alchemy - university of california, irvine 4 2015 @ uci uc irvine math ceo contents meeting 6...

November 4 2015 @ UCI

UC IRVINE MATH CEO

http://www.math.uci.edu/mathceo/

Contents

Meeting 6 Student’s Booklet

Alchemy

1 The Magic Flute

2 Transmutation

3 A Perfect Balance

4 Magnus Opus

5 Growth potion

8:00 pm

9:00 pm

10:00 pm

UCI Math CEO • Meeting 6 (NOV 4, 2015)

Continue the pattern. Draw the magic flute at each missing hour.

1 The Magic Flute

6:00 pm 7:00 pm

4:00pm 5:00 pm3:00 pm

2:00 pm1:00 pm1 The Magic Flute

a

We start building the magic flute at noon and we add a new piece every hour. Can you find out what piece was added at 7 pm? 8 pm? 9 pm?

yes this is the right version of the file

UCI Math CEO • Meeting 6 (NOV 4, 2015) 1 The Magic Flute

Time Numberof dots

Number of blank squares

Ratio DOTS : BLANK SQUARES

Ask your own question:: _________________________

Length of flute

4 pm

5 pm

6 pm 12 6

... ... ... ... ... ...

N pm

b Complete the following table:

We may create new events by using combinations. Here are some examples:

Alchemists were interested in transforming materials. Suppose that we start with a perfect balance of materials: 10 units of copper, 10 units of silver and 10 units of gold (these units may be grams, pounds or any other type). We are doing math, so we can and will allow negative units of material, which is kind of funny!

● 3g : 3 units of silver become gold.● g + 2c : 1 unit of silver becomes gold,

and 2 units of gold become copper.● - g : 1 unit of gold becomes silver.

s

g

c

s

g

afte

r:

Example:the event 3g

befo

re:

g : 1 unit of silver turns into gold.s : 1 unit of copper turns into silver.c : 1 unit of gold turns into copper.

The letters g, c, s will denote the following basic transmutation events, produced by playing notes in the magic flute:

New events

c


2 TransmutationChange theme

2 Transmutation

GOLD

SILVERCOPPER

copper silver

gold

10 7

13

c

s

g

copper silver

gold

10 10

10

Continues...

The Magic Flute can change different objects. But you have to combine the right melodies...

a Let y be the following event: y = 3c + 4s + 5g. Complete the picture for the event y:

d Express the following events using only the letters c and s :

● g + c

● -3g + 2c - s

● -900g

Answer:

Answer:

Answer:

Eliminating g

Event 1

UCI Math CEO • Meeting 6 (NOV 4, 2015) 2 Transmutation

b Consider the following event z: “5 units of copper were turned into gold”.How can you write this event formally (using some of the letters c, s, g)? Can you find more than one answer?

From copper to gold

c

s

g

copper silver

goldc

s

g

copper silver

gold

10 10

10

Answer: z =

c

s

g

copper silver

goldc

s

g

copper silver10 10

The key fact is the following:

g + s + c = 0.This is because if one unit of gold becomes copper, one unit of copper becomes silver and one unit of silver becomes gold, then this is the same as if nothing had happened at all!

From here we can deduce, for example, that g = -s - c. This means that we can always eliminate one letter and express it in terms of the other two

gold10 af

ter:

befo

re:

after

:

befo

re:

C Let z be the following event: z = 3(g + c + s + 4c + 4g). Complete the picture for the event z:

Event 2

c

s

g

Copper Silver

Gold

UCI Math CEO • Meeting 6 (NOV 4, 2015) 2 Transmutation

4 8


This is a scale in perfect balance. Some objects have their weights labeled. Each cube of silver weighs 10. All copper moons weigh the same. All gold coins weigh the same.

Can you find the weight of each gold coin?

3 A Perfect BalanceSilver cube Gold coinCopper moonThe left and right plates share the following

objects: 2 cooper moons, one silver cube and one gold coin. Thus we can remove them and keep the balance.

So we have the following equation (where x represents the weight of a gold coin):

2x + 4 = 10 + 8so x =7.

3 A Perfect Balance

The left and right plates share the following objects: 2 cooper moons, one silver cube and one gold coin. Thus we can remove them and keep the balance.

So we have the following equation (where x represents the weight of a gold coin):

2x + 4 = 10 + 8so x =7. =

UCI Math CEO • Meeting 6 (NOV 4, 2015) 3 A Perfect Balance

Think of equality as a scale: each number represents the weight of a coin. We place coins on the two plates of the scale, so that we have a perfect balance.

=1 75

3

Equation (quick way to express the balance): 1 + 7 = 5 + 3. Or: 1 + 7 = 3 + 5.

We can also use objects of mysterious weight, and place of the scale in a way that leads to a perfect balance.

=1 7

6

1

1

=5

3 8

Equation: 1 + + 7 = + 6 + 1 + 1.

Or, using the letter N for the weight of

1 + N + 7 = N + 6 + 1 + 1.

Working BackwardsSuppose that María knows that each weighs 9, but she does not tell Edwin. Instead, she shows him this scale:

=274

4

Equation: N + N + N + 4 = 27 + 4


Equality and Comic Solve Another example with a mystery object

Explanation: Equality and Comic Solve

Ex1:

Ex2:

Equation: + + 5 + 3 = + + 8.

Or, using a letter: 2K + 5 + 3 = 2K + 8, where 2K means K + K.

Ex3:

Ex4:

3 5

7

1

14

46

1 7

5

3 8

27

1

To solve this problem, Edwin will use the method of Comic Solve. His goal is to solve the mystery of finding the weight of each . He will write a comic in which he tells a story, and in the end he must discover the weight of each .

Here is how Edwin does it:

=274

4

=27

Three cubes plus 4 weighs the same as 27 plus 4

Remove the “4” from both plates keeps balance

= 9

Break the 27 into three equal pieces (there are three cubes)

9 9

=

So each cube must weight 9

9

Verification step: Replace each with 9 in the first scale to see if it is balanced

=274

49 9 9

UCI Math CEO • Meeting 6 (NOV 4, 2015) Explanation: Equality and Comic Solve

4

999

94

999

4 2727

274

Maria has 1 box and 16 coins. Edin has 3 boxes and 2 coins. Both have the same number of coins.

Each person spends 2 coins and we keep balance

=

Thus, each box must have 7 coins.

Verification step: Replace each box with 7 in the first scale to see if it is balanced

7

=14

=16

2

Suppose that all the boxes have the

same number of (gold) coins. Maria bought one box and already had 16 coins at home. Edwin bought three boxes, and already had 2 coins at home. Later they discovered that they had the same total number of coins. How many coins were in each box?

Each person gifts 1 box., and we still have balance

=14

=16

2

7 7 7


Ex5:

7

7777

2

2

16

16

14

14

box and 16 coins. Edin has 3 boxes and 2 coins. Both have the same number of coins.

=

Answer:, each box must have 7 coins.

Verification step

=

All the boxes have the same number of tennis balls. Each can has 3 balls. Suppose that Alan buys 2 boxes and 6 cans, and Betty buys 4 boxes and 1 cans (none had tennis balls before). After this, Alan has 1 ball less than Betty. How many balls are in each box?

Each person gifts 1 box., and we still have balance

=

=


d 16 coins. Edin has 3 boxes and 2 coins. Both have the same number of coins.

=

Now create your own Comic Solve!

Ex6:

Among other properties, The Philosopher’s Stone allowed its owner to live forever. Meet the great (and very, very old) alchemists Agathodaemon, Trevisan and Fulcanelli.

These three alchemists claim that they found the Philosopher’s Stone. They seem old enough to support this assertion. Agathodaemon has lived 3 times as much as Trevisan, while Trevisan is 200 years younger than Fulcanelli. However, Fulcanelli points out that he is 50 years older than Agathodaemon.

Can you find the age of each alchemist? To do this, follow the steps in the next pages...

4 Magnum Opus UCI Math CEO • Meeting 6 (NOV 4, 2015)

4 Magnum OpusLet x be the value of a normal potion.

We have the following potions:

x-1, x+2, x, x.Adding we have:4x + 1

Since this quantity fills one quarter of the cauldron and the cauldron holds 100 oz., we conclude that:

4x+1 = 25. Thus 4x = 24 and x=6.

We have 75 oz. more to fill using 6 oz. potions, so we will need 12.5 potions, which means we have to buy 13 potions.

Continues...

Alchemical allegorical glyph of the mathematical process “squaring the circle”. In Paracelsus philosophy, it was related with the maximum alchemists’ quest: The Magnum Opus. This was the Quest of discovering the Philosopher’s Stone.


Let x be the value of a normal potion.




4x+1 = 25. Thus 4x = 24 and x=6.


Step 1: The youngest of the pack

Although all of them being pretty old, is it possible to identify with the given information who is the youngest?

_________________________________________________

Step 2: Pairing

Use the stripes below to assign to each alchemist a symbolic length of their age. In other words:, which alchemist corresponds to each stripe? Put their initials on the blanks.

_______

_______

Recall

Agathodaemon has lived 3 times as much as Trevisan, while Trevisan is 200 years younger than Fulcanelli. However, Fulcanelli points out that he is 50 years older than Agathodaemon.

_______

Continues...


Let x be the value of a normal potion.




4x+1 = 25. Thus 4x = 24 and x=6.


Step 3: A Link to the Past

Find the the length of the each part with a question mark.

Step 4: Solving the problem

Finally, use the represen-tations to find the ageof each alchemist.

Funny bonus: Find which of the alchemists is not old enough to drive a horse carriage...

Recall

Agathodaemon has lived 3 times as much as Trevisan, while Trevisan is 200 years younger than Fulcanelli. However, Fulcanelli points out that he is 50 years older than Agathodaemon.

Wizard Mariano bought the following three growth potions at the “Night Cat” store, each having a certain unknown amount of ounces of a precious liquid known as Bigade. He discovered that one potion had one ounce less of Bigade than usual, while other potion had 2 ounces more than usual. The other two were fine. Next time Mariano will shop at the “Happy Owl”, where all growth potions are guaranteed to have the same amount.

Mariano poured the three potions into a small 100 oz. cauldron to and filled one quarter of it.

How many more potions does he need to buy in order to fill in the cauldron?

5 The Growth Potion UCI Math CEO • Meeting 6 (NOV 4, 2015)

5. The Growth PotionLet x be the value of a normal potion.




4x+1 = 25. Thus 4x = 24 and x=6.


alchemy - university of california, irvine 4 2015 @ uci uc irvine math ceo contents meeting 6...

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