There's No "Squared" Here
A = ( 0 , 0 ) , B = ( 1 , 3 ) , C = ( 3 , 4 ) , D = ( 5 , 1 )
Minimize the sum of the lengths of P A + P B + P C + P D .
This value can be written as X + Y , where X and Y are positive integers; submit your answer as X + Y .
The answer is 25.
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2 solutions
In a convex quadrilateral the shortest path that connects all vertexes is the sum of the two diagonals.
These are AC and BD.
A
C
=
3
2
+
4
2
=
5
,
a
n
d
B
D
=
(
3
−
1
)
2
+
(
1
−
5
)
2
=
2
0
.
S
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m
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(
P
A
+
P
C
)
+
(
P
B
+
P
D
)
=
A
C
+
B
D
=
5
+
2
0
=
X
+
Y
.
S
o
X
+
Y
=
5
+
2
0
=
2
5
.
N
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.
The first line is true for a convex quadrilateral. For a concave quadrilateral, the answer is quite different.
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Thanks. I have corrected.
What would be the minimum for a concave quadrilateral?
If the four points A , B , C , D form a convex quadrilateral, we see that, for any point P A P + P C ≥ A C B P + P D ≥ B D with equality when A , P , C and B , P , D are collinear. Thus we deduce that A P + B P + C P + D P ≥ A C + B D with equality when P is the point of intersection of A C and B D .
In this case, the point of intersection is ( 1 1 2 1 , 1 1 2 8 ) , and the shortest sum of distances is 5 + 2 0 , making the answer 5 + 2 0 = 2 5 .