Semi-circle is all set to blow your mind !

Let the radius of the semicircle be x. Thus, $OB = OP = x$

AC and AB are tangents to the semicircle at P and C respectively. So, $AP = AB = 3$ .

Thus, $PC = AC - AP = 2$

On joining O and P, in triangle OPC, angle P is right angled.

Thus, $OP^{2} + PC^{2} = OC^{2} = (BC - OB)^{2}$

$=> x^{2} + 2^{2} = (4 - x)^{2}$

$=> x = 3/2$

Thus, a+b = 5

Let $r$ be the radius of circle $O$ . Then $\overline{OB} = \overline{OC} =\overline{OP} = r$ .

The area of $\triangle{ABO} = \frac{1}{2}(\overline{AB})(\overline{BO}) = \frac{1}{2}(3)(r) = \frac{3r}{2}$ .

The area of $\triangle{AOC} = \frac{1}{2}(\overline{AC})(\overline{OP}) = \frac{1}{2}(5)(r) = \frac{5r}{2}$ .

The area of $\triangle{ABC} = \frac{1}{2}(\overline{AB})(\overline{BC}) = \frac{1}{2}(3)(4) = 6$ .

$Area_{\triangle{ABO}} + Area_{\triangle{AOC}} = Area_{\triangle{ABC}}$ .

$\Rightarrow \frac{3r}{2} + \frac{5r}{2} = 6$ .

$\Rightarrow r = \frac{3}{2}$ .

$a+b=\boxed{5}$ .

$OP$ and $AC$ are perpendicular. $OPC$ and $ABC$ are similar. Setting up the appropriate ratios gives

$\frac{3}{r}=\frac{5}{4-r}$

and $r=3/2$ , thus the answer is $3+2=\boxed{5}$ .

WE have $\angle ABO = 90^{\circ}$ and $OP \perp AC$ and $AB = AP$ [ tangents to the same circle from a single point] Hence in $\triangle ABO$ and $\triangle AOP$ , $AB=AP$ , $\angle ABO = \angle OPA = 90^{\circ}$ and $OB = OP$ . Hence $\triangle ABO \cong \triangle APO$ . Therefore, $\angle OAB = \angle OAP$ . Thus $OP$ is the internal bisector of $\angle A$ . Hence by the angle bisector theorem in $\triangle ABC$ , we get

$\dfrac{AB}{AC} = \dfrac{BO}{OC}$

$\implies$ $\dfrac{3}{5} = \dfrac{BO}{OC}$

$\implies$ $5BO=3OC$ and we already have $BO + OC = BC = 4$

Solving, we get $OB=\dfrac{3}{2}$ and hence $a+b = \boxed{5}$

By equal tangents, we have that $AP = 3$ . By LoC, we have that $BP^2 = \frac{36}{5}$ . We can set up another LoC equation in terms of the radius for $BP$ , using cyclic quadrilaterals to determine $\angle BOP = 180^{\circ} - \angle BAP$ , and solving to get $r= \frac{3}{2}$

OB=OP=r

angBAC= cos(angBAC)= $\frac{3}{5}$

angABC=arcos0.6=53.13°

angBAO=53.13°/2=26.56°

Tan(26.56°)=r/3

r=3*Tan(26.56°)

1)r=1.49966

PC=AC-AP

AP=AB=Tan a la semicircunferencia

PC=5-3=2

AB/OP=BC/PC

3/r=4/2

2)r=3/2

1 y 2 son iguales

a=3 y b=2

a+b= 3+2 = 5

Angle BAC can be found out using trigonometry.Join A and O.Angle BAO is half of angle BAC because BAP is an isosceles triangle.Now we can find r as r/3=tan Angle BAO.r comes out to be 1.5, so the ans is 3+2=5

since OB = OP = r

and OB + OC = BC = 4

thus OC = 4 - OB = 4 - r

thus (CP)² + r² = (4 - r)² = 16 - 8r + r²

thus (CP)² = 16 - 8r

thus CP = 2√(2(2-r))

and since AP, AB are tagnets

thus AP = AB = 3

But AP + CP = AC = 5

thus CP = 5 - AP = 5 - 3 = 2

thus 2√(2(2-r)) = 2

thus √(2(2-r)) = 1

thus 2(2-r) = 1

thus 2 - r = 0.5

thus r = 1.5 cm = 3/2 cm

but r = a/b

thus a = 3, b = 2

thus a + b = 5

Notice that semicircle $O$ is tangent to $\overline{AB}$ at $B$ and $\overline{AC}$ at $P$ . So we have that $\angle ABO = \angle APO = 90^\circ$ . Also, $BO = PO$ since they are radii of the same circle and $AO = AO$ , so $\triangle ABO \cong \triangle APO$ by HL congruence. Thus, $\angle BAO = \angle PAO$ so $\overline{AO}$ lies on the angle bisector of $\angle BOC$ . So by the angle bisector theorem, we have $\frac{3}{BO} = \frac{5}{CO} \implies 3\cdot CO = 5 \cdot BO$ and $BO + CO = 4.$ Solving this system gives us the solution $BO = \frac{4}{3}$ .