Nice Angles, But Not-So-Nice Answer

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Draw the quadrilateral ABCD.

$m\angle ABC=45^\circ\text{ and }m\angle BCD=60^\circ$

From point $\bf{A}$ , draw a line perpendicular to $\overline{BC}$ at point $\bf{E}$ .

From point $\bf{D}$ , draw a line perpendicular to $\overline{BC}$ at point $\bf{F}$ .

From point $\bf{A}$ , draw a line perpendicular to $\overline{DF}$ (and parallel to $\overline{BC}$ ) at point $\bf{G}$ .

$\text{The triangles }\Delta ABE\text{ and }\Delta DFC\text{ are special triangles.}$

That means $\boxed{AE=BE=\displaystyle\frac{\sqrt{2}}{2}}$ .

$FC=CD\cos{60^0}\rightarrow\boxed{FC=\displaystyle\frac{3}{2}}$

$DF=CD\sin{60^0}\rightarrow\boxed{DF=\displaystyle\frac{3}{2}\sqrt{3}}$

$BC=BE+EF+FC$ and $BC=2$

$2=\displaystyle\frac{\sqrt{2}}{2}+\color{#0000ff}{EF}+\frac{3}{2}$

$\boxed{EF=\displaystyle\frac{1-\sqrt{2}}{2}}$

$AGFE\text{ is a rectangle. Therefore, }AG\text{ is parallel and equal to }EF\text{ and }AE\text{ is parallel and equal to }GF$

$AE=\boxed{GF=\displaystyle\frac{\sqrt{2}}{2}}\text{ and }EF=\boxed{AG=\displaystyle\frac{1-\sqrt{2}}{2}}$

$DF=DG+GF$

$\displaystyle\frac{3}{2}\sqrt{3}=DG+\displaystyle\frac{\sqrt{2}}{2}$

$DG=\displaystyle\frac{3\sqrt{3}-\sqrt{2}}{2}$

Consider the triangle $\Delta AGD$ . $AD$ can be measured using the $\bf{Pythagorean Theorem}$ .

$AD=\sqrt{AG^2+DG^2}$

$AD=\sqrt{\left(\displaystyle\frac{1-\sqrt{2}}{2}\right)^2+\left(\displaystyle\frac{3\sqrt{3}-\sqrt{2}}{2}\right)^2}$

$AD=\sqrt{8-\displaystyle\frac{1}{2}\sqrt{2}+\displaystyle\frac{3}{2}\sqrt{6}}$

$\text{It is said in the problem that AD can be written in the form }\sqrt{a+\sqrt{b-\sqrt{c}}}$

$\sqrt{a+\sqrt{b-\sqrt{c}}}=\sqrt{8-\displaystyle\frac{1}{2}\sqrt{2}+\displaystyle\frac{3}{2}\sqrt{6}}$

$\text{Square both sides of the equation to get }$

$\color{#ff0000}{a}+\sqrt{b-\sqrt{c}}=\color{#ff0000}{8}+\displaystyle\frac{3}{2}\sqrt{6}-\displaystyle\frac{1}{2}\sqrt{2}$

We can now conclude that $\boxed{a=8}$

Since $a=8$ , we cancel it to get the new equation $\sqrt{b-\sqrt{c}}=\displaystyle\frac{3}{2}\sqrt{6}-\displaystyle\frac{1}{2}\sqrt{2}$

Square both sides.

$(\sqrt{b-\sqrt{c}})^2=\left(\displaystyle\frac{3}{2}\sqrt{6}-\displaystyle\frac{1}{2}\sqrt{2}\right)^2$

$\color{#ff0000}{b}-\sqrt{c}=\color{#ff0000}{14}-3\sqrt{3}$

We conclude that $\boxed{b=14}$ .

That leaves us $\sqrt{c}=3\sqrt{3}$ or $\sqrt{\color{#ff0000}{c}}=\sqrt{\color{#ff0000}{27}}$ .

Therefore, $\boxed{c=27}$

$a+b+c=8+14+27=\boxed{49}$

Let us put this problem on the complex plane. Let $C$ be the origin, and let $B=-2$ .

Clearly, $D=3\text{cis}(240^{\circ})$ and $A=\text{cis}(45^{\circ})-2$ for $AD$ to be maximum.

All we need to do now is to compute $AD=\left|(3\text{cis}(240^{\circ}))-(\text{cis}(45^{\circ})-2)\right|$ .

Note that $\text{cis}(240^{\circ})=-\dfrac{1}{2}-\dfrac{i\sqrt{3}}{2}$ and $\text{cis}(45^{\circ})=\dfrac{\sqrt{2}}{2}+\dfrac{i\sqrt{2}}{2}$ . Substituting these values in, we have $\begin{aligned}AD&=\left|3\left(-\dfrac{1}{2}-\dfrac{i\sqrt{3}}{2}\right)-\left(\dfrac{\sqrt{2}}{2}+\dfrac{i\sqrt{2}}{2}-2\right)\right|\\&=\left|\left(-\dfrac{3}{2}-\dfrac{3i\sqrt{3}}{2}\right)-\left(\dfrac{\sqrt{2}}{2}+\dfrac{i\sqrt{2}}{2}\right)+2\right|\\&=\left|\dfrac{1-\sqrt{2}}{2}-\left(\dfrac{3\sqrt{3}+\sqrt{2}}{2}\right)i\right|\\ &= \sqrt{\left(\dfrac{1-\sqrt{2}}{2}\right)^2+\left(\dfrac{3\sqrt{3}+\sqrt{2}}{2}\right)^2}\\ &= \sqrt{\dfrac{32+6\sqrt{6}-2\sqrt{2}}{4}} \\&= \sqrt{8+\dfrac{3\sqrt{6}-\sqrt{2}}{2}}\end{aligned}$

However, this isn't the format that the problem asked for. How can we change it to it the format?

We conjecture that $\dfrac{3\sqrt{6}-\sqrt{2}}{2}=\sqrt{b-\sqrt{c}}$ . We square $\dfrac{3\sqrt{6}-\sqrt{2}}{2}$ to get $\left(\dfrac{3\sqrt{6}-\sqrt{2}}{2}\right)^2=14-\sqrt{27}$ . Now squarerooting again we arrive at the conclusion that $\dfrac{3\sqrt{6}-\sqrt{2}}{2}=\sqrt{14-\sqrt{27}}$ .

Therefore $AD=\sqrt{8+\sqrt{14-\sqrt{27}}}$ and our desired answer if $8+14+27=\boxed{49}$ .

First, we form a triangle with vertexes $B$ , $C$ and $D$ , so we can find $\overline{BD}$ applying cosine's law: $\overline{BD}^{2}=\overline{CD}^2+\overline{BC}^2-2(\overline{CD})(\overline{BC})(cos \angle BCD)$

$\overline{BD}^{2}=(3)^{2}+(2)^{2}-2(3)(2)(cos 60°)$

$\overline{BD}^{2}=13-12(\frac{1}{2})$

$\overline{BD}^{2}=13-6$

$\overline{BD}^{2}=7$

$\overline{BD}=\sqrt{7}$

Now, we observe that $\angle ADB + \angle CBD = \angle ABC = 45°$ , so $\angle ABD = 45° - \angle CBD$ , now we find the angle $\angle CBD$ , again, by cosine's law: $cos \angle CBD = \frac{\overline{BD}^{2}+\overline{BC}^{2}-\overline{CD}^2}{2\overline{BD}\overline{BC}}$

$cos \angle CBD = \frac{(\sqrt{7})^{2}+(2)^{2}-(3)^2}{2(\sqrt{7})(2)}$

$cos \angle CBD = \frac{7+4-9}{4\sqrt{7}}$

$cos \angle CBD = \frac{2}{4\sqrt{7}}$

$cos \angle CBD = \frac{1}{2\sqrt{7}}$

Now, we find $\angle ABD$ , by applying cosine to both sides of: $\angle ABD = 45° - \angle CBD$ : $cos(\angle ABD) = cos(45° - \angle CBD)$

$cos(\angle ABD) = cos(45°)cos(\angle CBD) + sin(45°)sin(\angle CBD)$

$cos(\angle ABD) = \frac{\sqrt{2}}{2}cos(\angle CBD) + \frac{\sqrt{2}}{2}sin(\angle CBD)$

$cos(\angle ABD) = \frac{\sqrt{2}}{2}(cos(\angle CBD) + sin(\angle CBD)$ )

See that we have $cos(\angle CBD)$ , now we have to find $sin(\angle CBD)$ : $sin(\angle CBD)=\sqrt{1-cos(\angle CBD)^{2}}$

$sin(\angle CBD)=\sqrt{1-(\frac{1}{2\sqrt{7}})^{2}}$

$sin(\angle CBD)=\sqrt{1-\frac{1}{28}}$

$sin(\angle CBD)=\sqrt{\frac{28-1}{28}}$

$sin(\angle CBD)=\sqrt{\frac{27}{28}}$

$sin(\angle CBD)=\frac{3\sqrt{3}}{2\sqrt{7}}$

Replace these values to find $cos(\angle ABD)$ :

$cos(\angle ABD) = \frac{\sqrt{2}}{2}(\frac{1}{2\sqrt{7}}+\frac{3\sqrt{3}}{2\sqrt{7}})$

$cos(\angle ABD) = \frac{\sqrt{2}+3\sqrt{6}}{4\sqrt{7}}$

Finally, we form a triangle with vertexes $A$ , $D$ and $B$ , and we apply cosine's law to find $\overline{AD}$ : $\overline{AD}^{2}=\overline{AB}^{2}+\overline{BD}^{2}-2(\overline{AB})(\overline{BD})(cos(\angle ABD))$

$\overline{AD}^{2}=(1)^{2}+(\sqrt{7})^{2}-2(1)(\sqrt{7})(\frac{\sqrt{2}+3\sqrt{6}}{4\sqrt{7}}$ )

$\overline{AD}^{2} = 1+7-\frac{\sqrt{2}+3\sqrt{6}}{2}$

$\overline{AD}=\sqrt{8-\frac{\sqrt{2}+3\sqrt{6}}{2}}$

$\overline{AD}=\sqrt{8-\sqrt{(\frac{\sqrt{2}+3\sqrt{6}}{2}})^{2}}$

$\overline{AD}=\sqrt{8-\sqrt{\frac{54+12\sqrt{3}+2}{4}}}$

$\overline{AD}=\sqrt{8-\sqrt{\frac{56+12\sqrt{3}}{4}}}$

$\overline{AD}=\sqrt{8-\sqrt{14+3\sqrt{3}}}$

$\overline{AD}=\sqrt{8-\sqrt{14+\sqrt{27}}}$

Hence, $a=8$ , $b=14$ , and $c=27$ . So, $a+b+c=\boxed{49}$