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  /Title (Problem Set 4)
}

\title{Problem Set 4}
\author{Enoch Yu}
\date{May 2025}

\begin{document}

\section*{2023 AMC 12B Problem 16}
In Coinland, there are three types of coins, each
worth $6, 10,$ and $15$. What is the sum of the digits
of the maximum amount of money that is impossible to have?

$\textbf{(A) } 8 \qquad \textbf{(B) } 10 \qquad
\textbf{(C) } 7 \qquad \textbf{(D) } 11
\qquad \textbf{(E) } 9$

\textbf{Solution I: During Test}

\textbf{Key Word} Pattern Recognition

The method is to utilize the fact that when a number
could form with 6, 10 and 15, numbers formed by adding
multiples of 6 to the initial number also could form.
Number "6" was chosen to minimize the range of period
of the pattern.

Choose an arbitrary integer 20.

$20+6k$ is valid.

$21+6k$ is valid.

$22+6k$ is valid.

$23$ is not valid.

$24+6k$ is valid.

$25+6k$ is valid.

It is likely that $23 + 6k$ produces the maximum
amount of money that is impossible to have. 29 is
impossible to have. However, from 35, moneys could form.
Therefore, $2 + 9 = \boxed{11}$.


\textbf{Solution II: Generalizing Method}

\textbf{Key Word} Forming a Generalized Relationship

Let $a$, $b$, and $c$ be the number of coins that worth
6, 10 and 13 respectively. If $6a + 10b + 15c = k$, then
$6 (a + 1) + 10 (b - 2) + 15 (c + 1) = k + 1$. The
following represents the relationship between an arbitrary
triple of $(a, b, c)$.
\[
\begin{array}{ccccccccccccc}
    (1, 6, 2) & \rightarrow & (2, 4, 3) & \rightarrow
    & (3, 2, 4) & \rightarrow & (4, 0, 5) & \leftarrow
    & (0, 10, 0) & \leftarrow & (1, 8, 1) & \leftarrow
    & (2,6,2) \\
    96 & & 97 & & 98 & & 99 & & 100 & & 101 & & 102
\end{array}
\]
By induction, it is evident that all natural numbers
greater than or equal to 96 may form. Because a limit
is given, same method used in Solution I may be utilized
to obtain $\boxed{\textbf{(D) } 11}$.

\newpage
\section*{2023 AMC 12B Problem 18}
Last academic year Yolanda and Zelda took different
courses that did not necessarily administer the same
number of quizzes during each of the two semesters.
Yolanda's average on all the quizzes she took during
the first semester was $3$ points higher than Zelda's
average on all the quizzes she took during the first
semester. Yolanda's average on all the quizzes she took
during the second semester was $18$ points higher than
her average for the first semester and was again $3$
points higher than Zelda's average on all the quizzes
Zelda took during her second semester. Which one of the
following statements cannot possibly be true?

$\textbf{(A)}$ Yolanda's quiz average for the academic
year was $22$ points higher than Zelda's. \\
$\textbf{(B)}$ Zelda's quiz average for the academic
year was higher than Yolanda's. \\
$\textbf{(C)}$ Yolanda's quiz average for the academic
year was $3$ points higher than Zelda's. \\
$\textbf{(D)}$ Zelda's quiz average for the academic
year equaled Yolanda's. \\
$\textbf{(E)}$ If Zelda had scored $3$ points higher
on each quiz she took, then she would have had the same
average for the academic year as Yolanda.
\begin{solution}

\textbf{Key Word} Property of Fraction

Let
\begin{align*}
    \text{Yolanda's 1st Semester Average}
    &= \frac{a_1+\dots+a_n}{n} \\
    \text{Yolanda's 2nd Semester Average}
    &= \frac{a_1'+\dots+a_n'}{n'} \\
    \text{Zelda's 1st Semester Average}
    &= \frac{b_1+\dots+b_m}{m} \\
    \text{Zelda's 2nd Semester Average}
    &= \frac{b_1'+\dots+b_m'}{m'}.
\end{align*}
According to the problem, $\frac{a_1 + \dots + a_n}{n} + 18
= \frac{a_1' + \dots + a_n'}{n'}$. Moreover,
$\frac{a_1 + \dots + a_n}{n} = \frac{b_1 + \dots + b_m}{m} + 3$
and $\frac{a_1' + \dots + a_n'}{n'} =
\frac{b_1' + \dots + b_m'}{m'} + 3$ are true. In another words,
\[
\frac{a_1 + \dots + a_n}{n} =
\frac{a_1' + \dots + a_n' - 18n'}{n'} =
\frac{b_1 + \dots + b_m + 3m}{m} =
\frac{b_1' + \dots + b_m' - 15m'}{m'}.
\]
The relationship leads to
\[
\frac{a_1 + \dots + a_n + a_1' + \dots + a_n' - 18n'}
{n + n'} =
\frac{b_1 + \dots + b_m + b_1' + \dots + b_m' + 3m - 15m'}
{m + m'}, \text{or}
\]
\[
\frac{a_1 + \dots + a_n + a_1' + \dots + a_n'}
{n + n'} - \frac{18n'}{n' + n} - \frac{3m}{m + m'}
+ \frac{15m'}{m + m'} =
\frac{b_1 + \dots + b_m + b_1' + \dots + b_m'}{m + m'}.
\]
For choice $(A)$ to be true, $-\frac{18n'}{n' + n}
- \frac{3m}{m + m'} + \frac{15m'}{m + m'}$ must be $-22$.

Let $n' = kn$ and $m' = k'm$.
\[
-\frac{18n'}{n' + n} - \frac{3m}{m + m'} +
\frac{15m'}{m + m'} = -\frac{18kn}{n (k + 1)} -
\frac{3m}{m (k' + 1)} + \frac{15k'm}{m (k' + 1)} =
-\frac{18kn}{n (k + 1)} + \frac{3m (5k' - 1)}{m (k' + 1)}
\]
$n, n', m, m', k \text{ and } k'$ must be non-zero
integer to satisfy the condition in the question.
While the equation above is a multiple of three, 22 is not.
Therefore, choice $\boxed{(A) \text{ cannot be true.}}$
\end{solution}

\newpage
\section*{2023 AMC 12B Problem 20}
Cyrus the frog jumps $2$ units in a direction,
then $2$ more in another direction. What is the
probability that he lands less than $1$ unit away
from his starting position?

$\textbf{(A)}~\frac{1}{6} \qquad \textbf{(B)}~\frac{1}{5}
\qquad \textbf{(C)}~\frac{\sqrt{3}}{8} \qquad
\textbf{(D)}~\frac{\arctan \frac{1}{2}}{\pi}\qquad
\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}$

\begin{solution}

\textbf{Key Word} Law of Cosines, Inverse Trigonometric Identities

WLOG, the probability that Cyrus lands less than 1 unit away
from the starting position is 
\[
\frac{\text{Length of Red Curve}}
{\text{Length of Red Curve + Length of Blue Curve}}.
\]
\begin{center}
\begin{tikzpicture}[scale=0.7]
    \coordinate (O) at (0,0);
    \coordinate (A) at (1.41421356237,1.41421356237);
    \coordinate (B) at (-0.50788, 0.86143);
    \coordinate (C) at (0.86143, -0.50788);

    \draw (O) circle (1);
    \draw (O) circle (2);
    \draw[color=blue, thick] (A) circle (2);

    \pgfmathsetmacro{\angleB}{atan2(0.86143 - 1.41421356237,
        -0.50788 - 1.41421356237)}
    \pgfmathsetmacro{\angleC}{atan2(-0.50788 - 1.41421356237,
        0.86143 - 1.41421356237)}

    \draw[red, thick] (A) ++(\angleB:2) arc[start angle=\angleB,
        end angle=\angleC, radius=2];

    \draw (A) -- (B);
    \draw (A) -- (C);
    \draw (A) -- (O);

    \node at (O)[circle,fill,inner sep=0.4pt]{};
    \node at (A)[circle,fill,inner sep=0.4pt]{};

    \draw pic[draw=black, radius=0.5cm, "$\theta$",
        angle eccentricity=1.4] {angle=B--A--O};
\end{tikzpicture}
\end{center}
The length of blue curve and red curve is the circumference
of a circle with radius of 2, or $4\pi$. The length of the
red curve may be found by computing the value of $\theta$,
where $\theta$ is in radians for convenience. Using Law of
Cosines, $\cos\theta = \frac{2^2 + 2^2 - 1^2}
{2 \cdot 2 \cdot 2} = \frac{7}{8}$. Using a right triangle
with lengths $7, 8 \text{ and } \sqrt{15}$, it is evident
that $\sin\theta = \frac{\sqrt{15}}{8}$.
\begin{align*}
    \text{Probability}
    &= \frac{ 4 \pi \cdot \frac{2\theta}{2\pi} }{4\pi} \\
    &=\frac{\theta}{\pi}\\
\end{align*}
Assume $(E)$ is the correct answer. If
$\arcsin{\frac{\sqrt{15}}{8}} = 2 \arcsin{\frac{1}{4}}$
is true, choice $(E)$ is the answer.
\begin{align*}
    \arcsin{ \frac{\sqrt{15}}{8} }
    &= 2 \arcsin{\frac{1}{4}} \\
    \sin\left( \arcsin{ \frac{\sqrt{15}}{8}} \right)
    &= \sin\left( 2 \arcsin{ \frac{1}{4} } \right) \\
    \frac{\sqrt{15}}{8} &= 2 \cdot
    \sin\left( \arcsin{ \frac{1}{4} } \right) \cdot
    \cos\left( \arcsin{ \frac{1}{4} } \right) \\
    \frac{\sqrt{15}}{8} &= 2 \cdot \frac{1}{4} \cdot
    \sqrt{1 - \left( \frac{1}{4} \right)^2} \\
    \frac{\sqrt{15}}{8} &= \frac{\sqrt{15}}{8}
\end{align*}
Therefore, the probability that Cyrus lands less
than 1 unit away from his starting position is
$\boxed{\textbf{(E)}~\frac{2\arcsin \frac{1}{4}}{\pi}}$.
\end{solution}

\end{document}

