Proof without words

[jay_shark]

A dog is standing at the center of a circular yard. A rabbit is at the edge. The rabbit runs round the edge at constant speed v. The dog runs towards the rabbit at the same speed v, so that it always remains on the line between the center and the rabbit. Show that it reaches the rabbit when the rabbit has run one quarter of the way round.

[jay_shark]

bump

[Borodog]

Something seems wrong. For the dog to reach the rabbit, the dog's trajectory must cross the rabbit's trajectory at some angle. But this is impossible, since the dog must be travelling in uniform circular motion when he reaches the rabbit. It seems like the dog will never reach the rabbit.

However, I have not worked the problem.

[Borodog]

I see where my physical intuition went wrong. 2nd derivative.

r(t) = Rsin(vt/R)

Cool.

[djames]

Anyone care to show a little detail in this one? I'm not getting pi/2, but I'm likely not expressing the position of the dog at time t correctly.

[Siegmund]

I am looking forward to the proof without words of this one. It took me a sheet of scratch paper to convince myself they met at all, let alone met at that particular point.

[Borodog]

If someone doesn't beat me to it, I'll post the solution later tonight. Off to the movies.

[jay_shark]

Ok I lied , you need a few words but nothing too difficult .

Label the points A(0,0), B (0,R ), C (-R,0) and a smaller circle with center D( -R/2, 0) .Take a line from A which intersects the small circle and the large circle at points E and F respectively . I guessed that the dog should run along the arc path of the smaller circle (arc AC ) .It also turns out that the arc length of AE and FB are always equal for any line you choose through the origin .

Notice angle EDA is twice the angle of FAB since it is the center of the smaller circle which uses the fact that AEC is 90 degrees . Therefore

arc(DE)/<EBD = 2pieR/2pie hence arc (DE) = R*<EBD
and arc(CB)/<2*EBD= 2pie*R/2/2pie , so arc(CB) = R*<EBD

QED

[Borodog]

That is so NOT the way I worked it. /images/graemlins/wink.gif

In fact, I have no idea what your unintelligable mathspeak even says.

I worked it like this:

The rabbit runs in a circle of radius R at speed v, and hence with angular velocity omega = v/R. Since the dog is always found on the radial containing the rabbit, the dog's angular velocity is also omega, and his azimuthal velocity will be v_theta = r(t)*omega, where r(t) is his radius. Since his speed is constant and equal to v:

dr/dt = v_r = (v^2 - v_theta^2)^(1/2) = (v^2 - omega^2*r^2)^(1/2)

Or, dr/dt = v(1 - r^2/R^2)^(1/2)

I substituted variables:

r' = r/R, dr = Rdr'

(1 - r'^2)^(-1/2) dr' = (v/R)dt

Integrating the left yields arcsin(r'), so

r' = sin(vt/R),

so,

r(t) = Rsin(vt/R)

Which has its first maximum when vt/R = pi/2, or at time:

t = pi*R/(2v) = (2*pi*R)/(4v)

Since it takes a time:

T = 2*pi*R/v

to go around the entire circle, time t occurs at 1/4 of the circuit.

[jay_shark]

I thought my solution was elegant which doesn't even require calculus .


Take the time and read it over again , you'll be pleased :P

[Borodog]

Here's where I don't buy it:

[ QUOTE ]
I guessed that the dog should run along the arc path of the smaller circle (arc AC ).

[/ QUOTE ]

How do you justify this without refering to the physics?

Specifically, what would change about your solution if the dog ran at 2v rather than v? I see no reference to the speed of the dog in your solution.

In other words, I think you got lucky. But I'm willing to listen.

[jay_shark]

I just had a really strong intuition about this problem . I knew right away that the dog should travel along a semi circle , and then I conjectured that the arc lengths were always equal . All that matters is that the dog and rabbit are traveling at the same speed which is equivalent to the arcs of the two circles always being the same length .

If you draw the picture out using the coordinates I gave , you can see how this may possible be true . Then all you have to do is prove it which I already did .

[Borodog]

I still don't buy it. Your entire setup for the problem depended on being given the answer ("Show that . . ."). If you hadn't been told that the correct solution was 1/4 of the circle, would you have honestly placed point C where you did? How would you justify this a priori? In fact, how is your placement of ABC not simply assuming your conclusion?

Not trying to jump on you, by the way. Just trying to understand.

[Duke]

[ QUOTE ]
I still don't buy it. Your entire setup for the problem depended on being given the answer ("Show that . . ."). If you hadn't been told that the correct solution was 1/4 of the circle, would you have honestly placed point C where you did? How would you justify this a priori? In fact, how is your placement of ABC not simply assuming your conclusion?

Not trying to jump on you, by the way. Just trying to understand.

[/ QUOTE ]

I think textbooks do this all the time, as it's much more difficult to do what you did and figure out what it actually is.

[Neuge]

[ QUOTE ]
I still don't buy it. Your entire setup for the problem depended on being given the answer ("Show that . . ."). If you hadn't been told that the correct solution was 1/4 of the circle, would you have honestly placed point C where you did? How would you justify this a priori? In fact, how is your placement of ABC not simply assuming your conclusion?

Not trying to jump on you, by the way. Just trying to understand.

[/ QUOTE ]
His solution depends on a leap of faith of sorts. If you assume a solution then prove that solution to be true, it's still a solution. However, it's obvious flaw is that it provides no methodology for solving a similar problem (i.e. rabbit and dog run at different speeds).

Yours is obviously a more formal proof and provides mathematical rigor in solving this class of problems, rather than the specific case of same speeds.

[jay_shark]

I definitely would have guessed that the dog should travel along a circular path . You must understand that solving math problems requires a lot of guessing and speculating which is what I've done in this problem . I made the right conjecture and then I proved my claim .

I was very pleased with this solution considering that it was one of the final questions of an old Olympiad . Problems like this one usually requires techniques that are foreign to solutions found in math textbooks . This is why I didn't lean towards a calculus solution .

[thylacine]

[ QUOTE ]
Ok I lied , you need a few words but nothing too difficult .

Label the points A(0,0), B (0,R ), C (-R,0) and a smaller circle with center D( -R/2, 0) .Take a line from A which intersects the small circle and the large circle at points E and F respectively . I guessed that the dog should run along the arc path of the smaller circle (arc AC ) .It also turns out that the arc length of AE and FB are always equal for any line you choose through the origin .

Notice angle EDA is twice the angle of FAB since it is the center of the smaller circle which uses the fact that AEC is 90 degrees . Therefore

arc(DE)/<EBD = 2pieR/2pie hence arc (DE) = R*<EBD
and arc(CB)/<2*EBD= 2pie*R/2/2pie , so arc(CB) = R*<EBD

QED

[/ QUOTE ]

Assuming vector point from the center of the circle:

Is it the dog's position vector of the dog's velocity vector that points towards the rabbit. (Look at Jay's problem and solution.)

Jay? Boro?

[Borodog]

[ QUOTE ]
[ QUOTE ]
Ok I lied , you need a few words but nothing too difficult .

Label the points A(0,0), B (0,R ), C (-R,0) and a smaller circle with center D( -R/2, 0) .Take a line from A which intersects the small circle and the large circle at points E and F respectively . I guessed that the dog should run along the arc path of the smaller circle (arc AC ) .It also turns out that the arc length of AE and FB are always equal for any line you choose through the origin .

Notice angle EDA is twice the angle of FAB since it is the center of the smaller circle which uses the fact that AEC is 90 degrees . Therefore

arc(DE)/<EBD = 2pieR/2pie hence arc (DE) = R*<EBD
and arc(CB)/<2*EBD= 2pie*R/2/2pie , so arc(CB) = R*<EBD

QED

[/ QUOTE ]

Assuming vector point from the center of the circle:

Is it the dog's position vector or the dog's velocity vector that points towards the rabbit. (Look at Jay's problem and solution.)

Jay? Boro?

[/ QUOTE ]

Position. The dog remains on a radial that connect the rabbit to the center.

[Borodog]

[ QUOTE ]
I definitely would have guessed that the dog should travel along a circular path . You must understand that solving math problems requires a lot of guessing and speculating which is what I've done in this problem . I made the right conjecture and then I proved my claim .

I was very pleased with this solution considering that it was one of the final questions of an old Olympiad . Problems like this one usually requires techniques that are foreign to solutions found in math textbooks . This is why I didn't lean towards a calculus solution .

[/ QUOTE ]

I still don't see how your solution demonstrates that the dog is always on a radial that connects the center to the rabbit.

[jay_shark]

I appreciate your solution Borodog which certainly allows you to solve a larger class of similar problems . In that case ,
vector calculus is the way to go !!

[jay_shark]

Just draw ANY line from the center of the large circle which intersects the small circle and the large circle . The dog is traveling along the arc path of the smaller circle which I thought was very obvious to me at the time .

Now just go back to the part where I showed that the two arc lengths are always the same . This proves that the dog is running along this path which is traveling at the same speed
as the rabbit .

What happens if the rabbit is traveling twice as fast ?
hmmm...

Perhaps the dog would be traveling along some elliptical path .

[jason1990]

Put the dog and rabbit in the complex plane, set the rabbit's velocity at 1, and the dog's at c. The rabbit follows r(t) = e^{it} and the dog follows d(t)=f(t)e^{it}, for some real-valued function f. We know that |d'| = c, so this gives

|f' + if| = c, or (f')^2 + f^2 = c^2.

The solution to this DE with f(0) = 0 is f(t) = c*sin(t). This means the dog travels in a circle of radius c/2 centered at ic/2, or (0,c/2) in the xy-plane. If c > 1, he reaches the rabbit at time t = sin^{-1}(1/c). If c < 1, he never reaches the rabbit.

[Borodog]

[ QUOTE ]
If c < 1, he never reaches the rabbit.

[/ QUOTE ]

Physically, this is because a maximum velocity less that v limits the radius of the circle the dog can run in while chasing the rabbit to less than R.

[jay_shark]

[ QUOTE ]


Notice angle EDA is twice the angle of FAB since it is the center of the smaller circle which uses the fact that AEC is 90 degrees . Therefore

arc(DE)/<EBD = 2pieR/2pie hence arc (DE) = R*<EBD
and arc(CB)/<2*EBD= 2pie*R/2/2pie , so arc(CB) = R*<EBD

QED

[/ QUOTE ]

I have my letters mixed up but the solution still holds .

It should be arc (fb) /<Fab =2PieR/2Pie arc fb=<fab*R
And arc EA/(2<Fab)=(2pie R/2)/2pie so arc EA= <fab*R

[djames]

[ QUOTE ]
Since the dog is always found on the radial containing the rabbit

[/ QUOTE ]

While this is intuitive, it seems like it is something that should be proven and not simply assumed. Having said that, I haven't yet been able to prove this. If this is simple physics knowledge, perhaps one of you can share a proof of this for those of us that haven't touched physics in 10+ years.

To be clear, I'm not sure why the dog must lie on the ray eminating from the origin and extending through the rabbit.

Thanks!

[Borodog]

[ QUOTE ]
[ QUOTE ]
Since the dog is always found on the radial containing the rabbit

[/ QUOTE ]

While this is intuitive, it seems like it is something that should be proven and not simply assumed. Having said that, I haven't yet been able to prove this. If this is simple physics knowledge, perhaps one of you can share a proof of this for those of us that haven't touched physics in 10+ years.

To be clear, I'm not sure why the dog must lie on the ray eminating from the origin and extending through the rabbit.

Thanks!

[/ QUOTE ]

Uh, it's given in the OP?

/images/graemlins/confused.gif

[Borodog]

[ QUOTE ]
A dog is standing at the center of a circular yard. A rabbit is at the edge. The rabbit runs round the edge at constant speed v. The dog runs towards the rabbit at the same speed v, so that it always remains on the line between the center and the rabbit. Show that it reaches the rabbit when the rabbit has run one quarter of the way round.

[/ QUOTE ]

[djames]

Hehheh, sorry. In translating this to a colleague, I garbled the original post and phrased the question as "so that it always runs towards the rabbit." I need to think about how/if the question changes with the phrase in bold removed from the question.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
A dog is standing at the center of a circular yard. A rabbit is at the edge. The rabbit runs round the edge at constant speed v. The dog runs towards the rabbit at the same speed v, so that it always remains on the line between the center and the rabbit. Show that it reaches the rabbit when the rabbit has run one quarter of the way round.

[/ QUOTE ]

[/ QUOTE ]

I agree with djames. The primary condition is that "The dog runs towards the rabbit at the same speed v". It's not clear that this produces the conclusion stated in the problem, "so that it always remains on the line between the center and the rabbit". A satisfying solution should double check that those two conditions are indeed equivalent.

PairTheBoard

[Borodog]

You are making too narrow an interpretation of the word "toward" which clearly does not jibe with the intent of the problem.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]


Notice angle EDA is twice the angle of FAB since it is the center of the smaller circle which uses the fact that AEC is 90 degrees . Therefore

arc(DE)/<EBD = 2pieR/2pie hence arc (DE) = R*<EBD
and arc(CB)/<2*EBD= 2pie*R/2/2pie , so arc(CB) = R*<EBD

QED

[/ QUOTE ]

I have my letters mixed up but the solution still holds .

It should be arc (fb) /<Fab =2PieR/2Pie arc fb=<fab*R
And arc EA/(2<Fab)=(2pie R/2)/2pie so arc EA= <fab*R

[/ QUOTE ]

I think I figured out which angles you meant in the first post. It's clear that if angles a=2b then the arc length a cuts on a circle of radius R/2 equals the arc length b cuts on a circle of radius R. My problem is I can't see any geometric argument showing that one angle is half the other. I see the right angle in there. But it's not helping me.

PairTheBoard

[jay_shark]

OK I'll explain some more .

Using the coordinates I gave , <cea = 90 degrees since ac is the diamater of the smaller circle with radius r/2 .We also know that < fab+ < cae = 90 degrees . Also , < cae + <eca=90
Therefore <eca= < fab

Now we use the fact that <eda = 2*<eca =2*<fab

[holmansf]

Two comments:

1) The problem stated here is not the same as the problem where the dog runs straight toward the rabbit at all times. In that case I don't think the dog would ever catch the rabbit. An interesting question might be to ask how the close the dog gets to the rabbit in that case. I don't know the answer.

2) I had to draw a picture to understand Jay Shark's solution. To prove the one angle is twice the other I used the fact that a certain triangle is isoceles since two of its sides are radii of the smaller circle.

[PairTheBoard]

[ QUOTE ]
OK I'll explain some more .

Using the coordinates I gave , <cea = 90 degrees since ac is the diamater of the smaller circle with radius r/2 .We also know that < fab+ < cae = 90 degrees . Also , < cae + <eca=90
Therefore <eca= < fab

Now we use the fact that <eda = 2*<eca =2*<fab

[/ QUOTE ]

ok, got it. Good job. It took me a while to see why <eda = 2*<eca. Boy is my Geometry rusty.

PairTheBoard

[PairTheBoard]

[ QUOTE ]
You are making too narrow an interpretation of the word "toward" which clearly does not jibe with the intent of the problem.

[/ QUOTE ]

I think jason1990's solution proves that it does jibe.

PairTheBoard

[PairTheBoard]

[ QUOTE ]
1) The problem stated here is not the same as the problem where the dog runs straight toward the rabbit at all times. In that case I don't think the dog would ever catch the rabbit. An interesting question might be to ask how the close the dog gets to the rabbit in that case. I don't know the answer.


[/ QUOTE ]

I think jason1990's solution proves that they are the same problem and that if the dog is always running toward the rabbit at the same speed he stays on the smaller circle constructed by jayshark. Furthermore, if he runs slower than the rabbit he stays on yet a smaller circle which never intersect the larger one, thus never catching the rabbit. If he runs faster he stays on a larger circle and catches the rabbit sooner.

PairTheBoard

[jason1990]

No, it does not jibe. I assumed only that |d'| = c. That is, I assumed only that his speed was some given constant. Guided by what seemed to be the standard interpretation in this thread, I threw away the "toward" part, which would require assuming that d' is a positive real multiple of r - d. You cannot keep both the "toward" part and the "on the line" part. They are inconsistent.

[PairTheBoard]

Well then so far as my last two posts go. Nevermind.

PairTheBoard

[holmansf]

[ QUOTE ]
I think jason1990's solution proves that they are the same problem and that if the dog is always running toward the rabbit at the same speed he stays on the smaller circle constructed by jayshark.

[/ QUOTE ]

No. Draw the circles. Take a line from the center of the larger one. This line is only tangent to the smaller one at the start.