# An equilateral triangle is drawn in a circle with one of its vertices on the diameter. What is $x$ in the figure?

$$O$$ is the center of the arc $$AEC$$; $$ABD$$ is an equilateral triangle

$$\angle ACB = 45^o$$; $$|BO|= 6$$ cm

Find $$|DC|=x$$

I tried completing the square, drawing radii to the intersection points, but I can't figure out how to solve this.

This is a problem from a high school geometry test which allows only ~2 minutes to solve each problem without a calculator.

How do I solve this problem?

In triangle $$CAF$$ $$\angle AFC=\angle FCA=45^\circ$$, hence $$AO\perp FC$$ and

\begin{align} \triangle ABO:\quad a&=\frac 6{\cos 75^\circ} =6(\sqrt2+\sqrt6) ,\\ x&=a\sin15^\circ\cdot \sqrt2 = 6(\sqrt2+\sqrt6)\cdot\tfrac{\sqrt2}4\,(\sqrt3-1)\cdot \sqrt2 =6\sqrt2 \approx 8.485281372 . \end{align}

• How did you make this illustration? Is this by any chance directly integrated into LaTex? – 10012511 Mar 25 '19 at 12:26
• @mar_cel: Yes, Asymptote is deeply integrated with LaTeX. – g.kov Mar 25 '19 at 12:45
• Why is $\angle AOC=90$? – Aqua Apr 2 '19 at 17:01
• @Maria Mazur: The arc is extended to make a semicircle, $|FC|$ is the diameter, $O$ in the middle of it, $\angle CAF=90^\circ\Rightarrow \angle AFC=45^\circ$ $\Rightarrow |AF|=|AC|$... – g.kov Apr 2 '19 at 18:18
• OK................ – Aqua Apr 2 '19 at 18:50

Denote $$AB=AD=BD=a$$ and $$OA=OC=r$$. Note that $$AC=\sqrt 2r$$, so $$a=\sqrt 2 r-x$$. Consider $$\triangle ABC$$. By law of sines we have: $$\frac{AB}{\sin 45^\circ}= \frac{AC}{\sin 75^\circ}= \frac{BC}{\sin 60^\circ},$$ i.e. $$\frac{\sqrt 2r-x}{\sin 45^\circ}= \frac{\sqrt 2r}{\sin 75^\circ}= \frac{r+6}{\sin 60^\circ}.$$ Recall $$\sin 45^\circ=\frac{\sqrt 2}{2}$$, $$\sin 60^\circ=\frac{\sqrt 3}{2}$$ and $$\sin 75^\circ=\frac{\sqrt 2+\sqrt 6}{4}$$. Now first calculate $$r$$ from the equality of the second and the third term, and then calculate $$x$$.

I found a trigonometric approach but I am unable to find some pure geometric solution.

Connecting two points $$O$$ and $$A$$, we get a right angled isosceles triangle because $$O$$ is the center of arc $$AEC$$. Let denote the length of equilateral $$\triangle ABD$$ as $$a$$.

Thus we get the hypotenuse of the right angled $$\triangle AOC$$, $$|AC| = (a + x)$$ and $$|AO| = |OC| = \frac{(a + x)}{\sqrt2}$$. From all required information, we get

$$\angle ABO = 75^\circ$$ and from $$\triangle ABO$$,

$$\frac{AO}{OB} = \tan 75^\circ$$

$$\frac {AO}{OB} = \frac{\sqrt3 + 1}{\sqrt3 - 1}$$

$$AO = 3(\sqrt3 + 1)^2$$

And, $$\frac {OB}{AB} = \cos 75^\circ$$

$$AB = \frac {6}{\frac {\sqrt6 - \sqrt2}{4}}$$

Hence, $$AB = 6(\sqrt6 + \sqrt2)$$

Now, $$\frac{(a + x)}{\sqrt2} = AO = 3(\sqrt3 + 1)^2$$

Therefore, $$(a + x) = 3\sqrt2(\sqrt3 + 1)^2$$

$$\implies x = 3\sqrt2(\sqrt3 + 1)^2 - a$$

$$\implies x = 3\sqrt2(\sqrt3 + 2)^2 - 6(\sqrt6 + \sqrt2)$$

$$x \approx 8.48528$$