KöMal 04
Point
P
is located inside or on the perimeter of a unit square. If the maximum value of its distances from the vertices of the square (
w
+
x
+
y
+
z
) is
a
+
b
, where
a
and
b
are positive integers and
b
is square-free, then find the value of
a
+
b
.
The answer is 4.
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1 solution
- First we observe that the square has symmetrical properties and so we can ignore three quadrants of the square, this leaves us only to analyse the point P lying between 0 ≤ x ≤ 0 . 5 and 0 ≤ y ≤ 0 . 5 .
- If the point P was to leave the square the combined lengths of w,x,y,z would be greater than if P was contained within the square. We can therefore conclude the point P must lie on the edge of the square for the added lengths of w,x,y,z to be a maximum. This case has symmetrical properties too so we can assume it either lies on the y axis of the square or the x axis. For simplification let it lie on the x axis. We are now dealing with points between 0 ≤ x ≤ 0 . 5
- Solving for the total lengths we reach the equation x + ( 1 − x ) + x 2 + 1 + ( 1 − x ) 2 + 1 f o r 0 ≤ x ≤ 0 . 5
- Taking the derivative d x d y ( x + ( 1 − x ) + x 2 + 1 + ( 1 − x ) 2 + 1 ) = x 2 − 2 x + 2 x − 1 + x 2 + 1 x
- Solving for d x d y = 0 we get x = 2 1
- This is a minimum and can be shown by a sign diagram or observing the graph.
- Therefore as x decreases the total length increases.
- As 0 is the lowest value we substitute this into the original equation giving 0 + ( 1 − 0 ) + 0 2 + 1 + ( 1 − 0 ) 2 + 1
- This gives us 2 + 2
- Therefore a + b = 4