Brain Teasers ? or 50 Years On........... ?

Trig quadratics - yuk !

Solve        2sin²β + 5cosβ = 4         for 0° ≤ β ≤ 360°

This one shouldn't "need" Math Type, so the field is still open to all..............

Don

I fully agree with your comments on Ken's work or art. Makes me feel very shy to put anything in print.

Anyway I'll have a stab at this one.:

60 and 300 degrees

No need to be shy Peter, 60 and 360 are spot-on !

I used sin²β + cos²β ≡ 1

plus a substitution to make the quadratic factorisation a bit easier to visualise.

Did you know that the Tool Bar in the Post Reply box has an Ω sign that gives access to a small, but useful range of mathematical symbols ? means you can quote 60° and 360° quite easily.

But not as elegantly as ken's Math Type

Also, Word and Power-Point have a cut-down version of Math Type that ken pointed out and I am beginning to play with. Its quite useful, but i'm still finding it a bit clumsey to use. Practise, practise....!

i hope i am not putting you guys off.

in fact i think i should hand-write the formulae -- much more 'real' and post this as a picture here. Probably faster too -- dunno.

enjoy

ken

Not at all Ken. Just admiration.

I think I must be suffering from an acute form of "formulae envy" ? And note acute in this context doesn't mean < 90° (Thanks Don !!)

Regards,

Peter

Looking at the last diagram, a roller will need to places at the front of the stone when the stone overhangs the black roller by 1m (IE. when it is perfectly balance on that roller )

Because the roller is rolling in the direction of the stone, this will occur when the stone has moved 2m (at 998m to go). Therefore 499 to complete the job.

Summarising...............

A couple of dozen rollers, placed repeatedly at not more than 2m intervals, each retrieved as the block moves forward and leaves that roller behind. The "roller" gang carrying retrieved rollers forward so that the "pulling" gang can keep up their momentum and don't have to stop.

Anybody need a new pyramid to be built ? fatcat could be your man

Now, i'm not sure whether I have set this one in the hope it inspires a yet more elegant solution of the two intersecting circles (r = L and R = 2L), or whether it's to test ken's ability with Math Type

Find the area enclosed by the curves y = x² + 1 and y = 9 - x²

one can just intergrate to the point of intesection and subtract, but this one also has some interesting symmetry which leads to adding 2 much simpler areas. what that symmetry is, i leave as an exercise for now but i believe this is easier to spot than the previous L/2L teaser 

enjoy

ken

Using Don's proposed method of integration and subtraction I get an area of 64/3. Basically I have solved for x in the 2 equation points which gives x = +/- 2. This gives the limits for the integration of the 2 original functions.

As for the symmetry method, I'm not certain that I have got this right ? Essentially the 2 intersecting curves for a type of ellipse and are symmetrical about both the x and y axis. I have translated the top right quadrant such that it passes through the point (2,0)

This is equivalent to translating the y = 9-x^2 line 5 units down the y-axis (since the point of intersection is (2,5). The equation of this line becomes:

y = 4-x^2

This can be integrated between the limits of x=2 and x=0 to give the area of a quadrant. This works out to be 16/3.

You then need to multiply by 4 to get the total area, which again works out to be 64/3.

I'm not sure whether this was the ,method that Ken was alluding to ? It still requires integration - but only of a single function rather than two ?

Apologies that embedded graphics are still beyond my capability !

Regards,

Peter

yes, Peter, this is the xformation i had in mind. coincidentally because of the nature of the two parabolas, we end up with them being mirror images after transformation via origin shift. and then you end up just integrating one function as you said Peter. Problems like this are nice -- but this is pure luck in this case -- different parabolas would not necessary yield the same simplification.

enjoy

ken

Thanks Ken

I have just googled "area under a parabolic segment" and apparently it is simply 2/3  * width * height. So this seems to work.

Mind you Don had suggested that I needed to brush up on my integration - so I'm glad I didn't take the easy way out

Regards,

Peter

Peter, i like the short cut! Great...

enjoy...

ken

my turn:

prove that the product of 4 consecutive integers is one less than a perfect (integer) square

enjoy

ken

Nice question Ken. I found many blind alleys - but eventually hit upon which numbers to multiply first (I hope ?)

I started with the four consecutive numbers: (n-1), n, ((n+1), (n+2)

Now multiply (n-1)*(n-2) to give n^2+n-2

Multiply the remaining 2 numbers to give n*(n+1) = n^2 +n

Now let (n^2+n) = x

The product of the 4 consecutive numbers becomes (n^2+n-2) * (n^2+n) = (x-2)*x

Now (x-2)*x  = X^2 - 2x which can be expressed as (x-1) ^2 -1 and hence the desired result.

Long winded....but got there.

Regards,

Peter

Whoops...

Just read my reply and noticed a typo. The first part of the multiplication should read:

Now multiply (n-1)*(n+2) to give n^2+n-2

Sorry for the confusion !

no problem -- and no confusion here -- i read past the typo.

peter, this is a very elegant solution!

i attacked it by brute force and so my solution is not that neat.

By the way Peter, i had a maths teacher at A level who would give ZERO marks for my solution and 100% for yours - how cruel is that!!! 

enjoy

ken

Answers to 4 sig figs please..........................

as you can see, i'm still trying to get the hang of this Math Type (Microsoft Equation) thing..

y're doing very well Don...

enjoy

ken

integration after substutution of x=2sinθ is one way, but there is a much simpler way way exploiting that y^2 = 4-x^2 is the equation of a circle centre (0,0) and radius 2. so the area in question is simply the area of the 1st quadrant, i.e. π for which you ar free the choose umber of decimals

i know someone who can recall this to 100 decimals !!! erhmmm..., life's too short!

enjoy

ken

ps: slightly harder, but still interesting, if the limits were (0,1). any takers? Peter?

Ken

That is extremely elegant - and I'm sure your maths teacher would have been very proud of you

I will try and rise to your challenge of working the area for limits (0,1). Using your method of geometry rather than integration I end up with the area of a triangle plus the area of a sector.

My final result is Π/3 + (3 ^ 0.5) / 2 which is about 1.913

Regards,

Peter

ps - The integration route looks very ugly - but I'm sure there are some elegant tricks to make it easier.

Very neat solutions ken and Peter. And well spotted about the circle radius 2

When I find a "difficult" integration, I resort to the "brute force and ignorance" route ie the Trapezium Rule or, if i'm feeling a bit more egalitarian, Simpson' Rule.

Using the Trapezium Rule and an interval of 0.01, I get 3.141 and 1.913 respectively. I brutally used Excel to do the tedious arithmetic.

Using an interval of 0.1, I get 3.1283 and 1.9127 respectively.Pretty brutal.....................

see my last post..............................

i know our Don likes the cosine rule 

judicious application of pythagoras theorem is one way to prove it, but it is a bit tedious.

but there happens to be a very nice/elegant proof of it using vectors and cross products.

any takers?

enjoy

ken

Whilst I do like the cosine rule, I prefer the sine rule even more .

It's usually more useful and certainly easier to remember. But alas, you do need both, together with the area rule, to sort things out.

Anyway ken, you were looking for a vector proof of the cosine rule. I will need to think about that a bit more.

However, I do recall from 50 years back, that our proof at school included a Pythagoras based proof, but also a slight variation based on laying out the triangle on a set of Cartesian axes x,y as shown below. Now I suppose I could claim that the three sides of the triangle are "Position Vectors" since they are tied to a point "A" at the origin, But I won't. I'm also not sure that I have actually recalled the Cartesian proof absolutely correctly, 50 years ago is a long time !