HMMT-Nov 2017 gen, Problem 9

Find the minimum possible value of \[ \sqrt{58-42 x}+\sqrt{149-140 \sqrt{1-x^{2}}} \] where $-1 \leq x \leq 1$.

Solution: Substitute $x=\cos \theta$ and $\sqrt{1-x^{2}}=\sin \theta$, and notice that $58=3^{2}+7^{2}, 42=2 \cdot 3 \cdot 7,149=7^{2}+10^{2}$, and $140=2 \cdot 7 \cdot 10$. Therefore the first term is an application of Law of Cosines on a triangle that has two sides 3 and 7 with an angle measuring $\theta$ between them to find the length of the third side; similarly, the second is for a triangle with two sides 7 and 10 that have an angle measuring $90-\theta$ between them. "Gluing" these two triangles together along their sides of length 7 so that the merged triangles form a right angle, we see that the minimum length of the sum of their third sides occurs when the glued triangles form a right triangle. This right triangle has legs of length 3 and 10, so its hypotenuse has length $\sqrt{109}$. $\fbox{\sqrt{109}}$.

Answer: \sqrt{109}