A quadrilateral $ABCD$ is inscribed in a circle of center $O$. It is known that the diagonals $AC$ and $BD$ are perpendicular. On each side we build semicircles, externally, as shown in the figure.
a) Show that the triangles $AOB$ and $COD$ have the equal areas.
b) If $AC=8$ cm and $BD= 6$ cm, determine the area of the shaded region.
manutd03 wrote:
Do you know its website?
They do not have one webpage. Search in internet for : olimpiada matematica CPLP.
edit: all problems from this contest (which started in 2011) have been translated in English, look here
ps. CPLP stands for Community of Portuguese Language Countries
$a)$ $\angle AOB=2\angle ACB=2(90^{\circ}-\angle DBC)=180^{\circ}-\angle COD$
$[AOB]=\frac{R^2}{2}\sin\angle AOB=\frac{R^2}{2}\sin\angle COD=[COD]$
$b)$ Let $\angle AOB=x, \angle BOC=y, \angle COD=z, \angle DOA=w$ , $S$ be the shaded area, $R$
the radius, and denote by $\bigcirc XY$ the circle with diameter $XY$.
$S=\frac{1}{2}([\bigcirc AB]+[\bigcirc BC]+[\bigcirc CD]+[\bigcirc AD])-([\overarc{AB}]+[\overarc{BC}]+[\overarc{CD}]+[\overarc{DA}])=$
$=\frac{1}{2}([\bigcirc AB]+[\bigcirc BC]+[\bigcirc CD]+[\bigcirc AD])-([(ABCD)]-[ABCD])$
$[\bigcirc AB]=\pi\cdot (\frac{AB}{2})^2$
$AB^2=R^2+R^2-2R\cdot R\cdot\cos x=R^2(2-2\cos x)\Rightarrow [\bigcirc AB]=\frac{\pi}{2} R^2(1-\cos x)$
$[(ABCD)]=\pi R^2$
$[ABCD]=\frac{AC\cdot BD}{2}=24cm^2$ (Because the angle between $AC$ and $BD$ is $90^{\circ}$)
$S=\frac{1}{2}\cdot \frac{\pi}{2} R^2(4-(\cos x+\cos y+\cos z+\cos w))-\pi R^2+24cm^2$
But $x+z=y+w=180^{\circ}$, as proved earlier. Thus $\cos x=-\cos z$, $\cos y=-\cos w\Rightarrow$
$\Rightarrow S=\pi R^2-\pi R^2+24cm^2=24cm^2$
That means the shaded area is $24cm^2$.
