Tricky Triangle

Draw a line perpendicular from $B$ to line $X$ and to line $Z$ , name the intersection points $D$ and $E$ ; and draw a line perpendicular from $C$ to line $X$ , name the intersection point $F$ . Clearly a rectangle is formed. Name $L$ the side lenght of the triangle, $a=\overline{DA}$ and $b=\overline{AF}$ . The dimentions of the rectangle are: height of $16$ and base $x+y$ . Apply Pythagorean Theorem to triangles $\triangle BDA$ , $\triangle BEC$ and $\triangle AFC$ : $10^2+x^2=L^2$ $6^2+(x+y)^2=L^2$ $y^2+16^2=L^2$ Solve for $x$ and $y$ from first and third equations and substitute in the second one: $x=\sqrt{L^2-100}$ $y=\sqrt{L^2-256}$ $36+(\sqrt{L^2-100}+\sqrt{L^2-256})^2=L^2$ $36+L^2-100+2\sqrt{(L^2-100)(L^2-256)}+L^2-256=L^2$ $2\sqrt{L^4-356L^2+25600}=320-L^2$ $4(L^4-356L^2+25600)=(320-L^2)^2$ $4L^4-1424L^2+102400=102400-640L^2+L^4$ $3L^4-784L^2=0$ $3L^2=784$ $L^2=\dfrac{784}{3}$ Finally, the area of an equilateral triangle of side $L$ is $A=L^2 \dfrac{\sqrt{3}}{4}$ : $A=\dfrac{784}{3} \times \dfrac{\sqrt{3}}{4}$ $A=\dfrac{196}{\sqrt{3}}=\dfrac{14^2}{\sqrt{3}}$ Hence, $a=14$ , $b=3$ and $a+b=\boxed{17}$ .

First we draw the line parallel to $XZ$ through $B$ and the perpendiculars from $A$ and $C$ to this line. The first one intersects the line at $P$ , the second at $O$ . If we call the length of each side of the triangle $l$ and the angle $\angle BOC=\alpha$ , we have $\angle BAP = 60-\alpha$ and get the following equations:

$\left\{\begin{array}{l}l\sin(\alpha)=OB=6,\\l\sin(60-\alpha)=BP=10.\end{array}\right.$

We now manipulate the second equation, using the angle sum formula for sines: $l\sin(60-\alpha) = l(\sin(60)\cos(\alpha)-\cos(60)\sin(\alpha))=l(\frac{\sqrt{3}}{2}\cos(\alpha)-\frac{1}{2}\sin(\alpha).$ Since we know $l\sin\alpha$ , we can fill this in to obtain $l\cos\alpha=\frac{26}{\sqrt{3}}$ . We thus have $l^2=l^2(\sin^2\alpha+\cos^2\alpha)=\ldots=\frac{784}{3}$ . Since the area of an equilateral triangle given a side $l$ equals $\frac{\sqrt{3}}{4}l^2$ , it follows that $A=\frac{\sqrt{3}}{4}\cdot\frac{784}{3}=\frac{14^2}{\sqrt{3}}$ and thus the answer is $14+3=17$ .

Let $B$ be the origin of the complex plane then C is the complex number $t-6i$ . We rotate $C$ round $B$ and get $A$ , that is

$A=C(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=(t-6i)(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=(\frac{t}{2}+6\sqrt{3})+(\frac{t\sqrt{3}}{2}-3)i$

Because the $y$ coordinate of $A$ is 10, that means $\frac{t\sqrt{3}}{2}-3=10$ , then $t=\frac{26\sqrt{3}}{3}$

The area of the $ABC$ is $\frac{BC^2\sqrt{3}}{4}$ , so we have

$S_{ABC}=\frac{(t^2+36)\sqrt{3}}{4}=((\frac{26\sqrt{3}}{3})^2+36)\frac{\sqrt{3}}{4}=\frac{13^2+27}{\sqrt{3}}=\frac{14^2}{\sqrt{3}}$

Let $s$ be the sides of the triangle. Make a vertical line passing through vertex $B$ .

From vertex $B$ , let

$\theta$ be the angle between the upper vertical line $\overline{XY}$ to the side $\overline{BA}$ .

$\alpha$ be the angle between the lower vertical line $\overline{YZ}$ to the side $\overline{BC}$ .

$\theta + 60 + \alpha = 180 ; \theta + \alpha = 120$

$\cos{\theta} = \frac{10}{s}$

$s = \frac{10}{\cos{\theta}}$

$\cos{\alpha} = \frac{6}{s}$

$s = \frac{6}{\cos{\alpha}}$

$\frac{\cos{\theta}}{10} = \frac{\cos{\alpha}}{6}$

$\frac{\cos{(120 - \alpha)}}{10} = \frac{\cos{\alpha}}{6}$

$\cos120cos\alpha + \sin120\sin\alpha = \frac{5\cos\alpha}{3}$

Simplifying can arrive

$\tan\alpha = \frac{26}{6\sqrt{3}}$

$\tan\alpha = \frac{26}{6\sqrt{3}} = \frac{26/\sqrt{3}}{6} = \frac{\overline{ZA}}{\overline{YZ}}$

$s = \sqrt{(\overline{ZA})^2 + (\overline{YZ})^2}$

$s = \sqrt{(26/\sqrt{3})^2 + (6)^2}$

$s = \frac{28\sqrt{3}}{3}$

Solving for the area of the triangle $A$

$A = \frac{1}{2}s^2\sin60$

Substituting values:

$A = \frac{392}{3}.\frac{\sqrt3}{2} = \frac{14^2}{\sqrt3}$

$a + b = 14 + 3 = \boxed{17}$