Corners of a square - Oxford Mathematics Admissions Test 2020

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  • Опубліковано 23 січ 2025

КОМЕНТАРІ • 6

  • @Fate-rh1ob
    @Fate-rh1ob Місяць тому +4

    Another way to derive the other corner points is to find the vector of the other diagonal and attach it to the center. It is easy to see that for the given diagonal vector (5,3) - (1,5) = (4,-2), a perpendicular vector of the same length is given by (2,4). Then the other corner points are (3,4) ± 0.5 * (2,4), resulting in (4,6) and (2,2).

    • @mathoutloud
      @mathoutloud  Місяць тому +1

      That’s a much faster way to do it than what I did I’m sure!

  • @slytherinbrian
    @slytherinbrian Місяць тому +1

    I did it a different way, since I didn't remember the vector stuff... I recognized that the slope of the line going through the center and the given corner is m=-1/2, then used that to find an equation for the line, then found the distance sqrt(5) between the center and the corners. That gave me the opposite corner. Then, I came up with the line through the other two corners, knowing that the perpendicular line has slope -1/m, or 2. And then used that to find the two points on that line at distance sqrt(5) from the center. Maybe not the simplest way, but it got to the right answer in the end.

    • @mathoutloud
      @mathoutloud  Місяць тому +1

      I end up doing things in not so simple ways all the time here. It matters that you figured it out!

    • @RajSandhu-gm8iz
      @RajSandhu-gm8iz Місяць тому +1

      I sort of did it that way as well, but, sort of cheated since they gave you options. Just by a rough sketch it looks as if answer will be (2,2), hence the other diagonal. As you said this has gradient 2 and passes through (3,4), so equation of line is y=2x-2. Only coordinate that works is (2,2).
      Another possible way is use the technique of shifting the coordinate axes so that the centre is at the origin.
      To move the centre to the origin is a vector move of (-3,-4). That means the given corner becomes (-2,1). Rotate this 90° anti-clockwise to get one of the other vertices (-1,-2). Now shift back to original centre by vector (3,4), gives (2,2). If had not worked could have done same with other diagonal you found (5,3).

  • @dan-florinchereches4892
    @dan-florinchereches4892 Місяць тому

    I would say (1,5)-(3,4)=(-2,1)
    I will use the property that a squares diagonals halve eachother and are perpendicular on one another.
    (3,4)-(-2,1)=(5,3) another corner
    Considering (-2,1) as a vector represented in complex plane would be -2+i. Rotate 90 degrees by multiplying by i:
    ( -2+i)*i=-1-2i
    So another corner is (3,4)+(-1,-2)=(2,2)
    And last corner (3,4)-(-1,-2)=(4,6)
    Solution D) and wonder about your method