This problem seems incorrect. Apply barycentric coordinates. Let $P=(\alpha,\beta,\gamma)$ so that $A_1=(0:\beta:\gamma)$. Then \[\frac{[PBA_1]}{[ABC]}=\begin{vmatrix} \alpha & \beta & \gamma \\ 0 & 1 & 0 \\ 0 & \beta & \gamma \end{vmatrix}=\alpha\gamma\]and similarly $\frac{[PCB_1]}{[ABC]}=\beta\alpha$ and $\frac{[PAC_1]}{[ABC]}=\gamma\beta$. So the problem condition is equivalent to $\beta\gamma+\gamma\alpha+\alpha\beta=\frac{1}{2}$ where $\alpha+\beta+\gamma=1$. But \[(\alpha+\beta+\gamma)^2\geq3(\beta\gamma+\gamma\alpha+\alpha\beta)\]which is a contradiction. So the equality in the problem condition can never hold.

oskarbohme34 wrote: This problem seems incorrect. Apply barycentric coordinates. Let $P=(\alpha,\beta,\gamma)$ so that $A_1=(0:\beta:\gamma)$. Then \[\frac{[PBA_1]}{[ABC]}=\begin{vmatrix} \alpha & \beta & \gamma \\ 0 & 1 & 0 \\ 0 & \beta & \gamma \end{vmatrix}=\alpha\gamma\]and similarly $\frac{[PCB_1]}{[ABC]}=\beta\alpha$ and $\frac{[PAC_1]}{[ABC]}=\gamma\beta$. So the problem condition is equivalent to $\beta\gamma+\gamma\alpha+\alpha\beta=\frac{1}{2}$ where $\alpha+\beta+\gamma=1$. But \[(\alpha+\beta+\gamma)^2\geq3(\beta\gamma+\gamma\alpha+\alpha\beta)\]which is a contradiction. So the equality in the problem condition can never hold. $A_1=(0:\beta:\gamma)$ this is not homogenized so your area is wrong

Indeed you are correct. Let me fix now. Instead, we have $\frac{[PBA_1]}{[ABC]}=\frac{\alpha\gamma}{\beta+\gamma}$ and likewise, so \[\frac{\alpha\gamma}{\beta+\gamma}+\frac{\beta\alpha}{\gamma+\alpha}+\frac{\gamma\beta}{\alpha+\beta}=\frac{1}{2}\]where $\alpha+\beta+\gamma=1$. The left hand side factors as \[\frac{(\alpha+\beta+\gamma)(\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta+\alpha\beta\gamma)}{(\beta+\gamma)(\gamma+\alpha)(\alpha+\beta)}\]so \[0=2(\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta+\alpha\beta\gamma)-(\beta+\gamma)(\gamma+\alpha)(\alpha+\beta)=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)\]so two of $\alpha,\beta,\gamma$ are equal. This amounts to $P$ being on a median.