Parallelogram Problem

From the properties of parallelogram, we can conclude that triangle $APR\sim{CQR}$

$AP:PB=3x:2x$ and $CQ:QD=7y:3y$

Since the 2 opposite sides of a parallelogram are equal, add up the ratios: $3x+2x=7y+3y$ and therefore $x=2y$

That means that $AP=3x=6y$

Since the 2 triangles are similar, $AP:QC=6:7$ , therefore $AR=6$ and $AC=6+7$

$These\:numbers\:not\:the\:actual\:lengths\:they're\:just\:ratios$

$AR:AC=6:6+7=6:13=a:b$ . Thus, $a+b=\boxed{19}$

Due to the properties of proportionality, we can treat the parallelogram as a square and solve the problem without affecting the answer.

Therefore, let $ABCD$ be a unit square and their coordinates be $A(0,1)$ , $B(1,1)$ , $C(1,0)$ and $D(0,0)$ , then $P(0.6,1)$ and $Q(0.3,0)$ .

Then the lines AC and PQ are:

$\begin{cases} AC: \dfrac {y-1}{x-0} = \dfrac {0-1}{1-0} & \Rightarrow y = 1-x & ... (1) \\ PQ: \dfrac {y-0}{x-0.3} = \dfrac {1-0}{0.6-0.3} & \Rightarrow 0.3y = x-0.3 & ... (2) \end{cases}$

Substitute $y = 1 -x$ in equation 2, we get the $x$ -coordinate of $R$ , $x_R$ :

$\Rightarrow 0.3(1-x_R)=x_R-0.3\quad \Rightarrow 1.3x_R=0.6 \quad \Rightarrow x_R =\frac {6}{13}$

If $x_C$ is the $x$ -coordinate of $C$ , then we note that $\dfrac {AR}{AC} = \dfrac {x_R}{x_C} = \dfrac {\frac{6}{13}}{1} = \dfrac {6}{13}$

$\Rightarrow a + b = 6 + 13 = \boxed{19}$

since its just ratios, we can let AP,PB, DQ, QC be 6,4,3,7 respectively. and then applying similarity on APR and CRQ we get AR/(AC -CR) = 6/7 or AR/AC = 6/7*2 = 12/7 => answer is 12 + 7 = 19

we can easily do the question if we take AB=CD=gcd(2+3,7+3) then AP=18x CQ=21x the solution is now very trivial