We are finding the dimensions of the largest object that can fit in another geometric shape. **Largest Circle in a Square**

Known: The radius, which is represented by line segment

Goal: Find the dimension of one side of the square. Let the measure of the line segment AC be represent the measure of the side.

Let:

represent the line segment which length is the diameter of the circle.

By the Pythagorean Theorem:

AC^2 + CB^2 = AB^2

Since AC = CB and AB = 2 AN,

2 AC^2 = (2 AN)^2

AC^2 = 2 AN^2

AC = AN √2

Take the positive root, as the negative root has negligible meaning.

The measure of the square's side is the radius times √2. (s = r √2)

**Largest Circle in a Square**

Known: Radius of the circle, R.

Goal: Find the length of a side of the square, S.

Observe that the largest circle will fit when it's center is in the square's center. As the picture above indicates, some of the edge of the circle touches the square.

Therefore, S = 2 R**Largest Triangle in a Square**

Known: A square with each side measuring S.

Goal: The dimensions of the largest triangle.

If we want the largest triangle to fit in the square, it would make sense to make one of the sides to be of length S. It would also make sense to have the vertex of the triangle's two other sides at the other side of the square. This can be accomplished by a 45°-45°-90° triangle, not an equilateral triangle. (See the diagram above)

To solve for B, use trigonometry.

Observe that sin 45° = cos 45° = √2 / 2

Then:

√2 / 2 = B / S

B = (S √2) / 2

So our triangle would have the dimensions S × (S √2) / 2 × (S √2) / 2 **Largest Square in an Equilateral Triangle**

Known: An equilateral triangle with sides of length B.

Goal: The length of the inscribed square, S.

Observe, by inspection, that the inscribed square is the largest when one of it's sides is touching the triangle's base. This allows for the largest S possible.

The measurement of the base is (B - S)/2 + S + (B - S)/2 = B.

We can find S by working with one of the two "mini right triangles" formed at the bottom of the triangle. A diagram of one of the "mini right triangles" is shown below:

Using trigonometry:

tan 60° = S / ((B - S) / 2)

√3 = (2S) / (B - S)

B √3 - S √3 = 2S

S = (B √3) / (2 + √3)

which is the desired result. **Largest Circle in an Equilateral Triangle**

I admit I have some difficulty with this one. In order for the circle to fit, the following have to be true:

1. The area of the triangle must be greater than the area of the circle.

2. The length of the triangle's side must be greater than twice the circle's radius.

3. The height of the triangle must be greater than twice the circle's radius.

Even obeying these rules did not guarantee an answer.

For me this is an open question. This may be one where numerical methods must be used. Any ideas, as always, are welcome.

Until then,

Eddie

This blog is property of Edward Shore. © 2012

Nifty examples Eddie! For the last one, circle inside triangle, all I can come up with is this: 1/2 of the side times the tan of 1/2 Delta is equal to a circle that is tangent to all sides if the center of the circle is coincidental with center of the triangle. R=½side*tan½ 60° thoughts??? Alternately stated as the radius is equal to .577350269 times ½ the side.

ReplyDeleteHey Eddie, I think the area for the largest circle that can be placed in an equalaterial triangle is π*s^2/12 where 's' is the length of the side of the triangle.

ReplyDeleteJason and Mike, you guys are amazing. Thank you so much. Eddie

ReplyDeleteThis topic is really very interesting like in circle a square can be fit and in square a maximum figure a circle can be fit.These form some special type of questions in maths.

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Manor, I read that questions like this appear on some math exams.

ReplyDeleteThis post was inspired by a contest I saw at a shopping mall the other day. The contest asked how many dice fit in a bowl for a trip to Las Vegas.

Eddie