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Congruent Incircles


Incircles

Given a triangle, draw a Cevian to one of the bases that divides it into two triangles having congruent incircles. The positions and sizes of these two circumcircles can then be determined by simultaneously solving the eight equations

x_1=(tan(1/2theta_(12)))/(tan(1/2theta_(11))+tan(1/2theta_(12)))d_1
(1)
x_2=(tan(1/2theta_(22)))/(tan(1/2theta_(21))+tan(1/2theta_(22)))d_2
(2)
a=(tan(1/2theta_(11))tan(1/2theta_(12)))/(tan(1/2theta_(11))+tan(1/2theta_(12)))d_1
(3)
a=(tan(1/2theta_(21))tan(1/2theta_(22)))/(tan(1/2theta_(21))+tan(1/2theta_(22)))d_2
(4)
h=(tantheta_(11)tantheta_(12))/(tantheta_(11)+tantheta_(12))d_1
(5)
h=(tantheta_(21)tantheta_(22))/(tantheta_(21)+tantheta_(22))d_2
(6)
d=d_1+d_2
(7)
pi=theta_(12)+theta_(21)
(8)

for the eight variables d_1, d_2, theta_(12), theta_(21), a, x_1, x_2, and h, with theta_(11), theta_(22), and d given. Generalizing to n congruent circles gives the 4n equations

x_i=(tan(1/2theta_(i2)))/(tan(1/2theta_(i1))+tan(1/2theta_(i2)))d_i
(9)
a=(tan(1/2theta_(i1))tan(1/2theta_(i2)))/(tan(1/2theta_(i1))+tan(1/2theta_(i2)))d_i
(10)
h=(tantheta_(i1)tantheta_(i2))/(tantheta_(i1)+tantheta_(i2))d_i
(11)

for i=1, ..., n,

 theta_(i2)+theta_(i+1,1)=pi
(12)

for i=1, ..., n-1, and

 d=sum_(i=1)^nd_i
(13)

to be solved for the unknowns d_i and x_i (n of them), theta_(i1) and theta_(i2) (n-2 of each for i=2, ..., n-1), and theta_(12), theta_(n1), a, and h, a total of n+n+2(n-2)+4=4n unknowns.

Given an arbitrary triangle, let n-1 Cevians be drawn from one of its vertices so all of the n triangles so determined have equal incircles. Then the incircles determined by spanning 2, 3, ..., n-1 adjacent triangles are also equal (Wells 1991, p. 67).


See also

Congruent Incircles Point, Incircle

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References

Wells, D. The Penguin Dictionary of Curious and Interesting Geometry. London: Penguin, 1991.

Referenced on Wolfram|Alpha

Congruent Incircles

Cite this as:

Weisstein, Eric W. "Congruent Incircles." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/CongruentIncircles.html

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