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

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

Find $|DC|=x$

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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?

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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}

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

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

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

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

$$x \approx 8.48528$$

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