perplexus.info :: Shapes : Two Square Settlement

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

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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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

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

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