by **Guest** » Sun Jun 23, 2019 3:07 pm

I presume that by "around the circumference" you mean that four vertices all lie on the circumference of a circle. At first it wasn't at all clear to me that there could be such a trapezoid!

Here's how I analyzed the first problem:

Since AD is perpendicular to AB, we can set up a coordinate system with origin at A, x-axis along AB and y-axis along AD. Take B to be (a, 0) and D to be (0, b). A circle passing through A, B, and D must have center at (a/2, b/2). A circle with that center and radius 1 has equation [tex](x- a/2)^2+ (y- b/2)^2= 1[/tex]. Taking (x, y)= (0, 0), we must have [tex](0- a/2)^2+ (0- b/2)^2= a^2/4+ b^2/4= 1[/tex] so [tex]a^2+ b^2= 4[/tex]. You can check that (a,0) and (0,b) lead to that same equation.

Dropping a line from the fourth point perpendicular to the x-axis gives a right triangle with height b and angle opposite that side 30 degrees. Since tangent is "rise over run", and tan(30)= 1/2, letting "u" be the length of the side of that triangle along the x-axis we have tan(30)= 1/2= b/u so u= 2b. The fourth vertex has coordinates (a- 2b, b). Does that point satisfy the equation of the circle? (Does it lie on the circle?)

[tex](a- 2b- a/2)^2+ (b- b/2)^2= (a/2- 2b)^2+ (b/2)^2= a^2/4- 2ab+ 4b^2+ b^2/4[/tex]. Since [tex]a^2/4+ b^2/4= 1[/tex] that reduces to [tex]1- 2ab+ 4b^2= 1[/tex] or [tex]4b^2- 2ab= 2b(2b- a)= 0[/tex]. b cannot be 0 so we must have a= 2b. Then [tex]a^2+ b^2= 4b^2+ b^2= 5b^2= 4[/tex]. [tex]b= \frac{\sqrt{5}}{2}[/tex] and [tex]a= \sqrt{5}[/tex]

The fourth vertex, then, is at [tex](a- 2b, b)= (\sqrt{5}- \sqrt{5}, \sqrt{5}/2)= (0, \sqrt{5}/2)[/tex].

No, there is no such trapezoid because that "fourth vertex", C, lies on the same line as A and D.