Equidistant Points 2

Directions: Using the digits 1 to 9 at most one time each, fill in the boxes to create two points that are equidistant from (4,-1).

Hint

Which methods are available to determine answers to this problem? What shape is defined by all of the points that are equidistant from (4, -1)?

Answer

There are many correct answers to this problem including: (2, 4) and (9, 1) as well as (2, 3), and (8, 1)

Source: Bryan Anderson

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5 comments

  1. We worked this problem out and only found 4 possible answers. Do you think there are more than that?

    Our other answers were (1,4), (9,2) and (1,3), (8,2).

    Thanks!

    • there are many more, if you think in terms of slope, any that create perpendicular lines will work; all the points to the left or right two and up one will be the same as going left or right one, then up two. etc

      • If remove the limits of only being able to use the digits 1 through 9 at most once, then there are infinite possibilities, but with that limit it place I think Andrew is right. I can only find the two pairs listed above. Any other pairs of points will require repeating a digit or using 0, negatives, or fractions.

  2. I just gave this to my students and here are some of the answers that I have verified as correct. There may be more if you count the triangles on the grid, like I did. I imagine there are several more that could be correct.
    (1,3) & (8,2)
    (2,3) & (8,1)
    (8,1) & (6,3)
    (6,2) & (7,1)
    (3,7) & (8,6)- this one was very creative! Distance is sqrt(65)
    (1,4) & (9,2)

  3. Here are all solutions to this problem, I determined this by using pythagorean theorem
    Moves: 2-3 distance = sqrt 13 (7,1) (6,2)
    Moves: 2-4 d = sqrt 20 (8,1) (6,3)or(2,3)
    Moves: 2-5 d = sqrt 29 (9,1) (6,4)or(2,4)
    Moves: 3-4 d = sqrt 25 (8,2) (1,3)or(7,3)
    Moves 3-5 d = sqrt 34 (9,2) (7,4)or(1,4)
    Moves 4-5 d = sqrt 41 (9,3) (8,4)

    Additionally: Moves 4-7 (8,6) and 1-8 (3,7) both have a distance of sqrt 65
    I believe this is all possible solutions, but I am continuing to test possible solutions

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