Butterfly Effect

Relevant wiki: Two Secants

If we draw a radius joining point $E$ and center $O$ , we will obtain the center angle of $\angle COE = 2\angle BAC$ according to Center Angle Theorem . Thus, $\angle COE = \angle ABC$ , and since $\angle ACB = \angle ECO$ , the triangles $ABC$ and $CEO$ have the same internal angles and are, therefore, similar, as shown below:

Now suppose $BC = x$ and $CE = y$ .

Then by similarity, $\dfrac{y}{10-x} = \dfrac{3}{x}$ .

And according to Intersecting Chords Theorem , $3y = x(20-x)$ .

Thus, $3(10-x) = \dfrac{x^2(20-x)}{3}$ .

$90-9x = 20x^2 - x^3$

$x^3 - 20x^2 - 9x + 90 = 0$

$(x-2)(x^2 - 18x -45) = 0$

Thus, $x = 2$ because the other positive root exceeds $20$ , which is the diameter length, while the other one is negative.

Then $3y = 2\cdot 18 = 36$ . Thus, $y = 12$ .

As a result, the side ratio of $CE$ : $AC$ = $12:3$ = $4:1$ .

Hence, the area ratio of $\triangle CEO$ : $\triangle ABC$ = $16:1$ .

Then, since $\triangle CEO$ and $\triangle ODE$ share the same height with respect to the diameter chord, where $CO$ : $OD$ = $8:10$ = $4:5$ , the area ratio of $\triangle CEO$ : $\triangle ODE$ = $4:5$ .

That is, $\triangle CEO$ : $\triangle ODE$ : $\triangle ABC$ = $16:20:1$ .

Finally, the area ratio of $\triangle CDE$ : $\triangle ABC$ = $36:1$ .

In $\triangle AOC$ law of sines will give us $\frac{3}{sin(180-4\alpha)}=\frac{10}{sin(3\alpha)}$ . This has only one positive solution under $45^\circ$ , namely $\alpha =41.41^\circ$ .

From $\triangle ABC$ using law of sines we get $BC=\frac{3 sin(\alpha)}{sin(2\alpha)}=2$ .

$CD=20-BC=18$ , so the ratio of sides between the similar triangles $ABD$ and $DEC$ is $18:3=6:1$ and the ratio of areas is $36:1$ .

Let the centre of circle be $O$ and $BC=x$ . Using the sine rule on $\triangle ABC$ , we have $\dfrac {\sin \theta}x = \dfrac {\sin 2\theta}3$ , $\implies \cos \theta = \dfrac 3{2x}$ . Using the cosine rule on $\triangle ACO$ , we have:

$\begin{aligned} (10-x)^2 & = 3^2 + 10^2 - 60\cos \theta \\ 100 - 20x + x^2 & = 109 - 60 \times \frac 3{2x} \\ x^2 - 20x - 9 + \frac {90}x & = 0 \\ x^3 - 20x^2 - 9x + 90 & = 0 \\ (x-2)(x^2 - 18x-45) & = 0 \\ \implies x & = \begin{cases} \color{#3D99F6}2 & \color{#3D99F6} \text{accepted} \\ \color{#D61F06} 20.225 & \color{#D61F06} > 20 \text{, rejected} \\ \color{#D61F06} -2.222 & \color{#D61F06} < 0 \text{, rejected} \end{cases} \end{aligned}$

Let the height of $\triangle ABC$ be $h$ . We note that $\triangle EOC$ is similar to $\triangle ABC$ . Since $\dfrac {OC}{BC} = \dfrac 82 = 4$ , then the height of $\triangle EOC$ which is also the height of $\triangle CDE$ , $H=4h$ . Then, we have ratio of areas, $\dfrac {A_{CDE}}{A_{ABC}} = \dfrac {\frac 12 CD \cdot H}{\frac 12 BC\cdot h} = \dfrac {18\cdot 4h}{2h} = \boxed{36}$

We can also solve by finding cos(theta) which comes out to be 3/4.then the apply formula for area.