Oldest unsolved puzzle

I don't understand the question. It's the so called Vitruvian man. And the question is ......? Please tell me it's not the Holy Grail, or anything Dan Brown related.

Has anybody tried to translate the words that go with the Vitruvian man?

Now you've confused me Mr Brain ! Vitruvius was a Roman architect who held that buildings should be based upon the most perfect symetry- that of man. Leonardo supposedly did a nice picture. He used mirror writing, but I'm not aware of any undeciphered texts of his. Am I missing some thing in the question here ?

Have read your Vitruian man pointer to the drawing by Leonardo da Vinci, yes that was what I was referring to. I have alway regarded this drawing as an unsolved problem in geometry. Namely that there is no geometric proof that a square can be constructed form a circle and can be proven to be of the same area. I read this a long time ago, and to the best of my knowledge I think it is true? Wikipedia refers to the drawing as Leonardo's depiction of the Proportions of man? My interpretation could be wrong, what do others think?

Regards JD.

I think you're referring to the 'squaring the circle problem' and Euclidean type problems. This kind of thing has cropped up on CR4 before ( usually in relation to Indiana Pi). There are people here far better qualified than I am to elaborate on this, so I'll hold back for a while. Meantime have a google at;

Duplication of the Cube.

Squaring the Circle.

Trisection of an arbitrary angle.

You may need to understand Euclidean construction, and also the meaning of irrational and transcendental numbers. The understanding will require this. For extended fun you may like to investigate a device called a 'Tomahawk' for trisecting angles. None of the mentioned 3 problems can be solved by Euclidean method, and this can be proven (I think) in all cases.

Yes you are correct in your summation, I'll have to look closer at it, and learn some thing along the way.

Regards JD.

Find that Tomahwak one - you'll love it ! Somewhere out there is a paper headed ' Conferssions of a weekend Trisector'. It's both brilliantly funny and educational. Feel free to mail me direct - exploring the impossible is the mark of an enquiring mind. More power to you ! I learnt most of my mathematics by asking the basic question - " Why". No fancy teachers, just a blank sheet of paper. I may not be in the higher leagues, but I certainly asked the questions.

Thank you Kris, found the tomahawk on the internet ok. Many many years ago when I was a young Marine engineer, one of the senior engineers come up with this mathematical problem. If you put a box 4ft long by 3ft high against a wall, so it fits into the corner, then place a 12ft ladder against it so that it just touches the corner of the box, how high up the wall will the ladder be? I used to know the answer but for the life of me I can't remember how to work it out, it breaks down into a quadratic, done by some substitution which then allowed some elimination of the harder bits, if that makes any sense? It started by extending the hypotenuse of the box and and marking the length off with a compass, centre point at the top of the ladder, cant remember from where. The ladder was then divided into two parts, above and below the box corner, "x" below and "12-x" above. Does any one know how to do it.

Regards JD.

Hi JD, I like a good puzzle and this has the making of one. However, I don't understand the scene you describe. Is it 2D or 3D etc ? If you can describe it more, or perhaps scan in a quick sketch to illustrate, it would be fun to look at. Dotted around CR4 are several similar brain-teaser problems. If you've not alrerady seen it, you may enjoy the 'Pyramid' one that was posted by 'Physicist?' as one of the Challenge Questions recently. It requires intuition rather than higher maths skills ( the setter would leave most of us 'dead in the tracks' if minded to do so. That is the problems beauty). The eventual answer given is annoyingly good and simple, yet nobody within CR4 seemed to spot it.

Yes did see Fyz puzzle, was in UK on holiday at the time and did not have time to give it the necessary thought. I respect Fyz and the comments he gives, and I have learnt a lot on the way.

Regards JD.

Does this help. Regards JD.

Thanks for posting the drawing JD. On quick inspection I don't think it has any nice integer solutions. That ladder could be 'slid' along the floor while keeping contact with box and wall. I'm guessing that the problem you refer to has some kind of integer solution, but the numbers don't add up to one. The triangles hint at 3:4:5 etc, but I don't think it works - somewhere there's going to be a non-integer number. I'm probably going to end up with egg on my face when somebody doodles a quick sketch and gives an answer !

My son teaches maths, and has shown this problem to others, their reply is that there is not enough info, But I do remember that there was a solution. And it has haunted me for a long time, may be there is some one out there up to the task? and due the respect that goes with it.

Regards JD.

I'm still baffled.

a² + b² = c²

a=2mn , b=m²-n² , c=m²+n²

Ive been baffled for a long time, I have a computer with endless drawing without a solution. Please some one put me at rest.

regards JD.

I get 10.627 and 4.705 (using Inventor)

can't help with the maths though!

Drew

Of course the American Indians learned this years ago as follows

There are three squaws, one slept on a bear hide , the second on a buffalo hide and the third on a Hippopotamus hide. The squaws on the bear and Buffalo hides had one boy child each. The third squaw had twins.

From this the Indian chief deduced that the product of th squaw on the hippopotumus is equalt to the sum of the product of the squaws on the other two hides!

Hey! That's GOOD! I hadn't heard it before...

Dick

<groan>.

The version I heard..all the ofspring were boys..so it was 'the sons of the squaw... ..is equal to the sons of the squaws on the other etc'

it's funny either way.

That ladder could be 'slid' along the floor while keeping contact with box and wall.

... but only if the floor surface is an arc in cross section, or you move the box out away from the wall.

I know what I meanted even if I didn't say it ! I meanted the ladder could be placed in another position and touch the box and wall. I keep a box of eggs on my desk, just in case though.

Kris -

I don't think the ladder can be moved. It has to be exactly 12 feet long; it has to touch the box, the floor, and the wall. There should be only one solution.

Hi AstroNut,

I haven't taken the care I perhaps should to look at the question, but I think it has 2 solutions. The drawing presented earlier looks to scale. Tear off a strip of paper (ladder length) and place it against your monitor, and I think you will find 2 positions that meet the criteria. The long way is the algebraic ( see lower down) , but it is something of a mind-**^*. A reasonable calculator will do it with the input value, but it's more fun to look for intuitive and hand calculated solutions. This one is 'messy' because of the decimal places involved, but trying to do it on paper is great mental exercise ( if you survive it - I've taken a break from it, I was getting so scrambled). It has to be 'reduced' in a number of stages.

By definition there will be two solutions....just swap the wall and floor.

True dat.

The problem as originally posted (post #8) was: "If you put a box 4ft long by 3ft high against a wall, so it fits into the corner, then place a 12ft ladder against it so that it just touches the corner of the box, how high up the wall will the ladder be?"

There are 4 possible answers to that question, as illustrated here:

Dick

I'm learning something along the way, not only do I misinterpret post but I seen to post questions without realising there are multiplet interpositions, its good to see your involvement, and hope it is enjoyable to you as is me, but must admit to it also being frustrating. Regards JD.

You're right! Even after reading the original challenge (post #8) several times, (standard practice for me), I Just thought of it as a 3' x 4' box - the 3' high didn't register 'till you brought it to my attention! Thanks!

When you said 'Swap the ceiling and floor', my mental picture was to rotate the whole room 90°, which of course rotates the box as well. Now I have to change that to leave the box in the original position.

I originally rotated the box in my mind, to get the larger angle with the floor and reduce the bending you spoke of. Since I have the impression that the question, when asked in the days before computers and graphic calculators, was intended as a mental exercise, I still think it was originally a 10 foot ladder, testing to see if the student knew about the 3,4,5 right triangle and similar triangles.

No spanking ever thought of... and NOT off-topic (as long as post #8 and all the responses to it can be considered on topic).

10' 7 3/8"

I'm wondering if you might have remembered the length of the ladder incorrectly. Make it a 10 foot ladder, and rotate your box so the 4' dimension is vertical; then you have a couple of 3,4,5 right triangles. The 10' ladder touches the wall 8' above the ground.

thats a noval approach, change the question that matchs the answer.

Yeah, but it does create a tidy, intuitive puzzle with a single answer!

I like it.

me too, i'm looking to apply that to different situations also.

And if I they can't give me the right question to match my answer, I can respond with your not asking the right questions. boy theres alot of possiblities with that.

that alone is management material.

The name 'jdretired', together with his referring to 'many,many years ago' implies that he (like me) is an experienced person. Unfortunately, for most of us who are 'experienced', at least a bit of that experience tends to fade somewhat..., so it is entirely possible that he might not remember every detail of the original problem correctly.

not quite sure what was intended from your response, I for one did not intend to questioned his experience or his character. Wealth from practical experience is why I am on CR4.

Only was making light humor with it no insult was intended and if taken that way, please except my apologises

humbled,

phoenix911

I was just trying to say in a nice way that jd and I are old guys by many people's standards, and people of our age (I don't know how old jd is!) tend to forget some things from time to time, sort of like telling a joke and not remembering the punch line!

Dick

70 yrs old, this puzzle was put to me when I was 21yrs old, working my way thought Ried's, The maths book for every marine engineer. Did not sit for my ticket, but got a diploma late in life. I'm still thinking about the answer, hope I come good.

Regards JD.

I'll be there in less than three years...

I gather you are pretty sure the 12 was the correct value.

Please let us know if you come up with something.

Same Regards Dick

Maybe I'm missing something here.

Treat it as 2 triangles. A squared + B squared = C squared.

9+16=25 small triangle side =5 12' ladder -5=7 left

16+ b squared =49 49-16=33 = 5.74+ 3 =8.74 feet up the wall.

You'll have to forgive me I haven't done math on paper long hand in ages. So My #s may be screwed up.

You assume that the ladder and the diagonal 'c' are parallel, when in fact they are not! I've done a scaled down drawing so you can see the error of your ways! This drawing was courtesy of Paint, The Professionals Choice! As for my 'post a picture' skills! I try OK!

Unfortunately yes, you are missing something. The 9+16=25 is correct for the box and its diagonal, but only a 10 foot ladder can simultaneously touch the floor, box, and wall AND be parallel to the diagonal of the box. This is why I suggested a 10 foot ladder earlier...

A 12 foot ladder can't be parallel to the box diagonal and still touch the box and both floor and wall, so neither part of the ladder can be 5 feet.

Here are the two possible solutions for both a 10' and 12' ladder, depending on the position of the box (done to scale in CAD, with the positions of the 12' ladder determined by successive approximation - otherwise known as trial and error):

Of course there are two other sets of solutions, where the long portion of the ladder is on the floor side of the box, which is why the mathematical solution is a quartic instead of a quadratic. As Ken (I think) indirectly pointed out, ladders are not designed to be bridges, so I ignored those other two solutions!

Dick

Bingo!

Nice solution(s). Of course, one solution yields a position of the ladder that would put it under a little more bending load than would usually be the case.

Fairly simple Geometry of similar triangles and solving an equation with a 4th power in it.

The result is 2 trivial answers and two that work.

The answers are X=10.627 ft and X= 4.705 ft

Following shawnbrier's notation:

Using the similarity of the first & second triangles: 3/x=y/4 or x=12/y.

Inserting this into (y+3)^2 + (4+x)^2 = 12^2 we have:

(y+3)^2 + (4+12/y)^2 =12^2.

or:

y^2 + 6y +96/y + 144/y^2 = 119.

This is a quartic equation which should have an algebraic solution. Hopefully tomorrow...luckily it is not a quintic!

Intuitively: it's true that the ladder can be slid but I think there are only 2 positions satisfying the definitions. Solving the quartic should prove it.

After messing up a classical solution of a quartic, I found a lovely site with a readymade calculator for these monsters.

One solution is 10.62724, the other is 4.70472.

For whoever needs more significant figures I recommend www.1728.com/quartic.htm

I'm getting a very perverse feeling that I need to work this through on paper. Make a substitution and reduce one term out of it etc. I have a loose grip on reality as it is, so doing this may not inflict much damage. First of all I must decide if using a calculator for any part of this would be cheating. Not checking your derived equation first, could add to my sick enjoyment of this. When I have my hands free for lunch-time meds, I may see if I can do this.

I wish you good luck and enjoyment. Perverse is right. This can and has been done using a cm (inch?) ruler and a one square angle triangle. The 4 complete formulas for the roots of a quartic (without shortcuts and tricks) are really and truly horrible!

I may need more insanity time on this ! By substition ( t= x - 1.5) , I mangaged to elliminate a term ;

t⁴ - 132.5 t² + 480 t - 4527/16

This yields ( by cheating) t = 9.1272... and 3.2047... Taking off the 1.5 gives solutions ( I think - can't remember, my brain has collapsed). I was going to progress and shove another substitution into all the above to make an easier cubic, but gave up (!). If this hadn't got all those nasty decimal places to fart around with....OK, I'm just making excuses. And the floor is knee-deep in post-it notes !

Yes, eliminating the 3d power is considered to be the right start. You are left with a cubic (which is no joke either...)

For full perverse enjoyment look at the site I've mentioned a few posts ago. Somebody there was foolhardy enough to present the correct completely explicit formulas for all 4 roots. Just looking at them makes you bless the inventors of MathCad and the other computerised algebra wizards.

I could convert the previous to a 3rd order, but lost the plot ( not that I fully understood how the solution to 3rd orders really works anyway !). Sometimes it's fun to follow this stuff through ( even if not understanding) just to be able to do it without recourse to an on-line calculator. Solutions for 3rd order can be found that are reasonably OK to understand, this one ( ie 4th) I've not yet twigged, and am a bit disinclined since it seems that they occur rarely in practice. With so many substitutions required ( ie 4th to 3rd to solution, I kind of lost it somewhere. When time allows I'll look again. Small thing like solving square roots by iteration are just amusing - it keeps the grey matter alive.). I seriously doubt I'll ever be stuck in outer Mongolia, needing to solve such a thing and having no calculator - it's just fun to know how to do it with paper and pen. I'll go back and look at the site you mentioned dovy. Cheers.

My thanks for the site for solving the quartic... We came up with two non-trivial solutions, but what of the trivial solutions??

I just couldn't leave well enough alone.

For x=-11.57, one end of the ladder is against the wall 11.57 feet below the floor. The other end is at the floor 3.1836 feet from the wall. If the ladder were extended, it would pass through the corner of the box at (3,4).

For x=2.24, one end of the ladder is located 7.619 feet outside the wall at floor level. It passes through the wall at 2.24 feet, and the other end of the ladder is at the corner of the box at (3,4).

My math did not come out precise, because I did alot of rounding off in the calculations... doing them by hand, and with a stupid calculator. I deemed it "Good enough for government work".

I also bookmarked the site for solving quartics. Much thanks. It might also work for solving 3rd order equations. Just plug in 0 as the coefficient for x⁴... but I will have to play with this.

Thanks

Bill

There are of course two solutions. The diagrams we have seen show the ladder against the wall. The other solution is where the ladder is laid with the long portion on the floor and the short end against the wall.

Bill

The solution is simple.

The Vitruvian man was a study of positions a man can be drawn in. ( standing with the left arm out and both feet on the ground, the right foot extended and the right arm extended, etc.) It has nothing to do with math or arcitecture. Da Vinci sketched 16 seperate positions of a man standing in only one drawing. ( count them) It's all about art!!

It's a concept drawing by DaVinci inspired by the ideas of the Roman architect Vitruvius. The only puzzle there is in the minds of man.

I think I could have a solution...

it corressponds to solve the equation :

x⁴ - 6x³ - 119x² + 864 x -1296 = 0

Unfortunately, I only get Excel to graphically solve it (sorry for next to):

Sol 1 : (negative) next to -11.58

Sol 2 : next to 2.25 (but not OK with the height of the box)

Sol 3 : next to 4.705

Sol 4 : next to 10.627

I believe it symbolizes the geometry of a "GOLDEN RECTANGLE" , a rectangle from which if you remove a perfect square, you are left with a GOLDEN RECTANGLE. from this equation the Parthenon and other classical Greek structures are based.If I remember correctly, this was, or is still ,considered the most sound geometry in architecture, based on the symetry of the "perfect human body".

I used the method of similar triangles along with the Pythagorean theorem.

base of one triangle divided by adjacent = base of other triangle divided by other adjacent. solve for adjacent leg and substitute into Pythagorean and solve.

4 solutions two of which are feasible the same two the others have gotten 10.63 and 4.7

I meant i solved for the base then plugged into Pythagorean to solve for the adjacent leg or height on wall.

Thank you all, it will take me time to work throught the posts. having truoble with some of them already.

Regards to all JD.

My solution?

Fig 1 is of the puzzle triangle, and using previous information it can be simple solved by trial and error, 4/SinA + 3/CosA = 12. But the following is my attempt to solve it mathematically. This is not the original solution but my own, got fed up with x = x, as an answer.

Fig 2 is simular to the original solution, a circle of diameter AC is scribed and line BD extended to intersect it a E, and a triangle AEC constructed with height EF, where AE = 9.6, EC = 7.2, and EF = 5.76. As EF is off set from the centre, I bog down when trying to solve it by a quadratic. This led me to develop the following hypothesis out lined in Fig3 and 4.

If you draw a line BE at a slight angle to line AC, which lies on the diameter, and drop lines BG and FE to touch line AC at 90deg, then we have a Z shaped outline, and you will find that BG * DE = FE * BD. If this is considered as a constant, and this constant is maintained as points G and F are moved together they will meet at a line HH, who's length is the square root of the constant (R min) * 2. We can then reference line HH as being angle R1 from the diameter JR, and using the constant the angle of R2 can be calculated, where the constant = length (R max) * radius.

Now the hypothesis goes like this: There is an ideal diameter circle where the difference in length between (R min) and (R max) is directly proportional to the difference in the angles R1 and R2. This represented by line XZ in Fig4, and line XY is the given diameter K1 to K2 is the difference correction.

Worked solution.

Constant = 3/5 * 4/5 * 12 * 5 = 28.8.

R min = square root of 28.8 = 5.3666.

R max = constant/radius = 4.8.

Difference between R min - R max = 0.5666

Line HH = R min * 2 = 10.733.

Angle R1 = (90-Sine HH/12)/2 = 13.2825deg.

Angle R2 = Diameter * cosR - radius/cosR = R max. Therefore R2 = 20.7966deg.

Ideal difference between R min and R max = R2/R1-1 = 0.5657.

K1 correction factor = actual diff/ideal diff = 0.5666/0.5657 = 1.00159.

K2 = (5- 4.8)/1.00159 correction = 0.199968.

Corrected angle R = R2/(K2+1) = 17.3351deg.

Therefore Angle A = 45 - 17.3351 = 27.665deg.

Answer A = 27.665deg and x = 12 cos27.3351 = 10.628.