My first attempt led to a lot of numerical errors that I corrected with Geometer's sketchpad so I'll present it first.
Construct two perpendicular lines, call their intersection R. Create a movable point on one of the lines, call it Q. Rotate the line by 45 degrees about Q and call the new intersection P.
Create circles of radius 11, 6, 7 centered at P, R, Q respectively.
Slide Q so that the intersections of these three circles coincide.
When this happens PR=12.01653 (with some uncertainty in the final digits.)
(II) 4 triangles way
Call length PR=QR=x so PQ=x√2
Wherever U is there are 3 triangles formed, their areas sum to that of ΔPRQ.
The four triangles haves sides
Using Heron's formula yields
While it is possible to simplify this, doing so would yield a 16th degree polynomial. No thanks. But a quick graph gives an approximate x=12.016529
x²=144.3969696...=47651/330 (or so my calculator says)
The area sum becomes
43361/1320 + 19601/1320 + 49/2 = 47651/660
(III) 3 circle equations
Suppose R at the origin, Q at (q,0), P at (0,q).
Since UR=6 we have x²+y²=36 (i)
Since QU=7 we have (x-q)²+y²=49 (ii)
Since PU=11 we have x²+(y-q)²=121 (iii)
Solving (i) and (ii) for x we get x=(q²-13)/(2q)
Solving (i) and (iii) for y we get y=(q²-85)/(2q)
Plug these back into (i) gives the bi-quadratic
This is implies PQ=√(85±42√2)
But 85±42√2 is irrational and so it cannot be written as X/Y
The given task is impossible.
Posted by Jer
on 2016-06-11 12:06:08