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It should be clear that one vertex of the triangle of maximum area will coincide with one end of the circle's diameter, say the right side, and that the vertex $P$ at the right angle of the triangle will then lie to the left of the center $O$ of the diameter of the semicircle. Letting $|OP| = x$ , and letting the vertical leg of the triangle extend up from $P$ to intersect with the semicircle, the height of the triangle will then be $\sqrt{1 - x^{2}}.$ The area of the triangle is then

$S = \dfrac{1}{2}(1 + x)\sqrt{1 - x^{2}}.$

Now differentiate with respect to $x$ and set the derivative equal to $0$ :

$\dfrac{dS}{dx} = \dfrac{1}{2}[\sqrt{1 - x^{2}} - \dfrac{x(1 + x)}{\sqrt{1 - x^{2}}} = 0$

$\Longrightarrow 1 - x^{2} = x + x^{2} \Longrightarrow 2x^{2} + x - 1 = 0$

$\Longrightarrow x = \dfrac{-1 \pm \sqrt{1 + 8}}{4} = \dfrac{-1 \pm 3}{4}.$

Since we want $x \gt 0$ we find that $x = \dfrac{1}{2}$ , and thus

$S = \dfrac{1}{2}(1 + \dfrac{1}{2})\sqrt{1 - \dfrac{1}{4}} = \dfrac{3\sqrt{3}}{8},$

and so $a + b + c = 3 + 3 + 8 = \boxed{14}.$

(Note that it should also be clear that any critical points found in the differentiation process will yield a maximum, since we could let $x = 1$ to get a minimum of $S = 0$ .)

I took a short cut to the problem, knowing that the triangle equilateral inscribed in a circunference yields the biggest area, half of such an equilateral triangle will do so in half circle. So the asking area S equals 3*sqrt(3)/8

I took one vertex as (-1, 0), and one vertex on the circumference as (cost, sint) and the remaining is (cost, 0)

Draw the right-angled triangle in the semicircle where $\angle ABC = 90$ , BC lies on the diameter and D is the centre of the semicircle. Now draw radius AD and let $\angle BDA = x$ . Now we have triangle ABD, where $\large BD = cos x \ and \ AB = sin x$ . For triangle ADC, $\large ADC = 180 - x$ . Now using area of a triangle = $\frac{1}{2}$ absin(x) (or $\frac{1}{2}$ base $\times$ height for triangle ABD) $area\ = \frac{1}{2}sinxcosx + \frac{1}{2}sin(180-x)$ $= \frac{1}{2} (sinxcosx + sinx) = \frac{1}{2}(\frac{1}{2}sin(2x) + sin(x)) = f(x) \ (from \ sin2x = 2sinxcosx)$ Differentiating the equation and equating it to zero to find a stationary point: $f'(x) = \frac{1}{2}(cos2x + cosx) = 0\Rightarrow\ cos2x = -cosx \Rightarrow\ 2cos^{2}x -1 = -cosx$ $\Rightarrow\ (2cosx -1 )(cosx+1) \Rightarrow \ cosx = \frac{1}{2} \ as \ 0 < x < 90$ $\Rightarrow\ x = 60$ Now differentiating again to find the nature of this stationary point: $f''(x) = -\frac{1}{2}(2sin2x + sinx) \Rightarrow\ f''(x) < 0 \Rightarrow\ max \ point$

$\large \therefore\ max[f(x)] = \ f(x=60) = \frac{3\sqrt{3}}{8}$