Medals Are For Stars

Since the circle is placed right the center of the square and the figure is symmetrical, then the red straight line $AD$ passes through the center of the circle and has a length equal to the side length of the square, $AD = 1$ .

With $A$ and $D$ passes through the midpoints of the side lengths of the square, then $ED = 1/2$ and $\angle ECD = \dfrac{120^\circ}2=60^\circ$ . So $\tan (\angle ECD) = \dfrac {ED}{CD} \Rightarrow {\sqrt3} = \dfrac{1/2}{CD} \Rightarrow AB= CD = \dfrac1{2\sqrt3}$ .

Thus, the diameter of the circle is $d = AD - AB - CD = 1 - 2 \left( \dfrac1{2\sqrt3} \right) = 1 - \dfrac1{\sqrt 3}$ .

Hence, the area of the circle is $\pi r^2 = \pi \left( \dfrac d2\right)^2 = \dfrac\pi4 \left( 1 - \dfrac1{\sqrt 3}\right)^2 = \dfrac{2-\sqrt3}6 \pi$ .

Our answer is $2\times6=\boxed{12}$ .

Let us first label the points $A$ to $G$ as shown above.

We want to find the area of the circle that passes through the points $E,F,G$ and $H$ .

Since the figure is symmetric about the lines $AC$ and $DB$ , then $AD = AE = AF = FB = DH = HC = CG = GB$ . And $EF = FG = HG = EH$ .

For simplicity sake, let $t = DE = AE > 0$ and $u = EF = FG > 0$ .

Looking at the triangle $AED$ , and applying cosine rule gives

$(DE)^2 + (AE)^2 - 2(DE)(AE) \cos(120^\circ) = AD^2 \quad \Rightarrow \quad 2t^2 (1 - \cos120^\circ) = 1 \quad \Rightarrow \quad t = \dfrac1{\sqrt3} \; .$

Because the triangle $ADE$ is isosceles, then $\angle ADE = \angle DAE = \dfrac{180^\circ - 120^\circ}2 = 30^\circ$ . Similarly, $\angle FAB = 30^\circ$ , so $\angle EAF = 90^\circ - 30^\circ - 30^\circ = 30^\circ$ .

Looking at the triangle $AEF$ and applying the cosine rule once more gives

$(AE)^2 + (AF)^2 - 2(AE)(AF) = (EF)^2 \quad \Rightarrow \quad 2t^2 (1 - \cos 30^\circ) = u^2 \quad \Rightarrow \quad u^2 = \dfrac{2-\sqrt3}3 \; .$

Finally, let $r$ and $d$ denote the radius and diamater of the circle in question. The area of the circle is

$\pi r^2 = \pi \left( \dfrac d2 \right)^2 = \dfrac \pi4 d^2 = \dfrac\pi4 [ (EF)^2 + (FG)^2 ] = \dfrac\pi4 (2u^2) = \dfrac\pi2 \cdot \dfrac{2-\sqrt3}3 = \dfrac{2-\sqrt3}6 \pi\; .$

The penultimate step follows from Pythagorean theorem . Our answer is $2\times 6 = \boxed{12}$ .

If we draw a line (from side to side in the diamond) through the marked vertex and the opposite point (vertex of the opposite triangle) on the circle, we will find (due to symmetry), that the line is parallel to 2 sides of the rhombus (and therefore, its lenght is 1 (same as the side lenght of the diamond), it goes through the centre of the circle (the part inside the circle is a diameter), and its parts outside the circle form the altitudes of the two isosceles triangles.

These altitudes (h) can be calculated easily (e.g. from the 30-60-90 half triangles):

$h = 0.5 × tan 30° = \frac {\sqrt{3}}{6}$

Radius of the circle (r):

2h + 2r =1

r = 0.5 - h

$r = \frac {3 - \sqrt{3}}{6}$

Area of the circle:

$r^2 \pi = \frac {12 - 6 \sqrt{3}}{36} \pi = \frac {2 - \sqrt{3}}{6} \pi$

Hence, a = 2 , b = 6 and our answer should be

$ab = 2 × 6 = \boxed {12}$

Find the base of the isosceles triangle ( one of the hand of the star) and due to symmetry the quadrilateral inscribed in the circle is a square then find the radius and tpye your answer. Phewwwww.....😢