Four Circle Theorem

jonbenedick shared this question 4 years ago
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

Comments (15)

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Hi,


script solution... ? (to search)

https://ggbm.at/565245

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Hi,


an other solution...(without script)

https://ggbm.at/565247

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Very nice solution, Patrick!

Its core is represented by the root of function f(x), point R, as per att'd screenshot.

Cheers

Philippecbc1599caba5cfe88bbb59f33ca7407a

Files: 01.png
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Here is my solution to your problem with explanations (spanish written and without scripts):


http://matematicainteractiv...

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Nice explanation, cfleitas.


jtico

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Is there an explanation in English?

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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.

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Let 0<R_1 <=R_2<=R_3

Then: (R−R_3)/(R+R_3) +(R-R_1)/(R+R_1) * (R-R_2)/(R+R_2)= sqrt(1-(R-R_1)^2 /(R+R_1)^2 ) * sqrt(1- (R-R_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

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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 a-priori choice and this creates the said drawback.

Cheers

Philippe17e87737a3e595e59231c910b91f9333

dff796f97967684b11dc3b50d02c5f77

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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" ...

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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.

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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 so-called 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

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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 b-4, c > 0.36

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Hi,


Let 0<R_1 <=R_2<=R_3

Then: (R−R_3)/(R+R_3) +(R-R_1)/(R+R_1) * (R-R_2)/(R+R_2)= sqrt(1-(R-R_1)^2 /(R+R_1)^2 ) * sqrt(1- (R-R_2)^2/(R+R_2)^2) (http://matematicainteractiva.com/problema-triangulo-y-circunferencias-cuatro-circunferencias-inscritas-y-tangentes-entre-si)


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


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

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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...

Superb! The appropriate constraints have reached the goal, as shown underneath.

Best regards

Philippe8d0014cf72a485ef2751e2deae6f3ff8

Files: 21.png
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