- Additionally,

if consecutive side lenghts of

cyclic quadrilateral are

a, a+k, a+2k, and a+3k, then

Area = k^2 tan(x),

Zak

--- Zakir Seidov <seidovzf@...> wrote:> Yes, Mustafa,

http://mathworld.wolfram.com/CyclicQuadrilateral.html

> you are right!

>

> I've just checked that:

>

> > The area of ABCD is independent of a and is equal

> to

> > tan(x).

>

> with Mathematica,

> and using formulas from

>

>

>

__________________________________

> though I yet don't know simple geometric proof

> of your result.

>

> Also I don't know is your result new...

> Hopr triangle gurus know answer...

> Zak

>

> --- mustafa yagci <yagcimustafa@...> wrote:

> > ABCD is a cyclic quadrilateral with consecutive

> side

> > lenghts.

> > |AB|=a

> > |BC|=a+1

> > |CD|=a+2

> > |DA|=a+3

> > Let the measurement of the angle between the

> > shortest and the longest sides be (x).

> > The area of ABCD is independent of a and is equal

> to

> > tan(x).

> > I found this surprising result by some easy

> > calculations.

> > I called surprising because i don't remember that

> > i've seen it before.

> > Curiously, is it right?

> > Does anyone know?

> >

> > MUSTAFA YAGCI,Turkey

> >

> >

> > ---------------------------------

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> >

> >

>

>

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http://sbc.yahoo.com - For Andy Talmadge:

Thanx for your warning but I didn't wan't to mean that, i meant that i found an equality without writing any 'a'. Sorry for my English. Of course it depends on 'a'.

Also i want to ask another question..

I use the word 'consecutive' for x,x+1,x+2,...

but x,x+2,x+4,... are consecutive even numbers,

2x+1,2x+3,2x+5,... are consecutive odd numbers,etc.

But if you double the lengths then they won't be consecutive...

�f i am false in the usage of the word consecutive, could you please warn me again?

For Zakir Seidov:

Let ABCD be consecutive sided cyclic quadrilateral as the side lengths are as i gave before...draw the diagonal BD. say 'm' for its' length. Write cosine rules in the triangles BCD and BAD. Equalize 'm's. Then write the areas of BCD and BAD by using (1/2).p.q.sinx, you will see the result by very easy calculations.

Andy Talmadge <atalmadge1@...> wrote:

You can't mean that the area of the quadrilateral is independent of the value of "a". (Take one such inscribed quadrilateral inscribed in a circle and then double the lengths of all the sides--the angles remain the same and the area quadruples).

Brahmagupta's Formula gives (interestingly)

Area^2 = a(a+1)(a+2)(a+3),

not independent of "a".

mustafa yagci wrote:

ABCD is a cyclic quadrilateral with consecutive side lenghts.

|AB|=a

|BC|=a+1

|CD|=a+2

|DA|=a+3

Let the measurement of the angle between the shortest and the longest sides be (x).

The area of ABCD is independent of a and is equal to tan(x).

I found this surprising result by some easy calculations.

I called surprising because i don't remember that i've seen it before.

Curiously, is it right?

Does anyone know?

MUSTAFA YAGCI,Turkey

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[Non-text portions of this message have been removed] - (Saturday, July 12, 2003 8:15 PM)

mustafa yagci <yagcimustafa@...> escribió:

> ABCD is a cyclic quadrilateral with consecutive side lenghts.

By Brahmagupta formula,

>> AB|=a

>> BC|=a+1

>> CD|=a+2

>> DA|=a+3

> Let the measurement of the angle between the shortest and the longest

> sides be (x). The area of ABCD is independent of a and is equal to

> tan(x).

> I found this surprising result by some easy calculations.

> I called surprising because i don't remember that i've seen it before.

> Curiously, is it right?

> Does anyone know?

S^2 = (s - a)(s - (a + 1))(s - (a + 2))(s - (a + 3))/2 =

= (a + 3)(a + 2)(a + 1)a = a^4 + 6a^3 + 11a^2 + 6a

By cosine th. applied to triangle ABD and BCD,

a^2 + (a + 3)^2 - 2a(a + 3)cos(x) = (a + 1)^2 + (a + 2)^2 + 2(a + 1)(a +

2)cos(x) ==>

cos(x) = 1/(a^2 + 3a + 1) ==>

tg^2(x) = 1 + 1/cos^2(x) = S^2

As x < pi/2, because cos(x) > 0, S = tg(x), but it isn't independent of a.

For any value of a there is a value of x and S,

S = sqrt(a^4 + 6a^3 + 11a^2 + 6a)

x = acos(1/(a^2 + 3a + 1))

But it is certainly curious ...

Saludos,

Ignacio Larrosa Cañestro

A Coruña (España)

ilarrosa@... - (Saturday, July 12, 2003 8:15 PM)

mustafa yagci <yagcimustafa@...> escribió:

> ABCD is a cyclic quadrilateral with consecutive side lenghts.

And also, if y is the angle, acute, between the sides of length a + 2 and a

>> AB|=a

>> BC|=a+1

>> CD|=a+2

>> DA|=a+3

> Let the measurement of the angle between the shortest and the longest

> sides be (x). The area of ABCD is independent of a and is equal to

> tan(x).

+ 3, then

tan(y) = S/s

where s is the semiperimeter.

Saludos,

Ignacio Larrosa Cañestro

A Coruña (España)

ilarrosa@...