Quadrilateral problem
 

A simple quadrilateral is inscribed in a semi-circle so that the diameter of the semi-circle is one side of the quadrilateral. (Simple just means that no two sides cross each other, i.e, no self-intersections.) If the other three sides of the quadrilateral have lengths 3, 4, and 5 units, what is the diameter/radius of the semi-circle?

[SPOILER]Using the fact that we have two pythagorean triangles (with the diameter as hypotenuse) and Ptolemy's equation (cyclic quadrilateral <=> AB x CD + AD x BC= AC x BD (where AC and BD are diagonals)) we stumble upon the following equation:

r^3-50r=120
[/SPOILER]

:newcat:

PS: forgot again the way to hide, seem to have not be posting for a while :blush:

I get the same equation, but with the diameter d rather than r.

One of the amusing things about this problem is the variety of ways to solve it. I have three solutions, all different from your interesting and elegant approach. However, I am still looking for a method of solution which elucidates the form of the answer.

For anyone who works out an algebraic solution, consider the generalization where 3, 4, and 5 are replaced by the variables a, b, and c. (I discovered the hard way that my algebra skills are actually better than my arithmetic skills!)

[QUOTE=philmoore;97344]I get the same equation, but with the diameter d rather than r.

One of the amusing things about this problem is the variety of ways to solve it. I have three solutions, all different from your interesting and elegant approach. However, I am still looking for a method of solution which elucidates the form of the answer.

For anyone who works out an algebraic solution, consider the generalization where 3, 4, and 5 are replaced by the variables a, b, and c. (I discovered the hard way that my algebra skills are actually better than my arithmetic skills!)[/QUOTE]

:smile: You are right philmoore and there are a variety of ways to solve the problem by cosidering the value of angles from the cosine formula and then proceeding or as Keyes Has mentioned by the Cyclic quad formula which she has rightly called Ptolemys Theorem.

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:

silly me :blush:
of course it should have been the equation for the diameter, but I decided to use the confusing variable r for diameter...

General equation should be

D^3-(a^2+b^2+c^2)D=2abc

where D (diameter) and a,b,c are the sides. Given the symmetry of the equation it does not matter in which order we label the sides.

PS: someone kindly hid my solution, but did not give me a clue as to how.
I vaguely remember having to put bars [] around something like hide and \hide
but the exact sequence escapes me

:cat:

[QUOTE=Kees;97364]PS: someone kindly hid my solution, but did not give me a clue as to how.
I vaguely remember having to put bars [] around something like hide and \hide
but the exact sequence escapes me

:cat:[/QUOTE]


To hide it use: [ SPOILER ] [ / SPOILER ] (without the spaces).

Dear Malcolm,

Kees (although it could be used for a girls name), refers to a boy here...
A rapid look at the equation reveals that 8<D<9

Approximately D = 8.05

Fiddling by moderator
 

[QUOTE=Kees;97364]silly me :blush:
of course it should h

PS: someone kindly hid my solution, but did not give me a clue as to how.
I vaguely remember having to put bars [] around something like hide and \hide
but the exact sequence escapes me

:cat:[/QUOTE]

:wink: Take a look at the bottom line of your post 2 and you will know who hid your Half baked solution. As is common in this thread once again it was itself half baked without the tip off like ATH's

Mally :coffee:

[QUOTE=philmoore;97283]A simple quadrilateral is inscribed in a semi-circle so that the diameter of the semi-circle is one side of the quadrilateral. (Simple just means that no two sides cross each other, i.e, no self-intersections.) If the other three sides of the quadrilateral have lengths 3, 4, and 5 units, what is the diameter/radius of the semi-circle?[/QUOTE]

diameter/radius = 2 for any circle ;)

Mistaken identity
 

[QUOTE=Kees;97367]Dear Malcolm,

Kees (although it could be used for a girls name), refers to a boy here...
A rapid look at the equation reveals that 8<D<9

Approximately D = 8.05[/QUOTE]

:smile:
Im sorry Kees as I deduced from your smiley and the avatar you have chosen both of which are decidedly feminine that you were of the opp. sex.
I am a full blooded male and a blush on either sex could be arousing to me though I am not that way inclined ! :wink:

Mally :coffee:

[QUOTE=mfgoode;97363]In all cubics one root is rational the other two roots are imaginary.[/QUOTE]This is not correct : it should be at least one root is a real number. For instance :
x[sup]3[/sup]-6*x[sup]2[/sup]+11*x-6 =0 has three real solutions, i.e; 1, 2 & 3
x[sup]3[/sup]-4*x[sup]2[/sup]+5*x-2=0 has two real solutions, i.e; 1 & 2 (2 being a double solution)
x[sup]3[/sup]-3*x[sup]2[/sup]+3*x-1=0 has one real solution, i.e; 1 (it is a triple solution)

Of course if the coefficients are randomly chosen reals, most of the time there will be only one real solution and two more can be found by using imaginary numbers.

Jacob

My avatar is a very grumpy cat called Heinz, definitely not a female cat.
As there is no Heinz smiley (well, probably copyright reasons) I had to use another one.


:cat:

[QUOTE=mfgoode;97369]:wink: Take a look at the bottom line of your post 2 and you will know who hid your Half baked solution. As is common in this thread once again it was itself half baked without the tip off like ATH's

Mally :coffee:[/QUOTE]

Mally,
Your comments are uncalled for.
Anyone who would like to see exactly how I hid his response need only "quote" the message in question and look at the resulting text in the edit window.

Real and rational
 

[QUOTE=S485122;97375]This is not correct : it should be at least one root is a real number. For instance :
x[sup]3[/sup]-6*x[sup]2[/sup]+11*x-6 =0 has three real solutions, i.e; 1, 2 & 3
x[sup]3[/sup]-4*x[sup]2[/sup]+5*x-2=0 has two real solutions, i.e; 1 & 2 (2 being a double solution)
x[sup]3[/sup]-3*x[sup]2[/sup]+3*x-1=0 has one real solution, i.e; 1 (it is a triple solution)

Of course if the coefficients are randomly chosen reals, most of the time there will be only one real solution and two more can be found by using imaginary numbers.

Jacob[/QUOTE]

:smile:

Not quite so Jacob.
I'm afraid you are mixing up real and irrational numbers with rational numbers.
The real numbers consist of both rational and irrational, and complex numbers.

My quote is from the text book 'Higher Algebra' by Hall and Knight 2001 edition updated thru several editions and first published in 1887. That was about the time when these differences were rigorously defined and still stands good at the present time.

Moreover most irrational numbers are transcendental, unlike the rationals, which like algebraic, are countable.

However, thank you for bringing this up as its a good observation on your part.

Not being content with my text book I had to get on the Net and this is what I found.

[url]http://en.wikipedia.org/wiki/Real_number[/url]

If you still disagree I am open to discussion as numbers have always fascinated me from childhood and I am willing to learn from you.

Mally :coffee:

[QUOTE=Wacky;97380]Mally,
Your comments are uncalled for.
Anyone who would like to see exactly how I hid his response need only "quote" the message in question and look at the resulting text in the edit window.[/QUOTE]

:rolleyes:

Well I often do that myself when Im in doubt.

But a straight question deserves a straight answer not an anticipated revelation.

It is only for the glory of God who has the right to hide things :not us mortals

Mally :coffee:

[QUOTE=mfgoode;97391]If you still disagree I am open to discussion[/QUOTE]

If every cubic equation has a rational root, what is the rational root of

x[sup]3[/sup]-2 = 0

[QUOTE=mfgoode;97392] But a straight question deserves a straight answer[/QUOTE]

Where was there a question? In English, such queries customarily end with a particular punctuation.

I read Kees [B]statement[/B] as an apology that he had felt that it was appropriate to obscure his response, but that he had failed to do so.

As a moderator, I obliged him for the benefit of the other readers.

When, at message #5, he made a request for details, the answer was directly given. Since both inquiry and answer occurred while I was asleep, I saw no reason to elaborate.

[QUOTE=mfgoode;97391]I'm afraid you are mixing up real and irrational numbers with rational numbers.
The real numbers consist of both rational and irrational, and complex numbers.[/QUOTE]In mathematics, the real numbers may be described informally as a number that can be given by an infinite decimal representation, such as 2.4871773339…. The real numbers include both rational numbers, such as 42 and −23/129, and irrational numbers, such as π and the square root of 2, and can be represented as points on an infinitely long number line. No mention of complex numbers !
As pointed out by Wblipp x[sup]3[/sup]=2 has no rational solution, but it has a irrational (no ratios) solution meaning the solution is part of the real numbers.[QUOTE=mfgoode;97391]My quote is from the text book 'Higher Algebra' by Hall and Knight 2001 edition updated thru several editions and first published in 1887. That was about the time when these differences were rigorously defined and still stands good at the present time.[/quote]I'm afraid you misquoted.[QUOTE=mfgoode;97391]... most irrational numbers are transcendental, unlike the rationals, which <snip> are countable.[/quote] I fully agree.[QUOTE=mfgoode;97391]Not being content with my text book I had to get on the Net and this is what I found [url]http://en.wikipedia.org/wiki/Real_number[/url][/QUOTE]And that is where I found the definition I gave above ;-)

Jacob

The discriminant will tell whether there are three real roots or one real and two complex roots. In this case, the fact that the arithmetic mean of three positive numbers (at least two of which are distinct) is greater than the geometric mean implies that the discriminant is negative and all three roots are real. Now using the fact that the three roots must add up to zero, we can show that two roots are negative and that therefore the desired solution is the positive root.

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. This gives us the solution (I am having trouble getting the spoiler tags to work with tex markup):

[tex]d =\frac {10\sqrt6}{3}\cos(\frac{1}{3}\cos^{-1}\sqrt{\frac{486}{625}})[/tex]

Or more generally:

[tex]d =2\sqrt{\frac{a^2+b^2+c^2}{3}}\cos\Big(\frac{1}{3}\cos^{-1}\sqrt{\frac{27a^2b^2c^2}{(a^2+b^2+c^2)^3}}\Big)[/tex]

Can anyone see any particular significance to the angle that is being trisected in this formula?

To answer Wacky, it was indeed an apology and I am grateful for the spoilertags placed over my solution.

I see no reason why trisection should come into play in this problem or the solution.

A Googly!
 

[QUOTE=wblipp;97394]If every cubic equation has a rational root, what is the rational root of

x[sup]3[/sup]-2 = 0[/QUOTE]

:surprised

Well wblipp you've got me by a googly. I have to admit the roots are irrational so I can safely say that the roots are such that at least one is real as has been rightly pointed out by S485122.

However Hall and Night have an answer for this as roots have to be taken by pairs yz so the result is a rational, so they say, but I'm not sure right now.
I will quote them tomorrow as its too late now and the quote is quite lengthy.
I understand that if you take them separately one should get 9 roots for a cubic but this is not the case and the roots have to be paired to get only 3 as per the theory of equations. But don't take my word for it and lets wait and see.

Mally :coffee:

Trisection of angle.
 

[QUOTE=Kees;97447]To answer Wacky, it was indeed an apology and I am ...

I see no reason why trisection should come into play in this problem or the solution.[/QUOTE]

:smile: I presume that to get the roots of a cubic is as good as trisecting an angle.

This is the argument and proof to the assertion by the Greeks of antiquity that an angle (not all but most) cannot be trisected by compass and unmarked ruler only. I have this proof in my library but will have to locate it to post it here. Well we can also get it form the Net.

[url]http://mathworld.wolfram.com/AngleTrisection.html[/url]

This partly answers Philmoore's question but I would like to know his reason and thanks to your post phil as it was very enlightening.

Mally :coffee:

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:

I say more!
 

[QUOTE=davieddy;97847]I say no more.[/QUOTE]

I think you misunderstood my question of 'Any comments' ?

I meant on the Minkowskian spacetime equation and not a criticism of honest statements as I am out of touch with the cubics and bi-quads. for quite a while now. Four decades ago The 'Vietian' equations were old hat for me, but no longer. Still I did not expect a grunt from you.

However we can still be friends even though we dont see eye to eye, or equation for eqn.

Let bygones be bygones if I have offended you in any way.

Mally :coffee:

[QUOTE=mfgoode;97827]FYI I skipped your post.
[/QUOTE]

Too bad, because he gave a good way to reconstruct the formula for solutions to a cubic equation when three real roots are involved. I solved the equation starting from Cardan's formula, but this method was definitely simpler.

[quote=philmoore;97861]Too bad, because he gave a good way to reconstruct the formula for solutions to a cubic equation when three real roots are involved. I solved the equation starting from Cardan's formula, but this method was definitely simpler.[/quote]

THX.
Furthermore it was a derivation from first principles.

My typo arose because I was thinking:

C = x[sup]3[/sup] - Bx = a[sup]3[/sup]cos(3t)/4

(where x=a cos(t) and a=SQR(4B/3) as before)

David

Not so bad.
 

[QUOTE=philmoore;97861]Too bad, because he gave a good way to reconstruct the formula for solutions to a cubic equation when three real roots are involved. I solved the equation starting from Cardan's formula, but this method was definitely simpler.[/QUOTE]

:rolleyes:
Thank you for your concern, philmoore.

I do not rely on posters giving erroneous and misleading results:I dare say with even malevolent intentions.

When I said I skipped the post this was not to be taken literally. What I meant was that I checked the bottom line. It definitely did not tally and make sense so I did not waste my precious time with paper and pencil.

As I've said before and repeat again that I study posters and all their posts before I become hostile. If I find they are frauds I'm not afraid to expose them.

To me they are mathematical charlatans and there are quite a few amongst us whom I have brought to book. These guys are not interested in the TRUTH. Neither are they interested in learning. They glee in confusion and spreading mental chaos-just a pastime whilst their computers are reeling off digits. Some of our posters have even sent me PM's warning me of the same.

Well I have with me silent genuine friends Hall and Knight- and Bernard and Child who have written excellent books on 'Higher Algebra'. B&C are more advanced and meant for mathematical scholarships.

In them is all I need for my research work. At this juncture I wont say more.

Thank you,

Mally :coffee:

Right name.
 

:smile:

Philmoore I was not allowed to edit my previous post and had to log in again.
Well that does not daunt me in sending a further post which would have been a foot note to my previous post.
Please dont anglicise the name Cardano. The full name is Gorolamo Cardano and he published his formula in 1545 in his booK Ars magna. Being a teacher in Maths give credit where credit is due and please don't distort or cut short.

Mally :coffee:

[QUOTE=mfgoode;97914]I do not rely on posters giving erroneous and misleading results...[/QUOTE]
I am grateful, Mally, that you never do this, and that you are probably the only poster here who never makes mistakes.

Exaggeration !
 

[QUOTE=philmoore;97928]I am grateful, Mally, that you never do this, and that you are probably the only poster here who never makes mistakes.[/QUOTE]
:rolleyes:
Now, Now, Now, my dear philmoore, this is what is termed a hyperbole.

I have never claimed all what you say. I am fallible like all the rest of us. I am the first to admit my mistakes and thanked whoever has pointed them out always.

Yes I have made many mistakes during my tenure with this forum but I have never made them to mislead others or delude them. Many a time I have followed up posts painstakingly to learn some truth in them. Alas I have gone
after many well meaning problems and found I have been taken on a wild good chase so I have got wary.

No I am not the only one who never makes mistakes. I am proud to say there are many genuine posters beside myself who try to avoid mistakes as far as possible and don't attribute them to typographical errors.

Well philmoore I have said enough. There is coming a day and very soon too that you and others will be proud to claim that you have even been privileged to correspond with me but I'm not the one to count my chickens before they are hatched.

So with that I bid you adieu, but I have something to say for your other posts currently running.

Mally :coffee:

Exaggeration
 

[QUOTE=philmoore;97928]I am grateful, Mally, that you never do this, and that you are probably the only poster here who never makes mistakes.[/QUOTE]

This is what is called a pure hyperbole

Mally :coffee:

ERROR
 

[QUOTE=mfgoode;97961]This is what is called a pure hyperbole

Mally :coffee:[/QUOTE]

Wacky please do the needful and delete this post #41 as above I have already elaborated on it
THX
Mally :coffee:

Conjecture
 

If we relax the "simple" stipulation that AB,BC,CD,DA do not intersect
we get two "figures of 8". In one, AB crosses CD, in the other BC crosses
the diameter DA.

Perhaps the two negative solutions give the diameter in these cases?

David

Glad to see that your hyperbole detector is working, Mally. But it wasn't just the remarks on errors, it was the imputation of "malevolent intentions" that struck me as really very strange. I suspect your malevolence detector is working in overdrive.

[QUOTE=davieddy;97975]If we relax the "simple" stipulation that AB,BC,CD,DA do not intersect
we get two "figures of 8". In one, AB crosses CD, in the other BC crosses
the diameter DA.

Perhaps the two negative solutions give the diameter in these cases?

David[/QUOTE]

One negative solution definitely seems to give such an interpretation, but the other negative solution is less than 3 (in absolute value), so I really do not understand whether or not this solution has a straightforward geometric interpretation.

[quote=philmoore;97977]One negative solution definitely seems to give such an interpretation, but the other negative solution is less than 3 (in absolute value), so I really do not understand whether or not this solution has a straightforward geometric interpretation.[/quote]

Pity.
Another erroneous post for Mally to gloat about :(((