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2023 SSMO Relay Round 1 Problems/Problem 3: Difference between revisions

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==Problem==
==Problem==
Let <math>T=TNYWR</math>. Find the number of solutions to the equation  
Let <math>T=TNYWR</math>. Find the number of solutions to the equation  
<cmath>\sec^{N} (Nx) - \tan^{N}(Nx) = 1</cmath>
<cmath>\sec^{T} (Tx) - \tan^{T}(Tx) = 1</cmath>
such <math>0 \le x \le \pi</math>
such <math>0 \le x \le \pi</math>


==Solution==
==Solution==
We have <math>T = 2022</math>. Let
<cmath>\begin{align*}
S = \sum_{i=0}^\infty a_i &= a_0+a_1+a_2+\sum_{i=3}^\infty a_i\\
&= 0+1+2022+\sum_{i=3}^\infty \left(a_{i-1}-\frac{a_{i-3}}{8}\right)\\
&= 2023+\sum_{i=3}^{\infty}a_{i-1}-\sum_{i=3}^\infty\frac{a_{i-3}}{8}\\
&= 2023+\sum_{i=2}^\infty a_i-\frac{\sum_{i=0}a_i}{8}\\
&= 2023+\left(\sum_{i=0}^\infty a_i-a_0-a_1\right)-\frac{S}{8}\\
&= 2023+(S-0-1)-\frac{S}{8}\\
&= 2022+\frac{7S}{8}.\\
\end{align*}</cmath>
We have
<cmath>\begin{align*}
S &= 2022+\frac{7S}{8}\implies\\
\frac{S}{8} &= 2022\implies\\
S &= 8\cdot2022 = \boxed{16176}.
\end{align*}</cmath>
~pinkpig

Revision as of 10:23, 15 September 2025

Problem

Let $T=TNYWR$. Find the number of solutions to the equation \[\sec^{T} (Tx) - \tan^{T}(Tx) = 1\] such $0 \le x \le \pi$

Solution

We have $T = 2022$. Let \begin{align*} S = \sum_{i=0}^\infty a_i &= a_0+a_1+a_2+\sum_{i=3}^\infty a_i\\ &= 0+1+2022+\sum_{i=3}^\infty \left(a_{i-1}-\frac{a_{i-3}}{8}\right)\\ &= 2023+\sum_{i=3}^{\infty}a_{i-1}-\sum_{i=3}^\infty\frac{a_{i-3}}{8}\\ &= 2023+\sum_{i=2}^\infty a_i-\frac{\sum_{i=0}a_i}{8}\\ &= 2023+\left(\sum_{i=0}^\infty a_i-a_0-a_1\right)-\frac{S}{8}\\ &= 2023+(S-0-1)-\frac{S}{8}\\ &= 2022+\frac{7S}{8}.\\ \end{align*} We have \begin{align*} S &= 2022+\frac{7S}{8}\implies\\ \frac{S}{8} &= 2022\implies\\ S &= 8\cdot2022 = \boxed{16176}. \end{align*}

~pinkpig

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