From e73b7959cc34a8cf2562dde234b0b750c6f4cd5f Mon Sep 17 00:00:00 2001
From: craisin <kahkaicyap@gmail.com>
Date: Fri, 30 Jan 2026 22:12:45 -0800
Subject: [PATCH] vault backup: 2026-01-30 22:12:45

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 Petals in a Rose Curve.md | 32 ++++++++++++++++++++++++++++++++
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+#Math #Trig #Algebra 
+# Question
+Show there are $n$ petals in a rose curve with odd $n$ and $2n$ petals in a rose curve with even $n$.
+# Solution
+WLOG the rose curve has the form $r = \cos n\theta$.
+Then there are 2 cases to get a maximum point on a petal: either $\cos n\theta = 1$ or $\cos( n(\theta + \pi)) = -1$
+## Case 1
+$$
+\cos n\theta = 1
+$$
+For all integers $i$:
+$$
+n\theta = 2i\pi
+$$
+$$
+\theta = \frac {2i\pi} n
+$$
+## Case 2
+$$\cos( n(\theta + \pi)) = -1$$
+For any arbitrary integer $i$:
+$$n(\theta + \pi) = 2i\pi + 1$$
+$$\theta = \frac {(2i + 1)\pi - n\pi} n$$
+For $\theta \in [0, 2\pi)$ both sequences create 
+## Even Case
+Factor out $\frac \pi n$ from both sides. Case 1 has $\frac \pi n$ multiplied by an even factor, while case 2 has $\frac \pi n$ multiplied by an odd factor. Therefore the sequences are unique, creating $2n$ unique petals.
+## Odd Case
+Take case 2:
+$$\theta = \frac {(2i + 1)\pi - n\pi} n$$
+$$\theta = \frac {2i\pi + (1 - n)\pi} n$$
+Since $n$ is odd, $1 - n$ is even, and therefore the numerator is an even factor of $\pi$.
+$$\theta = \frac {2i\pi} n$$
+Therefore case 2 produces the same $\theta$ as case 1, resulting in $n$ petals.
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