Difference between revisions of "2024 SSMO Relay Round 5 Problems/Problem 3"

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==Solution==
 
==Solution==
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Note that <math>k = 13.</math> Now, <math>\sin^2x+\cos^2x = 1,</math> so we are seeking to find the number of solutions to <math>\sin x^2+\cos x^2 = 1.</math> The only solutions are when <math>x^2 = 2k\pi,\frac{\pi}{2}+2k\pi,</math> for integer <math>k.</math> Now, for <math>-13\le x\le 13,</math> we have <math>x^2\le 169.</math> Now, note that <math>52\pi+0.5\le 169< 54\pi.</math> So, there are <math>54</math> possible values for <math>x^2,</math> each giving two unique solutions, except <math>x^2=0.</math> In conclusion, the answer is <math>54\cdot2-1 = \boxed{107}.</math>
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~SMO_Team

Latest revision as of 14:48, 10 September 2025

Problem

Let $T = TNYWR.$ Let $k$ be the maximum prime factor that divides $T.$ How many values of $x$ satisfy both $\sin x^2+\cos x^2=\sin^2 x + \cos^2 x$ and $-k \le x \le k?$

Solution

Note that $k = 13.$ Now, $\sin^2x+\cos^2x = 1,$ so we are seeking to find the number of solutions to $\sin x^2+\cos x^2 = 1.$ The only solutions are when $x^2 = 2k\pi,\frac{\pi}{2}+2k\pi,$ for integer $k.$ Now, for $-13\le x\le 13,$ we have $x^2\le 169.$ Now, note that $52\pi+0.5\le 169< 54\pi.$ So, there are $54$ possible values for $x^2,$ each giving two unique solutions, except $x^2=0.$ In conclusion, the answer is $54\cdot2-1 = \boxed{107}.$

~SMO_Team