can you please help me to Find ... (see attachment)
5:30am here; need 2 coffees (one for each eye) before I can do anything...
so I'll just make an observation, hoping it makes sense:
make G the top of green triangle , so that we have right triangle GDC
make a=DG, b=DC and c=CG; then r = ab / (a + b + c); also, b = 2R
extend BA to left to point E, CD to left to point F, such that the resulting
rectangle EBCF has diagonal CE, G being on CE
make e=EF=BC, f=BE=CF and g = CE; then R = ef / (e + f + g)
right triangles GDC and EBC are similar
That's it for now...
Ok, did some work; my idea to create rectangle EBCF helps some, but not much:
by inserting the larger circle on left (tangent to EF, DF and EC), it is earier to see
that by similarity, e / (2R) = a / (2r) : remember that e=EF and a = DG.
With r = ab / (a + b + c) and b = 2R, lots of messy stuff (as interesting as a
root canal performed by a drunk dentist) leads to a = r(2R - r) / (R - r) and
to e = R(2R - r) / (R - r). This then makes these areas:
triangle CDG = Rr(2R - r) / (R - r) and rectangle ABCD = 2R^2(2R - r) / (R - r)
Makes for a neat ratio areaCDG : areaABCD = r : 2R
HOWEVER, all this does not take in consideration the smaller circle seen at corner C,
according to your diagram.
I noticed that your diagram "intends?" to show circle radius R; but it does not:
what appears is an ellipse! Seems to be only way to accomodate that smaller circle.
So could you CHECK that out please.
As far as I'm concerned, it it nor possible to have that 2nd smaller circle fit in.
I'll believe it is ONLY if YOU can show an example with dimensions.
I've given you plenty for the answer:
"This then makes these areas:
triangle CDG = Rr(2R - r) / (R - r) and rectangle ABCD = 2R^2(2R - r) / (R - r)"
Let J = Rr(2R - r) / (R - r) and K = 2R^2(2R - r) / (R - r)
Green area = J - pi r^2
Blue area = K - J - pi R^2
As I told you, I'm only using the smaller circle inscribed in the triangle.
You didn't answer my questions in last post, so I'm done with this.
Perhaps someone else here can help you further.
I can show that the right triangle is $\displaystyle 7-24-25 $ and $\displaystyle r:R = 1:4 $ .
Let me include my approach to this problem ( at first I thought the solution was so long but actually it wasn't ! )
Name the last 'unnamed' vertex of the right angled triangle , call it $\displaystyle E $ .
Let $\displaystyle \angle BCE = 2\theta $ then we have
$\displaystyle \frac{R-r}{R+r} = \sin(\theta) ~~~ --(1)$
Consider the right triangle $\displaystyle CDE $ , by using the formula calculating the inradius $\displaystyle r = \frac{S}{p} $ , we obtain $\displaystyle r = \frac{ 2R^2 \sin( \angle DCE ) }{ R( \sin(\angle DCE) + \cos(\angle DCE) + 1 )} $
$\displaystyle = \frac{ R( \sin(\angle DCE ) + \cos(\angle DCE) - 1)}{\cos( \angle DCE) } $
But $\displaystyle \angle DCE = 90^o - 2 \theta $ so
$\displaystyle r = \frac{ R( \sin(2\theta ) + \cos(2\theta ) - 1)}{\sin( 2\theta ) } $
$\displaystyle \frac{r}{R} = \frac{ \sin(2 \theta ) + \cos(2\theta ) - 1}{\sin( 2\theta ) } $
$\displaystyle \frac{R-r}{R} = \frac{ \sin( 2\theta ) - ( \sin(2\theta ) + \cos(2\theta ) - 1)}{ \sin( 2\theta ) } = \frac{ 1 - \cos(2 \theta )}{\sin( \theta )} = \tan( \theta ) $
From $\displaystyle (1) $ , we have
$\displaystyle \frac{R-r}{2R} = \frac{ \sin( \theta )}{1 + \sin( \theta )} $ so we have
$\displaystyle \frac{ \tan( \theta )}{2} = \frac{ \sin( \theta )}{1 + \sin( \theta )} $
$\displaystyle 2\cos( \theta ) = 1 + \sin(\theta ) $
$\displaystyle \sin(\theta ) = \frac{3}{5} $
so we can find the results I mentioned .
Ahhhh.....I see; agree SP.
The way the question was worded completely mislead me (that's my fault!).
Really, the question (as posed) can be answered this way:
let a = short leg of right triangle, b = other leg, h = rectangle height; P = pi;
ratio green : blue = (ab - 2Pr^2) : [b(2d - a) - 2P(r^2 + R^2)]