Incircle of $ABC$ touches $AC$ in $D$, $BC$ in $E$ and $AB$ in $K$. $J$ is the center of the excircle which touches the side $AB$. The circumcircle of $ADJ$ and $BEJ$ intersect in point $J$ and $T$. Prove that the circumcircle of $ATB$ and incircle of $ABC$ touch in one point.

I got that $T$ has to be on incircle of $ABC$. It also looks like that $K$ is on $JT$. One thing which could help is that $T$, center of the incircle of $ABC$ and center of the circumcircle of $ATB$ (labeled $M$ below) are colinear.



First, observe that $\angle DTE = \angle DTJ + \angle JTE = \left(\pi - \angle JAD\right) + \left(\pi - \angle EBJ\right) = 2\pi - \angle JAD - \angle EBJ = 2\pi - \left(\frac \pi 2 + \frac \alpha 2\right) - \left(\frac \pi 2 + \frac \beta 2\right) = \pi - \frac \alpha 2 - \frac \beta 2.$

On the other hand, $\angle EKD = \pi - \angle DKA - \angle BKE = \pi - \left(\frac \pi 2 - \frac \alpha 2\right) - \left(\frac \pi 2 - \frac \beta 2\right) = \frac \alpha 2 + \frac \beta 2$.

Thus $\angle DTE + \angle EKD = \pi$ which implies the fact you mentioned: $T$ lies on the incircle $o$ of triangle $ABC$.

Note that $\angle ATJ = \angle ADJ = \pi - \angle JDC = \pi - \angle CEJ = \angle JEB = \angle JTB$, so $TJ$ is the angle bisector of angle $ATB$.

Notice that $DK \perp AI \perp AJ$, so $DK \parallel AJ$. Moreover, $\angle TJA = \pi - \angle ADT = \angle TDC = \angle TKD$. Therefore $TK \parallel TJ$, so $K \in TJ$. Since $TJ$ is the angle bisector of angle $ATB$, we deduce that $\angle ATK = \angle KTB$.

Let $TK$ intersect circumcircle $\omega$ of triangle $ATB$ at points $T$ and $X$. Then $K$ is the midpoint of arc $AB$ of $\omega$ because $TK$ is bisector of angle $ATB$. Let $l$ be the tangent line to $\omega$ at $X$. Then $AB \parallel l$.

Consider homothety centered at $T$ which maps $X$ to $K$. Then $l$ is mapped to $AB$ so $\omega$ is mapped to a circle passing through $T$ which is tangent to line $AB$ at point $K$ - this circle is $o$. As a consequence, $\omega$ and $o$ are tangent to each other at point $T$.

  • $\begingroup$ A very good answer. $\endgroup$ – HeatTheIce Jan 27 '15 at 22:13
  • $\begingroup$ Tim, are you some IMO participant, as your geometry is very strong indeed? $\endgroup$ – Sawarnik Feb 18 '15 at 14:51
  • $\begingroup$ I participated at IMO a few years ago. Now I am a university student. I still like geometry and enjoy solving problems from olympiads. $\endgroup$ – timon92 Feb 18 '15 at 17:06

Since no one has come up with an answer yet, here is a rather brute-force computation on homogeneous coordinates to verify this fact. Without loss of generality I'll choose my coordinate system in such a way that the incircle becomes the unit circle, with $K=(1:0:1)$ at the $0°$ position. Then $D$ and $E$ can be described by one peal parameter each, using the half-angle formula. This gives us

\begin{align*} E&=(t^2-1:2t:t^2+1) & D&=(u^2-1:2u:u^2+1) \end{align*}

Everything else can be expressed in terms of $t$ and $u$. The unit circle has the diagonal matrix $U=\operatorname{diag}(1,1,-1)$, and multiplying that matrix with the vecotr of a point gives the tangent in that point. Computing the cross product between two lines gives their point of intersection. So you get

\begin{align*} A&=(U\cdot D)\times(U\cdot K)=(u:1:u) \\ B&=(U\cdot E)\times(U\cdot K)=(t:1:t) \\ C&=(U\cdot D)\times(U\cdot E)=(tu - 1 : t + u : tu + 1) \end{align*}

Next, we need $J$. I'd construct that by intersecting the line orthogonal to $AI$ through $A$ with the one orthogonal to $BI$ through $B$. To form a line connecting two points, you again compute the cross product. If you set the last coordinate of the resulting vector to $0$ you obtain a vector which describes a point infinitely far away in a direction orthogonal to the line. Connecting that with another point gives an orthogonal line. Dropping the last coordinate can be formulated by multiplication with the matrix $F=\operatorname{diag}(1,1,0)$.

\begin{align*} J &= (A\times(F\cdot(A\times I)))\times(B\times(F\cdot(B\times I))) = (tu - 1 : t + u : tu) \end{align*}

Next we need circles. I computed circles as conics through the ideal circle points $Q=(1:i:0)$ and $\bar Q=(1:-i:0)$. (Usually I'd call these points $I$ and $J$, but those letters are already taken in your problem statement.) You can find the matrix for the circle through three points $a,b,c$ by computing

\begin{align*} M_1 &= \det(a,c,Q)\cdot \det(b,c,\bar Q)\cdot (b\times Q)\cdot(a\times\bar Q)^T \\ M_2 &= (M_1 - \bar M_1)+(M_1 - \bar M_1)^T \end{align*}

Using this approach, your circles are described by

\begin{align*} \bigcirc_{ADJ}=\begin{pmatrix} 2 t u & 0 & 1 - t u - u^{2} \\ 0 & 2 t u & - t - 2 u \\ 1 - t u - u^{2} & - t - 2 u & 2 u^{2} + 2 \end{pmatrix} \\ \bigcirc_{BEJ}=\begin{pmatrix} 2 t u & 0 & 1 - t^{2} - t u \\ 0 & 2 t u & -2 t - u \\ 1 - t^{2} - t u & -2 t - u & 2 t^{2} + 2 \end{pmatrix} \end{align*}

Subtracting these matrices, you obtain a degenerate conic which factors into the line at infinity and the line connecting the two points of intersection. That line connecting the intersections is $g=(t + u : 1 : -t - u)$. It's point at infinity is $G=(-1:t+u:0)$. This allows you to find the point $T$ as the intersection of that line with one of the two circles:

\begin{align*} T&=G^T\cdot\bigcirc_{ADJ}\cdot G\cdot J - 2\cdot J^T\cdot\bigcirc_{ADJ}\cdot G\cdot G\\ T&=%(t^2 + 2tu + u^2 - 1, 2t + 2u, t^2 + 2tu + u^2 + 1) ((t+u)^2-1 : 2(t+u) : (t+u)^2+1) \end{align*}

One can already recognize the half angle format of this point, so it will lie on the unit circle. You need one last circle, $ABT$:

\begin{align*} \bigcirc_{ABT}=\begin{pmatrix} 4 t u & 0 & 1 - t^{2} - 2 t u - u^{2} \\ 0 & 4 t u & -2 t - 2 u \\ 1 - t^{2} - 2 t u - u^{2} & -2 t - 2 u & 2 t^{2} + 2 u^{2} + 2 \end{pmatrix} \end{align*}

Now you can verify that $T^T\cdot U\cdot T=T^T\cdot\bigcirc_{ABT}\cdot T=0$, so the point $T$ lies on both these circles. Furthermore, the tangent to the circle is the same in both cases as well, up to a scalar factor. So we have a single touching point.

\begin{align*} U\cdot T \sim \bigcirc_{ABT}\cdot T &= (t^2 + 2tu + u^2 - 1 : 2t + 2u : - t^2 - 2tu - u^2 - 1) \end{align*}

All through this post, I've cleared common denominators from the homogeneous coordinates and circle matrices to keep things simple. And I'm happy that I could write all of this without needing a single square root. I think about my homogeneous coordinates as column vectors, even though I wrote them as rows to save space. This explains which expressions I transposed in some computations and which I left as columns.

Both your assumptions are true, by the way. $K$ lies on $J\times T=(t+u:1:-t-u)$ and the center of the circle $ABT$, which is the point $M=((t+u)^2 - 1 : 2(t+u) : 4tu)$, lies on the line $T\times I=(-2(t+u) : (t+u)^2 - 1 : 0)$. $M$ does not lie on the incircle, even though it does look that way in the figure.

Seeing that the half angle tangent parameter of $T$ is $t+u$ tells me that the line $DE$ will intersect $AB$ in a point and that point connected to $T$ will intersect the incircle in another point, which also lies on $KI$. Not sure whether this is of any use to anybody.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.