Minnesota is the land of 10,000 lakes, but I know the answer is much smaller than that.Q:Take 15 coins. Arrange them in an equilateral triangle with one coin at the top, two coins touching below, three coins below that, then four, then five. Remove the three coins at the corners so you're left with 12 coins. Using the centers of the 12 coins as points, how many equilateral triangles can you find by joining points with lines?

Edit: My hint points to a shorter form of Minnesota, namely the abbreviation MN. That's also the abbreviation for Manganese (Mn) which has an atomic number of 25.

A:25 equilateral triangles total (see the video for details).

13 small triangles pointing up or down 4 medium triangles pointing up or down 6 medium triangles pointing left or right 2 large triangles at a slight angle

Here's my standard reminder... don't post the answer or any hints that could lead directly to the answer (e.g. via Google or Bing) before the deadline of Thursday at 3pm ET. If you know the answer, click the link and submit it to NPR, but don't give it away here.

ReplyDeleteYou may provide indirect hints to the answer to show you know it, but make sure they don't give the answer away. You can openly discuss your hints and the answer after the Thursday deadline. Thank you.

perhaps I'll know the answer for the first time on Tuesday. I am trying to go fishing in Minnesota

ReplyDeleteWhile I can't reveal my solution right now, I can say solving this puzzle is much easier if you drawn several coin patterns and then fill in the triangles according to their size on each fresh pattern. More clues later.

ReplyDeleteYou can also draw fresh coin patterns for different orientations of the triangle vertexes to make viewing all the possibilities easier to see.

ReplyDeleteWhat's with everyone's obsession with Minnesota? I think the lakes (and their surroundings) are just as scenic in South Dakota.

ReplyDelete@Tom W.: I think there are only two viewing orientations that work for this puzzle, but you do want to vary your horizon.

ReplyDeleteOr vertices, if you prefer!

ReplyDeleteWolfgang -> Yes, the symmetry makes it easier to see.

ReplyDeleteA nice counting puzzle, not as hard as the hats and pegs of a few months ago.

ReplyDeletePythagorean Hint: 1 + 27 = 16 + 12 = 25 + 3

Funny thing--as a child, I had a penpal by the name of Tom W., who lived in rural Pennsylvania. That was when you still wrote "real" letters.

ReplyDeleteNormally I'd get drunk for this, but unfortunately, I don't live in Cyprus.

ReplyDeleteI guess this puzzle is a bit of a composite this week.

ReplyDeleteMinnesota and Hong Kong have something in common with this puzzle.

ReplyDeleteEKW -> Perhaps the Law of Cosines as well?

ReplyDeleteDid my hint make sense to anyone else? Cause I'm not totally sure it's right.

ReplyDeleteThere are both equilateral and isosceles triangles between the points, so the wording of the puzzle is important.

ReplyDeleteTom W,

ReplyDeletePerhaps. There are lots of ways to approach this

problem. I favor analytic geometry. I did submit an answer in which I simply labeled all 12 points and then listed the vertices of all the equilateral triangles. But I think further discussion should wait until Thursday after 3 PM.

I went with the literal meaning of "equilateral" and didn't bother about Pythagoras or that Law of Cosines (which I wouldn't know to begin with). As Tom W. hinted at the importance of the wording, what I think is important is that the "centers of the coins" represent the corners of the triangles here.

ReplyDeleteClearly if the puzzle is asking for only equilateral triangles, then all three sides of all the triangles are equal in length, but not necessarily the same length, and all three interior angles are 60 degrees. But you can clearly see there are other triangles that can be drawn between the points that do not meet this criteria.

ReplyDelete7^0.5 or 7**0.5, That's more readable than this font's lame display of √7 that I used previously.

ReplyDeleteIf all else fails, try applying Morley's Trisector Theorem!

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteI'm pretty sure I know the answer, but dont understand the hints...

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteWhat the Puck, Wolfgang?

ReplyDeleteI had that same feeling about the hints when there were 11 comments, or later 19, or even 23 (the current count). It's just so ambiguous....

ReplyDeleteDoes ambiguous mean the puzzle will work either on land or water?

ReplyDeleteI have to disagree with Wolfgang. He said: "what I think is important is that the "centers of the coins" represent the corners of the triangles here."

ReplyDeleteThe puzzle says:"Using the centers of the 12 coins as points, how many equilateral triangles can you find by joining points with lines?"

To me that means the lines can intersect beyond the points, which would make for a larger number of total triangles.

Jim:

ReplyDeleteI think you missed the point.

Funny a "Skydiveboy" should ask that ;) It will work on both land or water, but not on either one without the other.

ReplyDeleteSo anyone got a good hint, or understand mine about a beer in Cyprus?

ReplyDeleteJim:

ReplyDeleteSkydiveboy is right; of course the lines can go right through a coin. Triangles can overlap, and a larger one can be completely atop another.

D35TROY3R:

I don't understand your Cyprus-beer hint; therefore, I won't make be the judge as to my hint(s) being "good."

First hint: The number of equilateral triangles equals my age in years;

ReplyDeleteSecond hint: u6FEAz;

Third hint: Minnesota Republicans in 2010;

Fourth hint: This post.

There are many definitions of “line.” The one that I’m using is the one I learned in high school math and is also found in some dictionaries: a line is a straight one-dimensional figure having no thickness and extending infinitely in both directions.

ReplyDeleteChuck

Come on, it isn't that hard. If we think of the coins as 1, 2, 3, ..., 13, 14, 15 then you remove 1, 11 and 15.

ReplyDeleteTake any 3 coins and connect their centers. As long as they are not in a single line you will will form a triangle (e.g. coins 4, 6, 15) But Will only wants equilateral triangles (all sides the same) so skip counting any other triagles. 4, 6, 15 won't be counted, but 2, 7, 9 should be.

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ReplyDeleteActually, the puzzle does not say you must have three or more points in the triangle, so if two points are acceptable, then there are an infinite number of equilateral triangles that connect any two points - but I don't think this is what what Will meant.

ReplyDelete(and strictly speaking they are line segments not lines).

Fifth hint: A little more than twelve percent of the noncollinear triangles that can be formed from the given array of twelve vertices are equilateral.

ReplyDeleteDaveJ:

ReplyDeleteImplicitly, connecting two points does not constitute an equilateral triangle as this is neither equilateral (no other side to compare against) nor a triangle.

I am assuming the points only have to be somewhere on the sides of the equilateral triangle, not necessarily at the vertices

ReplyDeleteDaveJ:

ReplyDeleteOh, I see what you meant. I would have understood better had you written "…whose sides pass through two arbitrary points" rather than "…that connect any two points."

However, not only would the answer for any coin setup go to infinity, but such an assumption would also be nonconstructive (as it would tell us nothing of how to find such triangles). Surely the more parsimonious interpretation of the question is to count only those equilateral triangles whose vertices are determined by, and coincide with, the centers of the coins.

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ReplyDeleteThis comment has been removed by the author.

ReplyDeleteTheRulerOfGolomb:

ReplyDeleteFrom the provided formula I have uniquely determined your answer, and while I initially thought it was wrong (as my answer is so much smaller), I'm now not so sure. (I've convinced myself that your answer fits Blaine's clue very nicely.) Therefore I deleted my above post (where I outright gave the number) and am now left to wonder whether my assumption about the centers of the coins determining the vertices of the triangles was correct. (The more I think about it, the more open-ended the question seems.) If this is the case, then all of my above clues need to be disregarded.

Damn, damn, damn, damn!

The phrase "joining points with lines" is a bit loose.

ReplyDeleteTo avoid the case of unlimited triangles (don't want issue with the infinite monkey's union !) I presume that each valid equilateral triangle must either have a point at each vertex OR at least one point on each side. Does that match other people's points of view (pun intended)

I agree with Blaine's interpretation. The vertex of each equilateral triangle must be at one of the

ReplyDeletetwelve points.

Blaine –

ReplyDeleteMay I assume from your comment above that you believe the puzzle demands each vertex must be at the center point of a coin?

Chuck

Come on folks! For once Will was very clear about a critical point (pun intended). You cannot solve the puzzle unless you understand that each line must begin at the CENTER of a coin and follow that stipulation. This is not rocket science. Now get with it as we are already about a quarter of the way there, time-wise.

ReplyDeleteCan anyone tell me if I'm on the right track here? Makes me think of the Sound of Music.

ReplyDeleteBlaine wouldnt that be 10,001 lakes?

ReplyDeleteHi PlannedChaos,

ReplyDeleteI didn't think anybody would try to brute force their way through that hint I provided. Sorry! I deleted it. Let me just say that I think there are a lot more triangles than most people here think there are.

Have fun!

D35 - I got your hint and I think you're correct.

ReplyDeleteRalph - I agree with your lake assessment, as would Jackie Robinson.

Movie Clue - William Holden?

ReplyDeleteI know nothing....

ReplyDeleteAwesome, I got this one in a little less than 20 seconds! Maybe I can get a triple crown!

ReplyDeletePlanned Chaos,

ReplyDeleteI think your hint five about percentages is slightly off. Perhaps you have come up with

an incorrect total for the number of

non-co-linear triangles one can draw using the twelve points. Or else your equilateral triangle count is not correct.

OK so now the vertex vortex is over, I get Lennon-McCartney-Buchan. Anyone else thinking along those lines ?

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ReplyDeleteTake 15 large Godfather pepperoni pizzas and arrange them in an equilateral triangle. Now see how many triangles can be made by drawing straight lines from the centers of each piece of pepperoni. The square root of the sum will equal the number of women who will come forward accusing Herman Cain of being a pervert.

ReplyDeleteOkay, I figured out a way of posting my answer that cannot be back-calculated.

ReplyDeleteIf you take the solution you get by allowing the lines to be of infinite length you get the answer m.

If you take the solution you get by only allowing the lines to start and stop at the coins, you get n.

m modulo n = 3.

The only lines that are infinite are the lines you find yourself in at the supermarket when you are in a hurry.

ReplyDeleteDaveJ - Your comment is way beyond compare

ReplyDeleteGuess it's time for the judgement day

ReplyDeleteHint: The prime factorization of the answer's square is the answer.

ReplyDeleteEdit: The prime factorization of the answer's square is the answer squared.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteTheRulerOfGolomb:

ReplyDeleteAssuming your m is the number I calculated earlier, and letting k=my count, then m = 7 (mod k). Obviously, k != n. I can't wait until Thursday.

PlannedChaos,

ReplyDeleteYup, that's my number. Did you get the same count, or are you sticking with k?

D35TR0Y3R:

ReplyDeleteYour statement is a tautology. I suspect you meant something else.

TheRulerOfGolomb:

ReplyDeleteI've yet to devise a counting argument to arrive at your number m.

PlannedChaos: That was the point of it, I think. I'm not sure what a tautology is, but what I said is what I meant.

ReplyDeleteTheRulerOfGolomb:

ReplyDeleteI suppose since you and I know m, I might as well divulge that my k = (m-7)/2, just so you know what k is. I counted up to k by assuming the vertices of the triangles coincided with, and were uniquely determined by, the centers of the coins. Now the question becomes, is this condition equivalent to, or more stringent than, the one you used to determine n? (Namely, one only allows the lines to start and stop at the coins.) If they're equivalent, then one of us is wrong, and if they're different then it reduces to what definition Mr. Shortz intended.

EKW:

To see if we're on the same page, my count for the number of collinear triangles that can be formed is equal to an even number that features prominently somewhere on every page of this website (depending on your screen resolution; YMMV).

D35TR0Y3R:

If what you wrote is what you meant, then you must have been making a joke and I apologize. I seem to become devoid of humor when my brain goes into analytic mode.

PlannedChaos,

ReplyDeleteJust to be clear. The answer that you know I have is my number "n" (see my previous post). The number I get when I allow the lines to extend infinitely far past the coins is my number "m" which is much larger. And m modulo n is 3.

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ReplyDeleteTheRulerOfGolomb:

ReplyDeleteOh...That changes things. In that case, k = (n-7)/2 (and thus n = 7 (mod k)). Did you submit n as your answer, or m?

I submitted the smaller number n. So all the lines I drew started and stopped on coins.

ReplyDeleteOkay, then I'm completely baffled as to how one could arrive at such a large number.

ReplyDeletePerhaps I assumed the lines had to start and stop exactly at the center of the coins, whereas your solution involves considering a line to start or stop on a coin so long as it comes to rest at some point on the coin. I concede that such a relaxing of what is considered "the center" of a coin could create a higher number of equilateral triangles, but I don't agree that such an assumption would match with the problem as stated.

I haven't been this baffled by a "math problem" since my senior yr in high school. Then again, I truly think this is quite the pedestrian puzzle: it's the total of all equilateral triangles within the 12-coin array (meaning, lines that would intersect outside the array do NOT make for a complete triangle). Also, per puzzle specs, triangles must point either up or down.

ReplyDeleteJust so we all know what I'm proposing, the question is whether we would consider the following equilateral triangle (with vertices resting on coins A, J and L) to be included in the count. I argue no, as this does not satisfy the condition that we use the centers of the coins as points. TheRulerOfGolomb: Is this the method you used to count up to n?

ReplyDeleteHi PlannedChaos. Nope, under those proposed conditions I think you'd be able to make an infinite number of equilateral triangles. Maybe I'm totally wrong. But I'll stay mum until Thursday.

ReplyDeleteThe puzzle clearly states that the centers of coins are to be used as points, so there's no debate there. But the line vs. line segment is obviously critical since the former would, for instance, allow the original triangle to be part of the tally.

ReplyDeleteGoing with the latter, I initially came up with a solution that matched Blaine's clue. But then I found more triangles. So now I can't decide whether Blaine's answer is incorrect or my interpretation of his clue is incorrect. Again, I'm going with the latter. :)

Wolfgang, if you consider the condition to hold only when one side is parallel with the horizon, I found some triangles that neither pointed up nor down.

ReplyDeletePlanned Chaos,

ReplyDeleteI owe you an apology. It is I who made the error,

not you. Your percentage estimate is correct.

TheRulerOfGolomb:

ReplyDeleteMy proposal was merely to find out if this was part of your thinking in determining n; as it was not, I think we can all agree that such a counting method is invalid.

Dan:

I agree with everything you've said. I suspect I know the initial solution you devised that "matches" Blaine's clue, as my counting went through a similar progression of first finding the initial triangles and then adding the ones that point neither up nor down. But I've also come up with a way that Blaine's clue can satisfy TheRulerOfGolomb's n count, adding to the frustration.

EKW:

No worries. But I guess this means your answer is the same as my k value from above? And perhaps the same as Dan's? Of course, majority doesn't rule and I've since realized the legitimacy of the argument that the vertices of the equilateral triangles needn't coincide with the centers of the coins—the lines need only be determined by the centers of the coins, not necessarily terminate at them. So this insight has added some to my value k—but not enough to make it equal Golomb's n.

I suspect this puzzle has in the past been expressed in terms of planting trees in a field instead of using coins. Perhaps this insight will help.

ReplyDeleteDid anyone happen to listen to the last Car Talk Puzzler? It has some similarities with this one. I got the answer as it was being read, and before the hint was given, but I never send in answers to their puzzles as I have no interest/use for their prize. This puzzle has been interesting and I would not pay attention to my earlier clues as I at first underestimated and then overestimated the number of acceptable triangles.

I changed my understanding of the puzzle rules to the apparently more prevalent view here that a vertex must fall on a coin center, recounted triangles and just submitted my answer.

ReplyDeleteAnyway, of all the would-be clues posted above only two really made sense to me. Conclusion: I know there’s at least two people here whose answer I agree with :) We’ll have to see if we’re right...

Chuck

Skydiveboy

ReplyDeleteI just submitted Cartalk Puzzler answer. Thanks for mentioning it. I agree about similarities to this puzzle.

Natasha:

ReplyDeleteDid you get it right away too? If we go on thinking outside the box all the time, we will not be able to get the simple answers to life's problems.

I'm not sure where all the discussion about

ReplyDeletewhich way the triangles "point" is coming from.

I do not see it in the statement of the problem. Some have a vertical side and therefore "point" left or right. Some have a horizontal side and therefore "point" up or down. And perhaps there are some others that have different orientations.

Chuck, whose clues make sense to you?

Planned Chaos,

ReplyDeleteMy answer could certainly be the same as your k

from above. I am taking the view that the sides of the equilateral triangles start and end at a "point", but may cross one or more other points in the process.

EKW:

ReplyDeleteThat sounds about like the assumption I initially made, although now I think it's more likely the intended value is somewhat larger (for the reasons already stated above).

I'm seeing some of these other triangles that people speak of pointing left or right but they don't appear to me to be equilateral but rather isosceles. I guess in a few days I'll know if my answer could be found in the pages of a teenage magazine.

ReplyDeleteRalph:

ReplyDeleteYes, there are exactly that many equilateral triangles that could be said to be pointing up or down and whose vertices coincide with the centers of the coins. But there are more that point neither up nor down, and more still whose three vertices do not coincide with the centers of three coins....

Ralph:

ReplyDeleteFinally! For once I actually get a hint! (Did you get mine?)

PlannedChaos:

I guess you can arrange the coins spatially in a way that also gives you neither-up-nor-down eq'l triangles, possibly even no up-or-down ones at all.

Oh, and when you drop the center-of-coin requirement, of course that gives you many more, possibly infinitely more, eq'l triangles (within the confines of each coin, for starters).

But I'd better stop talking now; as I hadn't even heard the term "isosceles" before, I feel I still need to come of age to keep up with all the math that's being thrown around here....

Maybe just one more thing...

ReplyDeleteDan:

How can an eq'l triangle not point up or down if one side is parallel with the horizon?

(Clarification of my question above: Assuming just two dimensions, that is.)

ReplyDeleteI'm saying there are some triangles that do not contain a single side which is parallel with the horizon. I was arguing that the "condition" of pointing up or pointing down would require that one side be parallel.

ReplyDeleteWolfgang, you suggested that "as per puzzle specs, triangles must point either up or down". I have been reading too much into this!

I have to assume that each vertex of a given triangle MUST lie in the center of a coin or you would have an infinite number of solutions.

ReplyDeleteThat being said, after reading the discussion here, I have located 110 more lakes! Check it out, Ralph.

Fans of the Beatles and teen mags may want to reconsider.

OMG! I think those of you saying there are some eq'l triangles *NOT pointing either up or down* have it right!

ReplyDeleteD35TROY3R:

I still don't get your Cyprus-beer hint. That may have to do with what I thought the answer was up until now. I'll try going through a sixpack and then some, maybe then I'll get it.

Wolfgang, please don't drink a whole case!

ReplyDeleteNow it's been ten thousand comments, puzzlers have tried a billion ways.... Tommy Boy, your warning of the Beatles and teen mags is dead-on (I've SO outgrown teen mags). And no worries, I won't drink a whole case of that Cyprus beer; thanks for wonderin' if I'm gonna be alive.

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ReplyDeleteGreg,

ReplyDeleteI think you have summarized this very well. I think the discussion has neglected the symmetries of the 12 points. Flipping them from left to right

leaves the pattern unchanged,

and so does a rotation of either 60 or 120 degrees. For people who are not comfortable

using coordinate geometry to compute the distances between the points, these symmetries help find the equilateral triangles and make sure they all get counted.

Keep counting guys! There's only a day left, and you're less than halfway there. :)

ReplyDeleteE. Friedman

ReplyDeleteOK everyone.

ReplyDeleteUsing the centers of the 12 coins as points, how many equilateral triangles can you find by joining points with lines?

1) if all lines must contain at least one point (joining points with lines)

and

2) a line need not end on a coin center (not specifically stated)

then

There is a difference of exactly one triangle in the solution.

Do we all agree that if 1) is not the case then there are infinite solutions?

and if 2) holds, is there only one additional triangle, namely the one constructed by "connecting" the centers of the missing coins?

Thoughts, please.

Steely Dan ±

ReplyDeleteOK. Based on the above, I just found another.

ReplyDeleteTom W. From your musical clue, you agree that premise 2) is true?

ReplyDeleteTB -> Yes, but I can't be more definitive at this time. This is an easy puzzle. I think people are making it more difficult than necessary. Now, back outside for me because it's another beautiful sunny day here on "The West Coast of Florida!"

ReplyDeleteFor the beer drinkers: Try a shot of Asian Jin.

ReplyDeleteIt that fails, try a Bloody Mary breakfast.

ReplyDeleteOK, I'm done. I have two vastly different answers depending on the "does a vertex have to

ReplyDeletecoincide with a coin center" conversation. I've created a graphic which I will post tomorrow.In the mean time, my new puzzle is ready.

Follow this link

I am going to just submit a guess - based on the "Minnesota clue". They never call me any way - even when I have the right answer to the puzzle.

ReplyDeleteI’ve exhaustively checked the possibilities and have determined that the answer is either

ReplyDeletek(as I calculated earlier) or 7k— 1 ≣p, depending on whether the triangle vertices must coincide with coin centers (the smallerk), or whether “joining points with lines” simply means any line used to form a triangle must have extended from the centers of (at least) two coins (the much largerp). I now believe my initial interpretation was too restrictive and so it is more likely that the latter was intended.In terms of

TheRulerOfGolomb’sn,p= 3n+ 3, sopcould conceivably equalGolomb’smsince we knowm= 3 (modn).What it comes down to is whether

Golomb’s condition of “only allowing the lines to start and stop at the coins” is less restrictive than “triangle vertices must coincide with coin centers.”From

DaveJ’s earlier comment, requiring at least one coin center reside on each triangle’s edge could be the in-between conditionGolombused to computen. This is just speculation as I have not done the analysis, nor do I really feel like doing so at this point—for if indeedp=mthen, betweennandm,mseems the more satisfactory answer, as arriving atnrequires imposing assumptions that are not stated in the puzzle.My reasoning is that with the coin centers, a pencil, and a straightedge (which needn’t go to infinity) we can create (a certain number of) patterns that feature lines at sixty degrees to one another, and whether or not any of the equilateral triangles we find in these patterns have the center of a coin residing on one of its sides should be immaterial when the lines were determined by the coin centers to begin with.

One final observation: Despite what a few posters have posited, allowing triangle vertices to differ from coin centers does

notresult in an infinite number of triangles, so long as we only consider lines to be legitimate if they are determined by (at least) a pair of coin centers.I’ll post the complete list of

ptriangles in set notation, along with some visuals, on Thursday.Right on, Plannedchaos! I also calculated the same p (=m), but then thought that the puzzle didn't really allow for infinite lines. The phrase "joining coin centers with lines" seemed to me to suggest you should just draw a line from one coin center to the other and then stop. Plus, this makes for more beautiful diagrams when you draw it up. So, the smaller number n seemed like the more satisfactory answer to me.

ReplyDelete

ReplyDeleteTheRulerOfGolomb:Good to know that our counts agree! However, I want to point out that the definition I'm using to count to

mdoes not require that the lines go to infinity, merely that some of them extend (a finite amount) past their endpoints to meet up with other extended lines. The infographic I whipped up is plenty beautiful.The only mystery that remains to me is how you arrived at

n...This puzzle and much of the discussion here reminds me of Gore Vidal's most recent memoir, POINT TO POINT NAVIGATION.

ReplyDeleteThere are 25 equilateral triangles:

ReplyDelete13 small triangles pointing up or down

4 medium triangles pointing up or down

6 medium triangles pointing left or right

2 large triangles at a slight angle

25

ReplyDeleteMy only good clue this week came from this post:

ReplyDelete"Come on folks! For once Will was very clear about a critical point (pun intended). You cannot solve the puzzle unless you understand that each line must begin at the CENTER of a coin and follow that stipulation. This is not rocket science. Now get with it as we are already about a quarter of the way there, time-wise."

The clue is in the last sentence where I use the word "quarter" as a hint to 25 cents (coins).

Saturday evening when I first read the puzzle I posted something about going and having a beer. This was not a hint at 21, as some may have thought, but it was a hint at 18, which is the legal age for beer drinking in some states. I then realized I missed some triangles and thought it was 21. Then I came up with 28 and hinted that it was a composite as 28 is a composite number, but I was still not convinced I had the right answer and I assigned numbers to each coin (actually I used chess pawns) and saw where I was finding three triangles twice. So finally I got it right. Good puzzle this time in my opinion.

I got 25, too, but had no confidence at all that I hadn't missed any. I found the huge mass of comments no help, but I have to admit that MN was a clever clue.

ReplyDeleteI'm glad for the explanation about Minnesota, which I couldn't figure. But, the only prime factors of 10,000 are 2 and 5, leading to 25, right? Or is that an accident?

ReplyDeleteI think the answer is 57. Can you guys see this picture?

ReplyDeletehttp://tinypic.com/view.php?pic=en0xj&s=5

Okay, here we go.

ReplyDeleteMy initial count was

k= 25. But then I noticed that this interpretation of the wording was too restrictive, and I came up withp=m= 174 triangles, which I now believe to be the correct answer; whether this is what Will intended or not, this is the answer that most accurately follows from the wording of the puzzle.Unaccounted for is

n= 57, for which I eagerly await an explanation fromTheRulerOfGolomb. The deleted original hint fromTheRulerOfGolombas to his countn, from which I derived it, was something like this:nis a two-digit number with first digitxand second digitywhere the sum of the digits of (n*x*y*(x+y)*(x—y))^(ymodx) results in another two-digit number ≣zthat is evenly divisible by (x+y).What follows is my NPR puzzle entry explaining everything, along with links to some infographics I made. The line in bold references the subset satisfying the more restrictive (but unstated) condition everyone here seems to be assuming that all triangle vertices must coincide with coin centers.

There are a total of 174 equilateral triangles to be found, which I list below using set notation, and are defined as follows:

ReplyDeleteWe label the twelve coins alphabetically A through L in the usual ordering from left to right and top to bottom. Then we write, for instance, {AB-AC-BE} to represent the equilateral triangle formed when one extends a line from coin center A to coin center B, and similarly for coin centers A to C and B to E.

Note that any triangle of the form {AB-AD-BD}, which permutes a maximum of three distinct coins, represents a triangle whose every vertex coincides with the coin centers of A, B and D; indeed, if a coin

Xappears twice in any listing then the represented triangle has a vertex that coincides withX's center.In total, 25 of the 174 triangles have all three vertices in correspondence with coin centers.To help visualize the problem, I made some graphics. A total of four distinct orientations of equilateral triangles can be formed, as seen here:

http://oi39.tinypic.com/91i36f.jpg

And here's three ways to visualize the triangles simultaneously:

http://i39.tinypic.com/6z7ihh.jpg

http://i42.tinypic.com/riyj2f.jpg

http://i41.tinypic.com/hv6he9.jpg

Below, the exhaustive list of all 174 equilateral triangles, sorted alphabetically.

{{AB-AC-BE}, {AB-AC-CG}, {AB-AD-BD}, {AB-AF-FJ}, {AB-AH-EH}, {AB-AL-IL}, {AB-BE-EH}, {AB-BG-CG}, {AB-BI-IL}, {AB-BJ-FJ}, {AB-CG-EH}, {AB-CK-IL}, {AB-EK-FJ}, {AB-FJ-IL}, {AC-AD-CD}, {AC-BE-CE}, {AE-AG-BC}, {AE-AG-DF}, {AE-AG-EG}, {AE-AG-HJ}, {AE-AG-IK}, {AE-BC-BH}, {AE-BC-CJ}, {AE-BC-EL}, {AE-BH-EG}, {AE-BH-HJ}, {AE-BH-IK}, {AE-CJ-DF}, {AE-CJ-EG}, {AE-CJ-HJ}, {AE-CJ-IK}, {AE-DF-DK}, {AE-DF-EL}, {AE-DK-EG}, {AE-DK-HJ}, {AE-DK-IK}, {AE-EL-HJ}, {AE-EL-IK}, {AF-AH-FH}, {AF-AL-JL}, {AF-BI-FI}, {AF-BI-JL}, {AG-BC-CH}, {AG-BC-FK}, {AG-BC-GL}, {AG-CH-EG}, {AG-CH-HJ}, {AG-CH-IK}, {AG-DF-DI}, {AG-DF-FK}, {AG-DF-GL}, {AG-DI-EG}, {AG-DI-HJ}, {AG-DI-IK}, {AG-EG-FK}, {AG-FK-IK}, {AG-GL-HJ}, {AG-GL-IK}, {AI-AJ-EF}, {AI-AJ-IJ}, {AI-BK-EF}, {AI-BK-IJ}, {AJ-CL-EF}, {AJ-CL-IJ}, {AK-BF-CI}, {AK-BF-FL}, {AK-CI-EJ}, {AK-EJ-FL}, {BC-BH-CH}, {BC-BH-DI}, {BC-BH-FK}, {BC-BH-GL}, {BC-CH-DK}, {BC-CH-EL}, {BC-CJ-DI}, {BC-CJ-FK}, {BC-CJ-GL}, {BC-DI-DK}, {BC-DI-EL}, {BC-DK-FK}, {BC-DK-GL}, {BC-EL-FK}, {BC-EL-GL}, {BD-BE-DE}, {BF-BL-CI}, {BF-BL-FL}, {BF-EJ-FL}, {BG-BI-GI}, {BH-CH-DF}, {BH-CH-EG}, {BH-CH-IK}, {BH-DF-DI}, {BH-DF-FK}, {BH-DF-GL}, {BH-DI-HJ}, {BH-DI-IK}, {BH-EG-FK}, {BH-EG-GL}, {BH-FK-HJ}, {BH-FK-IK}, {BH-GL-HJ}, {BI-BJ-JL}, {BK-CL-EF}, {BK-CL-IJ}, {BL-CI-EJ}, {CD-CF-FJ}, {CD-CG-DG}, {CD-DJ-FJ}, {CE-CK-EK}, {CE-CK-IL}, {CE-EK-FJ}, {CE-FJ-IL}, {CF-CG-FG}, {CF-CK-JK}, {CF-FJ-JK}, {CH-CJ-DF}, {CH-CJ-EG}, {CH-CJ-HJ}, {CH-CJ-IK}, {CH-DF-DK}, {CH-DF-EL}, {CH-DK-HJ}, {CH-DK-IK}, {CH-EG-EL}, {CH-EL-HJ}, {CJ-DF-DI}, {CJ-DF-FK}, {CJ-DI-EG}, {CJ-DI-HJ}, {CJ-DI-IK}, {CJ-EG-GL}, {CJ-FK-HJ}, {CJ-FK-IK}, {CJ-GL-HJ}, {CJ-GL-IK}, {DE-DH-EH}, {DE-DL-IL}, {DE-EI-IL}, {DF-DI-EL}, {DF-DK-FK}, {DF-DK-GL}, {DF-EL-FK}, {DF-EL-GL}, {DG-DH-GH}, {DI-DK-EG}, {DI-DK-HJ}, {DI-DK-IK}, {DI-EG-EL}, {DI-EL-IK}, {DJ-DL-JL}, {DK-EG-FK}, {DK-EG-GL}, {DK-FK-HJ}, {DK-GL-IK}, {EG-EL-FK}, {EG-EL-GL}, {EH-EI-HI}, {EI-EK-KL}, {EI-IL-KL}, {EL-FK-HJ}, {EL-FK-IK}, {EL-GL-HJ}, {EL-GL-IK}, {FG-FJ-GJ}, {FH-FJ-HK}, {FI-FJ-IL}, {FJ-HK-JK}, {FJ-IL-JL}, {GH-GK-HK}, {GI-GK-IL}, {GJ-GK-JK}, {GK-IL-KL}, {HI-HL-IL}, {HK-HL-KL}}

I wonder if anyone else thought the number was 17, which is exactly what Blaine posted minus the last 8 triangles whose vertices connected to points, with sides passing through open space instead of connecting with points. After reading the puzzle again, I can see that 25 is the correct answer. Great work Blaine (and others who got 25), and I'm sure Blaine's superb video will be mentioned on NPR this Sunday. My musical clues were based on 17: Steely Dan, Hey Nineteen; Asian Jin, Janis Ian, At Seventeen, and Bloody Mary Breakfast, Kings of Leon 17.

ReplyDelete^^^Ah. Beat me to it,

ReplyDeleteTheRulerOfGolomb! I can now unequivocally state that your result directly implies you are assuming the more correct interpretation of the wording of the puzzle that leads to the count of 174, while (in my opinion, arbitrarily) discounting the other 174-57=117 triangles.

ReplyDeleteTom W.:See the infographic from the post above at

http://oi39.tinypic.com/91i36f.jpg

to find the subset containing 17 triangles. Yes, I understand how many peiople could come to the conclusion that this is the answer, but even with the more rigorous interpretation of the puzzle it misses three of the four possible lattices.

Haha. No problem. We just have different interpretations of "joining points with lines." I think both answers are right. :)

ReplyDeleteHere’s some hyperlinked versions of the infographics I posted above. (Sorry; I forgot to modify my initial post appropriately.)

ReplyDeleteA total of four distinct orientations, or “lattices,” of equilateral triangles can be formed, as seen here:

http://i39.tinypic.com/91i36f.jpg

And here's three ways to visualize the triangles simultaneously:

http://i39.tinypic.com/6z7ihh.jpg

http://i42.tinypic.com/riyj2f.jpg

http://i41.tinypic.com/hv6he9.jpg

PC -> Yes, I see that now. I should also mention I was working with a rough sketch that was not to scale. Next time I solve a problem like this I will use a scaled diagram with dividers or try a computer graphics program. Nevertheless, it's been great fun, and I'm sure most agree! I people who have posted here for the first time continue to return weekly. I believe most word puzzles are easier to solve, and the clues are more definitive.

ReplyDeleteTom W.:

ReplyDeleteSame here; I thought 17 at first, but then I saw the light and arrived at 25, which I submitted. Still no call from WS %(

My first clues referred to 17: South Dakota (17th largest state); rural Pennsylvania, "real" letters (all ZIP codes beginning with 17 are rural PA); senior year in high school (when you are about 17); and "still need to come of age" (= not yet 18).

My clues referring to 25: "Now it's been ten thousand...tried a billion...gonna be alive" allude to the song "In the Year 2525"; and "a sixpack and then some" is adding 6+2 to 17 = 25.

I am SO not following the rationale for the 117 and 174 solutions; I hope WS wasn't targeting just the math geniuses....

Cool pictures! Thanks.

ReplyDeleteDang, missed Blaine's last two.

ReplyDeleteAn explanation of my initial hints, which all alluded to the incorrect value 25:

ReplyDeleteFirst hint: Yes, I am 25 years old. (Not terribly helpful.)

Second hint: bit.ly/u6FEAz was the first graphic I made as an attempt to solve this puzzle, while intentionally leaving it ambiguous as to how many triangles this created, in case anyone figured out the hint.

Third hint: In the November 2010 election, the Minnesota Republican Party gained 25 house seats.

Fourth hint: The overall post contained exactly 25 words. In fact, it had been my goal to make every one of my posts contain exactly this many words, which I succeeded at for this post, for hint five, and for my response to DaveJ, but once I realized the actual answer was much larger this exercise became moot.

Fifth hint: The calculations behind this went as follows: 25/((12 choose 3)—18) = 25/202 ≈ 12.38%. The denominator represents the number of ways to choose 3 distinct points from twelve, less the 6*3 = 18 that are collinear (and thus not triangles); the number 18 features prominently on every page of this website as part of the background image.

As to how I correlated Blaine’s hint to

TheRulerOfGolomb'sn= 57, I had convinced myself that it could be referring to Obama’s oft-cited gaffe that he had visited 57 states, which frankly seems as likely as anything given how oblique these clues usually are. I then also realized that Blaine’s hint contained exactly 17 words and that this could be another reference to the first 17 triangles that show up in the 0° lattice, of which Ralph Loizzo alluded with his “teen mag” reference.After reading PlannedChaos and TheRulerOfGolomb's posts, they make excellent points (no pun intended)and it all depends on how the puzzle is interpreted. They can't be ruled incorrect for their interpretations. In my opinion, they should be given extra credit for their solutions (A+). Now how about a n-dimensional version of this puzzle for next week? Are there any material science students out there (FCC, BCC, HCP etc.)

ReplyDeleteLennon-McCartney-Buchan =

ReplyDeleteWhen I'm 64 minus The 39 Steps = 25

For a coin diameter of 1", I counted:

ReplyDeleteTriangles..side length

13 1"

4 2"

6 1.732" = √3"

2 2.645" = √7"

for a total of 25, the smallest base 10 Friedman number.

I was all over the place, originally had 18, thus the reference to Tuesday first time voter. But then I had 33 but realized half of that were isosceles (sp?) so, Dave J, I went with an early Beatles hit "she was just 17" and the Julie Andrews song since I could only find 17. Wow, reminded now of why I was not a math major.

ReplyDeleteI’ve created an 87 second explanatory video that shows the formation of all 174 equilateral triangles, which you can see by clicking here.

ReplyDeleteI've made up an HTML file which divides all the triangles found by PlannedChaos into individual groups based on in which of the four lattices they belong and then by size, compared to the smallest unit for a triangle within that lattice. I found out that PC mislabeled one and also missed one ({AB-CK-EK}). My HTML can be found at http://users.az.com/~jwaters/Puzzles/2011-11-06.html

ReplyDeleteEnya_and_Weird_Al_fan:

ReplyDeleteThank you for the error checking. I've updated the video appropriately to reflect your corrections. Yes, {BL-EJ-FL} was mislabeled as {BF-EJ-FL}; however, {AB-CK-EK} was not missing per se, I simply did not follow my own notation by including it as {AB-CG-EH}.

Blaine –

ReplyDeleteGreat job on the video! You should definitely let Will know ASAP if you haven’t already.

Chuck

Wow - Now I want to know how that video was made.

ReplyDeleteBut this is why I hate contributing hints on Blaines. Because I was WRONG!!!!! And I'm sorry for anyone who followed my hints and were led astray.

I would remind Ralph Loizzo that ultimately we are each responsible for the answer we submit. It is a puzzle game designed for those who are so inclined to see if they can solve the given puzzle using their own wits. It is always a little annoying for me when a "winner" for the week is chosen who did not even solve the puzzle herself, but discloses that she obtained the answer from her son who lives in another part of the country. I always wonder what posses such a person to submit an answer obtained in such a manner. To me this is unethical. Fortunately this does not seem to happen often.

ReplyDelete@PlannedChaos, I'd like to know what tools you used for creating your graphics, both the original and the various lattice diagrams.

ReplyDeleteFor my "movie" I basically created several images and colored in the various triangles in each, then played them in sequence.

Blaine,

ReplyDeleteI crafted the 174 frames by hand from the Photoshop file I had made for the infographics, and then I compiled them as a 2 fps image sequence that I exported as a 1080 MP4 file in QuickTime.

(Thanks again to Enya_and_Weird_Al_fan for the error checking on my set notation.)

So, Blaine, are you convinced that the more faithful interpretation of the puzzle wording leads to 174 triangles rather than 25?

@PlannedChaos,

ReplyDeleteI agree that going with your alternate interpretation leads to 174 triangles and that number has been confirmed by you, me and others that have checked your counting.

However, given that Will is more of a word puzzler and since he likes to pick puzzles that you can be worked in your head or on a piece of paper usually, I think he intended for the triangles to strictly be those formed using the coin centers as their vertices. In his stating of the puzzle, I felt like he wanted us to think of points and lines not in the geometric sense but like one would in a "connect the dots" puzzle. Imagine the coin centers as dots. How many ways could you draw 3 lines (as in a segment joining two dots) to form an equilateral triangle? Using this way of thinking of the puzzle, I'm posting the answer as 25 triangles, but I do appreciate the effort you went through to enumerate all the equilateral triangles formed using the extension of line segments through centers.

And as I said before I really do like your diagrams. My image editor is a little less useful when it comes to getting exact placement of points and perfect circles, but I did my best. :)

Blaine,

ReplyDeleteI agree with your assessment that probably Will intended for the answer to be 25. However, whether this was his intention or not, it still remains that 174 is the most accurate interpretation of the puzzle as given, since the more rigorous condition of triangle vertices coinciding with coin centers cannot be assumed as it is worded—no matter how tempting it may be. As I’ve said, even I initially fell into this trap.

I would hope Will at least acknowledges the validity of the larger result as following directly from the given wording.

I think it might be well to keep in mind that someone in a position such as Will is, that he has concerns other than simply providing a puzzle for us to solve each week. What I mean is that he needs to protect his interests, which include keeping in the good graces of NPR and its' bosses. Therefore he will naturally be interested in generating as many positive email responses to NPR as he reasonably can. This will send a very strong message to those in charge that he and his puzzle segment are well attended to by the listeners. Because of these concerns he has nothing to gain, and everything to lose, by presenting obtuse puzzles that will only appeal to a very tiny minority of listeners. Because of this it seems obvious to me, Blaine (I believe) and others here that the more complicated puzzle is highly unlikely.

ReplyDeleteSDB:

ReplyDeleteI’m aware of the larger context of Will’s responsibilities and am under no delusion that he could ever regularly dedicate time equivalent to our collective hours of thought.

As to the heart of the matter, I am not claiming what is

likely. Rather, I am observing that the logical conclusion that follows from the puzzleas wordedresults in 174 equilateral triangles. Whether our intuitions want to assume an embedded assumption that does not exist with regards to the placement of the triangle’s vertices (for the sake of simplicity or otherwise) does not change the fact that no such assumption is stated, explicitly or even implicitly, in the wording of the puzzle as given.Just to quell suspicion that I’m engaging in flame bait here, let me just emphasize that I believe Will’s intent

ReplyDeletewasfor such an assumption to be embedded in the puzzle, and while it is very human of us to accept this belief as fact when solving the puzzle, it is a logical leap to do so.Oh what the hay. Since I messed up on this one I feltI had to redeem myself by at least getting the Car talk puzzle tho I didnot (could not? too late?) submit the answer. I just felt that was the least I could do to hang in with this prestigeous group. Also impressed by the number of blogs. Don't think I 've seen so many in the last couple of years since I joined in the fun.

ReplyDelete@RoRo, you are correct. This puzzle post has had the most comments to date. The last time we had a similar response was back in May 2009 with a deceptively difficult puzzle with numbers. That puzzle was one comment short of breaking the 3 digit mark.

ReplyDeleteIt’s my hope that the number of comments garnered here reaches at least, oh I don’t know, let’s say 174. (#156 here.)

ReplyDeleteNew puzzle just came up:

ReplyDeleteWhat number comes next in the following series: 2, 4, 6, 9, 11, 15, 20, 40, 60 and 90?

I think there is a number missing from the post.

ReplyDeleteAnd pray tell, what is that number?

ReplyDeleteis it not 43

ReplyDeleteFor the Triangle coin puzzle. The 6 mid sized R-L Triangle make up 18 more small sized equilateral triangles.

ReplyDeleteShould have read the post first. Started to count more and more. 174 it is.

ReplyDeleteMr. Sid,

ReplyDeleteIndeed! To clear this up for everyone,let me re-emphasize the video I made that demonstrates the solution. View it by clicking here.

Maybe word the problem so the triangles can not have points made of intersecting lines. or must share all three vertices with the center point of the remaining coins. Nice video, I did the same method of counting small then mid sized and so on.

ReplyDeleteEveryone tried so hard to overthink this one, with all the mathematical equations and theories. Did anyone think about just simply counting how many equilateral triangles you could find -like the instructions asked? I drew the coins in Illustrator, then drew each set of triangles, a different color each, then... wait for it... COUNTED THEM. High School education (barely) and common sense got this one. That is why we are in so much trouble now as a country and worldwide: Too many SMART people trying to be too clever. My clue also referred to the (In the Year) 2525 song, "Guess it's time for the judgement day"

ReplyDelete