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gwiyomi17

Optimization Problem -

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An isosceles triangle is inscribed in a circle of radius R. Find the value of theta that will mazimize the area of the triangle

post-91416-0-40831200-1383695881_thumb.png

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A chord forms the base of your triangle. Angle theta is the angle subtended from the ends of the chord to the circumference to form your triangle at its apex. I suggest you use 2 theta rather than theta and work on half the triangle.

 

Consider the angle subtended to the circumference on the other side of the chord to your isoceles triangle. this also forms an isoceles triangle.

 

The half chord is a side common to both triangles and allows calculation of the division of the diameter.

Edited by studiot

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You don't seem to have used my suggestion.

Your original idea has taken you away from the radius which is a constant for a given circle and given you an expression in terms of the half chord length, which is variable.

You really want to do the opposite ie work towards an expression that puts the area in terms of the circle constants and theta alone. Then you can differentiate and set to zero.

 

Here is some help towards this. My final expression can be further simplified, in particular look for a simplification of (1+tan2)

 

post-74263-0-17889500-1383737897_thumb.jpg

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hmmm - again I seem to have a different solution! I have not compared numerically nor made sure they are not identical.

 

I like the handwritten solutions and so much quicker than struggling with awful latex

 

post-32514-0-04714800-1383745403_thumb.jpg

 

 

 


 

and in case you don't fancy doing the maths - my solution gives an answer of alpha = pi/3 and thus theta = pi/6. this is the equilateral triangle which makes sense and I am pretty sure is the largest triangle that can be inscribed.

 

you can think of it heuristically this way.

 

area = half base times height

take any chord to be the base then the maximum height and therefore maximum area will be isoceles triangle with chord at base

turn triangle to make one of those edges the base and you have a non-isoceles - rinse repeat

you end up with an equilateral

 

 

 

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One thing Imatfaal didn't seem to mention is that his alpha is twice gwiyomi's theta as shown here.

 

post-74263-0-75970000-1383754129.jpg

 

Edit just found the hidden meaning in post#5 - spoiler.

Edited by studiot

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I tried solving this again yesterday, I changed my solution..

I found a different LONG way, but I still get the right answer..tongue.png

I also tried to do your solutions and both are also right,..

My solution was almost similar with imatfaal, i also used derivative to find the maximum area.

Thanks to both of yousmile.png

Edited by gwiyomi17

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I found a different LONG way, but I still get the right answer..tongue.png

 

Always good to do it yourownself. smile.png

 

Hope you also enhanced your knowledge of the properties of circles and how these properties all fit together as a result.

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Always good to do it yourownself. smile.png

 

Hope you also enhanced your knowledge of the properties of circles and how these properties all fit together as a result.

thanks a lot! your so kindwink.png

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I love the fact that there are two or three independent (or are they?) solutions to the same puzzle - as well as a heuristic explanation of why the answer must be so.

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hmm..

i need to indicate it's domain

so correct me if i'm wrong, the angle must be between 0 and 90 degrees(0°<Θ<90°) right?

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so correct me if i'm wrong, the angle must be between 0 and 90 degrees(0°<Θ<90°) right?

 

 

Yes that's the right idea for 2(theta)=90

 

When theta = 0 then the 'triangle' is a vertical diametral line (ST in my diagram)and the chord length is zero.

 

The longest possible chord (PQ in my diagram)is a diameter. A basic property of a circle is that the apex angle of any triangle based on a diameter is 90.

 

So shouldn't your domain be specified by greater than or equal to signs, rather than strictly greater than?

Edited by studiot

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