In a parallelogram $$ABCD$$, a point $$P$$ on the segment $$AB$$ is taken such that $$\frac{AP}{AB}=\frac{61}{2022}$$ and a point $$Q$$ on the 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.

Let $$\vec{AB}=\vec{v}$$ and $$\vec{AD}=\vec{w}$$. Take the origin at $$A(0,0)$$, so

$$B:\;\vec{v}, \qquad D:\;\vec{w}, \qquad C:\;\vec{v}+\vec{w}.$$

Because $$P$$ divides $$AB$$ in the ratio $$AP:PB=61:(2022-61)$$, its position vector is

$$\vec{P}=A+\frac{61}{2022}\,\vec{v}=\frac{61}{2022}\,\vec{v}.$$

Similarly, $$Q$$ divides $$AD$$ in the ratio $$AQ:QD=61:(2065-61)$$, giving

$$\vec{Q}=A+\frac{61}{2065}\,\vec{w}=\frac{61}{2065}\,\vec{w}.$$

Write the diagonal $$AC$$ in parametric form: every point on it is

$$\vec{T}=t(\vec{v}+\vec{w}),\qquad 0\le t\le 1,$$

where $$t=\dfrac{AT}{AC}$$. We need $$\dfrac{AC}{AT}=\dfrac{1}{t}.$$

The same point $$T$$ also lies on $$PQ$$, whose parametric equation is

$$\vec{R}(s)=\vec{P}+s(\vec{Q}-\vec{P}),\qquad 0\le s\le 1.$$

Equate $$\vec{T}=\vec{R}(s):$$

$$\frac{61}{2022}\,\vec{v}+s\!\left(\frac{61}{2065}\,\vec{w}-\frac{61}{2022}\,\vec{v}\right)=t\vec{v}+t\vec{w}.$$

Match coefficients of $$\vec{v}$$ and $$\vec{w}$$ separately.

1. Along $$\vec{w}:$$ $$s\left(\frac{61}{2065}\right)=t \;\;\Longrightarrow\;\; s=\frac{2065}{61}\,t.$$

2. Along $$\vec{v}:$$ $$\frac{61}{2022}-s\left(\frac{61}{2022}\right)=t.$$

Substitute the value of $$s$$ from step 1 into step 2:

$$\frac{61}{2022}-\frac{2065}{61}\,t\left(\frac{61}{2022}\right)=t.$$

Simplify:

$$\frac{61}{2022}-\frac{2065t}{2022}=t \;\;\Longrightarrow\;\; 61-2065t=2022t.$$

Combine like terms:

$$61=4087t \;\;\Longrightarrow\;\; t=\frac{61}{4087}.$$

Therefore

$$\frac{AC}{AT}=\frac{1}{t}=\frac{4087}{61}=67.$$

Rounded to the nearest integer (already an integer), the required value is 67.