This is a practice question for a math competition that I am lost on how to solve:
Los Angeles is located at (34°N, 118°W), and Osaka, Japan is at (34°N, 136°E). If a plane flies the great circle route from Osaka to Los Angeles, what is the highest latitude that the plane reaches during this flight?
A. 52.1°N
B. 45.5°N
C. 41.7°N
D. 48.3°N
The correct answer is D. Any guidance is appreciated.
Let $O$ be the center of the earth, $N$ the north pole, $K$ Osaka, and $L$ Los Angeles. Check that the difference in longitude between $K$ and $L$ is $\theta:=106^\circ$.
Luckily $K$ and $L$ lie at the same latitude. Let $\phi$ be the polar angle to this latitude, i.e., $\phi:=\angle NOK=\angle NOL=90^\circ-34^\circ=56^\circ$. The plane of the great circle through $K$ and $L$ passes through the center $O$ of the earth and contains the segment $KL$. By symmetry the airplane reaches highest latitude where the ray $OM$ intersects the surface of the earth, where $M$ is the midpoint of the segment joining $K$ and $L$. Hence the highest latitude is at polar angle $\alpha:=\angle NOM$.
How to calculate $\alpha$? Slice the earth with a horizontal plane at the latitude of $K$ and $L$. Let $P$ be the intersection of this plane with the north-south axis $ON$. Then $OP=\rho\cos\phi$ where $\rho$ is the radius of the (assumed spherical) earth. Look at isosceles triangle $\triangle PKL$, where $\angle LPK=\theta=106^\circ$, and $PK=PL=\rho\sin\phi$, and deduce for right triangle $\triangle MPK$ that $MP= PK\cos\frac\theta2=\rho\sin\phi\cos\frac\theta2$.
Now look at right triangle $\triangle OPM$ and deduce $$\tan\alpha = \frac {MP}{OP}=\frac{\rho\sin\phi\cos\frac\theta2}{\rho\cos\phi}=\tan\phi\cos\frac\theta2=\tan 56^\circ\cos53^\circ=0.89222$$ and so $\alpha=41.74^\circ$ and the latitude of $\alpha$ is $90^\circ-41.74^\circ=48.26^\circ$.
