Right triangle inscribed in a circle

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sk8ternite: Master | Next Rank: 500 Posts; Posts: 119; Joined: Sun May 10, 2009 7:46 pmFollowed by: 1 member; Thanked: 3 times

by sk8ternite » Mon Aug 24, 2009 7:49 pm

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

A) sqr(3)/4
B) 1/2
C) (pie)/4
D) 1
E) sqr(2)

This problem has been posted before, but I had a question regarding the fundamentals behind it. In order to maximize the area of a triangle, you set the two sides perpendicular to each other and make a 90 degree angle. But I thought a right triangle inscribed in a circle had to have a diameter as one of its sides? If you make two radiuses perpendicular to each other, can it be a right triangle?

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Source: Beat The GMAT — Problem Solving | Mon Aug 24, 2009 7:49 pm

fltingley: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Mon Oct 02, 2006 4:19 am; Thanked: 3 times

by fltingley » Tue Aug 25, 2009 7:51 am

I think your confusion lies in this statement:

"But I thought a right triangle inscribed in a circle had to have a diameter as one of its sides?"

Conceptualize just that that for a moment: imagine a circle with a line running the diameter. This means you have essentially cut the circle in half since the diameter has to run through the center of the circle.

Now, with one line running border-to-border through the circle, where does the right angle vertices go? Answer: it can't go anywhere since the two ends of your line are touching the edges of the circle. Any movement to try and create a right angle at either edge of the line will result in that perpendicular line running OUTSIDE of the circle.

Make sense?

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sk8ternite: Master | Next Rank: 500 Posts; Posts: 119; Joined: Sun May 10, 2009 7:46 pmFollowed by: 1 member; Thanked: 3 times

by sk8ternite » Tue Aug 25, 2009 7:59 am

fltingley wrote:I think your confusion lies in this statement:

"But I thought a right triangle inscribed in a circle had to have a diameter as one of its sides?"

Conceptualize just that that for a moment: imagine a circle with a line running the diameter. This means you have essentially cut the circle in half since the diameter has to run through the center of the circle.

Now, with one line running border-to-border through the circle, where does the right angle vertices go? Answer: it can't go anywhere since the two ends of your line are touching the edges of the circle. Any movement to try and create a right angle at either edge of the line will result in that perpendicular line running OUTSIDE of the circle.

Make sense?

Ok, I was thinking that the only right angle that could exist in a circle was one that had a diameter as one of its sides. But I guess other types of 90 degree angles can exist within a circle. Is this correct?

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