Geo

ricaototti · Posted: Thu Aug 28, 2008 6:00 am Post subject: Geo

what is the greated possible area of a triangular region with one vertex at the center of circle of radius 1 and the other two vertices on the circle?

OA is 1/2. But I really dont get it.

Thanks

codesnooker · Posted: Thu Aug 28, 2008 7:12 am Post subject:

I love Geometry, and in this question I am going to use more weird triangle that one can imagine for this question. See the attached image.

Area of Triangle: (1/2) * Base * Height

Now as shown in the image,

angle OAC = angle OBC (always because OA and OB are equal which make this triangle is isosceles triangle.)

Now let's say this angle OAC = x degree.

So applying trigonometry properties, lets find the height of the given triangle and base of the given triangle.

In triangle OCA,

OC/AO = Sin x

i.e. OC = AO * Sin x (Height of the triangle)

Now as AO = 1 (radius of the circle), therefore OC = Sin x

Now again in triangle OCA,

AC/AO = Cos x

i.e. AC = AO Cos x [1/2 base of the triangle)

Now as AO = 1 (radius of the circle), therefore AC = Cos x

therefore, AB = 2 * AC = 2 Cos x

Now apply the formula of area of the triangle.

area = (1/2) * base * height
area = (1/2) * OC * AB
area = (1/2) * Sin x * 2 * Cos x
area = (1/2) * Sin 2x (by the property Sin 2x = 2 * Sin x * Cos x)

Now finally area = (1/2) of Sin 2x

Now apply Maxima on the both sides,

Max (area) = Max ((1/2) * Sin 2x)
Max (area) = (1/2) * Max (Sin 2x)

(As we know that the maximum value of Sine of any angle = 1, therefore we can replace Max (Sin 2x) = 1)

Hence Max (area) = (1/2) * 1 = (1/2)

Hope this clear your doubt.

pepeprepa · Posted: Thu Aug 28, 2008 7:17 am Post subject:

I don't find the old post. Here is Ian's property solution:

Put your circle on a coordinate plan with the center of the circle at the origin. You circle is radius 1.
Let's take one side of the triangle, for example the side from (0;0) to (1;0). You know the formula (base*height)/2.
Let's say the segment you just draw is the base. Your goal is to get the biggest area, you will find it only if you get the highest "height". In your draw you can clearly see that to maximize the height your third one vertex must be either at (0;1) or at (0;-1).
So 1*1/2=1/2

Lot of words but that's simple.

kshankker · Posted: Thu Aug 28, 2008 7:28 am Post subject: Yeah OA is correct

In a circle....the biggest triangle will be ( 2 sides will be isosceles and 3rd one, which is connecting the two lines.) here the radius is 1. SO

1st = 2nd side = 1.(isosc. then the rule is X:X:Xsquart.2 )
3rd side will be (square rt 2.)..
then the height will be = 1^2 - (square rt 2/ 2)^ 2
= 1/ (square rt 2)

Therefore apply the triangle formula...A= 1/2 * Base * Height.
= 1/2 * 1/ (square rt 2)* (square rt 2)
= 1/2.......

Got it

pepeprepa · Posted: Thu Aug 28, 2008 7:43 am Post subject:

Codesnooker I am sure your reasoning is right but gmat questions can be solved without sin & cos. You don't need to study this.
But I would like to know, does it help you a lot for geometry questions or is it rare?

codesnooker · Posted: Thu Aug 28, 2008 8:13 am Post subject:

mayur00 · Posted: Thu Aug 28, 2008 8:53 am Post subject: Maxima/Minima anyone?

If the base of the isosceles triangle is assumed to be x then its height would be sqrt(1-x^2/4)

Therefore area of the triangle would be (1/2)*x*sqrt(1-x^2/4)

Differentiate this for maxima and equate to 0 which gives x= sqrt(2)

Plugging this value of x in the equation above gives the answer 1/2
.

I know this is more advanced maths, but if you are aware of this concept it'll give you a quick solution.

pepeprepa · Posted: Thu Aug 28, 2008 9:20 am Post subject:

What I inferred from this and previous posts is that cos/sin require many lines to solve a problem (I can be wrong). And I just wondered if I was going to study that.
I will take care if other persons use this in geometry questions.
Thanks

sumithshah · Posted: Sat Sep 06, 2008 11:02 pm Post subject:

Dont need cos and sines.

Just remember, the maximum area is achieved in a triangle when it is rt angled isoceles.

That is enough

Ian Stewart · Posted: Sun Sep 07, 2008 8:26 am Post subject: