M17-16

Question

Which of the following can be the perimeter of a triangle inscribed in a circle with a radius of 1?
 
I. 0.001
 
II. 0.010
 
III. 0.100

A. I only
B. III only
C. II and III only
D. I, II, and III
E. None of the above.

Bunuel · Answer

Official Solution:

Which of the following can be the perimeter of a triangle inscribed in a circle with a radius of 1?
 
I. 0.001
 
II. 0.010
 
III. 0.100

A. I only
B. III only
C. II and III only
D. I, II, and III
E. None of the above.

First of all, note that any three points on any circle form a triangle. If we imagine three points on a circle and move them closer to each other, we can decrease the perimeter of the triangle they form. Since we can position the points as close together as we want, we can make the perimeter of the triangle arbitrarily small, approaching zero. Thus, we can inscribe a triangle in any circle of any radius such that its perimeter is as small as desired. This implies that the lower limit of the perimeter of a triangle inscribed in any circle of any radius is zero, making answer D correct.
 
Furthermore, among all triangles inscribed in a given circle, the equilateral triangle has the largest perimeter. An equilateral triangle inscribed in a circle of radius 1 has a perimeter of  3\sqrt{3} . Hence, the range of perimeters inscribed in a circle of radius 1 is  0 < p \leq 3\sqrt{3} .

Answer: D

Klenex · Answer

Inscribed in a circle means that all the vertices of triangle must be on the circumference of the circle or some of them can be just inside?

Bunuel · Answer

The vertices must be on the circumference.

tejasa07 · Answer

High quality question. I took a pass at drawing.

pikolo2510 · Answer

Bunuel  - How to find out the max perimeter?

Bunuel · Answer

Among all triangles inscribed in a given circle, the equilateral one has the largest perimeter.

The radius of the circumscribed circle for an equilateral triangle is  R=side*\frac{\sqrt{3}}{3} . From this  side=\frac{3R}{\sqrt{3}} . Rationalize:  side=R\sqrt{3} . Perimeter =  side=3R\sqrt{3} .

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

rickyric395 · Answer

Hi  Bunuel  I tried this way, is this thought process correct?
Let's try to plot 3 points on a straight line.
If we have a straight line then, all points will be co-linear and we can't form any triangle.
A circle is a straight line curved in a round shape. It means  No two points can be co-linear .Thus we can have infinite number of such 3 point pair and therefore infinite number of triangles.

M17-16

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