#30 of March 2015 Grade 11 - Natural Sciences CSAT(Korean SAT) mock test

Let $A'(0,~-1).$

Then it's clear that

$\overline{AQ}+\overline{PQ}=\overline{A'Q}+\overline{PQ}\ge\overline{A'P}.$

And since there exists a point $Q$ such that $\overline{AQ}+\overline{PQ}\le6,$ we can say that

$\overline{A'P}\le6.$

Then point P should be in a circle that has a radius of $6$ and a center of $A'.$

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Let $x+y=k$ and you see that $y=-x+k.$

Then $k$ is the $y$ -intercept of $y=-x+k.$

Move the linear graph up and down and you see that $k$ is maximum when $y=-x+k$ contacts with the circle from above.

The distance from the center of the above circle from the linear graph $d=\dfrac{|-1-k|}{\sqrt{1^2+1^2}}$ should be the same with the radius of the circle( $6$ ).

$|-1-k|=\sqrt{2}\times6$

And since $k>0,$

$k=-1+6\sqrt{2}=p+q\sqrt{2}.$

Therefore $p=-1,~q=6.$