9.6. Inside an acute scalene triangle $A B C$, where $\angle A B C=60^{\circ}$, a point $T$ is marked such that $\angle A T B = \angle B T C = \angle A T C = 120^{\circ}$. The medians of the triangle intersect at point $M$. The line $T M$ intersects the circumcircle of triangle $A T C$ again at point $K$. Find $T M / M K$.

(A. Kuznetsov)
Answer: $1 / 2$.

First solution. Let $O$ be the center of the circumcircle $\Omega$ of triangle $ABC$. Since $\angle AOC = 2 \angle ABC = 120^{\circ}$, point $O$ lies on the circumcircle $\gamma$ of triangle $ATC$. Let the line $BT$ intersect the circle $\gamma$ again at point $X$, and the circle $\Omega$ at point $P$. Since $\angle ATX = \angle CXT = 60^{\circ}$, point $X$ lies on the perpendicular bisector of $AC$, so $OX$ is the diameter of $\gamma$. Therefore, $BT \perp OT$, which means $T$ is the midpoint of the chord $BP$ of the circle $\Omega$.

Finally, let point $K'$ be the point symmetric to point $P$ with respect to point $S$ - the midpoint of side $AC$. Since $\angle AK'C = \angle APC = 120^{\circ}$, point $K'$ lies on $\gamma$. Point $M$ lies on the median $BS$ of triangle $BPK'$ from vertex $B$ and divides it in the ratio $2:1$, counting from point $B$; therefore, $M$ is the centroid of triangle $BPK'$. This means $M$ also lies on the median $K'T$, so $K' = K$ and $KM:MT = 2:1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_88961667a024e1987dddg-04.jpg?height=707&width=555&top_left_y=585&top_left_x=203)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_88961667a024e1987dddg-04.jpg?height=703&width=525&top_left_y=585&top_left_x=769)

Fig. 2

Second solution. Let $AA_1$ and $CC_1$ be the medians in triangle $ABC$. Since $\angle TBC = 60^{\circ} - \angle TBA = \angle TAB$, triangles $ATB$ and $BTC$ are similar. Therefore, $\angle BC_1T = \angle TA_1C$ as corresponding angles (between a side and a median). This means points $C_1, A_1, B$, and $T$ lie on the same circle $\omega$.

Consider a homothety with center at point $M$ and coefficient $-2$. This homothety maps triangle $A_1BC_1$ to triangle $AB'C$, where $B'$ is the fourth vertex of the parallelogram $ABCB'$. Since $\angle AB'C = 60^{\circ} = 180^{\circ} - \angle ATC$, point $B'$ lies on the circumcircle $\gamma$ of triangle $ATC$. Therefore, under this homothety, circle $\omega$ maps to $\gamma$, so point $T$ maps to point $K$, and $TM / MK = 1 / 2$.

Third solution. Let $O$ and $Q$ be the centers of the circumcircles $\Omega$ and $\gamma$ of triangles $ABC$ and $ATC$, respectively, and let $H$ be the orthocenter of triangle $ABC$. Since $\angle AOC = 2 \angle ABC = 120^{\circ}$ and $\angle AHC = 180^{\circ} - \angle ABC = 120^{\circ}$, points $O$ and $H$ lie on $\gamma$. Since $\angle AQC = 360^{\circ} - 2 \angle ATC = 120^{\circ}$, point $Q$ lies on $\Omega$. As in the first solution, note that line $BT$ intersects $\gamma$ again at point $X$, which is diametrically opposite to point $O$.

Note that $OQ \parallel BH$ and $BO = OQ = QH$. Since $\angle QHB > 90^{\circ} > \angle OBH$, it follows that $BOQH$ is a rhombus. Therefore, $BH = OQ = OX / 2$.

Let $BX$ intersect $OH$ at point $L$; triangles $OLX$ and $HLB$ are similar with a ratio of $OX / BH = 2$. Therefore, $HL = HO / 3$. Recall that point $H$ maps to point $O$ under a homothety with center $M$ and coefficient $-1 / 2$, so $OM = OH / 3$, which means $OM = ML = LH$.

![](https://cdn.mathpix.com/cropped/2024_05_06_88961667a024e1987dddg-05.jpg?height=628&width=438&top_left_y=635&top_left_x=855)

Therefore, $TM$ is the median of the right triangle $OTL$, so $TM = MO$. Therefore, similar triangles $OMK$ and $TMH$ are equal, so $MK = MH = 2OM = 2TM$. This leads to the answer.

Remark. From the last solution, it is clear that $OTHK$ is an isosceles trapezoid.