Circles in a square!

Soln

r + x = 1/2

x = sqrt ((2*r)^2 - r^2)

x = sqrt(3)*r

r = 0.183

Let $r$ be the radius of the circle, $a$ be the short leg, and $b$ be the long leg. From the diagram, we can tell that $2r = b - a$ . There is a well known relation for the radius of a circle inside a right triangle: $r = \frac{1}{2}\left(a + b - c\right).$

We equate the two and discover that $a= \frac{1}{2}$ . From here, the solutions for $b$ and $r$ are obvious.

This is not as elegant as Michael Mendrin's proof without words, but it is short.

image

Let $r$ be the radius of the circle.

From a point we can draw two tangents to circle and both will be of same length.

From diagram, there are four Right angle triangles and all are congruent.

The Hypotenuse of the triangle is 1. Assume the length of the tangent from the top left point of the square to the left bottom circle is $y$ . the other part of the hypotenuse will be $1-y$

The sides of one triangle looks as in the figure attached.

The longest side can be $((1-y) + r + 2r )$ or $(y + r)$

i.e., $1 - y + 3r = y + r$

i.e., $y = \frac{1}{2} + r$

The longer side of the triangle is $\frac{1}{2} + 2r$ and shorter side is $\frac{1}{2}$

Applying pythagoras theorem for the right angle triangle, you will get below quadratic equation,

$8r^{2} + 4r - 1 = 0$

solving the equation, $r = 0.183$ or $-0.683$

Radius should be $> 0$

Therefore answer is $0.183$

Here's a "proof without words" graphic. After staring at this a bit, you can already make out the answer to this problem. No need for complicated algebra involving two variables, or even one.

Circles In Square

By looking at the diagram it is clear that the four triangles are congruent and and also right angled. It is given to us that the hypotenuse of each of these 4 triangle is equal to 1.

Let the longer side of each of the right triangle be 2c and the shorter side be 2b. Now clearly the length of each of the side of the smaller square in the center of the diagram is 2c - 2b which is also the diameter of each of the circle. That means that the radius of each circle can be represented as c-b.

Using the property that the tangents drawn from an external point to a circle are congruent , we can find that the lengths of all the remaining segments of the right triangle that are tangent to the circle and represent the hypotenuse of each of the right triangles as 4b.

But as it is given to us that each side of the large square is 1, therefore we get that:

4b =1

b = 1/4

2b = 1/2 ( shorter side of the right triangle in the given diagram)

We can use the pythagorean theorem to find that 2c is approximately equal to 0.866.

The diameter of the circle (2c - 2b) is therefore 0.866 - 0.5 = 0.366.

Finally dividing that be 2 we get the radius of the circle as 0.183.

Here is the solution applying inscribed circle and in-center rule.

(cos 30 (deg) X 1 = .8660254)-(sin 30(deg) *1 =.5) = .3660254 (sides of small square) / 2 =.1830 radius of circle

In the diagram we can observe two main values: $x$ and $y$ , which are distributed along all the triangles due to the tangential points properties. We can observe that $k=2x+y$ , and because $k+y=1 \implies x+y=\dfrac{1}{2}$ . Now we can also find that the remaining side length of the right triangle is $\dfrac{\sqrt{3}}{2}$ , which gives us the following system of equations:

$x+y=\dfrac{1}{2}$ $3x+y=\dfrac{\sqrt{3}}{2}$

Where we only solve for $x$ since it is the radius of the circle:

$x=\dfrac{\sqrt{3}-1}{4}\approx 0.183$

The triangles are 30-60-90 right triangles, which always have the same ratio: Since we know that the hypotenuse of the triangle is 1, we can say that the side opposite the $30^\circ$ angle is .5, and the side opposite the $60^\circ$ angle is $.5\sqrt{3}$

The length of the side of the smaller square + .5 = $.5\sqrt{3}$ $.5\sqrt{3}$ - .5 = .366 (the diameter of the circle)

.366 / 2 = .183

Let the radius of the circle be $r$ . By symmetry, (assume that) the triangles are equal. Let the length of the shorter leg be $l$ , which makes the length of the longer leg $l + 2r$ . From the right angled triangle, we get that

$l^2 + (l+2r)^2 = 1 .$

The area of an individual triangle is given by the inradius multiplied by the perimeter, divided by 2, which is $r \times ( 1 + l + l + 2r) \div 2$ . Thus, the area of the large square, viewed as the area of the 4 triangles plus the small center square, gives us

$1 = 2r ( 1 + 2l + 2r) + (2r)^2 .$

With these 2 equations, I couldn't figure out an easy way to deal with them, so I threw it into Wolfram. The only reasonable solution, where $0 < l < 1$ and $r > 0$ was $l = \frac{1}{2}, r = \frac{1}{4} ( \sqrt{3} - 1 ) \approx 0.183$ .

If anyone knows how to deal with those equations, please let me know!

• $2(θ+φ)=π/2$ or $θ+φ= π/4$
• $OH/DH=r/DH=tanφ$ or $DH=rcotφ$
• Similary $AG=DF=rcotθ$ because triangle ADM and CDE are congruent
• $DH-DF=rcotφ-rcotθ=2r$ or $cotφ-cotθ=2$ or $tanθ-tanφ=2tanθtanφ$
• $tanφ=tan(π/4-θ)=(1-tanθ)/(1+tanθ)$
• Substituting and solving will give $3tan^2θ-1=0$ or $tanθ=1/√3$ or $θ=π/6$
• Therefore the ADM is 30-60-90 right angled triangle (Note that angles are 2θ and 2φ)
• $ADsinπ/3=1/2=r(1+cotπ/6)$ or $r=1/(2(√3+1))=0.183$

We have inner circlle right triangle raduis $r=\frac {a+b-c}{2}$ (0) also we have four equal triangles and one square inside surface of bigger square so the sum of four triangles and little square is - $\frac{a \times b } {2}$ + $\frac{a \times b } {2}$ + $\frac{a \times b } {2}$ + $\frac{a \times b } {2}$ + $(2 \times r )^2$ = $c^2$ - and we know that $c=1$ and $c^2$ is area of bigg square so we have $2 \times a \times b + (2 \times r )^2 =1$ (1) then because triangles are rectangular we have $a^2 + b ^2 = c^2$ and when we change $c=1$ we get $a^2 + b ^2 = 1$ (2) then we eqal left sides of expresion (1) and expresion (2) (because right sides are equal to 1 ) then we get $2 \times a \times b + (2 \times r )^2 = a^2 + b ^2$ (3) then expresion (3) becomes $a^2 + b ^2-2 \times a \times b =(2 \times r )^2$ and further we have $(a-b)^2=(2 \times r )^2$ (4) so if we do square root of expresion (4) we get - $a-b=2 \times r$ (5) . Now we go back to expresion (0) $r=\frac {a+b-c}{2}$ and multiplicate it by 2 and we know $c=1$ so we get $a+b-1=2 \times r$ (6) now if we equal left sides of expresions (5) and (6) we get $a-b=a+b-1$ so $2 \times b=1$ and then $b= \frac {1}{2}$ now we can find (a) from $a^2 + b ^2 = 1$ so $a^2 + (\frac {1}{2}) ^2 = 1$ then $a^2 =\frac {3}{4}$ and finaly $a=\frac {\sqrt 3}{2}$ . We have $a=\frac {\sqrt 3}{2}$ , $b= \frac {1}{2}$ and $c=1$ so we can find ( r ) from $r=\frac {a+b-c}{2}$ so $r=\frac {\frac {\sqrt 3}{2}+\frac {1}{2}-1}{2}=0.183$

Did it with pythagoras and then I devided equal lines in 3 parts. I did this without angles. I took the side 1 and the other side as 1/2.

The radio is a simple function of the area of the small square (in the center). The area of each of the four triangles is simple to calculate. The smal square area is equal to the large square (1) minus four triangles.