3. … where one of the most significant
mathematics conferences is being held.
At the entrance of the hotel conference
room there is an inscription:
4. Let no one ignorant of Geometry
enter these doors
5. From the podium the famous
mathematician, Mr. X, states at the
closing of his speech:
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.
6. The work of the first day of the
conference is done.
Mr. X is alone in the dining room
reading.
7. We must
A waiter enters …
know the
truth
The truth is
we have to
close the
bar. Do you
need
anything?
8. Mr. X asks for a glass of water and the waiter
leaves to bring it.
The
mathematicians
are totally crazy
11. 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.
12. 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.
13. 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.
14. 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.
17. 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.
G Ε
S2
Η
Ν L J
Ρ Ζ
F
Κ
S3 Ο
S5 S4
V D
Ι
M
C Β
Α
S1
Q W U
18. M w2 − 9 D
w 2 − 2w − 20
KNOWN
3w − 1 EVIDENCE ELEMENTS:
C w +8 Β
Mr. Karl Friedrich (S1) states ΒC // DM
Α that he was at point A when ΑC=w+8
S1
the clock struck. ΒC=3w-1
CM=w2-2w-20
SOLUTION-ANSWER DM=w2-9
ΑC ΒC
The triangles ABC and ADM are similar so =
ΑM DM
w +8 3w − 1 w +8 3w − 1
= 2 or =
w 2 − w − 12 w − 9 (w − 4)(w + 3) (w − 3)(w + 3)
w + 8 3w − 1
or = or ( w + 8) (w − 3) = ( 3w − 1) (w − 4)
w −4 w −3
or 2w2 −18w + 28 = 0
and the root is w=7 (We don’t accept the root w=2 since it produces DM= -5)
ΑM=30m, so Mr. Karl Friedrich cannot be the murderer.
19. G Ε
3x-y-3 KNOWN
S2 EVIDENCE ELEMENTS:
Η
x+2y+10
2x+y+5
Mr. Constantin (S2) states that DΖ= x+6y-8
he was at point H when the
Ζ ΕΖ=x+2y+10
x+6y-8 clock struck.
ΗG=3x-y-3
M D
ΗM=2x+y+5
∧ ∧
SOLUTION-ANSWER Z = H = 90o
The triangles ΗGM and DΕΖ are equal so ΗG=DΖ and ΗM=ΕΖ
3x-y-3= x+6y-8 and 2x-7y= -5 and 2x-7y= -5 and
2x+y+5=x+2y+10 x-y=5 x=y+5
2(y+5)-7y= -5 and 2y+10-7y= -5 and -5y= -15 and
x=y+5 x=y+5 x=y+5
x=8 and y=3
ΗM=24m , so Mr. Constantin cannot be the murderer.
20. Ν b L b J
KNOWN
EVIDENCE ELEMENTS:
a
Κ S
F 12 Mr. Isaac (S3) states that he was at ΚΝ=ΚΙ
3 Ο point K when the clock struck. LΝ=LJ
a 12 ΟΙ=ΟJ=12
Ι M a<b
SOLUTION-ANSWER
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.
ΟΙ
We have KF = = FL since K, F and L are mid-points of NI, NO and NJ.
2
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= (2a) + (2b) from the Pythagorean Theorem in triangle MNI.
2 2
It is sufficient to prove that KI+IM>NM, or a + 2b > (2a) 2 + (2b) 2
3
We square both sides, (a + 2b)2 > (2a)2 + (2b)2 ⇔ a 2 + 4ab + 4b 2 > 4a 2 + 4b 2 ⇔ 4ab > 3a 2 ⇔ b > a ,
4
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.
21. 40 Ε
G
EVIDENCE KNOWN
ELEMENTS:
Η Mr. Leonhard (S4) states that
DM=GΕ=40
Ρ when the clock struck he was MΗ=24
Χ
24 in corridor MD at a point such ∧
Τ Y Η = 90ο
that the distance from corridor
ΡD plus the distance from
M R D corridor MΡ was equal to 24m.
The detective’s assistant cries out:
Mr. Leonhard was at point M, so he is the murderer!
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 RMX = TRM , since angles RMX = RDY (the triangle MPD is isosceles) and angles TRM = RDY
(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.
22. S5
V M EVIDENCE
Mr. 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
Q W 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!
SOLUTION-ANSWER
Let E be the actual area of the rectangle and E΄ the area of the similar rectangle
2
E d
then its true that =
E′ d′
since the ratio of the areas of two similar shapes is equal to the square of the ratio
of their sides.
2 2
E d 1 d 1 d
= or = or = a n d fin a lly, d = 3 0 m
4E 60 4 60 2 60
So, Mr. René cannot be the murderer.
23. EVIDENCE
Mr. 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
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.
SOLUTION-ANSWER
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= − = 25
4α
So, Mr. Pierre cannot have been the murderer since his minimum distance
from point M was 25m.
24. EVIDENCE
Mr. 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 detective’s assistant cries out:
The answer is 0 then Mr. Blaise was at point M, so he is the murderer!
SOLUTION-ANSWER
If d is the distance then it holds that:
5d + 10
= 2.5 or 5d + 10 = 2.5(2d + 4) or 5d + 10 = 5d + 10
2d + 4
or 0 ⋅ d = 0.
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.
25. 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.
26. 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
10-x so that: x2=10.(10-x). Then calculate the ratio x .
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 5( 5 − 1)
Then, 10 = 5 + 1 , which is the number φ of the Golden Ratio.
x 2
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.
27. EVIDENCE
Mr. 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.
28. 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!!!
29. 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”
30. If the above sentence is characterized as true,
then its meaning is corroborated. So the
conclusion is mathematics can’t prove a true
sentence.
31. However, if sentence A is characterized as
false that means mathematics can prove a
false sentence, which is not acceptable.
33. 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.
34. 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.
35. Mr. X dedicated his life tο the quest for
truth and
when it was revealed to him, it took his
own life.
37. The above story is thankfully… imaginary!
THE THEORY
OF INCOMPLETENESS,
as proven by Kurt Gödel in 1931,
is unfortunately…
REAL!!!
38. 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)
39. 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)
40. 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)
42. 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
43. 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 Inequalities
Fractional equations Identities
Quadratic equations Angles between parallel lines
Equality of triangles Isosceles triangles
Theory of parallelograms Areas ratio of similar shapes
System of equations Max-min points of parabolas
Theory of mid-points of triangles Indeterminate equations
Median of a right triangle The Golden ratio or Golden Mean
The Pythagorean Theorem Logic
Square roots A little… Math history
