Broken Hypotenuse

The area $A$ of the right triangle with inradius r may be expressed in two different ways:

$A$ = inradius x perimeter / 2 = $r * [(x + r) + (y + r) + (x + y)] / 2 = r * (x + y + r),$ and

$A$ = base x height / 2 = $(x + r) * (y + r) / 2 = r * (x + y + r) / 2 + x * y / 2 = A / 2 + x * y / 2.$

Hence $x * y = A$ .

So the required area $= 7*9 = 63$

Imagine a circle with unit radius in the first quadrant tangent to the axes, a point constrained to move on the circle and a line tangent to the circle through the point. Find the point that results in a tangent line that is divided in the ratio 7:9. Then find the parallel line with length 16. Definitely not as elegant a solution as that of Mr Ahmed. (But better than not getting a solution.)

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from math import sqrt
from sympy import Point, N
from scipy.optimize import minimize_scalar

tangentPoint=lambda x:Point(x,sqrt(x*(-x + 2)) + 1)
yIntercept=lambda x:Point(0, -sqrt(x*(-x + 2))*(-x + 1)/(-x + 2) + sqrt(x*(-x + 2)) + 1)
xIntercept=lambda x:Point(x*(-x + sqrt(x* (-x + 2)) + 2) / (sqrt(x* (-x + 2))*(x - 1)), 0)

def opt(x0):
    tP=tangentPoint(x0)
    return abs(7./9-N(yIntercept(x0).distance(tP)/tP.distance(xIntercept(x0))))

res = minimize_scalar(opt, bounds=(1., 2.), method='bounded')
print res

hypotenuse=yIntercept(res.x).distance(tangentPoint(res.x))+xIntercept(res.x).distance(tangentPoint(res.x))
scale=16./N(hypotenuse)

Area=0.5*scale**2*yIntercept(res.x).y*xIntercept(res.x).x
print Area

Output:

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  status: 0
    nfev: 19
 success: True
     fun: 1.84529327729166e-6
       x: 1.64183859367641
 message: 'Solution found.'
3.26942684062943
62.9999813162989