It has certainly been posted before. You can use Casey's theorem and the transversal theorem. Do you have a different solution, Pascual?

well i basically used caseys to find the ratio between incircle and the other tangent circle, but a friend of mine did it without caseys with a bash of trigonometry...I was wondering if there is a solution by inversion????

I have a different proof using the following Lemma: Lemma: For a circle $S_1$ which is internally tangent to $S_2$ at X, let $A,B\in S_2$ s.t. AB is tangent to $S_1$ at a point $C$. Then CX bisects the arc AB. From this lemma, it is enough to prove the following: Let D,E be the points on CA and CB respectively s.t. $DE\perp CI,I\in DE$. Let also lines AI and BI meet the circumcircle at another point A',B' respectively. Then it suffices to show $\angle IA'E+\angle IB'D=\angle C$. This is easy by considering the similarity between triangles IBA' and IAB'.

Solution by inversion... Let $o, i, j$ be respectively Circumcircle, Incircle, ThirdCircle. In inversion in $i$ lines $AB, BC, CA,$ and $o$ will give four congruent circles, call them $c', a', b', o'$. $o'$ passes through points of intersection of pairs $(a',b'), (b',c'), (c',a')$ distinct from $I$ (call them $K, L, M$ respectively). Angle bisector of $\angle ACB$ is perpendicular bisector of $DE$. $J$ (center of $j$) lies on this pernpedicular bisector. So it is enough to prove that $J$ lies on $DE$ or in other words $DI || IE$. Image of $j$ in this inversion is a circle $j'$ to which $a', b', o'$ are tangent internally. So $J'$ (center of $j'$) is point $K$. Let $(a'), (b')$ be centers of $a'$ and $b'$ and $X$ point of intersection of angle bisector of $\angle ACB$ and $(a')(b')$. Now because $D'(a') = (a')K$ we get $D'I || (a')X$ ($D'$ and $E'$ are images of $D$ and $E$) Similary $E'I || (b')X$ Thus $E'I || D'I$ And $EI || DI$, just as we wanted.

This problem has already been posted for at least 3 times. As far as I remember, there are solutions making use of Pascal, Ceva, etc.

Well-known lemma: If circle $K$ is tangent internally to the circle $L$ at point $F$ and the tangent to the smaller circle at $G$ cuts the bigger circle in $X, Y$, then the line $FG$ bisects the $\angle XFY$. So, let $F$ be the tangency point of the circle to the circumcentre of triangle $ABC$, and let $A', B'$ be the middle points of arcs $BC$ and $AC$ respectively (the arcs which don't contain $A$/$B$). Then, from the lemma the lines $FD$ and $FE$ cuts the circumcircle of $ABC$ in $B'$ and $A'$ respectively. Note also that $I$ is intersection of the lines $AA'$ and $BB'$, so from the Pascal Theorem for the hexagon $A, F, B, A', C, B'$ the points $D, E, I$ are collinear. Now, it's also obvious that $I$ is the midpoint of $DE$ since $CI$ is the angle bisector in the isosceles triangle $CDE$.

grobber wrote: It has certainly been posted before. You can use Casey's theorem and the transversal theorem. Do you have a different solution, Pascual?

Sorry to revive the thread, but I have some questions 1) How do you know that the homothecies actually hit the midpoints of the chords? 2) How does Pascal's imply that D,I,E are collinear? Doesn't pascal's just say that if you extend the sides the intersections are concurrent? o.O

Dear Mathlinkers, see also for this nice of Deprez, http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure p. 15 Sincerely Jean-Louis

I hope my solution is correct.... it is too easy so I feel queaky about this. Please rectify my errors.

Diagram: [geogebra]82cceafa5c8f38551c0b8f05ce6eab94a4fa8718[/geogebra]

It is not strange the topic be posted many times, is a classical result, one of the oldest references is an english problem book by Rev. Radford 1925. It has been also studied by Leo Bankoff in Crux 1983, pp.2-7 (A mixtilinear adventure).

Dear F. and Mathlinkers, the reference I gave in my article is Deprez E., Mathesis X (1890) 67-68. It is right ? Sincerely Jean-Louis

Dear J-L and Mathlinkers, yes, the reference is right. Mathesis 1890 p. 67-68 gives the solution by Deprez to a more complicated problem proposed by De Longchamps.

Potla wrote: I hope my solution is correct.... it is too easy so I feel queaky about this. Please rectify my errors. Let the circle drawn be denoted by $ (P)$. It is easily proven that $ I$ lies on $ DE$. Could you please explain how it is proven that $ I$ lies on $ DE$ ? Thanks.

Just for convenience, here's the official wording of the problem: 1993 G1 wrote: Consider the triangle $ABC$, its circumcircle $k$ with center $O$ and radius $R$, and its incircle with center $I$ and radius $r$. Another circle $k_c$ is tangent to the sides $CA,CB$ at $D,E$, respectively, and it is internally tangent to $k$. Show that the incenter $I$ is the midpoint of $DE$. Let circle $k_c$ have center $P$ and radius $t$ and let it touch circle $k$ at point $X$. Let the midpoint of $DE$ be $M$ and the midpoint of arc $AB$ be $N$. Clearly line $CMPN$ is the angle bisector of $\angle{C}$ and $O,P,X$ are collinear. From right triangle $\triangle{CDP}$, and the Law of Sines on $\triangle{CAN}$, we get (set $\angle{A}=2\alpha$, etc.) \[PC=\frac{t}{\sin\gamma},\qquad CN=2R\cos(\alpha-\beta).\]From the power of point $P$ with respect to circle $k$, we see that \begin{align*} t(2R-t)=R^2-(R-t)^2&=R^2-PO^2\\ &=PC\cdot PN=PC(CN-PC)=\frac{t}{\sin\gamma}\left(2R\cos(\alpha-\beta)-\frac{t}{\sin\gamma}\right), \end{align*}so \[2R\sin^2\gamma-t\sin^2\gamma=2R\sin\gamma\cos(\alpha-\beta)-t\]and \[t\cos^2\gamma=t(1-\sin^2\gamma)=2R\sin\gamma(\cos(\alpha-\beta)-\sin\gamma)=4R\sin\alpha\sin\beta\sin\gamma=r.\]It's simple to verify that \[CM=CI=\frac{r}{\sin\gamma},\]so we're done.

Consider the first problem of this. Two proofs have been provided. One by inversion and the other by using the lemma. If we consider the inversion proof, we can see that since $I$ maps to $I_A$, $M$ maps to $M'$, $N$ maps to $N'$ and we also know that $I_AM'=I_AN'$ which means $\frac{IM\cdot r^2}{IN\cdot AM}=\frac{IN\cdot r^2}{AI\cdot AN}\Longrightarrow IM=IN$ as required. Generalisation: Let $ABC$ be a triangle and consider the circle $\Omega$ which is tangent to the rays $AB, AC$ in the points $M, N$ respectively and which is externally tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the excenter of $\Delta ABC$ opposite to $A, I_A$ is the midpoint of $MN$. Proof is given here as a generalisation.

This is a very common and famous problem.The most common solutions are using homothety,but I shall present a solution using Casey. Let the smaller circle be tangent to the circumcircle at $X$.Then by Casey's theorem to the circles $A,B,BXE,C$ we have $AB*CE+AC*BD=AD*BC \Rightarrow \boxed{AD=AE=\frac{2bc}{a+b+c}}$.Thus if the midpoint of $DE$ is $I'$ we have $AI'=\frac{2bc}{a+b+c}\cos\frac{A}{2}$.It is also well known that $AI=\frac{2bc}{a+b+c}\cos\frac{A}{2}$ so $I \equiv I'$ as desired.

For the Sawayama's Lemma in $\triangle ABC $ and cevian " $ BC $ ". Then the chord $ ED $ passes through the incenter $ I $ of $ \triangle ABC \Rightarrow I$ is midpoint of $ DE $

Let the point $M$ to be the midpoint of $DE$. Suppose that the circle touches the circumcircle of $ABC$ in $P$. Prove that $MDCP$ and $MEBP$ are concyclic. Then we are done.

Does this approach work?

Can someone give me a hint on how to proceed? Thank you, book_learner

Above can't work because you are not using that the circle touch the circumcircle of $\triangle ABC$.

Hey can someone check my solution? The other solutions used complex stuff like Casey's theorem, some homothety (haven't learnt this), and some other inversion stuff. But I think my solution only uses elementary theorems which makes me really wonder if I even have it right Claim: $ID=IE$ Proof: Because $I$ is on the angle bisector of $\angle{ACB}$ we know that $ID=IE$. Claim: $\triangle{DWI} \cong \triangle{EWI}$ Proof: We have $\triangle{DIC} \cong \triangle{EIC}$ because of SSS. So we know that $\angle{WID}=\angle{WIE}=90^\circ$. Thus we have $\triangle{DWI} \cong \triangle{EWI}$ by SAS (or by SSS with Pythagorean Theorem). Claim: $I, D, E$ are all collinear Proof: Now, labeling $\angle{DWI}=\angle{EWI}=\alpha$ and $WI=x, ID=IE=y$ we have \[ \frac{ID} {\sin(\alpha)}=\sqrt{x^2+y^2} \implies ID=\sin(\alpha)\sqrt{x^2+y^2} \]by Law of Sines. Then doing Law of Sines on $\triangle{WDE}$ we have \[ \frac{DE}{\sin(2\alpha)}=\frac{\sqrt{x^2+y^2}}{\cos(\alpha)} \implies DE=\frac{\sqrt{x^2+y^2}}{\cos(\alpha)} \cdot \sin(2\alpha) \]Then because $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ we have \[ DE=2\sin(\alpha)\sqrt{x^2+y^2}=2ID=ID+IE \]which implies $I, D, E$ are all collinear. Because we have already proved that $ID=IE$ we have just proved that $I$ is the midpoint of $\overline{DE}$.

(Here’s an FYI to anyone who has solved this problem.) This handout utilizes a more natural approach to solve this question and prove other properties of the mixtlinear incircle.

lol i didn't even know what a mixtilinear thing was when i "solved" it. also bump, can somebody check my sol to this problem? its @2above

Cute problem. By symmetry we can assume that the sides are instead $AB$ and $AC$. Under $\sqrt{bc}$ inversion the $A$-mixtilinear incircle goes to the $A$-excircle. Then $D$ and $E$ go to the points of contact of the $A$-excircle with the sides $AB$ and $AC$ (call these $D'$ and $E'$). Finally, it's well known that $(A,M;I,I_A)=-1$ where $M$ is the midpoint of the arc $BC$ not passing through $A$ which implies that $I$ goes to $I_A$ (inversion induces harmonic bundles). Then $AD'I_AE'$ is cyclic so $I$ lies on $DE$. We are done. $\blacksquare$