Help! Math problem!!

<p>For those that have the NEW edition of the official blue book
it's page 485, problem 12</p>

<p>for those who don't have the book, it's gonna be hard cause there's an illustration, but here's the question.</p>

<p>in the figure above, rectangles PQRS and WXYZ each have a perimeter 12 and are inscribed in the circle. How many other rectangles with perimeter 12 can be inscribed in the circle?</p>

<p>a) one
b) two
c) three
d) four
e) more than four</p>

<p>Thanks guys, appreciate it</p>

<p>Because I haven’t had my cup of coffee yet I will not be able to answer your question in my own words…BUT I will give you this:</p>

<p>Choice (E) is correct. Let C be the center of the circle. Diagonal QS of PQRSis a diameter, since QRS is an inscribed right triangle. Similarly, diagonal PR is a diameter of the circle. Thus, both diagonals pass through C As PQRS is rotated about point C each rotated diagonal passes through C and thus is again a diameter of the circle. Thus, each rotation of rectangle PQRS about C yields a congruent rectangle that is inscribed in the circle. Thus there are an infinite number of rectangles congruent to PQRS inscribed in the circle. All of these rectangles have perimeter 12 Thus there are more than four rectangles other than PQRS and WXYZ that can be inscribed in the circle.</p>

<p>I just copied and pasted CB’s explanation to the solution.
Sorry for being lazy. </p>

<p>xD</p>

<p>lol i didn’t get it >.<</p>

<p>The diagonal of the rectangle is the same as the diameter. You can have infinite positions for the diameter. Thus you can have infinite amount of rectangles. </p>

<p>Another way to look at it is to imagine a rectangle rotating within the circle. It can be in any position. Thus more than 4.</p>