contest-matheuclidean-geometrygeometrytrianglestrigonometry
A very unique question featuring a composite diagram of two triangles, with a missing angle and two equal sides.
I am posting this here to see what kind of different approaches there could be to solve it. Please feel free to leave your own answers!
(I have posted my own approach as an answer below)
So this'll be my approach. I'm going to add a brief explanation too.
Here's how I did it:
1.) Mark the $\triangle ABC$ with all the appropriate angles and mark the midpoint of segment $AC$ as $D$.
2.) Connect point $B$ with midpoint $D$, because $D$ is circumcenter of $\triangle ABC$, $BD=AD=DC$. Locate point $F$ outside $\triangle ABC$ and join it with points $A$ and $B$ such that $AF=BD=AD=DC$ and $\angle BAF=15$. Thus implies that $\triangle ABF$ is congruent to $\triangle CED$ via the SAS property. This implies that $\angle AFB =\alpha$.
3.) Notice that $\angle FAD=60$. Connect $F$ and $D$, via $FD$, because $\triangle FAD$ becomes equilateral, this shows that $AF=BD=AD=FD=DC$. Notice also that $\triangle ADB$ is isosceles, implying $\angle ADB=30$. This also means that $\angle BDF=30$ and $\triangle BDF$ is isosceles as well congruent to $\triangle ADB$. This further implies that $\angle \alpha= \angle DFB-\angle DFA$. $\alpha=75-60$, therefore $\alpha=15$.
This is my own approach. I'll add a brief explanation too:
This is the procedure I followed:
1.) Label the triangle $ABC$ and mark all the appropriate angles and side lengths. Notice that since $\triangle ABC$ is obtuse, its circumcenter must lie outside of the triangle itself. Locate circumcenter of $\triangle ABC$ outside and call it point $E$. Connect all the vertices of $\triangle ABC$ to $E$ via $AE$, $BE$ and $CE$. Notice that this means $AB=BE=AE=CE=CD$ because $\triangle ABE$ is equilateral.
(for further explanation of above, point $E$ is a circumcenter of $\triangle ABC$, in that case, $\angle AEB$ must be twice the measure of $\angle ACB$ (inscribed angle theorem), which means it'll be $60$. Because $\angle AEB=60$ and know that $AE=BE$, it follows that $\triangle ABE$ is equilateral)
2.) Notice also that $\angle EAC=\angle ECA=6$. Connect point $D$ with $E$ via segment $DE$. Notice that because segment $CD=CE$ and $\angle ECD=36$, this implies that $\angle EDC=\angle DEC=72$. However, notice that $\angle EBD=96-60=36$, therefore, $\angle BED$ is also $36$. But this implies that segment $BD=DE$
3.) Lastly, notice that this means $\triangle ABD$ is congruent to $\triangle AED$ via the SAS property. This means $\angle BAD=\angle EAD=x$. This means that $2x=60$, therefore, $x=30$.
Best Answer
Yet another geometrical alternative. Since $\angle BAC + \angle BDC = 180^\circ$, we rotate and translate $\triangle BCD$ into $\triangle FBA$ as shown in the figure below:
(Note that $A$ lies on $CF$. However, we cannot not assume a priori that $D$ lies on $BF$, so we cannot simply observe that $\triangle BCF$ is equilateral.)
Now $$ \angle F = \angle CBD = 180^\circ - \angle BCD - \angle BDC = 75^\circ - x. $$ But $BC = BF$, so $$ \angle F = \angle BCF = \angle BCD + \angle ACD = 45^\circ + x, $$ so $$ 75^\circ - x = 45^\circ + x. $$ Therefore, $x = 15^\circ$.