Sorry, in part b) we have to determine values $\frac{c}{a}$.

Answer for a)

Answer for b) $\frac{c}{a}\in\left(\frac {3-\sqrt 5}{2}, 1 \right]$ We're waiting the picture and a complete solution.

From 2nd median theorem, it is $HM = \frac{{{b^2} - {c^2}}}{{2a}}$ and $HC = \frac{{{b^2} - {c^2}}}{{2a}} + \frac{a}{2} = \frac{{{b^2} + {a^2} - {c^2}}}{{2a}}$ From Menelaus theorem in $AMC$ with transversal $MGA$ we get: $\frac{{HM}}{{HC}} \cdot \frac{{CT}}{{TA}} \cdot \frac{{AG}}{{GM}} = 1 \Leftrightarrow $ $\frac{{{b^2} - {c^2}}}{{{b^2} + {a^2} - {c^2}}} \cdot \frac{a}{c} \cdot 2 = 1 \Leftrightarrow $ $\boxed{b = \sqrt {\frac{{c(2ac + {a^2} - {c^2})}}{{2a - c}}} }$

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What about b)?!

Reminder for b) part!!

If someone is quarantined they can complete the task.

Points $H$ and $T$ are marked respectively on the sides $BC$ abd $AC$ of triangle $ABC$ so that $AH$ is the altitude and $BT$ is the bisectrix $ABC$. It is known that the gravity center of $ABC$ lies on the line $HT$. a) Find $AC$ if $BC$=a and $AB$=c. b) Determine all possible values of $\frac{c}{a}$ for all triangles $ABC$ satisfying the given condition. (I. Voronovich)

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