recursal/Fanatic-Fandom huggingface dataset:
Two Squares Problem

## **Problem**

Suppose there are two adjacent squares: one red and one blue (as shown in Figure 1). A straight line connecting the lower right corner of the blue square with the upper left corner of the red square is of equal length with the diagonal of the blue square. Question: what is the ratio of the area of the red square to the area of the blue square?

## **Solution**

Let

x

{\displaystyle x}

 be the sidelength of the blue square and

y

{\displaystyle y}

 be the sidelength of the red square. The diagonal of the blue square is

2

x

{\displaystyle \sqrt{2} x}

. By the Pythagorean theorem, we get

(
x
y

)
2

y
2

=

2

x

2

x
2

2
x
y

y
2

y
2

=
2

x
2

2
(
x
y

y
2

)

=

x
2

.

{\displaystyle \begin{align}
(x+y)^2 + y^2 &= {\sqrt{2} x}^2 \\[5 pt]
x^2 + 2xy + y^2 + y^2 &= 2x^2 \\[5 pt]
2(xy + y^2) &= x^2.
\end{align}
}

Divide

x
2

{\displaystyle x^2}

 on both sides of the equation

2
y

x

2

y
2

x
2

=
1

{\displaystyle \frac{2y}{x} + \frac{2 y^2}{x^2} = 1 }

and define the ratio

# r

y
x

.

{\displaystyle r = \frac{y}{x}.}

Thus, we get the equation

2
r
2

r
2

=
1

2

r
2

2
r
−
# 1

{\displaystyle
\begin{align}
2r + 2r^2 = 1 \\[5 pt]
2r^2 + 2r -1 = 0.
\end{align}
}

There are two solutions:

# r

3

−
1

2

{\displaystyle r = \frac{\sqrt{3} - 1}{2} }

 and

# r

−

3

−
1

2

{\displaystyle r = \frac{-\sqrt{3} - 1}{2} }

. Of course, we cannot have a negative length, so only the positive solution

# r

3

−
1

2

{\displaystyle r = \frac{\sqrt{3} - 1}{2} }

 works.

The areal ratio of the red square to the blue square is equal to the square of

r

{\displaystyle r}

:

A

r
e
d

A

b
l
u
e

=

r
2

=

(

3

−
1

2

)

2

=

2
−

3

2

.

{\displaystyle
\begin{align}
\frac{A\_{red}}{A\_{blue}} &= r^2 \\[5 pt]
&= {\left(\frac{\sqrt{3} - 1}{2} \right)}^{2} \\[5 pt]
&= \frac{2 -\sqrt{3}}{2}.
\end{align}
}

Therefore,

A

r
e
d

:

A

b
l
u
e

=
2
−

3

:
2

{\displaystyle A\_{red}:A\_{blue} = 2 - \sqrt{3} : 2 }

.
Core Principles

1. Seek the way (道 do) and reason (理 li) in the cosmos.

2. Actions should be motivated by benefit (利 li) and righteousness (義 gyi).

3. Arrive at truth (實 sik/sit) through evidence (證 dzeng/dzing) and demonstration (明 myeng/mying).

4. Avoid or abstain from acts of vice (惡 ak/ok).

5. Honour is founded on trust (信 shin) and virtue (德 dek).

6. There is no fate (命 myeng/mying); only probable and improbable occurrences.
Variance with Integrals

## **Problem**

Begin with the integral definition of the variance

V
a
r
(
x
# )

∫

−
∞

∞

(
x
−
μ

)
2

p
(
x
)

d
x

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} (x-\mu)^2 \,p(x) \,dx }

where

μ

{\displaystyle \mu}

 is the mean, and

p
(
x
)

{\displaystyle p(x)}

 is the probability density function. Show that

V
a
r
(
x
# )

∫

−
∞

∞

x
2

p
(
x
)

d
x
−

μ

2

.

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} x^2 \,p(x) \,dx - {\mu}^2. }

## **Solution**

Begin with the integral definition of variance

V
a
r
(
x
# )

∫

−
∞

∞

(
x
−
μ

)
2

p
(
x
)

d
x

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} (x-\mu)^2 \,p(x) \,dx }

and expand the quadratic expression

V
a
r
(
x
# )

∫

−
∞

∞

(

x
2

−
2
μ
x

μ

2

)

p
(
x
)

d
x
.

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} (x^2-2\mu x + {\mu}^2) \,p(x) \,dx. }

In consequence, we get

V
a
r
(
x
# )

∫

−
∞

∞

x
2

p
(
x
)

d
x
−
2
μ

∫

−
∞

∞

x

p
(
x
)

d
x

μ

2

∫

−
∞

∞

p
(
x
)

d
x
.

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} x^2 \,p(x) \,dx - 2\mu \int\_{-\infty}^{\infty} x \,p(x) \,dx + {\mu}^2 \int\_{-\infty}^{\infty} p(x) \,dx. }

Since

# μ

∫

−
∞

∞

x
p
(
x
)
d
x

{\displaystyle \mu = \int\_{-\infty}^{\infty} x p(x) dx }

 (the expectation), and

∫

−
∞

∞

p
(
x
)
d
# x
1

{\displaystyle \int\_{-\infty}^{\infty} p(x) dx = 1 }

 (the total probability),

V
a
r
(
x
# )

∫

−
∞

∞

x
2

p
(
x
)

d
x
−
2

μ

2

μ

2

(
1
)

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} x^2 \,p(x) \,dx - 2{\mu}^2 + {\mu}^2 (1) }

V
a
r
(
x
# )

∫

−
∞

∞

x
2

p
(
x
)

d
x
−

μ

2

.

{\displaystyle Var(x) = \int\_{-\infty}^{\infty} x^2 \,p(x) \,dx - {\mu}^2. }
千秋夢（鄭氏中古漢語拼音）

**Transliterated by Tao Steven Zheng (鄭濤)**

## **千秋夢【tsen tsiuu myung】**

休 【hiuu】 賴 【lai】 人 【nyin】 間 【gan】 梧 【ngo】 桐 【dung】 輕 【kyeng】 ，

上 【dzyang】 棲 【sei】 有 【iuu】 鳳 【byung】 鳳 【byung】 自 【dzi】 鳴 【myeng】 。

銜 【gham】 得 【dvk】 鬢 【bin】 邊 【bien】 榴 【liuu】 花 【hua】 重 【dyung】 ，

名 【myeng】 冠 【guan】 長 【dyang】 安 【an】 城 【dzyeng】 。

千 【tsen】 秋 【tsiuu】 兮 【hei】 難 【nan】 知 【dhi】 禁 【gim】 中 【dyung】 事 【dzi】，

玉 【ngiok】 衣 【i】 兮 【hei】 一【it】 動 【dung】 紫 【tsi】 薇 【mwi】 知 【dhi】。

**Transliterated by Tao Steven Zheng (鄭濤)**
Reduced Row Echelon Form

## **Problem**

This example of row reduction is a step-by-step solution to the matrix constructed by Zhang Qiujian (*Zhang Qiujian Suanjing: Lower Scroll 14*).

[

3

2

1

300

4

6

3

600

2

2

3

300

]

{\displaystyle
\begin{bmatrix}
3 & 2 & 1 & 300 \\
4 & 6 & 3 & 600 \\
2 & 2 & 3 & 300
\end{bmatrix} }

## **Solution**

**Step 1: Subtract row 1 by row 3**

[

1

0

−
2

0

4

6

3

600

2

2

3

300

]

{\displaystyle
\begin{bmatrix}
1 & 0 & -2 & 0 \\
4 & 6 & 3 & 600 \\
2 & 2 & 3 & 300
\end{bmatrix} }

**Step 2: Subtract row 2 by two times row 3**

[

1

0

−
2

0

0

2

−
3

0

2

2

3

300

]

{\displaystyle
\begin{bmatrix}
1 & 0 & -2 & 0 \\
0 & 2 & -3 & 0 \\
2 & 2 & 3 & 300
\end{bmatrix} }

**Step 3: Subtract row 3 by two times row 1**

[

1

0

−
2

0

0

2

−
3

0

0

2

7

300

]

{\displaystyle
\begin{bmatrix}
1 & 0 & -2 & 0 \\
0 & 2 & -3 & 0 \\
0 & 2 & 7 & 300
\end{bmatrix} }

**Step 4: Subtract row 3 by row 2**

[

1

0

−
2

0

0

2

−
3

0

0

0

10

300

]

{\displaystyle
\begin{bmatrix}
1 & 0 & -2 & 0 \\
0 & 2 & -3 & 0 \\
0 & 0 & 10 & 300
\end{bmatrix} }

**Step 5: Divide row 3 by ten**

[

1

0

−
2

0

0

2

−
3

0

0

0

1

30

]

{\displaystyle
\begin{bmatrix}
1 & 0 & -2 & 0 \\
0 & 2 & -3 & 0 \\
0 & 0 & 1 & 30
\end{bmatrix} }

**Step 6: Add two times row 3 to row 1**

[

1

0

0

60

0

2

−
3

0

0

0

1

30

]

{\displaystyle
\begin{bmatrix}
1 & 0 & 0 & 60 \\
0 & 2 & -3 & 0 \\
0 & 0 & 1 & 30
\end{bmatrix} }

**Step 7: Add three times row 3 to row 1**

[

1

0

0

60

0

2

0

90

0

0

1

30

]

{\displaystyle
\begin{bmatrix}
1 & 0 & 0 & 60 \\
0 & 2 & 0 & 90 \\
0 & 0 & 1 & 30
\end{bmatrix} }

**Step 8: Divide row 2 by two**

[

1

0

0

60

0

1

0

45

0

0

1

30

]

{\displaystyle
\begin{bmatrix}
1 & 0 & 0 & 60 \\
0 & 1 & 0 & 45 \\
0 & 0 & 1 & 30
\end{bmatrix} }
Yale Babylonian Collection (YBC 7289) Clay Tablet

## **Problem**

This circular clay tablet from the Yale Babylonian Collection (Figure 1), called YBC 7289, was created around 1800 – 1600 BC. It shows the geometric diagram of a square, its diagonals, and some numbers etched beside the lines. The clay tablet is actually a mathematics problem (most likely a homework problem) that asks for the length of the square's diagonal given the side length of the square being 30. Since the Babylonians used a sexagesimal (base 60) numeral system, the author of the tablet expressed the square root of 2 to be

(
1
;
24
,
51
,
10
)

60

{\displaystyle {(1;24,51,10)}\_{60} }

.

**Part 1:** Calculate the length of the diagonal in base 60.

**Part 2:** Calculate the Babylonian value for

2

{\displaystyle \sqrt{2}}

 to 8 decimal places.

## **Solution**

**Part 1**

30
⋅

(
1
;
24
,
51
,
10
)

60

=
30
⋅
1

30
⋅
24

60

30
⋅
51

60
2

30
⋅
10

60
3

=
30

720
60

1530

60
2

300

60
3

=
30
12

25
60

30

60
2

5

60
2

=

(
42
;
25
,
35
)

60

{\displaystyle \begin{align}
30\cdot{(1;24,51,10)}\_{60} &= 30\cdot 1+\frac{30\cdot24}{60}+\frac{30\cdot51}{60^2} +\frac{30\cdot10}{60^3} \\
&= 30+\frac{720}{60}+\frac{1530}{60^2} +\frac{300}{60^3} \\
&= 30+12+\frac{25}{60} +\frac{30}{60^2}+\frac{5}{60^2} \\
&= {(42;25,35)}\_{60}
\end{align} }

**Part 2**

(
1
;
24
,
51
,
10
)

60

=
1

24
60

51

60
2

10

60
3

=
1

24
60

51

60
2

10

60
3

=
1.41421296

{\displaystyle \begin{align}
{(1;24,51,10)}\_{60} &= 1+\frac{24}{60}+\frac{51}{60^2} +\frac{10}{60^3} \\
&= 1 +\frac{24}{60}+\frac{51}{60^2} +\frac{10}{60^3} \\
&= 1.41421296
\end{align} }

This is very close to the exact value

2

=
1.4142135624

{\displaystyle \sqrt{2} = 1.4142135624 }

!
Simple Harmonic Oscillator

## **Problem**

Consider the position function

x
(
t
# )
A
cos
⁡

(
ω
t
ϕ
)

{\displaystyle x(t) = A \cos{(\omega t + \phi)} }

.

**Part 1:** Determine the value of

ω

{\displaystyle \omega}

 if this function solves the differential equation:

d
2

x

d

t
2

k
x

m

=

{\displaystyle \frac{d^2 x}{dt^2}+\frac{kx}{m} = 0. }

**Part 2:** Try to explain what each term of the above differential equation means.

## **Solution**

**Part 1**

Take two time derivatives:

d
x

d
t

=
−
ω
A
sin
⁡

(
ω
t
ϕ
)

{\displaystyle \frac{dx}{dt}= - \omega A \sin{(\omega t + \phi)} }

d
2

x

d

t
2

=
−

ω

2

A
cos
⁡

(
ω
t
ϕ
)

.

{\displaystyle \frac{d^2 x}{dt^2} = - {\omega}^{2} A \cos{(\omega t + \phi)}. }

Consequently,

−

ω

2

A
cos
⁡

(
ω
t
ϕ
)

k
m

A
cos
⁡

(
ω
t
ϕ
)

=
0

{\displaystyle - {\omega}^{2} A \cos{(\omega t + \phi)} + \frac{k}{m} A \cos{(\omega t + \phi)} = 0 }

(

k
m

−

ω

2

)

A
cos
⁡

(
ω
t
ϕ
)

=

{\displaystyle \left(\frac{k}{m}-{\omega}^{2}\right)A \cos{(\omega t + \phi)} = 0. }

Divide away

A
cos
⁡

(
ω
t
ϕ
)

{\displaystyle A \cos{(\omega t + \phi)} }

(

k
m

−

ω

2

)

=
0

{\displaystyle \left(\frac{k}{m}-{\omega}^{2}\right) = 0 }

and solve to the angular frequency to get

# ω
±

k
m

{\displaystyle \omega = \pm \sqrt{\frac{k}{m}} }

.

This quantity is the angular frequency of the wave. Mathematically speaking, it is the **eigenvalue** of the differential equation.

**Part 2**

 Multiply the entire equation by

m

{\displaystyle m}

:

m

d
2

x

d

t
2

k
# x

{\displaystyle m\frac{d^2 x}{dt^2} + kx = 0. }

Since acceleration is the second time derivative of position

# a

d
2

x

d

t
2

{\displaystyle a = \frac{d^2 x}{dt^2} }

−
k
# x
m
a
.

{\displaystyle -kx = ma. }

This is Newton’s second law with the spring force being the net force!
Newton's Method with Python

**By: Tao Steven Zheng (郑涛)**

```
'''Derivative function'''
def derivative(f,x):
    h = 0.00000001
    return (f(x+h) - f(x))/h

'''Newton's method function'''
def newton\_method(f, x):
    tolerance = 0.00000001
    while True:
        x1 = x - f(x)/derivative(f,x)
        t = abs(x1 - x)
        if t < tolerance:
            break
        x = x1
    return x

'''Test and results'''

def f1(x):
    return x\*\*3-1

def f2(x):
    return x\*\*2 - x - 1

root1 = newton\_method(f1, 1)
root2 = newton\_method(f2, 1.5)
print(root1) #RESULT: 1
print(root2) #RESULT: phi

```
Speed of a Cheetah

## **Problem**

The speed of a cheetah is 80 mph (miles per hour). If 1 mile = 1.609 km, calculate the speed of a cheetah in meters per second.

## **Solution**

First convert mph to km/h.

(

70

m
i
l
e

h
r

)

(

1.609

k
m

1

m
i
l
e

)

=

112.63

k
m

h
r

{\displaystyle \left(\frac{70 \; mile}{hr}\right) \left(\frac{1.609 \; km}{1 \; mile}\right) = \frac{112.63 \; km}{hr}}

Then convert km/h to m/s.

(

112.63

k
m

h
r

)

(

1000

m

1

k
m

)

(

1

h
r

3600

s

)

=

31.29

m

s

{\displaystyle \left(\frac{112.63 \; km}{hr}\right) \left(\frac{1000 \; m}{1 \; km}\right)\left(\frac{1 \; hr}{3600 \; s}\right) = \frac{31.29 \; m}{s}}

.
Therefore, the speed of a cheetah is approximately 31 m/s.
Rock-Paper-Scissors with Python

**By: Tao Steven Zheng (郑涛)**

## **Task**

   Write a program that initiates a game of rock-paper-scissors against the computer. The human player can type in three words (the weapons of the game): rock, paper, and scissors.

Things of include:

(1) Make sure the rules of the game makes "paper" defeat "rock", "rock" defeat "scissors", and "scissors" defeat "paper".

(2) Add a counter that keeps track of the number of victories made by the human player.

(3) The game ends when the human player is defeated by the computer.

(4) Don't forget to include the possibility that the game can result in a draw.

## **Code**

```
from random import choice

rules = {"rock": "paper", "scissors": "rock", "paper": "scissors"}
#paper beats rock, rock beats scissors, scissors beats paper
weapons = ["rock", "paper", "scissors"]
wins = 0

while True:
    human = input("\nPick your weapon:") # enter rock, paper, or scissors
    computer = rules[choice(weapons)]  # computer chooses a weapon randomly

    if human in rules:
        weapons.append(human)
        print("The computer played", computer, end = ": ")

        if rules[computer] == human:  # if what beats the computer's choice is the human's choice...
            print("You win! :D")
            wins += 1
        elif rules[human] == computer:  # if what beats the human's choice is the computer's choice...
            print("You lost...OTL")
            print("You won", wins, end = " time(s).")
            break
        else: print("It's a draw!")

    else: print("Pick rock, paper, or scissors")

```
Vector Norm

## **Problem**

The norm of a vector measures its magnitude (a.k.a length). For a vector in N-dimensioanl space

# A
⟨

x
1

,

x
2

,
.
.
.
,

x
n

⟩

{\displaystyle \boldsymbol{A} = \langle x\_1,x\_2,...,x\_n \rangle }

, the norm is defined as

A

=

x
1

2

x
2

2

.
.
.

x
n

2

{\displaystyle |\boldsymbol{A}| = \sqrt{{x\_1}^{2} + {x\_2}^{2} + ... + {x\_n}^{2}} }

.

**Part 1:** Show that the square of the norm of a vector in N-dimensional space is equal dot-product of the vector onto itself.

**Part 2:** Consider the 10-dimensional vector

# A
⟨
3
,
1
,
4
,
1
,
5
,
9
,
2
,
6
,
5
,
3
⟩

{\displaystyle \boldsymbol{A} = \langle 3,1,4,1,5,9,2,6,5,3 \rangle }

. Calculate the norm of this vector.

# A
⟨

x
1

,

x
2

,

x
3

,
.
.
.
,

x
n

⟩

{\displaystyle \boldsymbol{A} = \langle x\_1,x\_2,x\_3,...,x\_n \rangle }

A
⋅
A

=

x
1

⋅

x
1

x
2

⋅

x
2

.
.
.

x
n

⋅

x
n

{\displaystyle \boldsymbol{A \cdot A} = x\_1 \cdot x\_1 + x\_2 \cdot x\_2 + ... + x\_n \cdot x\_n }

A
⋅
A

=

x
1

2

x
2

2

.
.
.

x
n

2

{\displaystyle \sqrt{\boldsymbol{A \cdot A}} = \sqrt{{x\_1}^{2} + {x\_2}^{2} + ... + {x\_n}^{2}} }

Therefore,

A

2

=

A
⋅
A

{\displaystyle |\boldsymbol{A}|^2 = \boldsymbol{A \cdot A} }

.
**Part 2**

# A
⟨
3
,
1
,
4
,
1
,
5
,
9
,
2
,
6
,
5
,
3
⟩

{\displaystyle \boldsymbol{A} = \langle 3,1,4,1,5,9,2,6,5,3 \rangle }

A

=

3
2

1
2

4
2

1
2

5
2

9
2

2
2

6
2

5
2

3
2

{\displaystyle |\boldsymbol{A}| = \sqrt{3^2+1^2+4^2+1^2+5^2+9^2+2^2+6^2+5^2+3^2} }

A

=

207

{\displaystyle |\boldsymbol{A}| = \sqrt{207} }
Las Reglas de Ruina

מוויקיפדיה - Books in the Cthulhu Mythos :

"Las Reglas de Ruina (literally "the Laws of Ruin") first appeared in Joseph S. Pulver's novel Nightmare's Disciple. It is a tome written by Philip of Navarre in 1520, a Spanish friar of the 16th century. The book has been translated in English by Professors Theodore Hayward Gates and Pascal Chevillion in 1714 and describes the Great Old One Kassogtha, sister and incestuous bride of Cthulhu. The book also foretells of the coming of a messiah of destruction, who would be born in the western land of the red savage across the great ocean in Columbus' New World, a man that shall set the Great Old One free from her stellar prison. Livia Llewellyn elaborated on this, describing the violent sexual acts committed by Kassogtha worshipers".

הספר נכתב בשנת 1520 על ידי הנזיר המטורף פיליפ מנוורה Philip of Navarre הוא כתב וצייר את הספר תוך שימוש בדם ובצואה.
בפעם הראשונה שראיתי את הספר הוא הדיף ריח חזק של צואה וכך גם היה שנים רבות לאחר מכן, כאשר שוב החזקתי בספר, ריחו לא נחלש אף לא במעט.

למדתי על בשרי כי כל הקורא בספר דבק בו הריח הנתעב ורק רחצה ממושכת בחומץ עם מלח יכלה לריח הנורא.

   כל תמונה בספר גרמה בי לרעד בלתי נשלט ולאיבוד שלוות נפשי. בלילות הרבים שהיו לי לאחר מכן סיוטים אין ספור פקדו את חלומותי. סיוטים על דברים שלא אמורים להיות.

הטירוף ניכר בספר כמעט בכל פינה ופינה, חלק מדפיו חוזרים שוב ושוב, חלקים שלמים בו בלתי מובנים ובלתי קריאים כאילו נכתבו בשפה אשר אבדה בזמן וחלקים בו ברורים כאילו היו איים ספורים של שפיות בתוך הספר, איים בשפה הספרדית.

החלקים בספרדית מדברים על יישות אלוהית המכונה בארבעה שמות the Great Old One Kassogtha במקום אחד The Leviathan of Diseases במקום אחר, Eater-of-Souls במקום נוסף ו Bride of Cthulhu לפני סוף הספר.
Kassogtha מחפשה לטרוף את כל היצירה ולרעב שלה אין סוף. Kassogtha מקבלת את כוחה מהרעב, מחלות, חמדנות ותאווה אשר שם המפתח לכניסתה לעולם.

   הספר מתאר את משרתיה הרבים המכונים בשמות שונים ומשונים. אשר צועדים במקום המכונה "העולם שלייד". משרתים אשר ממתינים להגשים את רצונותיה התובעניים של Kassogtha. חלק מהספר אף כולל הוראות לזימון של אותם משרתים.

חלק מרכזי בספר קרוי בשם הבשורה של המלאך (אנג'ל) והוא מוצג כבשורה גנוסטית. המלאך מתאר שימוש ב"ידע" נסתר כאמצעי לגאולה. הרעיון שעומד מאחורי חלק זה הוא שהעולם הזה הוא עולם רע מיסודו, שנברא ונשלט על ידי ישויות אלוהיות-שטניות נחותות. האל העליון האמיתי הוא טרנסצנדנטלי, ומנוגד מטבעו לעולם הזה. כתוצאה מאסון קוסמי כלשהו נקרעו ניצוצות מהאל העליון ונכלאו בחלק מבני-האדם, שבויי העולם הזה. ניצוצות אלה מהווים את ה"רוח" של אותם אנשים. במצבה הרגיל נמצאת הרוח בתרדמה. על מנת להעירה יש לגלות את הידע, באמצעות התגלות חיצונית. בעזרת הידע תוכל הרוח, לאחר המוות, לפרוץ את מחסומי העולם הזה, ולהתאחד עם ה Eater-of-Souls.

בפרק האחרון בספר ששמו "הבשורה לפי פיליפ", מתאר הכותב כי המשיח המכונה בשם messiah of destruction אשר תוולד במערב בארץ אדומי העור הפראים מעבר לים הגדול. המשיח אשר תשחרר את ה Great Old One ממשכנה ומשנתה. המשיח שתביא לנו את הגאולה המיוחלת.

רק שנים מאוחר יותר למדתי את שם המשיח - וירג'יניה דייר .
Mimi

Full name: Mimi Abigail Rose

Born: March 2, 2004 (age 18)

## Bio

A girl who has 2 job she she likes to do. Mimi has worked in Color Me Crafty with her mom, who is the manager there. But more often, you will often see Mimi host one of her favorite childhood preschool shows of all time, Blue’s Clues. Mimi has always loved Blue’s Clues. She’s had her Blue plush, and would always tell her best friend, Lacey about it, whom she loves the show as much as Mimi! Not only does Mimi love Blue’s Clues, she has always dreamed of being the host of the show one day, and has always dreamed of meeting Blue, and the rest of the game. All of Mimi family, and friends are very supportive of her love for the show. And Mimi has not shown signs of stopping at all.

## Personality

Though Mimi can act like a tomboy sometimes, with her liking for mud, Mimi has shown to be a girly girl, who likes wearing a swimsuit around. Even if it is in an odd spot. But Mimi does try to be careful when wearing it. Much lik her pony friend, Rainbow Dash, Mimi is very loyal to her friends, and would never leave them hanging. She’d always be there to help her friends, or the Blue’s Clues cast. Though she isn’t a little bit like her Loud House friend, Fiona, for taking her job seriously (though Mimi tries not to take it too seriously An day wants to have fun), she is a little more like Rainbow Dash in that regard.

## Additional info

* Occupation: Arts and crafts studio member
* Height: 5'8
* Zodiac sign: Pisces
* Favorite colors: Just about all the colors of the rainbow
* Handedness: Right
* Belly button: Innie
* Ticklish level: 12
* Shoe size: 9
* Toe color: Blue

## Trivia

* Back when she was 5, Mimi and Percy were bitter enemies who did not get along. Since they were kids, they let a gross dare on each other get the best of them, and it wasn't until Lacey and Mitzie helped them realize how childish they were being that they became friends.
* She has been a part of school plays.
* Despite being a very minor character, her best Loud House character friend is Leni’s work friend, Fiona.
* Mimi absolutely loved Blue’s Clues when she was a little girl. In fact, she wants to do her own version of the Nick Jr. series, which would mark her the first ever female host.
Sig-Figs

## **Problem**

    Perform each calculation and express the answers in scientific notation with the correct number of significant figures.

**Part 1**

23.1
2.55
−
# 9.367
?

{\displaystyle 23.1 + 2.55 - 9.367 = ? }

**Part 2**

2500
# 13.5
?

{\displaystyle 2500 + 13.5 = ? }

**Part 3**

237050
×
# 3.41
?

{\displaystyle 237050 \times 3.41 = ? }

**Part 4**

435.00
÷
# 0.03100
?

{\displaystyle 435.00 \div 0.03100 = ? }

## **Solution**

**Part 1**

Line up the decimal places and include the decimal point.

2

3

.

1

2

.

6

−

9

.

3

6

7

{\displaystyle \begin{array}{cccccc}
& 2 & 3 & . & 1 & & \\
+ & & 2 &. &6 & & \\
- & & 9 &. &3 &6 &7 \\
\hline
\end{array} }

Draw a line to the right of the last significant digit in each number.

2

3

.

1

2

.

6

−

9

.

3

6

7

{\displaystyle \begin{array}{cccccc}
& 2 & 3 & . & 1| & & \\
+ & & 2 &. &6| & & \\
- & & 9 &. &3 &6 &7| \\
\hline
\end{array} }

Select the line farthest to the left and extend (disregard the other lines). In this problem we pick the line at the tenths place of 23.1 and 2.6. In other words, the least uncertain digit is the tenths digit. So round each number to the nearest tenths. Add or subtract the numbers.

2

3

.

1

2

.

6

−

9

.

4

1

6

.

3

{\displaystyle \begin{array}{cccc}
& 2 & 3 & . & 1| \\
+ & & 2 &. &6| \\
- & & 9 &. &4| \\
\hline
\end{array} }

The answer in scientific notation is

1.63
×

10
1

{\displaystyle 1.63 \times 10^1 }

.

**Part 2**

Line up the decimal places and include the decimal point.

2

5

0

0

1

3

.

5

{\displaystyle \begin{array}{cccccc}
2 & 5 & 0 & 0 & & \\
+ & & 1 & 3 &. &5 \\
\hline
\end{array} }

Draw a line to the right of the last significant digit in each number.

2

5

0

0

1

3

.

5

{\displaystyle \begin{array}{cccccc}
2 & 5| & 0 & 0 & & \\
+ & & 1 & 3 &. &5| \\
\hline
\end{array} }

Select the line farthest to the left and extend (disregard the other lines). In this problem we pick the line at the hundreds place of 2500. This is the least uncertain digit. So round each number to the nearest hundred. Add the numbers.

2

5

0

0

1

3

.

5

2

5

0

0

{\displaystyle \begin{array}{cccccc}
2 & 5| & 0 & 0 & & \\
+ & \, \, \,|& 1 & 3 &. &5 \\
\hline
2 & 5| & 0 & 0 & &
\end{array} }

The answer in scientific notation is

2.5
×

10
3

{\displaystyle 2.5 \times 10^3 }

.

**Part 3**

The number **23705**0 has five sig-figs. The number **3.41** has three sig-figs. So round the product to three sig-figs.

237050
×
# 3.41
808340.5

{\displaystyle 237050 \times 3.41 = 808340.5 }

237050
×
# 3.41
808
000

{\displaystyle 237050 \times 3.41 = \boldsymbol{808}000 }

The answer in scientific notation is

8.08
×

10
5

{\displaystyle 8.08 \times 10^5 }

.

**Part 4**

The number **435.00** has five sig-figs. The number 0.0**3100** has four sig-figs. So round the product to four sig-figs.

435.00
÷
# 0.03100
14032.25806

{\displaystyle 435.00 \div 0.03100 = 14032.25806 }

435.00
÷
# 0.03100
1403
0

{\displaystyle 435.00 \div 0.03100 = \boldsymbol{1403}0 }

The answer in scientific notation is

1.403
×

10
4

{\displaystyle 1.403 \times 10^4 }

.
Measurement of the Sphere

## **Problem**

Consider a sphere inscribed inside a cylinder of equal height and width.

**Part 1:** Show that the surface area of the sphere is equal to the lateral surface area of the cylinder.

**Part 2:** Derive the formula for the surface area of the sphere.

**Part 3:** Derive the formula for the volume of a sphere using its surface area.

## **Solution**

**Part 1**

Consider a cylindrical segment of height

z

{\displaystyle z}

. Let the length of the arc on the sphere be

w

{\displaystyle w}

. If the height

z

{\displaystyle z}

 and arclength

w

{\displaystyle w}

 are sufficiently small, the arc can be approximated as a tangent line. If we cast the height of the cylinder to the arc of the sphere (approximated as a tangent line), it forms a right triangle that is similar to the large right triangle of hypotenuse

r

{\displaystyle r}

 and half-chord

x

{\displaystyle x}

. Thus, the ratio of the height to the arclength can be determined by similar triangles:

x
r

=

z
w

.

{\displaystyle \frac{x}{r} = \frac{z}{w}.}

Consequently, by cross multiplication we get

w
# x
r
z
.

{\displaystyle wx = rz.}

The area of the cylindrical segment and spherical segment are

A

c
y
l
i
n
d
e
r
.
s
e
g

=
2
π
r
z

A

s
p
h
e
r
e
.
s
e
g

=
2
π
w
x

{\displaystyle \begin{align}
A\_{cylinder.seg} = 2\pi rz \\[5 pt]
A\_{sphere.seg} = 2\pi wx
\end{align}}

and since

x
# w
r
z

{\displaystyle xw = rz}

, the cylindrical segment and the area of the spherical segment are equal. By extending this relation through the entire height of the cylinder, we conclude that the surface area of the sphere must be equal to the lateral surface area of the cylinder.

A

c
y
l
i
n
d
e
r
.
l
a
t

=

A

s
p
h
e
r
e

{\displaystyle A\_{cylinder.lat} = A\_{sphere}}

**Part 2**

Determine the surface area of the sphere by calculating the lateral surface area of the cylinder

A

c
y
l
i
n
d
e
r
.
l
a
t

=
2
π
r
h

{\displaystyle A\_{cylinder.lat} = 2\pi rh}

 where

# h
2
r

{\displaystyle h = 2r }

:

A

s
p
h
e
r
e

=
2
π
r
×
2
r

=
4
π

r
2

.

{\displaystyle \begin{align}
A\_{sphere} &= 2\pi r \times 2r \\[5 pt]
&= 4\pi r^2.
\end{align}}

**Part 3**

The sphere can be transformed into a cone by unfolding the surface of the sphere and flattening it into a circle of area

4
π

r
2

{\displaystyle 4\pi r^2}

. Continue this process of flattening each infinitesimal layer of the sphere and stacking them on top of one another until reaching to the center of the sphere. The contour of this shape is linear (hence a cone) and the center point becomes the apex of the cone at a height

r

{\displaystyle r}

 away from the base. Therefore,

V

s
p
h
e
r
e

=

1
3

×

A

s
p
h
e
r
e

×
r

=

1
3

×
4
π

r
2

×
r

=

4
π

3

r
3

.

{\displaystyle \begin{align}
V\_{sphere} &= \frac{1}{3} \times A\_{sphere} \times r \\[ 5 pt]
&= \frac{1}{3} \times 4\pi r^2 \times r \\[5 pt]
&= \frac{4\pi}{3} r^3.

\end{align}}

This is the three-dimensional analogy to transforming a circle into a triangle of equal area.
Platonic Solids and the Euler Characteristic - Solutions

## **Part 2**

   Construct the following table. Count the number of vertices, edges, and faces for each polyhedron and record them in the table.

## **Part 3**

Let the **Euler characteristic

(
χ
)

{\displaystyle (\chi)}** be the constant that depends on

V
,
E
,

a
n
d

F

{\displaystyle V, E, \mathrm{and} \; F}

, where

V
,
E
,
F

{\displaystyle V, E, F}

 denotes the number of vertices, edges, and faces respectively. Verify that the formula

# χ
V
−
E
# F
2

{\displaystyle \chi = V - E + F = 2}

 holds true for the Platonic solids.

**Tetrahedron**

V
−
E
F

=
χ

4
−
6
4

=
2

{\displaystyle \begin{align}
V - E + F &= \chi \\[5 pt]
4 - 6 + 4 &= 2
\end{align}}

**Hexahedron**

V
−
E
F

=
χ

8
−
12
6

=
2

{\displaystyle \begin{align}
V - E + F &= \chi \\[5 pt]
8 - 12 + 6 &= 2
\end{align}}

**Octahedron**

V
−
E
F

=
χ

6
−
12
8

=
2

{\displaystyle \begin{align}
V - E + F &= \chi \\[5 pt]
6 - 12 + 8 &= 2
\end{align}}

**Dodecahedron**

V
−
E
F

=
χ

12
−
30
20

=
2

{\displaystyle \begin{align}
V - E + F &= \chi \\[5 pt]
12 - 30 + 20 &= 2
\end{align}}

**Icosahedron**

V
−
E
F

=
χ

20
−
30
12

=
2

{\displaystyle \begin{align}
V - E + F &= \chi \\[5 pt]
20 - 30 + 12 &= 2
\end{align}}
Sum of Squares

## **Problem**

In his work *On Spirals*, Archimedes (287 – 212 BC) derived the formula for calculating the sum of consecutive perfect squares. Figure 1 shows the geometric representation of the sum

1
2

2
2

3
2

4
2

5
2

{\displaystyle 1^2 + 2^2 + 3^2 + 4^2 + 5^2}

using Archimedes' method. He was able to derive the formula

∑

# k
1

n

k
2

=

n
(
n
1
)
(
2
n
1
)

6

{\displaystyle \sum\_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}}

Explain Archimedes’ derivation of the sum of consecutive perfect squares using modern algebraic notation.

## **Solution**

Figure 1 represents the equation

3
(

1
2

2
2

3
2

⋯

n
2

# )

n
2

(
n
1
)
(
1
2
3
⋯
n
)
.

{\displaystyle 3(1^2 + 2^2 + 3^2 + \cdots + n^2) = n^2 (n+1) + (1 + 2 + 3 + \cdots + n).}

Since

1
2
3
⋯
# n

n
(
n
1
)

2

{\displaystyle 1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}}

, it follows that

3
(

1
2

2
2

3
2

⋯

n
2

)

=

n
2

(
n
1
)
(
1
2
3
⋯
n
)

3
(

1
2

2
2

3
2

⋯

n
2

)

=

n
(
n
1
)

(

n

1
2

)

1
2

2
2

3
2

⋯

n
2

=

n
(
n
1
)
(
2
n
1

6

.

{\displaystyle \begin{align}
3(1^2 + 2^2 + 3^2 + \cdots + n^2) &= n^2 (n+1) + (1 + 2 + 3 + \cdots + n) \\[5 pt]
3(1^2 + 2^2 + 3^2 + \cdots + n^2) &= {n(n+1)}\left({n+ \frac{1}{2}} \right) \\[5 pt]
1^2 + 2^2 + 3^2 + \cdots + n^2 &= \frac{n(n+1)(2n+1}{6}.
\end{align}
}

Consequently,

∑

# k
1

n

k
2

=

n
(
n
1
)
(
2
n
1
)

6

.

{\displaystyle \sum\_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.}
TCM Algebra 5: Polynomial Equations

**By: Tao Steven Zheng (郑涛)**

## **Opening the Cube**

The study of cubic equations in ancient China began with the extraction of cube roots. The earliest extant cube root problems are found in Chapter 4 of the *Jiuzhang Suanshu*. Each problem involves a given volume in the shape of a cube and the question asks for the side length of the cube. Ancient Chinese mathematicians devised an ingenious algorithm for computing cube roots. the method is aptly called the *kai li fang shu* (開立方術, "method of opening the cube").

[1] 实：开方术的“实”是指被开立方根之数。

*Shi* means the radicand when computing cube roots.

[2] 借一算：又称“借算”，意思是借一枚算筹来表示未知数二次项的系数1。

Also called *jie suan*. From an algebraic point of view, the *jie suan* is a rod that represents the coefficient of the quadratic term.

[3] 超一等：隔一位。

Skipping every other place value (i.e. skip tens, thousands, etc)

[4] 议所得：商议根的第一位得数。

A first estimate of the cube root.

[5] 面：正方形面积的边长。

The side length of the cube.

[6] 通分内子：设整数部分为 ，分子为 ， 分母为 。

Let the whole part be , the numerator be , and the denominator be .

[7] 母不可开者：如果分母不是完全平方数，按照以下的计算方法进行。

If the denominator is not a perfect square, then proceed with the calculation method below.

### **Problem Study 1: Opening the Square**

答曰：一百二十三尺。

Answer: 123 *chi*.

**Solution for *Jiuzhang Suanshu* [4.19]**

The following solution gives a clear illustration of the *kai li fang shu* and its equivalent algebraic representation.

**(1)** Set up the initial problem

x
3

=
1860867

{\displaystyle x^3 = 1860867}

 on the counting board. The radicand (实 *shi*) is 1860867. The marker rod (借算 *jie suan*) is placed in the ones digit.

**(2)** Use the marker rod to make an estimate of the largest digit for the result. The largest digit for this problem is in the hundreds. For the algebraic interpretation, let

# x
100

x
1

{\displaystyle x = 100x\_1}

.

(
100

x
1

)
3

=
1860867

1000000

x
1
3

=
1860867

{\displaystyle \begin{align}
(100x\_1)^3 = 1860867\\[5 pt]
1000000x\_1^3 = 1860867

\end{align}}

**(3)**

**(4)**

**(5)**

**(6)**

**(7)**

**(8)**

**(9)**

**(10)**

**(11)**

**(12)**

**(13)**

**(14)**

**(15)**

**(16)**

**(17)**

**(18)**

**(19)**

## **Cubic Equations**

The ingenious method of the *kai fang shu* was extended to solve quadratic equations of the form

x
2

b
# x
c

{\displaystyle x^2 + bx = c}

.

### **Problem Study 2: Square Town**

答曰：二百五十步。
術曰：以出北門步數乘西行步數，倍之，為實。并出南門步數為從法，開方除之，即邑方。
Answer: 250 *bu*.
Method: Multiply the number of *bu* walked from the north gate by the number of *bu* walked west, then double this result. Let this be the radicand (实 *shi*). Add the number of *bu* walking from the south and north gates, and let this be the (從法 *cong* *fa*). Extract the square root to obtain the side of the city.
Muon Decay

## **Problem**

Muons are generated by interactions of cosmic rays with the upper atmosphere of the Earth.

**Part 1**

Determine the minimum energy needed for a muon to reach sea level. The thickness of the atmosphere is roughly

10
k
m

{\displaystyle 10 km }

. The lifetime in the muon’s rest frame is

2.2
μ
s

{\displaystyle 2.2 \mu s }

. The rest mass of a muon is

105.658
M
e
V

/

c
2

{\displaystyle 105.658 MeV/c^2 }

.

**Part 2**

Determine the speed of the muons as a multiple of the speed of light

c

{\displaystyle c}

.

## **Solution**

**Part 1**

Let

L

{\displaystyle L}

 be the thickness of the atmosphere according to the muon's moving frame.

Hence,

L
′

=
γ
# L
10000
m

{\displaystyle L' = \gamma L = 10000 m }

.
To calculate

L

{\displaystyle L}

, use the lifetime of the muon.

# τ
(
3
×

10
8

m

/

s
)
(
2.2
×

10

−
6

s
# )
660
m

{\displaystyle L = c\tau = (3 \times 10^8 m/s)(2.2\times {10}^{-6} s) = 660 m }

Divide the thickness of the atmosphere with the thickness according to the muon's frame to obtain the gamma factor.

# γ

10000
m

660
m

=
15.15

{\displaystyle \gamma = \frac{10000 m}{660 m} = 15.15 }

Therefore the minimum energy needed for a muon to reach sea level is

E
μ

=
γ

m
μ

c
2

=
(
15.15
)
(
105.658
M
e
V
# )
1601
M
e
V

{\displaystyle E\_\mu = \gamma m\_\mu c^2 = (15.15)(105.658 MeV) = 1601 MeV }

or

E
μ

=
1.6
G
e
V

{\displaystyle E\_\mu=1.6 GeV }

.
**Part 2**

# γ

1

1
−

(

v
c

)

2

{\displaystyle \gamma = \frac{1}{\sqrt{1 - {\left(\frac{v}{c}\right)}^{2}}}}

From **part 1** it is found that

# γ
15.15

{\displaystyle \gamma = 15.15 }

; therefore,

v
c

=

1
−

(

1
15.15

)

2

{\displaystyle \frac{v}{c} = \sqrt{1 - {\left(\frac{1}{15.15}\right)}^{2}} }

# v
0.997819186
c

{\displaystyle v = 0.997819186c }

<
The Holey Cube

**By: Tao Steven Zheng (郑涛)**

A fun problem I came up in my second year at UBC (2015).

## **Problem**

Imagine a cube with a hole drilled through the centers of each face. Each hole is equal in size. We will call this solid the *Holey Cube*.

**Part 1:** If the width of the cube is

s

{\displaystyle s}

 and the radius of each hole is

r

{\displaystyle r}

, determine the volume of the Holey Cube.

**Part 2:** What is the volume if the width of the cube is 10 cm, and the radius of each hole is 3 cm?

## **Hint**

(1) Read these two articles first before attempting this problem:

Volume of the Bicylinder

Volume of the Tricylinder

## **Solution**

**Part 1**

The volume of the Holey Cube is not that simple to determine. It requires a bit of set theory, more specifically the inclusion-exclusion principle.

When the holes are drilled through the center of each face, we can clearly see there are 3 cylinders taken out. But wait! What about the intersections of the cylindrical cavities? Since there are three cylinders (A, B, C), there are three bicylinders (AB, AC, BC) to patch up. But wait! What about the intersection of all three cylinders? Thus, we need to remove a tricylinder (ABC).

Hence, the volume of the Holey Cube is

# V

V

c
u
b
e

−
3

V

c
y
l
i
n
d
e
r

3

V

b
i
c
y
l
i
n
d
e
r

−

V

t
r
i
c
y
l
i
n
d
e
r

{\displaystyle V = {V}\_{cube} - 3{V}\_{cylinder} + 3{V}\_{bicylinder} - {V}\_{tricylinder} }

# V

s
3

−
3

(

π

r
2

s

)

3

(

16

r
3

3

)

−
8

(

2
−

2

)

r
3

{\displaystyle V = s^3 - 3 \left(\pi r^2 s \right) + 3\left(\frac{16r^3}{3} \right) - 8\left(2 - \sqrt{2} \right) r^3 }

# V

s
3

−
3
π

r
2

s
8

2

r
3

.

{\displaystyle V = s^3 - 3\pi r^2 s + 8\sqrt{2} r^3. }

**Part 2**

If

# s
10

c
m

{\displaystyle s = 10 \; cm }

and

# r
3

c
m

{\displaystyle r = 3 \; cm }

, then the volume of the Holey cube is

# V
(
10

)
3

−
3
π
(
3

)
2

(
10
)
8

2

(
3

)
3

{\displaystyle V = (10)^3 - 3\pi (3)^2 (10) + 8\sqrt{2} (3)^3 }

# V
457.2401

c
m

3

.

{\displaystyle V = 457.2401 \; {cm}^{3}. }
Standing Waves

## **Problem**

Consider two travelling waves

y

1

(
t
# )
0.3
sin
⁡
(
4
x
−
3
t
π
)

{\displaystyle y\_{1}(t)=0.3\sin(4x-3t+\pi )}

y

2

(
t
# )
0.3
sin
⁡
(
4
x
−
3
t
−
π
)

{\displaystyle y\_{2}(t)=0.3\sin(4x-3t-\pi )}

Determine the sum of the two waves.

**Hint:** Use the identity

sin
⁡
(
u
)
sin
⁡
(
v
# )
2
sin
⁡

(

u
v

2

)

cos
⁡

(

u
−
v

2

)

{\displaystyle \sin(u)+\sin(v)=2\sin \left({\frac {u+v}{2}}\right)\cos \left({\frac {u-v}{2}}\right)}

## **Solution**

y

1

(
t
)

y

2

(
t
# )
0.3
sin
⁡
(
4
x
−
3
t
π
)
0.3
sin
⁡
(
4
x
−
3
t
−
π
)

{\displaystyle y\_{1}(t)+y\_{2}(t)=0.3\sin(4x-3t+\pi )+0.3\sin(4x-3t-\pi )}

y

1

(
t
)

y

2

(
t
# )
0.3

[

sin
⁡
(
4
x
−
3
t
π
)
sin
⁡
(
4
x
−
3
t
−
π
)

]

{\displaystyle y\_{1}(t)+y\_{2}(t)=0.3\left[\sin(4x-3t+\pi )+\sin(4x-3t-\pi )\right]}

Use the identity

sin
⁡
(
u
)
sin
⁡
(
v
# )
2
sin
⁡

(

u
v

2

)

cos
⁡

(

u
−
v

2

)

{\displaystyle \sin(u)+\sin(v)=2\sin \left({\frac {u+v}{2}}\right)\cos \left({\frac {u-v}{2}}\right)}

.

y

1

(
t
)

y

2

(
t
# )
0.3

[

2
sin
⁡
(
4
x
−
3
t
)
cos
⁡
(
π
)

]

{\displaystyle y\_{1}(t)+y\_{2}(t)=0.3\left[2\sin(4x-3t)\cos(\pi )\right]}

y

1

(
t
)

y

2

(
t
# )
−
0.6
sin
⁡
(
4
x
−
3
t
)

{\displaystyle y\_{1}(t)+y\_{2}(t)=-0.6\sin(4x-3t)}
Thales' Circle Theorem

## **Problem**

Let there be a triangle inscribed in a circle such that the longest side of the triangle forms the diameter of the circle. Show that the triangle is a right triangle.

## **Solution**

Let

A
C

{\displaystyle AC}

 be the diameter of the circle. Construct triangle

A
B
C

{\displaystyle ABC}

 such that

A
C

{\displaystyle AC}

 is the longest side of the triangle and a point

B

{\displaystyle B}

 is placed on the perimeter of the circle such that it forms an inscribed angle that is to the arc of the circle.

Consider the diagram shown below.

Let the centre of of the circle be point

O

{\displaystyle O}

. This means the lines

O
A
,
O
B
,
O
C

{\displaystyle OA, OB, OC}

 are radii. This demonstrates there exists two isosceles triangles

A
O
B

{\displaystyle AOB}

 and

C
O
B

{\displaystyle COB}

. According to the diagram, it can be deduced that if

∠
# A
α

{\displaystyle \angle A=\alpha}

 and

∠
# C
β

{\displaystyle \angle C = \beta}

, then

∠
# B
α
β

{\displaystyle \angle B = \alpha + \beta}

.

Since the sum of the interior angles of a triangle is 180 degrees,

α
(
α
β
)
# β
180

{\displaystyle \alpha + (\alpha+\beta) + \beta = 180 }

2
α
2
# β
180

{\displaystyle 2\alpha + 2\beta = 180 }

2
(
α
β
# )
180

{\displaystyle 2(\alpha + \beta) = 180 }

α
# β
90

{\displaystyle \alpha + \beta = 90 }

This proves that the triangle is a right triangle.
6 Player Blokus

You need:

* One original Blokus set
* One Blokus Duo set (for the orange and purple pieces)

## How to play

1. The rules will be changed so that the score system is removed and the person with the least amount of unit squares in his or her pieces wins.
2. The 20x20 Blokus board from the original set will be used.
3. Each of the 6 players will take turns placing their colored pieces just like a normal 4 player game.
4. At the end of the game, each player will count the number of unit squares from their remaining pieces and the player with the least remaining unit squares wins.
Mean Density of Earth

## **Problem**

The mass of the Earth is

6
×

10

24

{\displaystyle 6 \times {10}^{24} }

 kg and its mean radius is 6400 km. Calculate the mean mass density of Earth in grams per cubic centimeter.

## **Solution**

Mass density is calculated by mass over volume:

# ρ

M
V

{\displaystyle \rho=\frac{M}{V}}

The Earth is approximately a sphere, so the volume of Earth is

# V

4
π
×

(
6400000

m
)

3

3

=
1.098
×

10

21

m
3

{\displaystyle V = \frac{4\pi \times {(6400000 \; m)}^3}{3} = 1.098 \times {10}^{21} \; m^3 }

Thus, the density is

# ρ

6
×

10

24

k
g

1.098
×

10

21

m
3

=
5464

k
g

m
3

{\displaystyle \rho = \frac{6 \times {10}^{24} \; kg}{1.098 \times {10}^{21} \; m^3} = 5464 \; \frac{kg}{m^3} }

Convert the density into grams per cubic centimeter.

5464

k
g

m
3

×

1000

g

1

k
g

×

1

m
3

10
6

c

m
3

=
5.464

g

c

m
3

{\displaystyle \frac{5464 \; kg}{m^3} \times \frac{1000 \; g}{1 \; kg} \times \frac{1 \; m^3}{10^6 \; cm^3} = 5.464 \; \frac{g}{cm^3} }

According to this calculation, the mean mass density is around the lower mantle level. This makes sense because the lower mantle is thickest portion of the Earth!
Zu Chongzhi's Value of π 祖冲之圆周率

**Translated by Tao Steven Zheng (郑涛)**

**《隋書·律曆志》**
古之九数，圆周率三，圆径率一，其术疏舛[1]。自刘歆[2]、张衡[3]、刘徽[4]、王蕃[5]、皮延宗[6]之徒，各设新率，未臻折衷。
宋[7]末，南徐州从事史[8]祖冲之，更开密法，以圆径一亿为一丈， 圆周盈数[9]三丈一尺四寸一分五厘九毫二秒七忽，朒数[10]三丈一尺四寸一分五厘九毫二秒六忽，正数在盈、朒二限之间。密率，圆径一百一十三，圆周三百五十五。约率，圆径七，周二十二。

***Book of Sui* (Treatise on the Calendar)**
According to the nine computational methods of antiquity, the ratio of the circumference to the diameter is 3 to 1. This method is inaccurate. Liu Xin, Zhang Heng, Liu Hui, Wang Fan, Pi Yanzong each found a new ratio, yet could not reach a consensus.
Towards the end of Song, the assistant prefectural governor of southern Xu Zhou, Zu Chongzhi, further devised a precise method. Taking a circle of diameter 10000000 to be 1 *zhang*, the excess value of the circumference is 3 *zhang* 1 *chi* 4 *cun* 1 *fen* 5 *li* 9 *hao* 2 *miao* 7 *hu*, and the deficit value is 3 *zhang* 1 *chi* 4 *cun* 1 *fen* 5 *li* 9 *hao* 2 *miao* 6 *hu*. The actual number lies between the excess and deficit values. The precise ratio is a diameter of 113 to a circumference of 355. The approximate ratio is a diameter of 7 to a circumference of 22.

**【注释】**

[1] 疏舛shū chuǎn：粗疏，因为太小而不准确。

Crude, inaccurate for being too small.

[2] 刘歆（公元前50年 – 公元23年）汉代经学家、天文学家。

Liu Xin (50 BC – 23 AD) scholar of classics and astronomy of the Han dynasty.

[3] 张衡（公元78年 – 139年）汉代文学家、天文学家、发明家。

Zhang Heng (79 – 139 AD) literary scholar, astronomer, and inventor of the Han dynasty.

[4] 刘徽（约公元225年 – 295年）魏晋时期数学家。

Liu Hui (c. 225 – 295 AD) mathematician of the Three Kingdoms and Jin dynasty.

[5] 王蕃（公元228 – 266年）三国时代吴国天文学家、数学家。

Wang Fan (228 – 266 AD) astronomer and mathematicians of the Three Kingdoms period.

[6] 皮延宗（公元5世纪）南北朝数学家。

Pi Yanzong (fl. 5th century AD) mathematician of the Northern and Southern dynasties period.

[7] 宋sòng：南朝宋（公元420年 – 479 年）。

The Liu-Song dynasty (420 – 479 AD).

[8] 从事史cóng shì shǐ：负责主众曹文书的官员。

An official in charge of compiling and managing archives.

[9] 盈数yíng shù：过大数值。

Excess number or excess value.

[10] 朒数nǜ shù：不足数值。

Deficit number or deficit value.

**【今译】**

据古代的九数，圆周与直径的比率是3:1。此算法粗疏。自刘歆、张衡、刘徽、王蕃、皮延宗，各设新圆周率，但未达到折衷。

宋末，南徐州的从事史祖冲之，重新发现精密的圆周算法，以圆10000000的直径为1丈，圆周盈数是3丈1尺4寸1分5厘9毫2秒7忽，朒数是3丈1尺4寸1分5厘9毫2秒6忽。正数落在盈数和朒数二限之间。密率的直径113比圆周355。约率的直径7比圆周22。
山居秋暝（鄭氏中古漢語拼音）

**Classical Chinese transliterated by Tao Steven Zheng (鄭濤)**

## **山居秋暝 【san guu tsiu myeng】**

**王維 【Uang Wi】**

空【kung】山【san】新【sin】雨【uu】後【hyou】，

天【tien】氣【ki】晚【muan】來【lai】秋【tsiuu】。

明【myeng】月【nguat】松【song】間【gan】照【dzio】，

清【tsyeng】泉【dzuen】石【sek】上【dziang】流【liuu】。

竹【dyuk】喧【huan】歸【gui】浣【huan】女【nyo】，

蓮【lien】動【dung】下【ha】漁【ngu】舟【tsiuu】。

隨【suui】意【i】春【tsuun】芳【pyang】歇【hyat】，

王【uang】孫【suon】自【dzi】可【ka】留【liuu】。

## **English Translation**

空山新雨後，kōng shān xīn yǔ hòu
天氣晚來秋。tiān qì wǎn lái qiū
明月松間照，míng yuè sōng jiān zhào
清泉石上流。qīng quán shí shàng liú
竹喧歸浣女，zhú xuān guī huàn nǚ
蓮動下漁舟。lián dòng xià yú zhōu
隨意春芳歇，suí yì chūn fāng xiē
王孫自可留。wáng sūn zì kě liú
After rain the empty mountain
Stands autumnal in the evening,
Moonlight in its groves of pine,
Stones of crystal in its brooks.
Bamboos whisper of washer-girls bound home,
Lotus-leaves yield before a fisher-boat —
And what does it matter that springtime has gone,
While you are here, O Prince of Friends?
Tuning Fork Vibration

## **Problem**

The tine of a tuning fork, when struck, has an amplitude of 0.13 cm. If the frequency of the fork is 210 Hz, what total distance will the tine travel in 1.00 minute?

## **Solution**

Given that the amplitude of the oscillation is

# A
0.13

c
m

{\displaystyle A = 0.13 \, cm }

 and the frequency of the oscillation is

# f
210

H
z

{\displaystyle f = 210 \, Hz }

, the distance traveled by the tine in one cycle is

d

=
4
A

=
4
(
0.13

c
m
)

=
0.52

c
m
.

{\displaystyle \begin{align}
d &= 4A \\[5 pt]
&= 4(0.13 \; cm) \\[5 pt]
&= 0.52 \; cm.
\end{align} }

The period of the oscillation is

T

=

1
f

=

1

210

H
z

=
0.004762

s
.

{\displaystyle \begin{align}
T &= \frac{1}{f} \\[5 pt]
&= \frac{1}{210\; Hz} \\[5 pt]
&= 0.004762 \;s.
\end{align} }

Hence, the number of cycle completed by the fork in 1 minute (60 seconds) is

n

=

60

s

0.004762

s

=

{\displaystyle \begin{align}
n &= \frac{60 \; s}{0.004762 \; s} \\[5 pt]
&= 12600.
\end{align} }

Therefore, the total distance will the tine travel in 1 minute

D

=
n
d

=
12600
×
0.52

c
m

=
6552

c
m
.

{\displaystyle \begin{align}
D &= nd \\[5 pt]
&= 12600 \times 0.52 \; cm \\[5 pt]
&= 6552 \; cm.
\end{align} }
Sums of Intercepts

## **Problem**

Find the sum of the x-intercept and y-intercept of any tangent line to the curve

x

y

=

r

{\displaystyle {\sqrt {x}}+{\sqrt {y}}={\sqrt {r}}}

, where

r

{\displaystyle r}

 is constant.

**Hint**

(1) Use implicit differentiation!

## **Solution**

Derive the implicit function with respect to

x

{\displaystyle x}

:

d

d
x

x

d

d
x

y

=

d

d
x

r

{\displaystyle {\frac {d}{dx}}{\sqrt {x}}+{\frac {d}{dx}}{\sqrt {y}}={\frac {d}{dx}}{\sqrt {r}}}

1
2

x

1

/

2

1
2

y

1

/

2

y
′

=
0

{\displaystyle {\frac {1}{2}}x^{1/2}+{\frac {1}{2}}y^{1/2}y'=0}

.
Solving for

y
′

{\displaystyle y'}

yields

y
′

=
−

y
x

{\displaystyle y'=-{\sqrt {\frac {y}{x}}}}

.
The equation of the tangent line for any point

(

x
0

,

y
0

)

{\displaystyle (x\_0, y\_0)}

 would be

y
−

y

0

=
−

y

0

x

0

(
x
−

x

0

)

{\displaystyle y-y\_{0}=-{\sqrt {\frac {y\_{0}}{x\_{0}}}}(x-x\_{0})}

.
To solve for the y-intercept, set

# x
0

{\displaystyle x = 0}

:

y
−

y

0

=
−

y

0

x

0

(
0
−

x

0

)

{\displaystyle y-y\_{0}=-{\sqrt {\frac {y\_{0}}{x\_{0}}}}(0-x\_{0})}

# y

x

0

y

0

y

0

{\displaystyle y={\sqrt {x\_{0}y\_{0}}}+y\_{0}}

.
To solve for the x-intercept, set

# y
0

{\displaystyle y=0}

:

0
−

y

0

=
−

y

0

x

0

(
x
−

x

0

)

{\displaystyle 0-y\_{0}=-{\sqrt {\frac {y\_{0}}{x\_{0}}}}(x-x\_{0})}

# x

x

0

y

0

x

0

{\displaystyle x={\sqrt {x\_{0}y\_{0}}}+x\_{0}}

.
The sum of the x-intercept and y-intercept is therefore

x
# y

x

0

y

0

2

x

0

y

0

{\displaystyle x+y=x\_{0}+y\_{0}+2{\sqrt {x\_{0}y\_{0}}}}

.
Notice that the square of the original implicit function is also

x
y
2

x
y

=
r

{\displaystyle x+y+2{\sqrt {xy}}=r}

; therefore, the sum of the x-intercept and y-intercept for this curve is just

r

{\displaystyle r}

.
Solution of a Modular Congruence

## **Problem**

**Part 1**

Show that if

a
,
m

{\displaystyle a,m }

 are coprime integers, there exists an integer

s

{\displaystyle s}

 that satisfies

a
s
≡
1

(
mod

m
)

{\displaystyle as \equiv 1 \pmod{m} }

. This integer is called the modular inverse of

(
a
,
m
)

{\displaystyle (a,m) }

.

**Part 2**

Show that if

a
,
m

{\displaystyle a,m }

 are coprime integers, then the solution of

a
x
≡
b

(
mod

m
)

{\displaystyle ax \equiv b \pmod{m} }

 is

x
≡
s
b

(
mod

m
)

{\displaystyle x \equiv sb \pmod{m} }

 where

a
s
≡
1

(
mod

m
)

{\displaystyle as \equiv 1 \pmod{m} }

.

The results in part 1 and part 2 are useful when solving the Chinese remainder problems. Check out Sun Zi's algorithm.

## **Solution**

**Part 1**

If

a
,
m

{\displaystyle a,m }

 are coprime, then

gcd
(
a
,
m
# )
1

{\displaystyle \gcd(a,m) = 1 }

.

Equivalently,

a
s
m
# t
1

{\displaystyle as+mt=1 }

a
# s
1
m
(
−
t
)

{\displaystyle as=1+m(-t) }

a
s
≡
1

(
mod

m
)
.

{\displaystyle as \equiv 1 \pmod{m}. }

**Part 2**

Rewrite the linear congruence

a
x
≡
b

(
mod

m
)

{\displaystyle ax \equiv b \pmod{m} }

 as

a
# x
m
y
b

{\displaystyle ax=my+b }

, where

y

{\displaystyle y}

 is an integer. Since

gcd
(
a
,
m
# )
1

{\displaystyle \gcd(a,m) = 1 }

, then

a
s
m
# t
1

{\displaystyle as+mt=1 }

 for integers

s
,
t

{\displaystyle s, t}

.

Therefore,

# s

1
−
m
t

a

.

{\displaystyle s=\frac{1-mt}{a}. }

Multiply

s

{\displaystyle s}

 to both sides in order to isolate

x

{\displaystyle x}

.

s
(
a
x
)
≡
s
b

(
mod

m
)

{\displaystyle s(ax) \equiv sb \pmod{m} }

(

1
−
m
t

a

)

(
a
x
# )
m
y
s
b

{\displaystyle \left(\frac{1-mt}{a}\right)(ax) = my+sb }

x
−
m
t
# x
m
y
s
b

{\displaystyle x-mtx=my+sb }

# x
m
(
t
x
y
)
s
b

{\displaystyle x=m(tx+y)+sb }

x
≡
s
b

(
mod

m
)
.

{\displaystyle x \equiv sb \pmod{m}. }