In a parallelogram $ABCD$, the point $P$ on segment $AB$ is taken such that $\frac{AP}{AB}=\frac{61}{2022}$ and a point $Q$ on segment $AD$ is taken such that $\frac{AQ}{AD}=\frac{61}{2065}$. If $PQ$ intersects $AC$ at $T$, find $\frac{AC}{AT}$ to the nearest integer.

Step 1: Let $\overrightarrow{AB}=\vec{u}$ and $\overrightarrow{AD}=\vec{v}$. Then the diagonal is $\overrightarrow{AC}=\vec{u}+\vec{v}$.

Step 2: We are given:

Step 3: The point $T$ lies on both $AC$ and $PQ$. Since $T\in AC$, we can write:

Since $T$ lies on the line segment $PQ$, by the section formula or collinearity of $P,T,Q$, there exists some $k\in [0,1]$ such that:

Step 4: Equating the components of $\vec{u}$ and $\vec{v}$:

Step 5: Add these two equations to eliminate $k$:

Step 6: The ratio $\frac{AC}{AT}$ is exactly $1/\lambda$: