HMMT February

Solution:

We will use the following notation: let $\omega$ be the circle of radius $49$ tangent to each of $\omega_{a}, \omega_{b}, \omega_{c}$. Let $\omega_{a}, \omega_{b}, \omega_{c}$ have radii $r_{a}, r_{b}, r_{c}$ respectively. Let $\gamma$ be the incircle of $ABC$, with center $I$ and radius $r$. Note that $DEF$ is the intouch triangle of $ABC$ and $\gamma$ is orthogonal to $\omega_{a}, \omega_{b}, \omega_{c}$ (i.e. $ID, IE, IF$ are the common internal tangents). Since $AD, BE, CF$ are concurrent at $T$, we have $K = AB \cap DE$ satisfies $(A, B; F, K) = -1$, so $K$ is the external center of homothety of $\omega_{a}$ and $\omega_{b}$. In particular, $K$ lies on $\ell$. Similarly, $BC \cap EF$ also lies on $\ell$, so $\ell$ is the polar of $T$ to $\gamma$. Hence $IT \perp \ell$ so if $L$ is the foot from $I$ to $\ell$, we have $IT \cdot IL = r^{2}$.

An inversion about $\gamma$ preserves $\omega_{a}, \omega_{b}, \omega_{c}$ and sends $\ell$ to the circle with diameter $IT$. Since inversion preserves tangency, the circle with diameter $IT$ must be $\omega$. Therefore $IT = 98$ by the condition of the problem statement. Letting $a, b, c$ be the radii of $\omega_{a}, \omega_{b}, \omega_{c}$ respectively and invoking Heron's formula as well as $A = r s$ for triangle $ABC$, we see that $\gamma$ has radius

$$r = \sqrt{\frac{r_{a} r_{b} r_{c}}{r_{a} + r_{b} + r_{c}}}.$$

We will compute this quantity using Descartes' theorem. Note that there are two circles tangent to $\omega_{a}$, $\omega_{b}, \omega_{c}$, one with radius $IT/2$ and one with radius $\infty$. By Descartes' circle theorem, we have (where $k_{a} := 1/a$ is the curvature)

$$k_{a} + k_{b} + k_{c} + 2 \sqrt{k_{a} k_{b} + k_{b} k_{c} + k_{a} k_{c}} = \frac{1}{IT/2}$$

and

$$k_{a} + k_{b} + k_{c} - 2 \sqrt{k_{a} k_{b} + k_{b} k_{c} + k_{c} k_{a}} = 0$$

which implies

$$\sqrt{\frac{r_{a} + r_{b} + r_{c}}{r_{a} r_{b} r_{c}}} = \sqrt{k_{a} k_{b} + k_{b} k_{c} + k_{c} k_{a}} = \frac{1}{2 IT}.$$

Therefore $r = 2 IT$, which means $IL = \frac{r^{2}}{IT} = 4 IT$ and $TL = 3 IT = 294$.

Solution:

Using the same notation as the previous solution, note that the point $T$ can be expressed with un-normalized barycentric coordinates

$$\left(\frac{1}{r_{a}} : \frac{1}{r_{b}} : \frac{1}{r_{c}}\right)$$

with respect to $ABC$ because $T$ is the Gergonne point of triangle $ABC$. The distance from $T$ to $\ell$ can be expressed as a weighted average of the distances from each of the points $A, B, C$, which is

$$\frac{1 / r_{a}}{1 / r_{a} + 1 / r_{b} + 1 / r_{c}} \cdot r_{a} + \frac{1 / r_{b}}{1 / r_{a} + 1 / r_{b} + 1 / r_{c}} \cdot r_{b} + \frac{1 / r_{c}}{1 / r_{a} + 1 / r_{b} + 1 / r_{c}} \cdot r_{c} = \frac{3}{1 / r_{a} + 1 / r_{b} + 1 / r_{c}}.$$

Note that there are two circles tangent to $\omega_{a}, \omega_{b}, \omega_{c}$, one with radius $49$ and one with radius $\infty$. By Descartes' circle theorem, we have (where $k_{a} := 1 / r_{a}$ is the curvature)

$$k_{a} + k_{b} + k_{c} + 2 \sqrt{k_{a} k_{b} + k_{b} k_{c} + k_{a} k_{c}} = \frac{1}{49}$$

and

$$k_{a} + k_{b} + k_{c} - 2 \sqrt{k_{a} k_{b} + k_{b} k_{c} + k_{c} k_{a}} = 0$$

so $k_{a} + k_{b} + k_{c} = 1 / 98$. The distance from $T$ to $\ell$ is then $3 \cdot 98 = 294$.

Solution:

As in the first solution, we deduce that $IT$ is a diameter of $\omega$, with $T$ being the point on $\omega$ closest to $\ell$. By Steiner's porism, we can hold $\omega$ and $\ell$ fixed while making $\omega_{c}$ into a line parallel to $\ell$, resulting in the following figure:

Let $R$ be the common radius of $\omega_{a}$ and $\omega_{b}$ and $r$ be the radius of $\omega$. Notice that $ABDE$ is a rectangle with center $T$, so $R = AE = 2 IT = 4r$. The distance from $T$ to $\ell$ is $IL - IT = 2R - 2r = 6r = 294$.