Area Problem 5

hint:

Going for the bonus

If we draw radii from the vertices to the centre, we get three identical isosceles triangles with sides R, R, 2, and three identical isosceles triangles with sides R, R, 4.

We therefore know that the angles at the centre of two different triangles add up to 120°.

To find out the individual angles, we start out with a unit circle centred at the origin, and choose points $A=(1,0)$ , $B=(-\frac{1}{2}, \frac{\sqrt{3}}{2})$ , so that they are 120° apart. We are looking for some point $P=(x,y)$ on the unit circle having $d(P,B)=2d(P,A)$

For shortness set $d_A = d(P,A)$ and $d_B = d(P,B)$ . We have

$x^2+y^2=1$

$d_A^2=(x-1)^2+(y-0)^2=x^2-2x+1+y^2 = 2-2x$

$d_B^2=(x+\frac{1}{2})^2+(y-\frac{\sqrt{3}}{2})^2=x^2+x+y^2-\sqrt{3}y+1=2+x-\sqrt{3}y$

$d_B=2d_A \Rightarrow 2+x-\sqrt{3}y = 4 (2-2x)$ $\Rightarrow 9x-\sqrt{3}y= 6 \Rightarrow y= (3x-2) \sqrt{3}$

So our point P lies on the line $y= 3\sqrt{3}x -2\sqrt{3}$

Plugging this expression for y into $x^2+y^2-1=0$ gives the equation $28x^2-36x +11=0$ .

which solves to $x = \frac{36 \pm \sqrt{36^2-4×28×11}}{2×28}=\frac{11}{14}$ . (The diagram above makes it clear that we should take the largest root.) Thus we find $P=(x,y)=(\frac{11}{14}, \frac{5}{14} \sqrt{3})$

Now that we have found $P$ , we can calculate the triangular areas, which is done easiest using cross products: $A_1=\frac{1}{2}|(\vec{P}×\vec{A})|=\frac{5}{28} \sqrt{3}$ and $A_2=\frac{1}{2} |(\vec{P}×\vec{B})|=\frac{2}{7} \sqrt{3}$ , so that $A_1+A_2= \frac{13}{28} \sqrt{3}$

Finally, we should scale up things by a factor $\frac{2}{d_A}$ . Doing so, the areas scale up by a factor $\frac{4}{d_A^2}=\frac{28}{3}$ . Furthermore, in order to count three pieces of each triangle area, multiply by 3. Thus we get: $A=\frac{28}{3}×3×(A_1+A_2)= 13 \sqrt{3}$ and

$A^2= (13 \sqrt{3})^2=\boxed{507}$

r=3.05505046330389 area squared=507. I first solved for the circle's radius (r) and then computed the altitudes to the two triangle types. That gave me the area of the figure, which squared to 507.

Let a and b be the central angles of the large triangle and small triangle respectively. Applying the Law of sines in each triangle, sin[(180 - a)/2]/r = [sin(a)]/4 and sin[(180 - b)/2]/r = [sin(b)]/2. Let b = 120 - a, equate each equation to r, and we get a non-linear equation for a, solved by Newton-Raphson. Approximate root is a =81.776 and b = 38.224. We then get r = 3.0545. Large triangle area = .5 r^2 sin(a) = 4.617. Small triangle area = .5 r^2 sin(b) = 2.886. Total area = 3*(large + small) = 22.51 and squared = 507.