Distance: CR4 Challenge (04/22/08)

This one is too easy, if you think geometrically instead of just numerically.

To simplify things, assume that cities A and B are at the same latitude, or use an azimuthal equidistant map. That means use can draw a triangle connecting the cities. You can also draw a circle with B at the center, and A and C on the circle. If the triangle is an equilateral triangle, then all 3 cities are 9000 miles from each other, and all the angles are 60°. You might be tempted to think that the sector of the circle created by this triangle represents the proportion of the circle where C could be, but we have to consider the whole circle. So, the proportion of where C could be is represented by a sector created by 2 equilateral triangles. This gives an angle of 120°. 120/360 = 1/3, or 33% probability. If C is located at any other point on the circle outside this sector means that C is farther from A than it is from B.

I would agree, sort of. The lines you have drawn are not actually straight lines, but should be rhumb lines. In the instances of the two chords you have drawn they would actually be the curved segments. You will not observe the curve in the lines that originate from city B, but they do exist, so the surface distance across the geoid is in fact 9,000 miles.

Such is the problem with working with a sphere and thinking in terms of 2-D.

However, I do not agree that the probability is as you stated. That would be true if the circumference of the globe was 4 times 9,000 miles or 36,000 miles, but it is not!

Such is the problem with working with a sphere and thinking in terms of 2-D.

Isn't an azimuthmal equidistant map a 2D projection of a the 3D spherical surface?

Now, if we're talking distances through the Earth, that's another matter.

Azimuthal equidistant maps are projections that maintain proportional distances along the azimuth. The minute you need measurements along any other direction these maps become irrelevant (or even misleading?) Take the example of a projection centred on the North pole, and look at a circle 1-mile radius centred on the south pole. That will look to have a circumference in the region of 80000 miles, instead of about 6-miles which is what it should be.

To get a feel for this I suggest:
Get an old baseball, put a pin through the ball to mark B. Use a thread centred on B that is 3/8 of the circumference of the ball to mark a number of spaced points on the ball that are 3/8 of the circumference from B (you will need to arrange each point separately, as the ball is not smooth). Now fill on the circle that contains these points. You will see that the radius (along the surface of the sphere) of this circle is 1/8 of the circumference of the sphere. Equivalents to A and C both lie on this circle; the maximum distance between them is when they are diametrically opposed - or 1/4 of the circumference of the sphere.

Assume the three cities are on planet Earth and distances between cities is measured along the surface. The Earth has a diameter of about 7890 miles and a circumference of about 24787 miles. If we measure straight line distances between cities, we couldn't be on Earth. All locations 9000 miles away from city B as measured along the surface lie on a circle with a diameter of 5980 miles. If city B were at the South pole, this circle would be at about 40.7 degrees Northern latitude. The circumference this latitude is 18,788 miles. Half of this is 9394 miles. The probability of city A being closer to city C than city B is 9000/9394 or 0.958.

Nice work!

Interesting but there is more to consider. A city by definition is a large metropolis and thus a city can not be located on water nor on the polar ice caps. This basically excludes 75 % of the Earth's surface. As most of the usable land mass sorta sits more to the north of the planet, this will greatly shift the odds that the A & C cities are indeed closer to each other than the 9000 mile reference.

I would even venture a guess of 95% or more!

Surely this is irrelevant, as the cities cannot all three be on the surface of the Earth and A and C be further apart than 7000 miles?

I see your point. A is alway closer to C than B. The probability is 1.

Thanks

I agree Fyz, you're bang on the money as usual.

Because I like visuals, here is my "not to scale" pictorial representation :-

Just to be very picky: I'm sure the circle containing the possible positions of C must go through A

And, I'm sure Fyz meant 24874 and 6874 rather than 6784 miles (from the average radius of the earth). Although strictly speaking if the cities are on the equator they could be 6901 miles apart.

Agreed - except to note that I gave my 6903 miles as a partial allowance for equatorial cities being above sea level... I'd also be inclined to add (curved) paths AB and BC to show the distance - but either way it's extraordinarily difficult to show spherical geometry on flat paper.

The drawings by yourself and MPM are both good, though I'd have stuck with angles (ie about 130^o) rather then distance. The notion of 130^o + 130^o leaving a maximum 100^o separation from A to C seems more intuitive to me.

Just trying to answer the question in the terms it was presented, perhaps... To my mind 3Doug's latest drawing (using more advanced techniques) is clearest.

I can't resist adding, in support of my angular view, the nautical mile is defined as........ Well, it would stop anybody arguing about which planet the cities are on. Maybe !

Nice thought, but I thought that Wikipedia got the present situation right:
. "The international definition is: 1 nautical mile = 1,852 metres exactly."
Then, of course, other planets might be dramatically more oblate than Earth, which could just about invalidate unity if city B was on the equator

I get propelled at a rate of nautical miles per hour so easily.

Yes it would seem that if you assume earth is the planet in question, you do come up a bit over 2000 miles shy of the 9000 mile goal.

Now it is time to explore the possibilities on other planets. Take Jupiter for instance, I see a huge probability of exceeding that 9000 mile target there. And why not, look at the gravity question a few weeks ago. There was certainly a lot of imagination that overflowed onto the floor there. And the question did not specify the planet in question.

Cities on a planet with such high gravity? Plus I don't know what is the diameter of the solid core (can you advise?).

Simpler to start with a differnet distance on the Earth, perhaps? Perhaps try 1/4 of the planet circumference for a relatively easy starting point.

Your picture makes it all the explanations clear, thanks.

The illustration is in reverse of the given order; B is " " from C, A is " " from B..

'The greatest possible distance between A and C is simply 24784-18000 = 6784 miles.'

...then you are obviously making assumptions somewhere that are incorrect, because the problem clearly states that the cities are 9000 miles apart. You can't simply throw away the only given data because it doesn't fit your line of reasoning and hope to arrive at the correct answer.

The statement you say contradicts the question is about the distance between A and C. The question gives the distances between A and B and between B and C respectively. The distance between A and C is what you are requested to estimate as part of the challenge. No data was ignored. OK?

Fyz

OK, part two of this question:

If city A is traveling west at 120 mph and city C is traveling east at 135 mph, where will the two cities collide?

I agree with 3DOUG's logic, my best friend Auto-Cad and I have come to nearly the same conclusion however the the distances between A to B and B to C must be considered as circles. And at the points where the distances between circle AB meet with circle BC creates a 120 degree circumferance segment. 360/120 equals 1/3, however at those exact points, they are not closer, they are the same and that is not the answer to the question. The answer to the question is the probability that city C is closer to city A that to city B is less than a 1/3 probability. BLAMO!

You are making the problem too hard. Imagine taking a ball that represents the Earth. Let the circumference around the ball represent 24,300 statute miles. Now drive a pin anywhere in the ball. This is city B. Next, cut a length of string that represents 9,000 statute miles relative to the ball's circumference. The string will be 90/243 of the ball's circumference.

Attach the string to the pin and the loose end to a pen. Now draw a circle on the ball around the pin with the string tight. Somewhere on this line is where cities A and C reside (both are 9,000 statute miles from B).

Next, relocate the pin somewhere on the line you just drew. This represents city A. If you try to draw another circle and you will see that city C, no matter where it might sit on the first drawn circle, will always be inside the reach of the string.

Why? 9,000 statute miles is greater than 1/3 the circumference of the Earth. Therefore the area swept out by the string attached to city B will be greater than 2/3 the surface area of the Earth. Any two points along the limit of the swept out area of the Earth will always be less than 1/3 the circumference of the Earth apart, which is significantly less than 9,000 statute miles.

Assuming the cities lie on a hypothetical infinite plane (flat) and not on the Earth's (or another sphere's) surface, then what is the solution?

When you think of it like that one might think, "well, are there hills, mountains, valleys? or might the said cities be underground?" unexpected variables that only infinateley complicate the problem. information has been left out of this problem, as to location topography. damn!

I will say the answer is 50%

Statistical point of view

probability = no of ways/total numbers

= 1/2

= 0.5 which is 50% of either being close to A or B

AC1 = r X θ (degree)

= r X (θ X π/180)

= 9000 X (θ X π/180)

d(AC1) = 9000 X (π/180) X d θ

As, d(AC1) < d(AB)

9000 X (π/180) X d θ < 9000

(π/180) X d θ < 1

d θ < (180/ π)

d θ < 57.3º

FOR TOTAL ARC LENGTH

2 X d θ < 114.6º

So, probability of AC1 always less than AB = 114.6/360 = 31.83%

Since question doesn't mention anything about surface I'll take it as a flat plane.

Hi,

my solution is 95.8%.

Assuming that the Earth is a sphere with a radius of R=3945miles.The distance between B and Ais 9000 miles so the angle between the two points is 9000/3945=2.28rad=130.7°.

the point A can be on a circle with radius r=R*sin(2.28)=2990miles. Also C has to be on the same circle. If the distance between A and C is 9000 miles the angle between them is 9000/2990=3.01rad=172.4°. The point C is closer to A than to B if the angle between A and C is inside the range -172.4° and +172.4°. The probability, thus, is 172.4°*2/360°=0.958..... 95.8%.

Your "distance between A and C" follows the circle on which they must lie. So your maximum "distance" is half the circumference of this circle. But the standard measure of distance would be the chord - longer than for a circle on a plane with the same circumference, but not all that long. Of course, the longest chord is the diameter - which is the circumference of the Earth minus twice 9000 miles. (Saves all the trig, too!)

I'm not agree...

The distance between two points on the earth surface is the arc length not the chord.

Guest appears to be using a standard abbreviation - in spherical geometry the shorter arc along the great circle is often referred to as the chord. The length of this "chord" is of course the distance between cities A and C.

You could try drawing a small circle on the sphere, so the surface appears approximately flat. That makes it more obvious what is the difference between great the circle and the one on which A and C must lie.

Guest appears to be using a standard abbreviation - in spherical geometry the shorter arc along the great circle is often referred to as the chord

ok... and why he said "the longest chord is the diameter"?The longest "chord" is half circumference, isn't it?

And so, what's wrong in my proof?

I'm sorry.... Now I get it!

.......the point A can be on a circle with radius r=R*sin(2.28)=2990miles. Also C has to be on the same circle.The maximum distance between A and C is R*(6.28rad-2*2.28rad)=6798 miles! So the probability is 100%.

To satisfy the requirement that city C will be closer to city A than city B the answer is slightly less than 33.33%.

To satisfy the requirement that city C will be equidistant or closer to city A than city B the answer is 33.33%.

Explanation: keeping city B in centre and BA & BC as two radii from B. Now keeping BA fixed and revolving BC around B. The condition can be satisfied only when <ABC is <= ±60º (since both sides are equal to 9000 miles and if the angle made between them is 60º then the third side will also be equal to 9000 miles, further, any angle smaller than 60º will make CA < CB) i.e. total internal angle when the condition will be satisfied is 120º out of 360º. So the probability for the said condition is 120 / 360 = 33.33%.

Are you by any chance a member of the "flat Earth" society? If so, why do you say "slightly less than" 33.33%?

Maybe the following explanation is clearer??

Call the longest circular path around the (approximately spherical) Earth "circumference".

Place point Y diametrically opposite city B. The distance between them (along the surface of the sphere) is circumference/2
And we can always route a circumference connecting B and Y so it passes through A or C
So YA = YC = (circumference/2 - YB)
AC ≤ AY + YC < 25000 - 18000 = 7000

So A and C are always closer together than A and B.
I.e the probability is unity.

There is a high probability. City C could be geographically located only a few miles from city A and still be 9,000 miles from city B.

geographycally this is impossible, maybe scientifically is possible

because the diameter of the earth is 7926.14 miles i.e including water.in conclusion the probability is 0% hence the distance from city A to city B is longer than the diameter of the earth, unless those cities are not on earth.

The diagram above shows city A at the center of a white circle with a 9000 mile radius.

City B is shown at a point on that white circle and in the center of a multicolor circle with a radius of 9000 miles.

The green arch represents the possible position of city C at any point where C is closer to A than B.

The red arch represents the possible position of city C at any point where C is further from A than B.

The red arc length is 37699.1118.

The green arc length is 18849.5559.

The red arc is twice as long as the green arc.

The probability that C is closer than B to C is 18849.5559 / 56548.6678 or 1/3

I'm not sure of the proper way to show a probability but I hope my diagram will adequately explain my theory.

I see some here assumed the measurements were taken on earth. I did not. I assumed a flat plain.

Best Regards,

Spencer

well i agree for diplomatic reason.

the arc might cover that length, but now you draw a straight line and consider that as the displacement from one point to another. what i'm trying to say is the actual distance u traveled from one piont to another is not exactly the distance from those place, considering the topography and other factors.

So the circumference of the globe is irrelevant?
OK, let us pretend it is 18000 miles exactly.
Now choose the location of city B as your reference. The distance between A and B is half the circumference of the globe, there is just one place that A can be - diametrically opposite to B. The same applies to C.

That says cities A and C are in the same location (different names for the same city?).

Obviously, the probability in this case is unity. Now make the globe a little bigger - say 25000 miles. A and C are now 3500 miles away from the point diametrically opposite B. So the largest possible separation between A and is ?? And the probability that they are closer than 9000 miles is??

Assuming for a flat earth the two cities A and C are both on a circle around B with a radius of 9000 miles. If we fix The City C and trace possible locations of A on the circle with a chord length (AC) smaller than the radius. We can allow only 60 degrees towards each side of the fixed city C.

Which means a central angle of 120 degrees at B. Total circle is 360 degrees

thus the possibility is 1/3

At an angle range between 120º and 240º, the distance between A and C are closer then a B to C. If you divide the portions in 1º you would get 118 places where it would be closer. Since the universe is 360º you have to divide 118 by 360 and get 0.3277. That´s what I think.

Since RsubA = RsubB, the circles intersect at points 120 degrees apart, so probability = 1/3.

Since the circumference of the earth is approximately 25,000 miles and the distance between the cities is 9000 miles, not 25 miles, then one can not assume a flat plane and assumes a spherical surface. Therefore, the probablility is 1.

my 2 cents:

if we assume an infinitely large flat space - Both A and C lie on circle (r = 9000 miles: C = 28274 miles). Determine an angle alpha where the cord length = 9000 miles

solve cl = (sin^2(alpha) + (1-cos(alpha))^2)^0.5 = 1 (from unit circle) it happens at alpha=60

now use (2 x alpha) / 360 for for a percentage

(2x60)/360 = 33%

on the other hand, if we assume earth (it is not stated) then

Let B arbitrarily lie on the North Pole (it could be anywhere -just rotate the reference frame)

Both A and C lie on a Latitude exactly 9000 miles south. Determine that latitude (at Earth C = ~25000 miles - it is not beyond the South Pole (which would eliminate any possibility) From that latitude - determine the Circumference - simple ratio to determine likelihood of them being within 9000 miles of each other.

Assuming a perfect sphere - it's not... but

NP to Equator is ~6250 - an additional 2750 beyond that is 44% of the way to SP. That's about 40º lat. The circumference of that line is ~19250.

A can lie anywhere on that circle. if C lies +/-9000 miles (an 18000 mile range) it satisfies condition.

therefore 18000/19250 ~= 93.5%

... one last thought - great circle route from 2 oposing pts on 40º Lat through the sout pole - is only 6900 or so miles - therefore - once considering that - odds are 100% that A and C will be closer.

So here's a trivia - at what distance smaller than 9000 miles do the odds fall below some threshhold - say 90%?

I can't easily answer your question. I can offer this.

On earth the farthest possible distance between A & C is appx. 6,260 miles.

Please explain. Are you taking straight line distances through the Earth? That would be a different way of measuring than used in the question (assuming that the cities are on Earth...)

If you know of a flat space that is larger than 18000 miles in diameter that has cities on it, I'd love to visit

No good answers so far. Doesn't seem that hard. Think: areas...as in areas which are the set of all radii in a circle. B at center of circle. A and B at points on circle. Calculate equilateral triangle containing A, B, C at smallest possible equal distances from each other. All radii, hence all paths from B to C will lie within the arc containing the equilateral triangle ABC; hence the ratio of the area of that that pie (pun?) sector to the the area of the circle will be the probability that the distance from A to C is equal to or less than that from A to B. Its complement, the ratio of the area of the circle outside the pie to the area of the circle is the probability that A is farther from C than it is from B.

Now someone go do the drawing and calculate the approximate .65 and .35 probabilities.

PS. no calculation for curvature...should not make significant/substantive difference.

Oops! Left out something. Double the area of the area of the "pie" sector...because an equilateral triangle (two in fact) would include paths from A to C on either side of the radius, AB.

Sorry if that threw anyone.

PS: the first guy was on the right track...just didn't make the leap from lengths to areas. Good work guy.

Also, A and C are points on circle, not B which is at center. Haste making waste.

50 percent

Bill

Okay, let me clean up thread #47 (posts 47, 48, 49, 52) to to make things more convenient....

No good answers so far. Doesn't seem that hard.

Think: areas...as in areas which are the set of all radii in a circle. B at center of circle. A and BC at points on circle. Calculate [area of] equilateral triangle containing A, B, C at smallest possible equal distances from each other [that's the distance given by challenge]. [Double that to account for an equilateral triangle lying on the other side of AB. All radii, hence all paths from B to C [that are equal or less than this shortest distance] will lie within the arc one of the arcs containing the [either] equilateral triangle ABC on either side of AB; hence the ratio of the area of that pie (pun?) sector [comprising the sectors on both sides of AB [(the doubled area mentioned above)] to the the area of the [entire] circle will be the probability that the distance from A to C is equal to or less than that from A to B. Its complement, the ratio of the area of the circle outside the [two-sector] pie to the area of the [entire] circle is the probability that A is farther from C than it is from B.

Now someone go do the drawing and calculate the approximate .65 and .35 (not confirmed yet) probabilities.

PS. no calculation for curvature...should not make significant/substantive difference

Hope this makes it easier.

The sphere makes all the difference.
[On a flat surface with A and C positioned independently of each other, the obvious answer of exactly 1/3 (or approximately 33.333%) is correct.]

As you have taken the trouble to present 5 similar posts on the topic, please make a semi-practical check before you walk away from the problem. Get yourself a sphere (any moderately stiff near-spherical ball will do), and mark a point on it as B. Then measure a distance of 1.3 times the sphere's diameter along the surface of the sphere (starting from B) and mark the end point as A.
Now measure the same distance away from B, in any direction you choose, and mark that end point as C. How far away from A is C (taking the shortest route along the surface of the ball)? Then, if you still believe it, come back and tell us all that you have found that Anonymous Hero's answer number 6 and my answer number 5 were "not good answers".

Fyz

Thanks. And, indeed, it depends on how one guesses the 9000 constituted the "trick" to the question. You aptly saw it as a (deceptively) inductive problem pertaining to Earth and great circles. I (possibly no less aptly? only the CR4 shadow knows) viewed the 9000 instead as something to throw guessers off from the concept of areas (sets [continuous probability spaces] of uncountable radii on a [practically--given the posited distances] planar surface)...on an ideal surface as opposed to Earth per se.

You are also right, your answer is GA deserving, as perhaps is mine (both in terms of insight), I believe.

Thank you for taking time to respond.

The other post, it was sufficent to view since I couldn't find a real sphere nearby.

Fascinating! ...fella complains, justifies his GA, ask for reconsideration. Respondent concedes point then give fella a GA--even promotes it to firm GA. Is rewarded by having own GA demoted by same fellas who asked and received GA promotion. Talk about fairplay? Guess it's only about collecting GA's? Guess being nice to a fella gets its comeuppance.

For me it was about clarification.

N.B. I didn't think guests could vote on the quality of answers. (If so, I could just log in from a number of places as guest and vote myself all the good answers I wished - mind you, with only marginally more trouble, I could log in under 2000 aliases and do the same)

Hmm. I suppose you could do that if you were of a mind to.

Fyz, et al,

I think that the trick here is to go with what is implied by the question. In this case the clues are cities and distance.

So far, the only known cities we have observed (without mind altering dugs and/or science fiction) are on Earth.

It is reasonable to assume the challenge speaks to cities on Earth.

Second, the normal method to measure distance on this planet is along the surface of the sphere, not through it. This has been the standard for thousands of years. We could be anal and use WGS84 geoid.

So, it is reasonable to assume that the distances between cities is the distance along the surface of the sphere.

One more thing, the challenge stated 9,000 miles between cities and there is plenty of slop in that number that even using a geoid with a circumference of 26,0000 miles you get the correct answer (100% probability). That means that we don't need to use a calculator, just our brains and some common sense.

So, in my opinion, not only has Fyz arrived at the correct answer first, but has done an excellent job explaining the answer to all. You get your second "Good Answer" from me.

Thanks - only timing and dupliction impeded your getting the good answer vote (though you had one from me). Personally, I would like to rate the illustration of #62 as a good answer - but I'm not willing to cast multiple votes

If city B is 9000 miles from city C, and city A is 9000 miles from city A, using a radius of 6874 for the earth, there is a 34% chance that city A is closer to city A than to city B.

Using a radius of 6,874 miles for Earth;

If city B is (at any given point on a radius) 9000 miles from city C, and city A is (at any given pont on a radius) 9000 miles from city B, there is a 34% chance that city A is closer to city C than to city B.

Error...Error...Error...That does not compute....

Trying to squeeze in too much during a break, which was full of interruptions.

I am having difficulty inserting my illustration that is complete with the calculations, however; Try this: Construct a globe with a radius of 6784 miles, and establish a location for city C, then mark a circle at 9000 miles away. Select a point on that circle and establish a location for city B and mark a circle at 9000 miles away. This should help to illustrate things in a way that is easy to understand. You will see that 50% of the possible locations for city A will be further from city C than city B, and 50% will be closer.

There is a 50% chance that City C will be closer to City A than to city B.

Well: I suppose if the world did double in size, it would solve lots of the current overpopulation and food and fuel shortages.

I think that the probability will be less than one third, or 33.33 per cent. A graphical explanation follows (Fig. 1): If the cities are A, B, C respectively, C will be closer to B than to A if C stays on the blue line, and C will be closer to A than to B if C stays on the red line, because now C will be into the black circumference of 9000 mile radius. The violet line divides the green line in two equal parts (4500 mi ea) - in fact, the "two circles" method is long ago used to divide a line in equal parts - then, the b angle will be arc cos 0.5, or 60 degrees, the full arc of two times b will be 120 degrees, and as the full circle has 360 degrees, the probability will be proportional to the ratio of 120 / 360, or 1/3.

Figure 1

the correct answer is 1/Pi, or a little less than 32%.

C is somewhere on a circular path with radius 9000 miles and B at the center. A is also on the same circular path. The total length of the path is Pi*diameter, or Pi*18000. The places where C is closer to A than to B are 9000 miles along the circle to one side of A, and 9000 miles along the path to the other side of A. Thats 18000 miles of path where C would be closer to A than to B. The only other places C can exist are on the circle, with a total length of 18000*pi , so the odds of C being closer to A than to B is the ratio of possible closer locations to total locations, which in this case equals 18000/Pi*18000 which equals 1/Pi which is roughly 31.8%.

Why would you measure the distance from A to C along the arc of the circle instead of "as the crow flies"?

If you can't assume that the problem is intended to be applied to the present day Earth then: you have no information at all about the structure which the cities are embedded in, so, (assuming only 3 dimensions and that the cities remain in fixed position relative to each other) the problem becomes one of the intersecting surfaces of two spheres and the probability is ¼.

Ok,my first attempt was too simple, but now I've got a better approach. This time I kept with the meaning of my screen name. I have also changed my position, and have an illustration to prove it.

First, I looked up the mean radius of the Earth, which Wikipedia gave as 6,371 km. This converted to roughly 3959 miles.

Next, I looked up a formula for arc length. The source this time was Pocket Ref. The formula, after some manipulation of a factor involving pi, is:

Arc length = 0.01745rΘ

I already knew arc length (9000 miles), and the radius (3959 miles). So, I manipulated to solve for Θ, and came up with an angle of roughly 130°.

What was I going to do with this information? I was expecting to create a somewhat dome-shaped "slice" of the Earth. The peak of the dome would be city B, and the edge would be a circle described by radius of 9000 miles from B along a great circle path. Cities A and C would then be on the edge. Instead of a dome, I came up with this:

The edge of the "crater" is where cities A and C will be. If the A and C where at opposite ends of the diameter, which is the fartherest apart they could be from each other, they would only be 6678 miles apart. So, they will be closer to each other that to B.

Now, the relevant question is: how many of you noticed that this resembles the Death Star of Star Wars?

'earth' is neither mentioned, nor implied in the original problem, and the numerical data actually exclude the earth being involved in all but the most abstract manipulations.

why do people feel compelled to fit the square peg in the round hole?

silly hominids.

You pseudo-aliens can be so irritating...

If you personally know of cities other than on Earth, you are within your rights to say that Earth is not implied. The measure "miles" also implies Earth, because no species other than homo insipiens uses it - and even most humans use km for theoretical measures outside the solar system

Distances between cities are generally measured along the surface of the Earth, so 9000 miles is quite possible. (E.g. central USA cities to Australian cities).

Finally, if you wish to include planets other than the Earth, you will presumably have to count all planets that have cities, and take all pairs of cities. Of those I can observe, only 3 have a circumference greater than 27000 miles, and they all have a single large continent whose largest dimension is less than 18000 miles.

For observable cities, the probability remains unity. Based on the available evidence, it would be very high indeed - greater than 99.99%.

The Alien

If cities are implied why not not imply Earth?

...City A is 9000 miles from city B, and city B is 9000 miles from city C. What is the probability that city C is closer to city A than to city B?...

Cities are not implied.

Since the answer is a percentage, requiring no units of measure, it doesn't matter if the distances are measured in miles, furlongs, decimeters, spans, or the long dimension of a 4x2 LEGO building block, the answer is still as I explained earlier: