Four distinct points are marked in a line. For each point, the sum of the distances from said point to the other three is calculated; getting in total 4 numbers. Decide whether these 4 numbers can be, in some order: a) $29,29,35,37$ b) $28,29,35,37$ c) $28,34,34,37$
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Tags: algebra proposed, algebra
Hello_Kitty
24.08.2014 00:04
their sum must be even, each distance being counted twice, answer is a)
chaotic_iak
24.08.2014 10:05
Nobody says the points are on integer points. Denote the points $A,B,C,D$ in order from left to right, with distances $AB = x, BC = y, CD = z$. Then the sum of distances from $B$ is $x + y + (y+z) = x + 2y + z$, and from $C$ is $(x+y) + y + z = x + 2y + z$. So we need two equal distances. Furthermore, the sum of distances from $A$ is $x + (x+y) + (x+y+z) = 2x + (x + 2y + z) \ge x + 2y + z$, and similarly from $D$ is $2z + (x + 2y + z) \ge x + 2y + z$. So we need two equal distances that are the lowest. Only option A works, and indeed we can obtain $x = 3, y = 11, z = 4$ as a working example for option A. Thus only option A works.
parmenides51
16.09.2021 10:38
solved from here
mavropnevma wrote:
Solution 1. Denote the distances between consecutive points, from left to right, by $x,y,z$. Then the four values are $3x+2y+z, x+2y+z, x+2y+z, x+2y+3z$, so two values are equal, which eliminates possibility b). Now, $(3x+2y+z) - (x+2y+z) = 2x > 0$ and $(x+2y+3z) - (x+2y+z) = 2z > 0$, which eliminates possibility b). Finally, taking for example $x=3$, $y=11$, $z=4$, we can have possibility a). Extremely easy.