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Four Circle Theorem
Answered
Let ∆XYZ be any triangle with a radius of incircle, r
Let a be the radius of the circle tangent to r, XY, and XZ
Let b be the radius of the circle tangent to r, XY, and YZ
Let c be the radius of the circle tangent to r, XZ, and YZ
Construct ∆XYZ if a = 1 cm, b = 4 cm, and c= 9 cm
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Hi,
script solution... ? (to search)
https://ggbm.at/565245
Hi,
an other solution...(without script)
https://ggbm.at/565247
Very nice solution, Patrick!
Its core is represented by the root of function f(x), point R, as per att'd screenshot.
Cheers
Philippe
Here is my solution to your problem with explanations (spanish written and without scripts):
http://matematicainteractiv...
Nice explanation, cfleitas.
jtico
Is there an explanation in English?
Hi all,
If you modify Patrick's nice solution by allowing slider c to range from 0 to 20 and then set a=2, b=2.5 and c=14, you get an erroneous solution. Simple reason being that there is no triangle ABC satisfying the requirements. The radius of the incircle must be greater than any of a, b and c; for these values R is not greater. I've been trying with no success to come up with a criterion on a, b and c so that the problem is solvable.
Let 0<R_1 <=R_2<=R_3
Then: (R−R_3)/(R+R_3) +(RR_1)/(R+R_1) * (RR_2)/(R+R_2)= sqrt(1(RR_1)^2 /(R+R_1)^2 ) * sqrt(1 (RR_2)^2/(R+R_2)^2) (http://matematicainteractiv...)
You get a cubic equation. Solutions are R=0, and R= sqrt(R_1 R_2) + sqrt(R_2 R_3) +sqrt( R_3 R_1) and another negative solution.
R= sqrt(R_1 R_2) + sqrt(R_2 R_3) + sqrt(R_3 R_1) makes sense only if R> R_i, which means problem solution condition is
sqrt(R_1 R_2) + sqrt(R_2 R_3) + sqrt(R_3 R_1) > R_3
Yes, the construction I previously defined as "nice solution" should be amended. In fact, the incenter should always reside inside the triangle, and in some circumstances it isn't; probably because the incenter I(0,0), or any coordinates, is an apriori choice and this creates the said drawback.
Cheers
Philippe
Hi,
:flushed: sorry for my English :cry:
I proposed just a "WAY" no a "solution"
But a way to view the limits ...
Thanks you :
sqrt(R_1 R_2) + sqrt(R_2 R_3) + sqrt(R_3 R_1) > R_3
the incenter should always reside inside the triangle
ways to search "solution" ...
cfleitis,
Bravo and thank you. Even knowing the answer, it took me a while to arrive at the solution. As you know, the main step was getting a cubic equation with a factor of R. But even from here the quadratic formula was a bit tricky for me. Again, bravo.
Now, last question. Does the solution have any geometric meaning? I can't see any geometric significance to the result.
Good Patrick!
Your way to search solution is highly suitable to afford the proposed problem, which involves (this is important from my point of view) a new field of investigation, the socalled Dynamic Geometry.
The human mind doesn't think in such terms, so what any CAS (computer aided software) performs may sometimes be far away from classical approaches.
Nevertheless, this is a forum devoted to this geometry and any solution should be discussed, like the one I've shown earlier (incenter outside the triangle).
Cheers
Philippe
Based on the formula for the radius of incircle, r given a, and b, and c, I think we can be able to determine the permissible value of c.
r=(ab)^0.5+(bc)^0.5+(ac)^0.5
since r must always greater than a, b or c, the maximum value of c must not exceed r. therefore
we can replace r by c
c<(ab)^0.5+(bc)^0.5+(ac)^0.5
solving for c, (plase verify)
c<(a + b + 4(ab)^0.5 + sqrt((a + b + 4(ab)^0.5)²  4ab)) / 2
if a=9 and b=4, c < 36
On the other hand, the minimum value of c must not be less than
((a(ab)^0.5)/(a^0.5+b^0.5))^2
if a=9 and b4, c > 0.36
Hi,
With cfleitas way , i amended my file :
cursors are limited a < b < c
and c limited by g(x) = sqrt(a b) + sqrt(b x) + sqrt(a x) and g(x) > x
x = c
c is the variable...
Thank you cfleitas !
And thank you, "Dynamic Geometry investigation" ...
https://ggbm.at/565373
Best regards
Philippe
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