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T O P I C    R E V I E W
TchrWill Posted - 03/17/2013 : 13:01:10
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?



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.

If anyone is interested in the whole story of the problem, I have the 1977 article which I would be happy to copy and send to anyone interested.

Enjoy












5   L A T E S T    R E P L I E S    (Newest First)
TchrWill Posted - 03/22/2013 : 18:05:07
quote:
Originally posted by someguy

After a little thought, I can come up with two methods but neither one is really satisfying. Both require a fair amount of symbol pushing (but don't use anything beyond highschool algebra).


One uses the law of cosines twice, the identity for the cosine of a sum, and the identity (sin theta)^2 + (cos theta)^2 = 1.

The other uses 3 applications of the Pythagorean theorem and similar triangles.

Both led to a speed of

(1260-8*sqrt(6461))/953 miles per minute

or

(75600-480*sqrt(6461))/953 miles per hour.

(approximately 38.843 mph)


I would be interested in seeing the article to see if there is a 'nice' method.



A simple approach. Mine was a tad longer.

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?


1--Let A stand for Ardsville, B for Bardville, C for Campton, and D for Davis.
2--Draw horizontal line DC, D on left, ~1 inch long and label it 10 miles.
3--Draw CB perpendicular to DC at C, ~1 inch long and label it 10 miles also.
4--Approximately 60 from DC draw line CA ~2 inches long.
5--Draw lines AB and AD.
6--Given the times it takes the commuter to travel each route at the unknown velocity V, the distances can be expressed as AB = 35V - 10, AD = 40V - 10, and AC = 30V.
7--Angle BCD = 90; let angle BCA = C1 and angle DCA = C2.
Using the Law of Cosines -
8--(35V - 10)^2 = (30V)^2 + 100 - 600Vcos(C1) or
....cos(C1) = (700 - 325V)/600.
9--Similarly, (40V - 10)^2 = (30V)^2 + 100 - 600Vsin(C1) or
....sin(C1) = (800 - 700V)/600.
Using sin^2x + cos^2x = 1
10--640,000 - 1,120,000V + 490,000V^2 + 490,000 - 455,000V + 105,625V^2 = 360,000 or
.....595,625 V^2 - 1,575,000V + 770,000 = 0 or
....V^2 - 2.64428V + 1.29275 = 0.
11--Therefore, V = [2.64428 +/-sqrt(6.9922 - 5.171)]/2 yielding V = .64738 miles/min. = 38.843 MPH.

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






Draw yourself a square and label it A, B, C, and D, starting with the upper left corner and proceeding clockwise. Locate a point representitive of 5 cm from A, 8 cm from B, and 13 cm from C, and label it P. Label angle ABP as "M" and angle BPC as " N". Label the side of the square as X. Given two sides, b & c, and the included angle, A, of a triangle, we know that cosA = (b^2 + c^2 - a^2)/2bc. Applying this to our two triangles containing our given lengths, we have for angle M, cosM = (8^2 + X^2 - 5^2)/16X and cos N = (8^2 + X^2 -13^2)/16X. Since cosN = cos(90 - M) = sinM, from sin^2 + cos^2 = 1, we have (4096+64X^2-1600+64X^2+X^4-25X^2-1600-25X^2+625) + (4096+64X^2-10816+64X^2+X^4-169X^2-10816-169X^2+28561) = 256X^2. Collecting terms and simplifying we have X^4 - 194X^2 + 6273 = 0. Through quadratic substitution we have X = [194X +/-sqrt(12544X^2)]/2X^2 = [194X +/-112X]/2X^2 from which 2X^3 = 306X or 82X from which X^2 = 153 or 41 X = 2.797. Then X = 12.369 cm or 6.403 cm. Clearly, only X^2 = 153 sq cm fits our given problem. The geometric significance of the X = 6.403 cm answer is that it is the solution for the case where the point P in question is outside the square altogether but still in the same plane.
These two solutions represent the two limiting cases where the point P is in the plane of the square. With X = 12.369, the point is inside the square and with X = 6.403, the point is outside the square. In between these two extremes, the problem could be considered three dimensional as point P rotates between the two extremes, about the edge AB, from inside the square to outside the square.







Admin Posted - 03/20/2013 : 09:46:32
quote:
Originally posted by TchrWill

Right on someguy.

I can get the article to you either by mail or as an E-Mail attachment.

Your call.

Will need an E-Mail address or a P.O. address.

Will post my solution later, for information purposes.



Please E-Mail me the complete article on Numbers.

Thank you,
Gisele
someguy Posted - 03/20/2013 : 01:57:31
Hi TchrWill, e-mail is good.

In the options menu under profile, I set "Allow Forum Members to Send you E-Mail?" to yes.
If you are logged in you should see an icon with an envelope on it above my posts that you
can click to send me an email.

Thanks.
TchrWill Posted - 03/19/2013 : 16:54:07
Right on someguy.

I can get the article to you either by mail or as an E-Mail attachment.

Your call.

Will need an E-Mail address or a P.O. address.

Will post my solution later, for information purposes.
someguy Posted - 03/19/2013 : 04:21:42
After a little thought, I can come up with two methods but neither one is really satisfying. Both require a fair amount of symbol pushing (but don't use anything beyond highschool algebra).


One uses the law of cosines twice, the identity for the cosine of a sum, and the identity (sin theta)^2 + (cos theta)^2 = 1.

The other uses 3 applications of the Pythagorean theorem and similar triangles.

Both led to a speed of

(1260-8*sqrt(6461))/953 miles per minute

or

(75600-480*sqrt(6461))/953 miles per hour.

(approximately 38.843 mph)


I would be interested in seeing the article to see if there is a 'nice' method.

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