Tangential circles inscribed within a square

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Discussion Overview

The discussion revolves around the problem of determining the radius of a smaller circle inscribed within a square that also contains four larger circles of radius 1, which are tangent to the sides of the square and each other. The context includes mathematical reasoning and problem-solving techniques related to geometry.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • One participant suggests that the radius of the smaller circle could be calculated as \( \sqrt{2} - 1 \), providing a method involving the diagonal of a square.
  • Another participant agrees with the proposed radius and offers a similar approach to derive it, emphasizing the relationship between the diagonal of the square and the radii of the circles.
  • There is a mention of dividing the larger square into four smaller squares to facilitate the calculation of the smaller circle's radius.
  • Some participants express uncertainty about the initial setup, asking for clarification on whether the problem was presented with a known answer or if it was purely exploratory.

Areas of Agreement / Disagreement

Participants generally agree on the method to derive the radius of the smaller circle, with some variations in the presentation of the solution. However, there is no explicit consensus on the problem's initial conditions or whether the proposed solutions are definitive.

Contextual Notes

Some assumptions about the arrangement of the circles and their tangential relationships may not be fully articulated, leading to potential ambiguity in the problem setup. The discussion also reflects varying levels of clarity regarding the geometric relationships involved.

Who May Find This Useful

Readers interested in geometry, mathematical problem-solving, or those preparing for standardized tests may find this discussion relevant.

dozzer
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In this diagram, a square is inscribed with four circles of equivalent size, all with the radius of 1. Each circle is tangent to two sides of the square, two circles, and the smaller circle. Obviously each side of the square is equal to 4, but what is the radius of the smaller circle?

I think the answer is something like 21/2 -1, or .414..., but how would you solve for it?

This is a problem from the SAT before the math section was reformed.
 

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Welcome to PF!

Hi dozzer! Welcome to PF! :smile:

I can't see your picture yet, but if you divide the square into four squares, doesn't that put the center of the small circle at one corner of each small square? :wink:
 
Were you provided the answer and don't know how to get there or are you asking something else?

I can't see your pic either, but from the description I came up with the same answer you supplied.
 
Divide the square into 4 even squares and you will have 4 squares, each with a circle of radius 1 circumscribed within.

Now, draw a line from the center of the small circle to one of the corners of the original large square. This line is a diagonal of one of the small squares and is equal to 2 * (R + r), where R is the radius of the large circles and r is the radius of the small circle.

We already know that R = 1, therefore the diagonal is equal to 2 * (1 + r).

The diagonal is the hypotenuse of a right triangle with legs of equal length (in this case, 2), so the length of the diagonal D = (2^2 + 2^2)^1/2, or
D = (4 + 4)^1/2
D = 8^1/2
D = 2 * (2^1/2)

From above, we know that D = 2 * (1 + r), so 1 + r = 2^1/2
and finally, r = 2^1/2 - 1
which is approximately equal to 1.4142 - 1, or 0.4142
 
Thank you for the explanations, everyone.
 
Just wanted to clean up my earlier post...

zgozvrm said:
Divide the square into 4 even squares and you will have 4 squares, each with a circle of radius 1 circumscribed within.

Now, draw a line from the center of the small circle to one of the corners of the original large square. This line is a diagonal of one of the small squares and is equal to 2(R + r), where R is the radius of the large circles and r is the radius of the small circle.

We already know that R = 1, therefore the diagonal is equal to 2(1 + r).

The diagonal is the hypotenuse of a right triangle with legs of equal length (in this case, 2), so the length of the diagonal D = \sqrt{(2^2 + 2^2)}, or
D = \sqrt{(4 + 4)}
D = \sqrt8
D = 2\sqrt2

From above, we know that D = 2(1 + r), so 1 + r = \sqrt2
and finally, r = \sqrt2 - 1
which is approximately equal to 1.4142 - 1, or 0.4142
 
Last edited:
Welcome to PF!

Hi zgozvrm! Welcome to PF! :smile:

Alternatively, instead of LaTeX, have a square-root to copy: √ and try using the X2 tag just above the Reply box :wink:
 

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