A, B are two points on a circle (O). There are two arcs AB and two circular segments AB but we choose the small one.
How to cut from this circular segment one right triangle with maximum area?

Best regards,
Bui Quang Tuan

>A, B are two points on a circle (O). There are two arcs AB
>and two circular segments AB but we choose the small one.
>How to cut from this circular segment one right triangle
>with maximum area?

This is a very interesting problem.
The solution exhibits significant interest, with many symmetries, alignments and circles worthy of further exploration.

Consider the minor arc APB of circle with centre O and diameter AC = 2.r. Let AB=2.h and let the perpendicular from O to S on BC =k. Let APQ be the right triangle within the minor segment, with Q on BC.
(see Fig 1).

Using Cartesian Coords with origin at A and x-axis along ASQB:

Equation of arc APB is

y = ((r^2-(x-h)^2)^1/2 + k

Area tAPQ = x.y/2 + y^3/(2.x)

Differentiate and set dA/dx = 0 and solve for x

The resulting equation is a sextic, so no easy closed form solution is likely.
However, for example, if h=5, k=3, then r=sqrt(34) gives
x=6.283 and theta=23.161 degrees.

A better alternative is to use Polar Coords, with the Pole at A, and theta=0 along x-axis. Then C=(2.r, -thetaC) where tan(thetaC) = k/h
and P=(rho,theta) with rho = 2.r.cos(theta+thetaC)

The structure of the solution is shown in Fig.2
Observe that 2.r.sin(theta+thetaC) = rho (from tAPC),
PC.cos(theta).sin(theta).2 = DE = rho (from kite EPDC),
rho = 2.t.cos(theta) (from tAPE),
PD = PE = EQ = QD = t (from rhombus EPDQ),
AE = EP (from tAPE),
ED = 2.t.cos(theta) = rho (from //gram APDE).

So, formally, in Polar Coords: Let AE = EQ = PD = t
. project APD onto AB so t.(1 + 2.(cos(theta)^2) = 2.h ...(1)
. from tAPC 2.r.cos(theta+thetaC) = 2.t.cos(theta) ...(2)
Solve (1) and (2) simultaneously for t and theta.

OR use rho.tan(theta+thetaC) = PC = rho/sin(2.theta)
.: tan(theta+thetaC) = 1/sin(2.theta)

The resulting equation is again a sextic, with either of the above approaches, but with only even powers present, it may be solved as a cubic, giving some advantage over the Cartesian approach. However the real advantage of the polar solution is the way in which it reveals the underlying structure of the solution geometry.

( Figures to follow, in a separate post ).

Regards, Peter Scales.

Peter Scales
Peter Scales