## 1. Mixtilinear in-circles

These are circles simultaneously tangent to two sides of a triangle ABC and to its circumcircle c. We distinguish between incircles, contained in the circumcircle, and excircles lying outside the circumcircle of ABC.

Below is described the construction of such an mixtilinear incircle contained in the angle A of triangle ABC.

As usual for such tangency problems, use an inversion. Here preferably to the circle A(2R), centered at A with radius 2R, R being the circumradius of ABC.
By this inversion the circumcircle of ABC maps to line DE, orthogonal to the diameter AA0 of the circumcircle.
The mixtilinear circle maps through the inversion to the tritangent circle of triangle DAE, which is similar to ABC.
The contact point A' of the mixtilinear with the circumcircle maps to A'', the contact point of the tritangent circle with side DE. Let M bet the intersection of AA' with BC.
In view of the similarity of triangles ABC and ADE, A''D/A''E=(s-c)/(s-b), s being the half-perimeter and {a,b,c} the side lengths opposite to {A,B,C} correspondingly.
Draw the parallel to DE from A, which is tangent to the circumcircle of ABC. Measuring the cross-ratio of the lines (AC,AB,AM,AG)=c through their intersections with lines DE and BC we find:
(i) On line DE, c = DA''/EA'' = (s-c)/(s-b).
(ii) On line BC c = (CM/BM):(CG/BG) = (CM/BM):(b2/c2), last equality shown below.
Equating we get CM/BM = (s-c)/(s-b)(b2/c2). This gives the position of M on BC, consequently the line AA' and its intersection point A' with the circumcircle.

Remark Since the external simulitide center X(56) of the circumcircle and incircle of ABC has trilinears:
(a/(b+c-a), b/(c+a-b), c/(a+b-c)),
the intersection point Q of AX with BC has CQ/BQ = (a+b-c)/(c+a-b)(b2/c2) (see TrilinearsRelatedToOthers.html ), thus coinciding with M. This gives another way of locating the contact point of the A-mixtilinear circle with the circumcircle.

## 2. Mixtilinear ex-circles

These are circles simultaneously tangent to two sides of a triangle ABC and to its circumcircle from outside.
Repeating almost verbatim the arguments of the preceding section we find the contact-point of the A-mixtilinear ex-circle as the intersection point A' of the line AX with the circumcircle. Here X is the triangle center X(55) coinciding with the inner similitude center of the incircle and circumcircle of ABC.

As in the previous section locate the position of M on BC and relate it to X(55). The A-mixtilinear ex-circle is circle d. Using the inversion with respect to the circle A(2R) this circle maps to the incircle of triangle ADE, line DE being the inverse of the circumcircle of ABC and similar to it.

Remark The previous results show that X(55) and X(56) are correspondingly intersection points of lines {AA',BB',CC'} each joining the vertex with the contact point of the corresponding ex/in-mixtilinear circle with the circumcircle.

### See Also

TrilinearsRelatedToOthers.html

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