source: #alcumus tags: #triangle #area #special-right-triangles course: [[Geometry]] status: #solved ### Problem Given that $m \angle A=60^{\circ}, B C=12$ units, $\overline{B D} \perp \overline{A C}, \overline{C E} \perp \overline{A B}$ and $m \angle D B C=3 m \angle E C B$, the length of segment $E C$ can be expressed in the form $a(\sqrt{b}+\sqrt{c})$ units where $b$ and $c$ have no perfect-square factors. What is the value of $a+b+c ?$ ![[alcumus-triangle-altitudes.svg]] #### Solution Writing the given information into the diagram ![[alcumus-triangle-altitudes-1.svg]] Using [[Triangle Sum Theorem]] on $\triangle AEC$ and $\triangle ADB$ to find $30\degree$ angles at $\angle ACE$ and $\angle ABD$ ![[alcumus-triangle-altitudes-2.svg]] Applying the [[Triangle Sum Theorem]] again on $\triangle EBC$ $ \begin{aligned}90\degree + \alpha\degree + (30+3\alpha)\degree = 180\degree\\ 4\alpha\degree&=60\degree\\ \alpha\degree&=15\degree\\ (30+3\alpha)\degree&=75\degree \end{aligned}$ $\triangle EBC$ is a [[Special Right Triangles#15-75-90]] so it's long leg $\overline{EC}$ is $\frac{\sqrt{6}+\sqrt{2}}{4}$ times the length of the hypotenuse $\overline{BC}$. $\frac{\sqrt{6}+\sqrt{2}}{4}\cdot12=3(\sqrt{6}+\sqrt{2})$ Comparing to the form mentioned in the problem: $3(\sqrt{6}+\sqrt{2})=a(\sqrt{b}+\sqrt{c})$ $a=3, \quad b=6, \quad c=2 $ $a+b+c=3+6+2=\boxed{11} $ #### Alternative Solution Start the same as previous solution, but use [[Special Right Triangles#45-45-90]] on $\triangle BDC$ to get $\overline{BD}=\overline{DC}=6\sqrt{2}$ Then use [[Special Right Triangles#30-60-90]] to find $\overline{AD}= 2 \sqrt{6}$ and $\overline{AB}=4\sqrt{6}$ ![[alcumus-triangle-altitudes-3.svg]] By the [[◊ Segment Addition Postulate]], $\overline{AC}=\overline{AD}+\overline{DC}=2\sqrt{6}+6\sqrt{2}$ The [[Area of a Triangle]] $\triangle ABC$ will be the same whether calculated using $\overline{AB}$ as the base or $\overline{AC}$ as the base. $\begin{aligned} \frac{\overline{AB}\cdot\overline{EC}}{2} &= \frac{\overline{AC}\cdot\overline{DB}}{2}\\ \\ {\overline{AB}\cdot\overline{EC}} &= {\overline{AC}\cdot\overline{DB}}\\ 4\sqrt{6} (\overline{EC}) &= (2 \sqrt{6} +6\sqrt{2})\cdot6\sqrt{2}\\ \overline{EC}&=(2 \sqrt{6} +6\sqrt{2})\cdot \frac{\sqrt{3}}{2} \\ &= \sqrt{18} +3 \sqrt{6} \\ &= 3\sqrt{2} +3 \sqrt{6} \\ &= 3(\sqrt{2}+\sqrt{6}) \end{aligned}$ Comparing to the form mentioned in the problem: $3(\sqrt{2}+\sqrt{6})=a(\sqrt{b}+\sqrt{c})$ $a=3, \quad b=2, \quad c=6 $ $a+b+c=3+2+6=\boxed{11} $