2 - u^2 - v^2)
b = 4e(uw^2 + vw^2 - u^3 - v^3)
c = 4e^2(u^2 + v^2 - w^2)

With u=35, v=40, w=30 and e=10 thrown in there: r = .647384295.... (as per Mark).
 
Denis said:
Made up a "general case" solution :
minutes going via B = u
minutes going via D = v
minutes directly to C = w
distance from B and D to C = e
speed per minute = r

r = [-b - SQRT(b^2 - 4ac)] / (2a) : the "+" solution produces negative distance, so n/a
where:
a = u^4 + v^4 + 2w^2(w^2 - u^2 - v^2)
b = 4e(uw^2 + vw^2 - u^3 - v^3)
c = 4e^2(u^2 + v^2 - w^2)

With u=35, v=40, w=30 and e=10 thrown in there: r = .647384295.... (as per Mark).
Click to expand...

Nice going Denis. I was on my way to determining the same thing. Why don't you submit it to Popular Science for its 70th birthday and give a new generation a chance to tackle it.
 
TchrWill said:
Why don't you submit it to Popular Science for its 70th birthday and give a new generation a chance to tackle it.
Click to expand...
Too lazy...you go ahead, since you would have got same thing!!
 
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