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Triangle inscribed in a circle

Expert replies
by getso » Fri May 15, 2009 1:20 am
Triangle ABC is inscribed in a circle with radius of 1; triangle DEF is within the circle and one of the vertexes is the center of the circle.what is the greatest possible area of triangle ABC minus the greatest possible area of triangle DEF?

The answer is ( 3*3^1/2-2)/4

Please help in solving the above problem.



Thanks
Last edited by getso on Fri May 15, 2009 2:08 am, edited 2 times in total.
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Source: — Problem Solving |

Re: Triangle inscribed in a circle

by dtweah » Fri May 15, 2009 1:49 am
getso wrote:Triangle ABC is inscribed in a circle with radius of 1; triangle DEF is within the circle and of
the vertexes is the center of the circle.what is the greatest possible are of trianble ABC minus
the greatest possible area of triangle DEF?

The answer is ( 3*3^1/2-2)/4

Please help in solving the above problem.

Thanks
What is this?" triangle DEF is within the circle and of
the vertexes is the center". Please edit and state problem clearly.
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Re: Triangle inscribed in a circle

by sanju09 » Fri May 15, 2009 3:22 am
dtweah wrote:
getso wrote:Triangle ABC is inscribed in a circle with radius of 1; triangle DEF is within the circle and of
the vertexes is the center of the circle.what is the greatest possible are of trianble ABC minus
the greatest possible area of triangle DEF?

The answer is ( 3*3^1/2-2)/4

Please help in solving the above problem.

Thanks
What is this?" triangle DEF is within the circle and of
the vertexes is the center". Please edit and state problem clearly.
The problem needs no edit work except writing vertices in place of vertexes, and it is stated clearly too.

What we need to do here is to maximize the area of both the triangles. The inscribed triangle will have the maximum area in case it is an equilateral one. Do a little construction and recall the medians' property of triangles (the medians of a triangle divide each other in the ratio 2:1, 2 parts towards the vertex and 1 part towards the side) to determine one altitude of the equilateral triangle as to be 3/2. We know that the area in this case will be given by (3/2) ^2/√3 = 3 √3/4; and the area of the other triangle whose one vertex is at the center of the circle will be maximum in case it is right angled at the center, and the maximum area will be ½ (1) (1) = ½.

Now, the difference = 3 √3/4 – ½ = (3 √3 – 2)/4. Take my answer.

Attached is the figure to make the understanding easier.
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Sanjeev K Saxena
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Thank you

by getso » Fri May 15, 2009 6:41 am
Hi Sanju,

Thanks a lot....Explaination was simple and easy to understand.

I could get only till that 2 triangles are equilateral and rt.angled...Was not able to arrive at height. U made it clear.

Will be posting more problems..
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