 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Two Square Settlement (Posted on 2016-02-09) Each of the sides of square PQRS has length 1.
Points E and F are on QR and RS, respectively, so that triangle PEF is equilateral.
A square with vertex Q has sides that are parallel to those of PQRS and a vertex on PE.
The length of a side of this smaller square is (x –√y)/z , where x, y, and z are positive integers and y is not divisible by the square of any prime.

Determine x,y and z

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
Angle QPE is (90-60)/2=15 degrees.
QE = tan 15 = 2-√3
Call T the corner of the small square on QE
Call U the corner of the small square on PE

QT=UT=x

Triangles PQE and UTE are similar so we can find TE = x*tan 15
Since QT+TE=QE we have
x + x*tan 15 = tan 15
or
x=tan15/(1+tan15)
which, with a bit of work simplifies to
(3-3)/6

 Posted by Jer on 2016-02-10 11:58:01 Please log in:

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