contest-matheuclidean-geometrygeometrysolution-verificationtrigonometry
As title suggests, the question is to find the measure of $\angle AKC$ in the triangle $\triangle ABC$, with an internal point $K$ and some given angles.
This is a pretty challenging problem, I'm going to post my solution below as an answer, but I'm not sure if my answer is correct and if my method is correct. Please share your own answers and approaches. I want to see if my answer is correct and if there are any other ways to arrive at an answer.
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 $\alpha=15$.
I think my solution is quite simple. Since $\triangle BDC$ is isosceles, it follows that $\angle BCD = \angle CBD = 2 \alpha$, and as such, $\angle BCA = 3\alpha$. We then compute that $$ \angle ABD = 180 - \angle CBD - \angle BCA - \angle BAC = 180 - 2\alpha - 3\alpha - 7\alpha = 180 - 12 \alpha. $$ Because also $\triangle ABD$ is isosceles, we decude from this that $$ \angle BAD = \frac{180 - \angle ABD}{2} = 6\alpha. $$ In particular, we find that $\angle CAD = 7\alpha - 6\alpha = \alpha$. This implies that also $\triangle CDA$ is isosceles; whence $|AD| = |DC| = |DB| = |BA|$. It follows that $\triangle BDA$ is even equilateral and as such, its angles are all $60^{\circ}$. Hence $6\alpha = 60^{\circ}$, and thus $\alpha = 10^{\circ}$.
Best Answer
We have $\angle ABC = 66^\circ$ and $\angle AKC = 150^\circ$. Assume $BC = 1$, then $AC = \frac{\sin 48^\circ}{\sin 66^\circ} = \frac{\sin 48^\circ}{\cos 24^\circ} = 2 \sin 24^\circ$ and $KC = \frac{\sin 12^\circ}{\sin 150^\circ} = 2\sin 12^\circ$.
Now $$AK^2 = AC^2 + KC^2 - 2AC \cdot KC \cdot \cos \angle ACK = 4\sin^2 24^\circ + 4\sin^2 12^\circ - 8 \sin 12^\circ \sin 24^\circ \cos 48^\circ.$$
Note that $\sin^2 12^\circ = \frac{(1 - \cos 24^\circ)}{2}$ and $\sin 12^\circ \cos 48^\circ = \frac{(\sin 60^\circ - \sin 36^\circ)}{2}$. Now everything is in $24^\circ$ and $36^\circ$ and the values can be calculated. The result is $\frac{(3 - \sqrt{5})}{2}$. So $AK = \frac{(\sqrt{5} - 1)}{2}$.
$$\cos X = \frac{AK^2 + KC^2 - AC^2}{2AK \cdot KC}$$
Calculate again and get $\cos X = -\frac{\sqrt{7 - \sqrt{5} - \sqrt{30 - 6\sqrt{5}}}}{4}$. Compare this with a cos table and get $X = 102^\circ$.