Question 978962
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I call bullshit on the "teacher isn't the type..." nonsense.  Your teacher gets paid a salary, said funds either derived from the tuition you pay to go to school or the property taxes your parents pay to run the public schools.  Your teacher therefore has a fiduciary responsibility to "explain things".  So cut the whiny "poor me" crap and use your energy to learn mathematics.


Be that as it may, on a coordinate system, create an arbitrary rectangle of dimensions *[tex \Large a] by *[tex \Large b].  Since coordinates can be translated and rotated arbitrarily, you can position the rectangle with one vertex at the origin and the sides parallel to the coordinate axes without loss of generality.  Hence, your vertices are (given *[tex \Large a] is the horizontal dimension), *[tex \Large (0,0), (a,0), (a,b), (0,b)].  Using the midpoint formulas:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ = \frac{x_1 + x_2}{2}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ = \frac{y_1 + y_2}{2}]


calculate the coordinates of each of the midpoints of the four sides of the rectangle, in terms of *[tex \Large a] and *[tex \Large b].  For example, the side between the points *[tex \Large (a,0)] and  *[tex \Large (a,b)] has a midpoint *[tex \Large (a,\frac{b}{2})].


Then, using the distance formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{(x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2}]


calculate the distance between each pair of adjacent side midpoints.  If the inscribed figure is a rhombus, all of the sides must be equal.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \