Perhaps the angle trisection doesn't have a straightforward interpretation in terms of the original problem and only occurs because of the cubic equation. However, I thought the nice form of the solution might lead someone to such a connection if it exists.

Kees' solution is elegant, but because Ptolemy's theorem is not so well-known, I thought I would sketch the other three solutions I came up with:

[spoiler]Solution 1:
(This probably occurred to me first because I am teaching trigonometry this term.) Use the relation arcsin(a/d) + arcsin(b/d) + arcsin(c/d) = pi /2. Now move one of the arcsin terms to the right-hand side, take the sine of both sides, and use angle addition to get an equation in terms of a, b, c, and d.[/spoiler]

[spoiler]Solution 2:
(I then wanted to find a solution not using trigonometry.) Consider the circle x^2 + y^2 = r^2 as the circle with the unknown radius. Find the intersections of this circle with (x-r)^2 + y^2 = a^2 and (x+r)^2 + y^2 = c^2. Equate the distance between the two solutions with y > 0 to b to get an equation in terms of a, b, c, and r.[/spoiler]

[spoiler]The problem with both solutions 1 and 2 is that the resulting equation in both cases, after getting rid of square roots, is of degree 6, actually of degree 3 in d^2. However, I did find that this equation factored into two cubics, and it was not hard to determine which factor was the one of interest. This cubic was the same one given above by Kees.[/spoiler]

[spoiler]Solution 3:
This one is my favorite so far, partly because it leads directly to a cubic equation rather than one of degree six, and partly because it only uses facts from trigonometry that many people actually remember! But it also uses a couple of other facts:
1) For any quadrilateral inscribed in a circle, the opposite angles must add up to pi (180 degrees). This follows because the arcs add up to 360 degrees, and any subtended angle is half the measure of the arc.
2) Any angle inscribed in a semi-circle subtends an arc of 180 degrees and is therefore a right angle. (This fact was also used in Kees' solution.)
Consider the segments from left to right to be of length a, b, and c. The measure of the angle between the segment of length a and the diameter of length d will be denoted t. Let x be the length of the diagonal that goes from the vertex between the segments of length a and b to the opposite end of the diameter. We now have:
cos(t) = a/d ,
x^2 = d^2 - a^2 ,
and from the Law of Cosines, x^2 = b^2 + c^2 - 2bc * cos(pi - t).
Now eliminate x^2 and use the fact that cos(pi - t) = - cos(t) = - a/d. [/spoiler]

Ptolemy
 

[QUOTE=philmoore;97583]Perhaps the angle trisection doesn't have a straightforward interpretation in terms of the original problem and only occurs because of the cubic equation. However, I thought the nice form of the solution might lead someone to such a connection if it exists.

Kees' solution is elegant, but because Ptolemy's theorem is not so well-known, I thought I would sketch the other three solutions I came up with:
QUOTE]

:smile:

Thank you philmoore for the many different solutions.

They were very interesting and instructive. This should be the aim of all 'puzzilists' to instruct by way of puzzles some deep mathematical truths required to solve the problems and alternative methods to tackle them.

In my school algebra text book, which I still cherish, Ptolemy's theorem is the 78th Theorem and a part of the exam syllabus, by far the last theorem, to master.

Then there are several constructions and problems based on the theorems.

Theorems not part of the syllabus such as Ceva's , Menelaus' Theorem, the nine points circle, and Simson's Line that I recall were included as supplementary reading for the geom enthusiasts who wished to explore further.

I look forward to many similar problems from you

Mally :coffee:

[QUOTE=mfgoode;97674]
In my school algebra text book, which I still cherish, Ptolemy's theorem is the 78th Theorem and a part of the exam syllabus, by far the last theorem, to master.[/QUOTE]

I found this quite interesting, as I don't believe I learned of Ptolemy's theorem until I began studying the history of mathematics. His theorem formed the basis for his construction of a "table of chords", a 1st century equivalent of a trigonometric table, where the theorem provides the geometric equivalent of the sine addition law:
sin(A+B) = sin A * cos B + sin B * cos A

In one sense, Kees' solution is more elementary, as it uses only plane geometry, although the trigonometric content of the solution becomes clear from the above connection.

[quote=philmoore;97415]

Now the problem is that even though all three roots are real, the cubic formula expresses them in terms of complex numbers. We can take a cube root of a complex number by writing it in polar form.

[/quote]

I have found a way to keep the solution of a cubic with three
real roots in the real domain:

x[sup]3[/sup] - Bx = C

Let x=a cos(t)
Then cos(3t) = 4 (x/a)[sup]3[/sup] - 3(x/a)

choose a s.t. 3/a = 4B/a[sup]3[/sup]
i.e. a = SQR(4B/3)

Then cos(3t) = Ca[sup]3[/sup]/4

David

[QUOTE=davieddy;97752]Then cos(3t) = Ca[sup]3[/sup]/4[/QUOTE]

Change this to 4C/a[sup]3[/sup] and you have a solution.

I like it because the supposition X = a cos(t) gives you an easy way to reconstruct it.

I believe that Viete was the first to solve cubics this way.

[quote=philmoore;97754]Change this to 4C/a[sup]3[/sup] and you have a solution.

I like it because the supposition X = a cos(t) gives you an easy way to reconstruct it.

I believe that Viete was the first to solve cubics this way.[/quote]

Thanks for the correction.
You get all three roots because you can add multiples of 2pi/3 to
t without changing cos(3t).

It also makes the condition for three real roots clear to derive.

Who and when was Viete? (Excuse my ignorance:)

David

François Viète (1540-1603) wrote "Introduction to the Analytic Art" and was one of the leading algebraists of his time. He is credited with popularizing the use of letters to represents constants as well as variables, as in y = mx + b, where m and b represent constants, x and y are variables. His work was an important influence on two generations of French mathematicians, including Fermat and Descartes.

F. Vie'ta
 

[QUOTE=davieddy;97762]Thanks for the correction.

Who and when was Viete? (Excuse my ignorance:)

David[/QUOTE]

The correction by Philimoore mentally noted for future posts by you (davie) as possible errors which could be misleading. FYI I skipped your post.

:smile:

Davieddy.

Your query on Viete reminds me of an intermediate grade howler submitted by a Tibetan co-ed at the weekly science/maths tutorial where reference books and notes could be referred too but the time limit was only 45 Min's for ten questions.

1st Ques: What is coal and its properties ?

Ans: It is black, burns easily and is used in making tea.

2nd Ques: Who was Viete?

Ans: You will find his picture on page 87 of the history book ‘A Concise history of mathematics’ :grin: :lol:

Philmoore: Thank you Phil for some history on the subject.

Well Francois wrote “In artem isagoge, 1591 which I take you have given the correct translation in to English.

He also wrote ‘logsitica speciosa’ where he uses letters as numerical coefficients and also the plus and minus signs , + and -.

Viete also expressed Pi as an infinite product viz:
2/pi = cos pi/4.cos p/8.cos pi/16…. in our notation.

Yes, he reduced Cardano's equation to a trigonometric one in which process the irreducible case lost its horrors by making the introduction of imaginaries unnecessary

Like the Hindu-Arabic numerals and Leibnitz in Calculus it exhibits a profound relation between content and form.

Of personal note, is his insistence on the Greek Principle of homogeneity. A product of two lines was necessarily conceived as an area and line segments could only be added to line segments, areas to areas, and volumes to volumes.

It makes me wonder and question the Minkowskian space time equation adopted by Einstein in his STR which had to be altered slightly to make it dimensionally correct.

Any comments ?

Mally :coffee:

[quote=mfgoode;97827]The correction by Philimoore mentally noted for future posts by you (davie) as possible errors which could be misleading. FYI I skipped your post.
Mally :coffee:[/quote]

Well go back and read it then, you *******.
You might just learn something and see that
your sneering was unjustified

David

[quote=mfgoode;97363]
The correct formula is d^3 -50d -120 = 0

where d is the diameter of the circle. But how about solving this cubic and giving us a near exact value ?

In all cubics one root is rational the other two roots are imaginary. So what is the rational value? Will it be plus or minus ?

I await with bated breath as its not as simple as it sounds.

Mally :coffee:[/quote]

I say no more.

Go Back
 

[QUOTE=davieddy;97845]Well go back and read it then, you *******.
You might just learn something and see that
your sneering was unjustified

David[/QUOTE]

:smile: Hey! hey! The time difference suggests to me you are somewhere In Gods good country-one which has barbarians in sheep's clothing - the majority at least, and you are betraying your breed.

Nevertheless! There's no point losing your shirt ( esp.if its freezing there !) and implying abusive language. Definitely not over a simple maths solution.

The father of psycho-analysis puts it down to losing an argument. Yes truth always hurts but don't be like rust which only corrodes the self.

Learn to take correction like a man and not a monkey as the Indian saying goes. Clean the inside of the cup first before you clean the outside. The sepulchre is white on the outside but filled with dead men's bones.

Please act your age and don't behave like a rebellious delinquent.

However I beg your pardon ( 'I never promised you a rose garden'), but as I have maintained I will be wary of your threads and posts as this is not the first time crossing swords with you.

So lets live and let live. Life is too short too quarrel !

Mally :coffee: