Circle's slightly harder challenge

[dontdisturbmycircles]

Alright, well seeing as my last challenge lasted a whole 30 minutes before Jimmy Snyder solved all my brain teasers, I dug deep to find more challenging ones.

Please use and and the closing tag [/ color] to make your text unreadable to those who don't want to see the solution, thanks.

If you recognize any of these puzzles from old threads please either ignore them or link to them and I may remove those puzzles from this thread to ensure that these are all new. Thanks.

Good luck!

1.Three distinct points with integer coordinates lie in the plane on a circle of radius r > 0. Show that two of these points are separated by a distance of at least r^(1/3).

2.What chemical compound is represented by the following: HIJKLMNO?

3. On his way to work each day, a man living on the fifteenth floor of an apartment building takes the elevator to the first floor from the fifteenth floor. On his return, he is forced to take it to the seventh floor and walk the remaining eight floors to his apartment. Why?

4.Three criminals just robbed a bank and go back to their hideout. They put the money behind a high tech security door. There are 3 locks on the door, each activated/deactivated by a button next to it. All locks are originally deactivated, and once a lock is activated it is impossible to tell whether it is activated or not. The three criminals want to work out a system so that any two of them can access the money but a single criminal cannot. The 2 criminals accessing the money must be assured that all locks are deactivated, otherwise an alarm will sound, and built-in lasers will shoot them. Also, each criminal may only give information about which locks he toggled to one other criminal. Figure out how their system will work.

5.Punctuate the following so it makes sense: Alice while Matthew had had had had had had had had had had had a better effect on the teacher.

 

[arunbg]

2. H₂O

lol...

 

[dontdisturbmycircles]

edit: Yup, :tongue:

 

[arunbg]

3. A probable answer would be that the man has his office also on the first floor of the building, and the office extends from floors 1 to 7 (say) and the elevator is privately owned by the office for the employees . Note that the first statement of the question doesn't say from which floor the elevator is boarded, only the destination (1st floor). So the man may well have his residence on the 15th floor, walk upto the 7th floor and then take the elevator to the 1st floor.
Who would construct such a building anyway ?

5. Can we delete spaces, so that we may combine all but the last two hads into a very big surname for Matthew

 

[dontdisturbmycircles]

Lol, nice try on both .

Highlight to see :


I am afraid that you are not quite right on either, but your thinking on Q#3 is on the right path, it is something quirky like that.


For Q#5, no you can't get rid of spaces, nice try

 

[arunbg]

Well, I only said that my answer to question 3 is only "possible", don't you agree. Maybe you could add a few more details so that you may contradict this case ?

Cheers


Arun

 

[dontdisturbmycircles]

True enough I suppose, I shouldn't have said you were incorrect. There is another answer though. heh. I'll post it in a few days.

 

[dontdisturbmycircles]

I also changed it so that he takes it from the 15th floor in the morning.

 

[Jimmy Snyder]

3. On his way to work each day, a man living on the fifteenth floor of an apartment building takes the elevator to the first floor from the fifteenth floor. On his return, he is forced to take it to the seventh floor and walk the remaining eight floors to his apartment. Why?

He is too short to reach the buttons above the 7.

5.Punctuate the following so it makes sense: Alice while Matthew had had had had had had had had had had had a better effect on the teacher.

Alice, while Matthew had had "had had" had had "had". "Had had" had had a better effect on the teacher.

I showed this to my wife and she gave a similar answer to this except that I while she had had had had had had had had had had had had had had had had had had had had had had been in the answer I posted.

 

[kant]

the first problem is hard

 

[dontdisturbmycircles]

He is too short to reach the buttons above the 7.

Alice, while Matthew had had "had had" had had "had". "Had had" had had a better effect on the teacher.

For Q#3, that is the "original answer" good job.

Correct on Q#5 as well, that is the answer (pretty close to "original" answer but its right anyway)

The "original answer" is::

Alice, while Matthew had had "had," had had "had had." "Had had" had had a better effect on the teacher.

 

[daniel_i_l]

#4:
the first two criminals activate or deactivate the first lock. then the first and third ones together either activte or deactivate the second one. then the second and third one either activate or deactivate the third lock. this way any given criminal knoes the position of only two locks.
am i missing something?

 

[Jimmy Snyder]

dontdisturbmycircles,
Given the difference between my answer and the original answer, you should be able to punctuate the sentence at the bottom of my post.

I showed this to my wife and she gave a similar answer to this except that I while she had had had had had had had had had had had had had had had had had had had had had had been in the answer I posted.

 

[dontdisturbmycircles]

"I showed this to my wife and she gave a similar answer to this except that I while she had had had had had had had had had had had had had had had had had had had had had had been in the answer I posted." - Jimmy Snyder

I can't see any other way to correctly punctuate "except that I while she"

:uhh:

 

[Jimmy Snyder]

I showed this to my wife and she gave a similar answer to this except that I, while she had had 'had had "had" had had "had had"', had had 'had had "had had" had had "had"'. 'Had had "had"' had been in the answer I posted.

By iterating in this fashion, any number of had's can be strung together and meaningfully punctuated.

 

[dontdisturbmycircles]

Yea definitely, it still takes some effort to do it though (although it is not a very creative process) I tried to select a few puzzles for different types of people. I already knew that any number of had's can be strung together in that fashion, it's not the kind of puzzle I enjoy doing but some may.

Although of course you still realize that your answer to your question is not grammatically correct.

 

[Jimmy Snyder]

Although of course you still realize that your answer to your question is not grammatically correct.

It seems ok to me. What's the problem?

 

[davee123]

1.Three distinct points with integer coordinates lie in the plane on a circle of radius r > 0. Show that two of these points are separated by a distance of at least r^(1/3).

This one's still bugging me. I've asked a few people I know, and so far everyone's stumped.

Just so I'm clear, though, can the question can be re-written as the following?

Three distinct points A, B, and C lie in the x,y plane, and each have exact integer coordinates. Each of the points also lies on a circle of radius R. Prove that at least one pair of these points is separated by a distance greater than or equal to the cube root of R.

Any hints on the method of solving it?

DaveE

 

[dontdisturbmycircles]

Had to ask a friend how to do this as I sort of forget how he showed me, use the points to form a triangle / circumcircle. If you need more help I have more hints. And yes I believe that your wording is fine.

What is the equation for the Circumradius if the sides are a,b,c, and D = largest distance between two points, and A = area?

 

[davee123]

Had to ask a friend how to do this as I sort of forget how he showed me, use the points to form a triangle / circumcircle. If you need more help I have more hints. And yes I believe that your wording is fine.

What is the equation for the Circumradius if the sides are a,b,c, and D = largest distance between two points, and A = area?

Alright. Uncle!

After trying this for a couple days, I asked a few of my more geeky friends, but nobody could figure it out. So I sent the problem to a math whiz at MIT, a college math professor, and someone who majored in math a long time ago. Nothing.

So far, I think the best working theory someone had was that there were only a few quasi-rational coordinates on the unit circle. Not that the coordinates are actually *rational* per se, but that the radius could be adjusted to a particular value that would *make* them be rational. For example coordinates sqrt(3)/2, 1/2, which, while irrational, can be made to *be* rational with a radius that's a multiple of sqrt(2).

Anyway, the assumption was that there's a fixed number of possible exact-integer coordinates on any given circle. And knowing that, you can calculate the circumradius of the points and compare it to the distance of the furthest points, and get an answer.

But otherwise, I dunno. The sticky wicket in all cases is somehow establishing equations wherein the coordinates of the three points are necessarily integers.

Could you post (or PM me) the proof?

DaveE

 

[dontdisturbmycircles]

Here it is. The question appeared on a putnam exam (Had to find hard ones!) :-)

We are given three points in the plane. Let D be the largest distance between any pair. If we view these points as vertices of a triangle, then we may view r as the radius of the circumcircle of a triangle. If the lengths of the sides are a,b, and c, and the area is A, then the circumradius is given by r = abc/(4A). Since D = max(a,b,c), we have r < D^3/4A, whence D > (4A r)^(1/3)

On the other hand, the area of the triangle may be determined from the coordinates of its vertices; it's half the area of a certain parallelogram, which is in turn computed by a determinant, showing that the area of such a triangle must be at least 1/2. This gives D > (2 r)^(1/3), a slightly stronger result than was asked for.

 

[davee123]

Here it is. The question appeared on a putnam exam (Had to find hard ones!) :-)

We are given three points in the plane. Let D be the largest distance between any pair. If we view these points as vertices of a triangle, then we may view r as the radius of the circumcircle of a triangle. If the lengths of the sides are a,b, and c, and the area is A, then the circumradius is given by r = abc/(4A). Since D = max(a,b,c), we have r < D^3/4A, whence D > (4A r)^(1/3)

On the other hand, the area of the triangle may be determined from the coordinates of its vertices; it's half the area of a certain parallelogram, which is in turn computed by a determinant, showing that the area of such a triangle must be at least 1/2. This gives D > (2 r)^(1/3), a slightly stronger result than was asked for.

Hm. Ok, I still don't quite get it.

I understand how we get to D > (4Ar)^(1/3), that's fine. Although, I don't quite see:

1) How do we know the area is at least 1/2? I may figure that one out, cuz... that looks like it might be something I could prove... But I'm still fuzzy on why that's true.

2) So, if the area is 1/2 or greater, I understand how we get a minimum value and it turns into D > (2r)^(1/3). But aren't we looking for D > r^(1/3)? Should the problem have been stated as "the cube root of the diameter" instead of "the cube root of the radius"?

DaveE

 

[dontdisturbmycircles]

if D > (2r)^(1/3) then D must also be greater than r^(1/3). right?

 

[gabee]

Here it is. The question appeared on a putnam exam (Had to find hard ones!) :-)

We are given three points in the plane. Let D be the largest distance between any pair. If we view these points as vertices of a triangle, then we may view r as the radius of the circumcircle of a triangle. If the lengths of the sides are a,b, and c, and the area is A, then the circumradius is given by r = abc/(4A). Since D = max(a,b,c), we have r < D^3/4A, whence D > (4A r)^(1/3)

On the other hand, the area of the triangle may be determined from the coordinates of its vertices; it's half the area of a certain parallelogram, which is in turn computed by a determinant, showing that the area of such a triangle must be at least 1/2. This gives D > (2 r)^(1/3), a slightly stronger result than was asked for.

Do we need to pay attention to the fact that the problem says the point coordinates are integers? I was trying to solve it a totally different way.

 

[dontdisturbmycircles]

Yes it is very important because the points become lattice points and Pick's theorem can be applied, and in general allows you to easily compute the minimum area of the triangle who's vertices are the 3 points.

Davee, you can use Pick's theorem to prove that the area of a triangle whose vertices have integer coordinates is at least 1/2.

Actually, the fact that a primitive triangle has an area of 1/2 can be used to prove Pick's theorem.

 

[dontdisturbmycircles]

edit: double post, sorry.

 

[davee123]

if D > (2r)^(1/3) then D must also be greater than r^(1/3). right?

Oh yeah, duh.

Davee, you can use Pick's theorem to prove that the area of a triangle whose vertices have integer coordinates is at least 1/2.

Huh-- I had never heard of this one before... Interesting.

DaveE

 

[dontdisturbmycircles]

Yea, I admit, it was a hard one... :P

 

[dontdisturbmycircles]

I probably should have said that you needed to know Pick's theorem or something, I hope you enjoyed trying to solve it none the less. Next time I post a problem that requires a mathematical theorem or something of the nature, I will put in a hint so that this doesn't happen again. My appologies, I wasn't really thinking.

 

[daniel_i_l]

I think that you can solve it without pick's theorem, here's how i did it:
-first put the first point on the circle, to make it easier let's assume that it's on the y-axis with the height of r which is an integer, this doesn't change anything cause i can always rotate the coordinate system so that it is on the y-axis.
-now we have to put down two more integer points, one with a positive x and one with a negative one (if you want to put them on the same side then just choose a different point as the "midpoint" so that there will be one point on the positive y-axis and two more on either side of it)
-since it's obvious that the longest line will be between the two points on either side of the middle we have to prove that that distance will always be bigger that r^1/3.
-it's also obvious that for the shortest line both points will have the same height and opposite x's ( (x,y) , (-x,y) ) because if this isn't true and one of the points has a smaller x that the other one than we can aways make the line shorter by making the point with the bigger x have the same value (with the opposite sign) as the smaller one.
-so the distance d = 2x where x is the distance of the points from the y-axis.
-so we have to prove that for every right-angle triangle with 3 integer sides, the smaller side (x) will always be bigger that 0.5r^1/3 where r is the diagonal.
-to prove this we use the following equation:
x^2 = 2ny + n^2 for every right angle integer triangle where x is the smaller side, y is the bigger side and n is r-y. it is easy to see that this is true by drawing r^2 as a square with a side the length of r and then to fill in it's bottom right corner with a square the size of y (the bigger side) all the area left is equal to x^2.
-using this it's easy to show that x > r^1/3 and then d > x > r^1/3
Q.E.D
what do you think?

 

[davee123]

-it's also obvious that for the shortest line both points will have the same height and opposite x's ( (x,y) , (-x,y) ) because if this isn't true and one of the points has a smaller x that the other one than we can aways make the line shorter by making the point with the bigger x have the same value (with the opposite sign) as the smaller one.

I don't quite follow this assumption-- Let's suppose you're dropping down points A, B, and C, and putting B directly on the Y axis as you state. Hence the coordinates for A will have a negative X value, and the coordinates for C will have a positive X value. I'm with you so far. But putting down A and C as you mention above seems like it's not covering all the cases.

Let's suppose that B = (0,10). Now you decide to drop point A at (-1,13). By your assumption, we should be putting C = (1,13). But then we've ruined the fact that the longest side is AC, and it's instead a tie between AB and BC. Instead, we could place point C = (1,6).

Now, in theory, you get the same circle if you place A at (-3,9) and C at (4,11), and yes, technically, the *distance* between A and C gets shorter if you place C at (3,9) instead of (4,11). But you've also decreased the radius of the circumcircle, which (in theory) may have suddenly crossed some threshhold.

I suppose if you can prove via calculus that the rate of decrease in the distance of AC is less than the rate of decrease of the cube root of the radius, and show what the minimal case is, you've got a solution.

DaveE

 

[daniel_i_l]

I think that the problem is that we're both looking at it from opposite perspectives.
in the solution that i posted above i first assumed that there was some circle centered at (0,0) with a point at B(0,b). now there're two more points to put down A(-x,a) and C(x,c). if both a and c are above the x-axis then AC is the longest line. if they're below then you can always "reflect" them up without changing the solution.
after looking this over a little more i saw that in my proof i never assumed that the coords were integers! the only restraint on the points needed to solve the problem is that none of the points are closer that sqrt(2) to each other which is true for integer points on a circle. because if this is true than the min distance on the y-axis (or the x-axis) is always atleast 1. with this restraint the proof is the same.

 

[davee123]

in the solution that i posted above i first assumed that there was some circle centered at (0,0) with a point at B(0,b). now there're two more points to put down A(-x,a) and C(x,c). if both a and c are above the x-axis then AC is the longest line.

after looking this over a little more i saw that in my proof i never assumed that the coords were integers! the only restraint on the points needed to solve the problem is that none of the points are closer that sqrt(2) to each other which is true for integer points on a circle. because if this is true than the min distance on the y-axis (or the x-axis) is always atleast 1. with this restraint the proof is the same.

Yeah, I think if you assume that the origin of the circle is on an integer coordinate, and that the middle point of the three points shares one coordinate with the origin of the circle (making the radius an exact integer), then I expect your proof holds.

I was tempted to try a similar strategy by assuming that two of the points were necessarily separated by a distance of 1, trying to reduce the overall distance to the 3rd point. But that didn't seem to be necessarily true, so I abandoned that one...

DaveE