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Algebra Difficulty 5.1 AIME, harder Find the answer

Find the minimum possible value of 5842x+1491401x2\sqrt{58-42 x}+\sqrt{149-140 \sqrt{1-x^{2}}} where 1x1-1 \leq x \leq 1

A number or a short expression. Spacing and $ signs are ignored.

Solution

Substitute x=cosθx=\cos \theta and 1x2=sinθ\sqrt{1-x^{2}}=\sin \theta, and notice that 58=32+72,42=237,149=72+10258=3^{2}+7^{2}, 42=2 \cdot 3 \cdot 7,149=7^{2}+10^{2}, and 140=2710140=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 θ\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 90θ90-\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 109\sqrt{109}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.