A fox, 90 rods due south of a greyhound, is pursued by the hound at the rate of 5 rods to 4 of the fox, the fox running a due east course. How far will the hound run to overtake the fox?
[Will leave this up for 1 2 days before showing submitted answers, thanks]
Math is definitely not my strength, so I doubt the accuracy of this, but I came up with 201.246 rods. This was reached by assuming the running path is in the shape of a triangle. If the fox is directly south of the hound, then running east puts him on a ninetydegree angle to the hound, whose running distance forms the slope of the triangle.
However, I know this would be inaccurate, if for no other reason than the hound’s path would instead be curved. If he actually pursues the fox from the start as the question states, then he would begin running directly south, and gradually be gaining an eastward heading as he closed distance on the fox running east. I would have factored all that in, but frankly, I don’t know how.
150 rods
The hound would run 150 rods.
The fox will run only 120 rods.
Assumptions:
1. By sheer luck, the greyhound took the most direct path, running in a straight line directly to the intercept point
2. By sheer lack of selfpreservation, the fox did not veer from its due east path as the greyhound approached
Set up the problem as a right triangle, with the right angle in the lower left corner at point F (Fox), and a leg extending vertically “upward” to point G (Greyhound).
To the right of and at 90 degrees to point F is point M (Meeting), such that line FM is perpendicular to line FG. Point M is connected to both points F and G, where FM and GM have as yet undetermined lengths.
The angle at FMG can be determined by observing that, for any such triangle:
inverse sine (FG/GM) = inverse tangent (FG/FM)
GM = 5x
FM = 4x
FG = 90
x = the amount of time run by both animals
so: inverse sine (90/5x) = inverse tangent (90/4x)
The angle at which the inverse sine equals the inverse tangent is found to be 36.87 degrees, with a value for “x” of 30 units of time.
Therefore, the greyhound ran 5x, or 150 rods.
Their paths will make a right triangle with the hypotenuse and the other side having a ratio of 4/5. Since any two sides have to exceed the value of the other side and the first side is 90, then the hypotenuse has to be 55 and the remaining side has to be 44. So the hound would have to run 55 rods.
This one is very easy, even for a 15 years old boy like me
So here’s my method, maybe I’m wrong
Fox: y=4x+90 (4 rods every ”turn” and starts 90 rods away from the hound)
Hound: y=5x (5 rods every turn and starts at 0 rods)
so y=y
4x+90=5x
90=x
So at 90 rods.
150 rods.
G = total greyhound running distance
F = total fox running distance
G/F=5/4
Draw a right triangle with length 90 rods and length F as the two legs, and length G as the hypotenuse. The pythagorean theorem gives you:
F^2 + 90^2 = G^2
Solve the system of two equations for F and G, and you get:
F = 120 rods
G = 150 rods
360?
It depends on how smart the hound is. If the hound is smart enough to judge the speed and direction of the fox, plus judge his own capability then he will be able to figure out where he will intercept the fox and head for that point immediately. In this case the hound will travel 150 rods.
If, however, the hound always runs toward the fox, then he will not travel in a straight line but a curved one. Too difficult for me to solve.
150 rods
450
i changed my mind///
hold up i didnt take east into consideration i was thinkin straight line. cancel that w.e too hard my easy mac is ready later.
ok so itd be… a pythagorian theorum right triangle 3 4 5. simple conversion once thats figured out.
ehh never mind i dunno its weird how do you know the course the hound would tkae? maybe itd be a curve prolly, quadratic? no cuz hed catch him before the curve expanded. im lost
I am assuming that the hound catches the fox and that the hound velocity vector points at the fox at all times.
Answers:
Capture time: 50 sec.
Hound arclength: 250 rods Hound



– s – “diagonal”

– …… Fox
Soln:
Form differentials. \ / \ dl
(1) Separation Distance: s => ds = dl – vd*dt _\ vd*dt / _\
dw vr*dt
(2) xaxis separation : r => dr = vr*dt – dw
(3) Similar Triangles : vd*dt ~ vr*dt
——– ——
dw dl
=> dw = dl*vd/vr
(4) Eliminate terms : => dr = vd*ds/vr + (vr^2 – vd^2)*dt/vr
rc 0 tc
/ / /
(5) Integrate :  dr = (vd/vr)ds + (vr^2 – vd^2)/vr/dt
/ / /
0 90 0
(6) Observe : rc = 0 (he got him!!!)
(7) denoucement : 0 = (vd/vr)*90 + (vr^2 – vd^2)*tc/vr
tc =vr*vd/(vr^2 – vd^2)*90/vr
=90*vd/(vr^2 – vd^2) Capture Time
= 90*5/(2516) = 50 sec
distance = tc*vd = 250 rods
works because dH/dt = vd*(RH)/rH
& arclength integral =
integrate( sqrt(dH/dt.dot.dH.dt) dt =
integrate( vd,dt) = vd*tc QED!!!
My calculus is a little rusty, not so my basic geometry.
If the hound ran directly to the nearest possible point of intersection, the fox would have travelled 120 rods, and the dog 150 (a 345 right triangle); but what hound would think it through that cleverly? (Actually, they do to a point, but for the sake of this problem, I won’t try to presume the dog’s ability.)
If the two were adjacent to a barrier, so that once the hound started toward the fox he had to run to the fox’s starting point, then turn left and chase along the fox’s path of travel, the fox would travel 360 rods and the dog 450.
So the answer is between 120 and 360, but the problem is that I will assume that the hound is always travelling directly towards the fox and because they are not both travelling along the same line, the angle between them is changing and the dog must run along a curve somewhat akin to a hyperbola, but a little different.
Let’s see, the instantaneous change in angle ? with respect to time…
Official answer: 250.
Shawn and Ed (as well as a few others) state how the problem would be solved and what the answer would be if the hound could precalculate the point of intersection.
However, is it possible to arrive at an answer of 250 without calculus?
sorry, meant to say Shawn & Ed (not James). now corrected
Nicely done James!
You can simplify an inverse x function such as this by assuming 2 straight paths. If the hyperbolic arc is symmetrical from its intersection points on the x and y axes, you can find the answer by measuring the distance covered by the hound if he ran half of the “vertical” distance due south, then cut a straight course to the intercept point.
So, the hound initially runs 45 rods south. At 5 rods/second (assume time is measured in seconds), this means he runs for 9 seconds. In this time, the fox travels 4rods/second x 9seconds = 36 rods east
The new triangle to solve is (using the same nomenclature as in my first post):
FG = 45
GM = 5x
FM = 4x+36 (fox already went 36 rods)
where x = the number of seconds left to run AFTER the greyhound’s initial 9 second southerly sprint
Even though the fox has moved, we can still use the right triangle created by the fox’s original starting point:
cosine FMG = adj/hyp = FM/GM = (4x+36)/5x
sine FMG = opp/hyp = 45/5x
The only values that satisfy these equations are:
FMG=12.68 degrees
x=41
So, the TOTAL amount of time run by the greyhound is 9 (due south) + 41 (southeast at 12.68 degrees), or 50 seconds.
If the greyhound runs at 5 rods/second for 50 seconds, he runs a total of 250 rods.
It is the classic 345 triangle. The fox travelv 120 rods the hound travels 150 rodd which makes the ratio to be 5 to 4. The correct answer is 150 not 250. Given the time lapsed to be the same, a simple algebra will give you the 150 answer.
Shawn & James are right, because dogs don’t know calculus, D.Q. or even the Pythagorean Theorem. However, here is a URL to the question about dogs knowing calculus:
Doggy Calculus
The problem with using this example for this puzzle, is that the ball is thrown at a much higher velocity and is stationary waiting to be retrieved for most of the time. Calculus just verifies that Elvis instinctively takes the best route. But Elvis (math dog) is chasing what is for all intents and purposes a stationary target, we must assume that OUR hound is always heading directly towards the fox, and that produces a curve and that’s why James & Shawn are my heroes!
Shawn and James great job and very nice explanations too.
Milocan you please double check the link/url for ‘Doggy Calculus? thanks
This is the URL a straight copy and paste from the page.
http://www.pen.k12.va.us/Div/W.....s/dog.html
Understood, Good Job.
Simple pythagoras (3,4,5 triangle) 3×30=90 on the short side and 3×40 =120 on the adjacent so the hound travels 3×50=150 on the hypotenuse. Fox travels 120 rods (660′ or 220yards) at rightangles to original strait line drawn between the two.
solution 250 rod, here is my explanation, non calculus based:
velocity is constant and are represented in linear vectors, you can used pytagoras for the velocity and get 5^2= 4^2+x^2, x=3–> t= velocity/distance, t=90/3–> 30 sec, 150 would be the optimal distance: 5*30sec= 150 rods by a “genius dog”, in the other hand a “non sense dog” would go south, so you go for a dog error: he run between the optimal route and the linear route thet is south because the initial postion is south:
so 3 velocity of linear route, 5 velocity of optimal route: 3/ 5= Doptmal/DReal–> 3*DReal =5*DOptimal –> 5*150/3=DReal–>250.
Another way is this:
how many seconds the dog takes to cover 150rods(optimal route) at 3 rods per sec(velocity of the linear route) ?
50 sec right–> thats the time that takes to the dog to reach the fox, 5*50 sec–> 250, thats part of laplace version of time and position theory: the dog route is the optimal but the time is if the dog goes at the suboptimal velocity and viceversa, wow!!! , the calculus version was very cool indeed, but i could not arrive to the answer in that way.
Even a running human being will not be able to choose the optimal path…But Calculus dogs do The result would be 150 rods (i never knew that a rod was a distance unit, about 5m)
Having read the puzzle, i understand that the hound always runs in the direction of the fox, hence following a curve, which i may post about perhaps in a future post.