Who Killed Mr

Who killed Mr. X?Who killed Mr. X?

THODORIS ANDRIOPOULOSTHODORIS ANDRIOPOULOS

The investigation of a crime by a The investigation of a crime by a

detective mathematician!detective mathematician!

The story takes place in 1900 in Paris …

… where one of the most significant

mathematics conferences is being held.

At the entrance of the hotel conference

room there is an inscription:

Let no one ignorant of Geometry Let no one ignorant of Geometry Let no one ignorant of Geometry Let no one ignorant of Geometry

enter these doorsenter these doorsenter these doorsenter these doors

Mathematics is the Absolute Truth.

Sooner or later, it can prove whether a theory is right or wrong or it can characterize a

sentence true or false.

From the podium the famous

mathematician, Mr. X, states at the

closing of his speech:

The work of the first day of the

conference is done.

Mr. X is alone in the dining room

reading.

A waiter enters … We must know the truth

The truth is we have to close the

bar. Do you need

anything?

Mr. X asks for a glass of water and the waiter

leaves to bring it.

The mathematicians are totally crazy

The waiter returns.

Mr. X is dead…

The waiter is interrogated by detective

Kurt. He describes the preceding scene

and adds that one second before he

entered the dining room for the second

time, he heard the central hotel clock

striking.

The detective confirms that the

murderer had 20 seconds at his disposal

to commit the crime, which is the time

between the waiter’s exit and his re-

entrance to the dining room.

Mr. Kurt talks with his assistant.

Let us assume that the murderer committed the crime immediately after the waiter left the dining room. That

gave him 20 seconds to move around until the clock struck.

I believe the murderer was walking as he left the dining room in order not to

look suspicious, so, estimating someone walks one meter per second, then the murderer could not have

been more than 20 meters away from the scene of the crime when the clock

struck.

Detective Kurt questions the suspects…

Here is the hotel ground plan

S3

S5 S4

S2

S1

M

Α

D

Β

Η

Ζ

C

Ρ

W

Ν

Ι

J

Ο

L

ΚF

V

Q

G Ε

U

After the first suspects S1, S2, S3, S4, and S5 testify, their positions are

placed on a diagram (a ground plan of the hotel), where M is the spot

where the murder took place.

The suspects could only move on the lines that represent the hotel

corridors shown below.

EVIDENCEMr. Karl Friedrich (S1) states that he was at point A when

the clock struck.

2w 9−

3w 1−

w 8+

2w 2w 20− −

S1

Α

ΒC

DM

SOLUTION-ANSWER

The triangles ABC and ADM are similar soC C

M DM

Α Β=

Α

2 2

w 8 3w 1

w w 12 w 9

+ −=

− − −

KNOWN

ELEMENTS:

ΒC // DM

ΑC=w+8

ΒC=3w-1

CM=w2-2w-20

DM=w2-9

ΑM=30m, so Mr. Karl Friedrich cannot be the murderer.

and the root is w=7 (We don’t accept the root w=2 since it produces DM= -5)

w 8 3w 1or

(w 4)(w 3) (w 3)(w 3)

+ −=

− + − +

w 8 3w 1or

w 4 w 3

+ −=

− −( ) ( )or w 8 (w 3) 3w 1 (w 4)+ − = − −

2or 2w 18w 28 0− + =

SOLUTION-ANSWER

EVIDENCEMr. Constantin (S2) states that

he was at point H when the clock struck.x+6y-8

3x-y-3

2x+y+5

x+2y+10S2

Ζ

D

Η

M

G Ε

ΗG=DΖ and ΗM=ΕΖ

3x-y-3= x+6y-8 and

2x+y+5=x+2y+10

x=8 and y=3

ΗM=24m , so Mr. Constantin cannot be the murderer.

KNOWN ELEMENTS:

DΖ= x+6y-8

ΕΖ=x+2y+10

ΗG=3x-y-3

ΗM=2x+y+5

oZ H 90∧ ∧

= =

2x-7y= -5 and

x-y=5

2x-7y= -5 and

x=y+5

2(y+5)-7y= -5 and

x=y+5

2y+10-7y= -5 and

x=y+5

-5y= -15 and

x=y+5

The triangles ΗGM and DΕΖ are equal so

Let us calculate the distance KF+FM.

KL//JΙ since K and L are the mid-points of the sides NI and NJ in triangle JNI.

F is the mid-point of NO since in the triangle ONI K is the mid-point of NI and KF//OI.

We have since K, F and L are mid-points of NI, NO and NJ.

NF=KF=LF since KNL is a right-angle triangle and NF is the median.

Therefore: KF+FM=NF+FM=NM=JΙ=24 (The diagonals of a rectangle are equal).

We must prove that the shortest distance from point K to point M is KF+FM=NM and not KI+IM.

NM= from the Pythagorean Theorem in triangle MNI.

It is sufficient to prove that KI+IM>NM, or

We square both sides,

which is a valid operation, since is given b>a.

The shortest distance that Mr. Isaac could have covered is KF+FM=24m, so he cannot be the murderer.

EVIDENCEMr. Isaac (S3) states that he was at

point K when the clock struck.

KNOWN

ELEMENTS:

ΚΝ=ΚΙ

LΝ=LJ

ΟΙ=ΟJ=12

a<b

SOLUTION-ANSWER

L

S3

b b

a

a

12

12

J

Ο

M

Ν

FΚ

Ι

2 2(2a) (2b)+

2 2a 2b (2a) (2b)+ > +2 2 2 2 2 2 2 2 3

(a 2b) (2a) (2b) a 4ab 4b 4a 4b 4ab 3a b a ,4

+ > + ⇔ + + > + ⇔ > ⇔ >

KF FL2

ΟΙ= =

G Ε

D

24

Η

M

40

Ρ

Χ

Τ

R

Y

KNOWN

ELEMENTS:

DM=GΕ=40

MΗ=24

90∧

οΗ =

The detective’s assistant cries out:

Mr. Leonhard was at point M, so he is the murderer!

EVIDENCEMr. Leonhard (S4) states that when the clock struck he was in corridor MD at a point such that the distance from corridor ΡD plus the distance from

corridor MΡ was equal to 24m.

SOLUTION-ANSWER

We assume point R on the side MD.

We take the heights RT to MH and RY to ΡD.

RY=TH (1), since RYHT is a rectangle.

We take the height RX to MP.

The triangles RMΧ and RTM are equal, since they are both right, they have a common side RM and

angles , since angles (the triangle MPD is isosceles) and angles

(corresponding angles on RT//DP).

So, RX=TM (2) and from (1) and (2) we have: RX+ RY=ΤM+ΤΗ=MH=24.

This means that the point R can be any point on the side MD, therefore we cannot conclude whether Mr.

Leonhard is guilty or innocent, since he could have been either to the right of the mid-point of MD or to

the left of the mid-point of MD.

RMX TRM∧ ∧

= RMX RDY∧ ∧

= TRM RDY∧ ∧

=

EVIDENCEMr. René (S5) states that when the clock

struck he was at point V and if the rectangle MVQW had an area equal to four times it’s actual area and remained similar to the initial rectangle, then the

distance from point M would have been 60m.

The detective’s assistant cries out: One fourth of 60 is 15, so Mr.

René was at a distance 15m from point M, so he is the murderer!

S5M

W

V

Q

SOLUTION-ANSWER

Let E be the actual area of the rectangle and E΄ the area of the similar rectangle

then its true that

since the ratio of the areas of two similar shapes is equal to the square of the ratio

of their sides.

So, Mr. René cannot be the murderer.

2 2E d 1 d 1 d

o r o r an d fin a lly, d = 3 0 m4 E 6 0 4 6 0 2 6 0

= = =

2E d

E d

= ′ ′

The function d(x) has a minimum value.

Let us calculate this value:

The discriminant is ∆= -200

The minimum value of d(x) is dmin= 254

∆− =

αSo, Mr. Pierre cannot have been the murderer since his minimum distance

from point M was 25m.

SOLUTION-ANSWER

The detective’s assistant cries out:

We can’t find the answer! We have to calculate the value of the function for an

infinite number of values of x.

EVIDENCEMr. Pierre (S6) states that his distance from point M when the clock struck

is given by the function d(x)=2x2-12x+43 for an appropriate value of x

5d 102.5

2d 4

+=

+

So, the equation is indeterminate.

Then, the distance could be 0 or any positive number.

Therefore, we cannot conclude whether Mr. Blaise is guilty or innocent.

SOLUTION-ANSWER

If d is the distance then it holds that:

or 5d 10 2.5(2d 4)+ = + or 5d 10 5d 10+ = +

or 0 d 0.⋅ =

The detective’s assistant cries out:

The answer is 0 then Mr. Blaise was at point M, so he is the murderer!

EVIDENCEMr. Blaise (S7) states: “You can find my distance from point M when the

clock struck, if you know that five times this distance increased by 10 and the whole thing divided by twice this distance increased by 4,

is equal to 2.5m.I know this statement is too long but I didn’t have time to make it shorter”

The assistant informs the detective

that an employee heard

Mr. Pheidias (S8) saying to someone:

“Do as I tell you, and

you will be rewarded with gold.”

The detective calls in Mr. Pheidias to

give evidence.

EVIDENCE

What Mr. Pheidias (S8) said was: “Separate a line segment 10cm

in length into two parts, one with length x and one with length

10-x so that: x2=10.(10-x). Then calculate the ratio

Do as I tell you, and you will be rewarded with gold.”

SOLUTION-ANSWER

The equation becomes x2+10x-100=0,

whose positive root is the number

Then, , which is the number φ of the Golden Ratio.

Any piece of artwork containing this number offers us the sense of harmony

and beauty.

Therefore, the gold that Mr. Pheidias promised was the “golden number” φ,

so we cannot consider him guilty.

5( 5 1)−

10 5 1

x 2

+=

10.

x

EVIDENCEMr. Evarist (S9) states that he knows who killed Mr. X.

He knows, because they both come from France.

“Mr. Pierre (S6) lied to you. He killed Mr. X!

I know the French very well, and they are all liars.”

SOLUTION-ANSWER

Assistant: If Mr. Evarist is telling the truth, then Mr. Pierre lied

to us and…

Detective: One moment. If Mr. Evarist is telling the truth that

the French always lie, then since Mr. Evarist is also French,

he is also lying that he knows the killer.

Assistant: So Mr. Evarist is lying.

Detective: If Mr. Evarist is lying about the French being liars,

then the French tell the truth and so Mr. Evarist, as a

Frenchman, is telling the truth.

Assistant: If Mr. Evarist is telling the truth, then… he is lying,

on the other hand, if he is lying, then… he is telling the truth.

Detective Kurt confirms the conclusion with a grimace and,

putting his hands to his temples, he thinks while staring out

of the window.

Assistant: After all mathematics

doesn’t have all the answers.

Detective: What did you say?

Assistant: I said Mathematics can’t

solve all problems.

Suddenly the detective’s face lights up

and he mumbles to himself as he

leaves…

That killed him!!!

Everyone is gathered in the conference room.

Detective Court explains:

In the last lines of his notes, Mr. X wrote:Sentence A: “Mathematics can’t prove

sentence A”

If the above sentence is characterized as true, then its meaning is corroborated. So the

conclusion is mathematics can’t prove a true sentence.

However, if sentence A is characterized as false that means mathematics can prove a

false sentence, which is not acceptable.

Conclusion: If sentence A is true mathematics can’t prove it.

Mathematics is the Absolute Truth.

Sooner or later, it can prove whether a theory is right or wrong or it can characterize a sentence true or

false.

It seems that Mr. X found out that

Mathematics is not complete,

in other words, that there will always be sentences or theories for which we cannot

determine whether they are true or false.

Mr. X dedicated his life tο the quest for truth and

when it was revealed to him, it took his own life.

The truth killed Mr. X!!!

THE THEORY

OF INCOMPLETENESS,

as proven by Kurt Gödel in 1931,

is unfortunately…

REAL!!!

The above story is thankfully… imaginary!

The Protagonists who took part unintentionally are:

Detective Kurt Kurt Gödel

Austrian Mathematician (1906-1978)

Mr. X David Hilbert

German Mathematician (1862-1943)

Suspect Karl Friedrich (S1) Karl Friedrich Gauss

German Mathematician (1777-1855)

Suspect Constantin (S2) Constantin Carathéodory

Greek Mathematician (1873-1950)

Suspect Isaac (S3) Isaac Newton

British Mathematician (1642-1727)

Suspect Leonhard (S4) Leonhard Euler

Swiss Mathematician (1707-1783)

Suspect René (S5) René Descartes

French Mathematician (1596-1650)

Suspect Pierre (S6) Pierre de Fermat

French Mathematician (1601-1665)

Suspect Blaise (S7) Blaise Pascal

French Mathematician (1623-1662)

Suspect Pheidias (S8) Pheidias

Greek Sculptor (498 BC – 432 BC)

Suspect Evarist (S9) Evarist Galois

French Mathematician (1811-1832)

THODORIS ANDRIOPOULOS

Project Design

This project is dynamic in design. The following

adjustable parameters can be manipulated in each

application by the teacher:

� Curricular Topics

� Level of Difficult

� Time Duration

� Application Area (Mathematics, Physics, …)

It can be easily adopted for all grade levels and

student abilities

This presentation constitutes a 4 hour review of the 3rd grade junior high school mathematics curriculum, employing a different and hopefully interesting way of teaching, as conducted at

the end of the academic year at the Anatolia College of Thessaloniki.

The topics that are included in this presentation are:

Similar triangles InequalitiesFractional equations IdentitiesQuadratic equations Angles between parallel linesEquality of triangles Isosceles trianglesTheory of parallelograms Areas ratio of similar shapesSystem of equations Max-min points of parabolasTheory of mid-points of triangles Indeterminate equationsMedian of a right triangle The Golden ratio or Golden MeanThe Pythagorean Theorem LogicSquare roots A little… Math history