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Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
Who Killed Mr
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Who Killed Mr

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  • @andriopoulos Some sample pages here:
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  • 1. Who killed Mr. X? The investigation of a crime by a detective mathematician! THODORIS ANDRIOPOULOS
  • 2. The story takes place in 1900 in Paris …
  • 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
  • 9. The waiter returns.
  • 10. Mr. X is dead…
  • 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.
  • 15. Detective Kurt questions the suspects…
  • 16. Here is the hotel ground plan
  • 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.
  • 32. Conclusion: If sentence A is true mathematics can’t prove it.
  • 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.
  • 36. The truth killed Mr. X!!!
  • 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)
  • 41. Suspect Pheidias (S8) Pheidias Greek Sculptor (498 BC – 432 BC) Suspect Evarist (S9) Evarist Galois French Mathematician (1811-1832) THODORIS ANDRIOPOULOS
  • 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

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