consider a semicircle with a diameter AB where the length is 4, and a point C on the circular arc.

Question

if the area of triangle ABC is a half of the maximum and point C is nearer to point A than point B then the angle CAB is? I have already found out that the maximum area of triangle ABC is 4.But I have no idea on how to continue.

Answer 1

Drop a perpendicular from point $C$ on $AB$ and call the feet of this perpendicular $D$. Let the centre of the semicircle be $O$.

Observe that, in right $\triangle CDO$, $CD=1=\frac {1}{2} 2=\frac {1}{2} CO$ and hence it is a $30-60-90$ triangle with $\angle COD=30^{\circ}$. Now find $\angle CAB$ taking advantage of the fact that $\triangle COA$ is isosceles.

Answer 2

Let AB=c=2R=4, AC=b and BC=a, we have:

$S_{ABC}=\frac 12 (\frac 12 \pi R^2)=\pi$

Now we use this formula :

$$\frac{a+b}{c=4}=\frac{\cos \frac{A-B}2}{\sin \frac C2=\sin \frac{90}2=\frac{\sqrt 2}2}$$

$\cos(\frac{A-B}2)=\cos \frac A2\cos \frac B2+\sin \frac A2 \sin \frac B2$

$\cos(\frac{A+B}2)=\cos \frac A2\cos \frac B2-\sin \frac A2 \sin \frac B2$

Summing these relations we get:

$2\cos \frac A2\cos \frac B2=\cos(\frac{A-B}2)+\cos \frac {\pi}4$

Using relation:

$\frac a{\sin A}=\frac b{\sin B}=2R=4$

we finally get:

$\cos\frac A2\cos\frac B2=\frac{\sqrt 2}4(1+\sin A+\sin B)$

Considering $A+B=90^o$, we get $A\approx 75^o$.

Answer 3

Let $O$ be the center of the semi-circle. Then $O$ is the midpoint of the diameter $AB.$ The maximal area for the triangle $ABC$ occurs when $C \equiv C_{max}$ so that $C_{max}O$ is perpendicular to $AB$. Then $ABC_{max}$ is right-angle isosceles triangle so $\angle \, BAC_{max} = 45^{\circ}$.

The triangle $ABC$ has half the area of $ABC$ when the distance of $C$ from $AB$ is half the distance of $C_{max}$ to $AB$, the latter being the length of $C_{max}O$. If $M$ is the midpoint of $C_{max}O$, then the perpendicular from $C$ to $AB$ is equal and parallel to $MO$, so $CM$ is perpendicular to $C_{max}O$. Therefore $CM$ is the orthogonal bisector of $C_{max}O$ and therefore the triangle $CC_{max}O$ is isosceles with $CO = CC_{max}$. But $CO$ and $C_{max}O$ are radii of the semi-circle, so $CO = CC_{max} = C_{max}O$, i.e. triangle $CC_{max}O$ is equilateral and therefore $\angle \, COC_{max} = 60^{\circ}$. Hence $\angle \, CAC_{max} = \frac{1}{2} \, \angle \, COC_{max} = 30^{\circ}$. Finally, $$\angle \, CAB = \angle\, CAC_{max} + \angle \, BAC_{max} = 30^{\circ} + 45^{\circ} = 75^{\circ}$$