Elements
Base is the square of given side length
with centroid at
Inscribed are the largest possible circle and right triangle of the same area.
General case
Segments in the general case
0) The side length of the base square: ![{\displaystyle a_{0}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/693ad9f934775838bd72406b41ada4a59785d7ba)
1) The radius of the inscribed circle:
, see Calculation 1
2) The side length of the inscribed right triangle:
, see Calculation 1
Perimeters in the general case
0) Perimeter of base square
1) Perimeter of the inscribed circle: ![{\displaystyle P_{1}=2\pi \cdot r_{1}={\frac {2\pi {\sqrt {2}}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 2.122\cdot a_{0}\quad }](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/d60e63cf9e3b687d7005817e6d9eb88c6246fb5b)
2) Perimeter of the inscribed triangle: ![{\displaystyle P_{2}=4\cdot a_{2}={\frac {4\cdot 2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 3.387\cdot a_{0}\quad }](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/ca8773691162d37dd0f77bafc8b22c034d0c09b8)
Areas in the general case
0) Area of the base square ![{\displaystyle A_{0}=a_{0}^{2}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/b61c9532e83c548bc8e2b82561874984edfd492b)
1) Area of the inscribed circle: ![{\displaystyle A_{1}=\pi r_{1}^{2}=\pi \left({\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/0d489838055bd85571f0db0589e8fbb1415283db)
2) Area of the inscribed triangle: ![{\displaystyle A_{2}={\frac {a_{2}^{2}}{2}}={\frac {1}{2}}\left({\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/7c9315a6ce902158d77cd08b05822fdbad0452a2)
Calculations
Equations of given elements and relations
(0) The area of the right triangle
has to be the same as the area of the circle with radius
around center point
(1)
since
is a square
(2)
since
is the diagonal of square
(3)
since
is a right triangle
(4)
since
is a square, because
and
are tangent points of the circle around
with the sides of square ![{\displaystyle \square ABCD}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/83f54153304ec3963272dedf0701611bf6633aeb)
Calculation 1
The radius
is calculated:
a) First the relation between
and
is determined from equation (0):
applying equation (0)
, definitions of areas of right triangles and circles
, multiplying both sides by two
, since negative length cannot be applied
b) Then the relation between
and
is determined from the following identity:
since
is a square
, because the three segments are forming the diagonal of the square
, since
is a right isosceles trinagle
, applying part a) of the calculation
, rearranging
, rearranging
, because E is a tangent point on the circle and a diagonal side of the triangle ![{\displaystyle \triangle AKH}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/af72fc66cd14e5b4744d8478b2cf49fac292517d)
, because
is a square with side length ![{\displaystyle r_{1}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/ea214f2b31fb3869344bb9311da41c5cc38a99e1)
, extracting
out of the bracket
, rearranging
c) Eventually the relation between
and
is determined by using part a) of the calculation:
, from part a) of the calculation
, using the result from part b) of the calculation
, rearranging
Calculation 2
The centroid of the inscribed circle is calculated in orientated expression:
![{\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}C}}+{\overleftarrow {CS_{1}}}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/9866dcabf6758324da0bfe2fcb2d92b18b37d0fe)
, applying definition of ![{\displaystyle S_{0}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/ebe0ac45a38c4437bd2689a14ec434cd499e7e49)
, applying the orientation principle
, rearranging
, since
and
are collinear
, since
is half the diagonal of square ![{\displaystyle \square {ABCD}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/581143de0293a8beaacce48aef32c3fc459e2680)
, since
is the diagonal of square
with side length ![{\displaystyle r_{1}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/ea214f2b31fb3869344bb9311da41c5cc38a99e1)
, rearranging
, rearranging
, applying result from Calculation 1
, extracting ![{\displaystyle a_{0}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/693ad9f934775838bd72406b41ada4a59785d7ba)
, rearranging
, rearranging
, rearranging
, rearranging
, rearranging
, rearranging