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

Geometry Level 3

A ( 0 , 1 ) A(0,~1) and P P that are points on a Cartesian coordinates system, satisfying these conditions.

  • A) Point P is on Quadrant 1.

  • B) For some point Q Q on the x x -axis, A Q + P Q 6. \overline{AQ}+\overline{PQ}\le6.

Let D D be the area shown by all possible locations of P . P.

A point ( x , y ) (x,~y) belongs to D . D. The maximum of x + y x+y is p + q 2 p+q\sqrt{2} where p , q p,~q are integers.

Find the value of p + q . p+q.


This problem is a part of <Grade 11 CSAT Mock test> series .


The answer is 5.

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1 solution

Boi (보이)
Aug 2, 2017

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

Then it's clear that

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

And since there exists a point Q Q such that A Q + P Q 6 , \overline{AQ}+\overline{PQ}\le6, we can say that

A P 6. \overline{A'P}\le6.

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


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

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

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

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

1 k = 2 × 6 |-1-k|=\sqrt{2}\times6

And since k > 0 , k>0,

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

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

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