Geo

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ricaototti: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Wed Feb 27, 2008 4:56 am; Thanked: 2 times

by ricaototti » Thu Aug 28, 2008 6:00 am

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

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Source: Beat The GMAT — Problem Solving | Thu Aug 28, 2008 6:00 am

codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Thu Aug 28, 2008 7:12 am

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.

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pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Thu Aug 28, 2008 7:17 am

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.

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kshankker: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Fri Jul 04, 2008 5:28 am; Thanked: 1 times

Yeah OA is correct

by kshankker » Thu Aug 28, 2008 7:28 am

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

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pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Thu Aug 28, 2008 7:43 am

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?

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Thu Aug 28, 2008 8:13 am

pepeprepa wrote: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?

Well I can't say about the level of help it provides as I studied all these stuffs during school days and applied most of complex geometrical concepts during programming years.

Secondly I think we should try to generalize the solution every time with the help of Math basic concepts (but again it depends upon individual to individual).

But about one thing I am 100% sure that it provides you an edge over the others, in case, if you are targeting high or highest score in GMAT.

I guess still one can stick to Ian's logic (that provided by you) as indeed it is a shortcut for the persons those having weak geo concepts. But certainly I feel the method that I posted as shortcut for me because to solve this I just need to draw the figure once and rest I can solve in my mind.

Well this is about me. As I am not an instructor, so I may be I am not the right person to guide. Please consult with some instructor.

Also Geometry questions are rare in GMAT, so you may skip the sin and cos concepts. And there so many to solve a single question, so it does not mean, one needs to learn everything. But there is no harm in knowing all the techniques, if one have time to spare to learn those techniques.

Edit: Added last line.

Regards,

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mayur00: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Mon Jun 23, 2008 1:41 pm; Thanked: 2 times

Maxima/Minima anyone?

by mayur00 » Thu Aug 28, 2008 8:53 am

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.

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pepeprepa: Legendary Member; Posts: 661; Joined: Tue Jul 08, 2008 12:58 pm; Location: France; Thanked: 48 times

by pepeprepa » Thu Aug 28, 2008 9:20 am

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

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sumithshah: Master | Next Rank: 500 Posts; Posts: 165; Joined: Wed May 28, 2008 11:25 pm; Thanked: 9 times; GMAT Score:730

by sumithshah » Sat Sep 06, 2008 11:02 pm

Dont need cos and sines.

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

That is enough

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Sun Sep 07, 2008 8:26 am

pepeprepa wrote: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.

Yes I posted that with slightly different wording in this thread:

www.beatthegmat.com/largest-area-of-of- ... 12221.html

This was the post:

Imagine the circle is in the co-ordinate plane, centre O at (0,0). You might as well let one of the points A be at (1,0) (you can rotate the circle to get it there if you need to). Consider OA to be the base of our triangle: b=1.

Now, if (c,d) is the third point in the triangle, then the height will be |d|. To get the largest area we need the largest height, and that clearly happens when (c,d) is (0,1) or (0.-1). So the maximum area is 1*1/2 = 1/2.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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