Ozymandias Posted September 6, 2020 #326 Share Posted September 6, 2020 (edited) Thanks for that, Anton. I forgot about the redundant subtraction of the zero should have inserted back into my solution. Here is my next problem based on geometry for a change. Problem: A rectangle whose long side is 2.4 times the length of its short side is inscribed in a circle of diameter 130m (see below). What is the area inside the circle but outside the rectangle? Edited September 6, 2020 by Ozymandias 1 Top Link to comment Share on other sites More sharing options...

ant0n Posted September 6, 2020 #327 Share Posted September 6, 2020 49 minutes ago, Ozymandias said: Thanks for that, Anton. I forgot about the redundant subtraction of the zero should have inserted back into my solution. Here is my next problem based on geometry for a change. Problem: A rectangle whose long side is 2.4 times the length of its short side is inscribed in a circle of diameter 130m (see below). What is the area inside the circle but outside the rectangle? (zero was originally missing in the official answer and I couldn't stand zero got skipped that way. So I added that *important* substraction. Whenever we're asked to count to 10, we're supposed to start from zero, right? ^^ I cherish zero. I even regularly list things from the zeroth: 0th, 1st, 2nd, etc) Link to comment Share on other sites More sharing options...

ant0n Posted September 6, 2020 #328 Share Posted September 6, 2020 (edited) 1 hour ago, Ozymandias said: Thanks for that, Anton. I forgot about the redundant subtraction of the zero should have inserted back into my solution. Here is my next problem based on geometry for a change. Problem: A rectangle whose long side is 2.4 times the length of its short side is inscribed in a circle of diameter 130m (see below). What is the area inside the circle but outside the rectangle? The diameter is one diagonal of the rectangle. Pythagoras: 130² = x² + (2.4x)² = 6.76x², therefore x² = 2500 m, therefore x (width) = 50 m. Rectangle area = 50 x 2.4 x 50 = 6,000 m² Disk area = pi x (130/2)² = 13,273 m² approximately. Therefore the difference area = 7,273 m² approximately. Edited September 6, 2020 by ant0n 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted September 6, 2020 #329 Share Posted September 6, 2020 33 minutes ago, ant0n said: The diameter is one diagonal of the rectangle. Pythagoras: 130² = x² + (2.4x)² = 6.76x², therefore x² = 2500 m, therefore x (width) = 50 m. Rectangle area = 50 x 2.4 x 50 = 6,000 m² Disk area = pi x (130/2)² = 13,273 m² approximately. Therefore the difference area = 7,273 m² approximately. Nice to see you back, Dan. Correct again, Anton. It's back to you for another problem. 1 Top Link to comment Share on other sites More sharing options...

ant0n Posted September 6, 2020 #330 Share Posted September 6, 2020 (edited) Great While in jail, John was realizing that if he had 3 cigarette ends, he could mix them together to get a full cigarette. He got a cigarette butt each time after smoking a full cigarette. One day, in his cell, John had a conversation with one of his two cellmates, Jim : « I’d really like to smoke 5 cigarettes today but I’ve got only 10 butts. I guess they won’t be enough. » - Jim asked: "Why don’t you steal some butts from our cellmate Mike?", pointing at that little pile of butts on their cellmate’s bed. Mike was out all day because of community service. - - I can’t, John replied. « Mike always carefully counts his remaining butts and he would literally kill me if he noticed I stole some. » However, upon reflexion, John realized there actually was a way to smoke 5 cigarettes without irritating Mike. What was his plan? Edited September 6, 2020 by ant0n Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #331 Share Posted September 7, 2020 15 hours ago, ant0n said: Great While in jail, John was realizing that if he had 3 cigarette ends, he could mix them together to get a full cigarette. He got a cigarette butt each time after smoking a full cigarette. One day, in his cell, John had a conversation with one of his two cellmates, Jim : « I’d really like to smoke 5 cigarettes today but I’ve got only 10 butts. I guess they won’t be enough. » - Jim asked: "Why don’t you steal some butts from our cellmate Mike?", pointing at that little pile of butts on their cellmate’s bed. Mike was out all day because of community service. - - I can’t, John replied. « Mike always carefully counts his remaining butts and he would literally kill me if he noticed I stole some. » However, upon reflexion, John realized there actually was a way to smoke 5 cigarettes without irritating Mike. What was his plan? My solution: John has 10 butts, that will allow him to make 3 cigarettes with 1 butt over. He smokes those 3 cigarettes giving him 3 new butts plus the one he already had = 4 butts. He makes a fourth cigarettes using 3 of these 4 butts and smokes it , leaving him with 2 butts. To make a fifth cigarette he borrows a butt from his cellmate's pile. That allows him to roll and smoke a fifth cigarette. When he is finished he has a butt left which he returns to his cellmates pile leaving him non the wiser. Job done! 1 Top Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #332 Share Posted September 7, 2020 (edited) 6 minutes ago, Ozymandias said: My solution: John has 10 butts, that will allow him to make 3 cigarettes with 1 butt over. He smokes those 3 cigarettes giving him 3 new butts plus the one he already had = 4 butts. He makes a fourth cigarettes using 3 of these 4 butts and smokes it , leaving him with 2 butts. To make a fifth cigarette he borrows a butt from his cellmate's pile. That allows him to roll and smoke a fifth cigarette. When he is finished he has a butt left which he returns to his cellmates pile leaving him non the wiser. Job done! Well done! I was still a smoker 3 3/4 years ago and I know that butt trick. Objectively, over-recycled butts must end up stinking quite bad! Your turn! Edited September 7, 2020 by ant0n Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #333 Share Posted September 7, 2020 (edited) 1 hour ago, ant0n said: I was still a smoker 3 3/4 years ago and I know that butt trick. Objectively, recycled butts must end stinking quite bad! I was a heavy smoker - 20 to 40 per day depending on what I was doing - until the millennium New Year's Eve, 31 Dec 1999. Haven't smoked since. Problem: 5 numbers are generated in sequence following the same rule. They are: X 192 Y 108 Z Find X, Y and Z. Edited September 7, 2020 by Ozymandias Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #334 Share Posted September 7, 2020 3 minutes ago, Ozymandias said: I was a heavy smoker - 20 to 40 per day depending on what I was doing - until the millennium New Year's Eve, 31 Dec 1999. Haven't smoked since. Problem: 5 numbers are generated in sequence following the same rule. They are: X 192 Y 108 Z Find X, Y and Z. 234 192 150 108 66 X Y Z I guess so. 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #335 Share Posted September 7, 2020 (edited) 10 minutes ago, ant0n said: 234 192 150 108 66 X Y Z I guess so. No, Anton. Not the answer I am expecting. How did you get your numbers? PS - I pressed the 'like' button by accident on post #338. Meant to click on 'confused'. Edited September 7, 2020 by Ozymandias Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #336 Share Posted September 7, 2020 1 minute ago, Ozymandias said: No, Anton. Not the answer I am expecting. How did you get your numbers? It is not the answer you're expecting HOWEVER my sequence is valid anyway: substracting 42 each time, from left to right. If you expect some other specific sequence, you have to add conditions. Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #337 Share Posted September 7, 2020 2 minutes ago, ant0n said: It is not the answer you're expecting HOWEVER my sequence is valid anyway: substracting 42 each time, from left to right. If you expect some other specific sequence, you have to add conditions. Yes, I should have stipulated that my progression was geometric. You solved the problem on the basis that it was an arithmetic progression. So, what is X, Y and Z if the sequence is geometric? Sorry about that. 1 Top Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #338 Share Posted September 7, 2020 4 minutes ago, Ozymandias said: Yes, I should have stipulated that my progression was geometric. You solved the problem on the basis that it was an arithmetic progression. So, what is X, Y and Z if the sequence is geometric? Sorry about that. No worries Multiply by 3/4 each step from left to right. 256 192 144 108 Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #339 Share Posted September 7, 2020 1 hour ago, ant0n said: No worries Multiply by 3/4 each step from left to right. 256 192 144 108 Yes, Anton. That's it: X = 256, Y = 144, and Z = 81. Your turn. 1 Top Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #340 Share Posted September 7, 2020 7 minutes ago, Ozymandias said: Yes, Anton. That's it: X = 256, Y = 144, and Z = 81. Your turn. Great Find the number 28 as the result of a calculation involving the numbers 2, 3, 4, and 5, each one of them being used only once (parenthese allowed). Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #341 Share Posted September 7, 2020 (edited) 28 = 7 x 4 = (10 - 3) x 4 = [(2 x 5) - 3] x 4 = 4[2(5) - 3] Edited September 7, 2020 by Ozymandias 1 Top Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #342 Share Posted September 7, 2020 11 minutes ago, Ozymandias said: 28 = 7 x 4 = (10 - 3) x 4 = [(2 x 5) - 3] x 4 = 4[2(5) - 3] Good! Your turn. Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #343 Share Posted September 7, 2020 (edited) Another geometry/trigonometry problem ... Problem: A circle of radius 10cm is inscribed inside a right-angled triangle whose base length is 35cm (see diagram). What is the area coloured black? Edited September 7, 2020 by Ozymandias Link to comment Share on other sites More sharing options...

ant0n Posted September 7, 2020 #344 Share Posted September 7, 2020 I hope I didn't make mistakes... - angle DGH = 90° therefore DGE = EGH = (360 - 90)/2 = 135° - EG = GD = 10 cm - angle FGE = 135/2 = 67.5° = FGD - angle FEG = 180 - ( 90 - 67.5) = 22.5° = FDG - cos22.5° = EF/10 = 0.9239 therefore EF = 9.24 cm = FD - triangle AFD: sin(FAD) = FD/35 = 9.24/35 = 0.264 therefore angle FAD = 15.31° = FAE - cos(FAD) = cos15.31° = AF/AD therefore AF = 0.965 x (35 - 10) = 24.13 cm - area AFD = 1/2 x AF x FD therefore area AED = 24.13 x 9.24 = 222.96 cm² - area FEG = 1/2 x FG x EF; sin(FEG) = FG/10 = sin22.5° therefore FG = 3.83 cm therefore area FEG + area FDG = 2 x 1/2 x 10 x 3.83 = 38.3 cm² - sector of the disk with angle EGD = pi x 10² x 135 / 360 = 117.81 cm² - portion of disk EFD = 117.81 - 38.3 = 79.51 cm² - black part = 222.96 - 79.51 = 143.42 cm² Link to comment Share on other sites More sharing options...

Ozymandias Posted September 7, 2020 #345 Share Posted September 7, 2020 3 hours ago, ant0n said: I hope I didn't make mistakes... - angle DGH = 90° therefore DGE = EGH = (360 - 90)/2 = 135° - EG = GD = 10 cm - angle FGE = 135/2 = 67.5° = FGD - angle FEG = 180 - ( 90 - 67.5) = 22.5° = FDG - cos22.5° = EF/10 = 0.9239 therefore EF = 9.24 cm = FD - triangle AFD: sin(FAD) = FD/35 = 9.24/35 = 0.264 therefore angle FAD = 15.31° = FAE - cos(FAD) = cos15.31° = AF/AD therefore AF = 0.965 x (35 - 10) = 24.13 cm - area AFD = 1/2 x AF x FD therefore area AED = 24.13 x 9.24 = 222.96 cm² - area FEG = 1/2 x FG x EF; sin(FEG) = FG/10 = sin22.5° therefore FG = 3.83 cm therefore area FEG + area FDG = 2 x 1/2 x 10 x 3.83 = 38.3 cm² - sector of the disk with angle EGD = pi x 10² x 135 / 360 = 117.81 cm² - portion of disk EFD = 117.81 - 38.3 = 79.51 cm² - black part = 222.96 - 79.51 = 143.42 cm² Again, Anton. Not the result I was expecting. I think, at the very beginning of your solution, your statement that the angle DGE = EGH = (360 - 90)/2 = 135° is not correct. Using your diagram the 'black' area is equal to twice the area of the triangle ADG (ADG + AEG) minus the circle sector swept out by the angle DGE. Link to comment Share on other sites More sharing options...

ant0n Posted September 8, 2020 #346 Share Posted September 8, 2020 8 hours ago, Ozymandias said: Again, Anton. Not the result I was expecting. I think, at the very beginning of your solution, your statement that the angle DGE = EGH = (360 - 90)/2 = 135° is not correct. Using your diagram the 'black' area is equal to twice the area of the triangle ADG (ADG + AEG) minus the circle sector swept out by the angle DGE. Oh... I thought DGE = EGH... I assumed it was the result of a general law for an inscribed circle in a rectangle triangle... Here, B1IC1 = 90° but B1IA1 =/= A1IC1 indeed: I'm trying to find another solution. Link to comment Share on other sites More sharing options...

Ozymandias Posted September 8, 2020 #347 Share Posted September 8, 2020 3 hours ago, ant0n said: B1IC1 = 90° but B1IA1 =/= A1IC1 indeed: This is correct. They are nor equal. They would only be equal if the sides AB and BC were equal. 3 hours ago, ant0n said: The area required is the are of A1-C-B1-I-A1 (an area twice the size of the triangle A1IB1) less the area of the circle sector swept out by the angle B1IA1. Link to comment Share on other sites More sharing options...

Ozymandias Posted September 8, 2020 #348 Share Posted September 8, 2020 (edited) The area required is the are of A-D-C-B-A (an area twice the size of the triangle ABC = ADC) less the area of the circle sector swept out by the angle DCA. Angle CAD = arctan(10/25) = 21.8deg >> angle ACD = angle ACB = 90deg - 21.8deg = 68.2deg Angle DCB = angle ACD + angle ACB = 2(68.2deg) = 136.4deg Area of Circle Sector swept out by the angle DCB = (136.4/360)(pi)(10^2) = 119 cm^2 Area of triangle ADC = (1/2)(base)(perp.height) = 1/2(25)(10) = 125cm^2 Area of A-D-C-B-A = 2(area triangle ADC) = 2(125cm^2) = 250cm^2 'Black' area = 250cm^2 - 119cm^2 = 131cm^2 Edited September 8, 2020 by Ozymandias Link to comment Share on other sites More sharing options...

ant0n Posted September 8, 2020 #349 Share Posted September 8, 2020 17 minutes ago, Ozymandias said: The area required is the are of A-D-C-B-A (an area twice the size of the triangle ABC = ADC) less the area of the circle sector swept out by the angle DCA. Angle CAD = arctan(10/25) = 21.8deg >> angle ACD = angle ACB = 90deg - 21.8deg = 68.2deg Angle DCB = angle ACD + angle ACB = 2(68.2deg) = 136.4deg Area of Circle Sector swept out by the angle DCB = (136.4/360)(pi)(10^2) = 119 cm^2 Area of triangle ADC = (1/2)(base)(perp.height) = 1/2(25)(10) = 125cm^2 Area of A-D-C-B-A = 2(area triangle ADC) = 2(125cm^2) = 250cm^2 'Black' area = 250cm^2 - 119cm^2 = 131cm^2 Oh no, I was just having the solution. You revealed it too early. Link to comment Share on other sites More sharing options...

ant0n Posted September 8, 2020 #350 Share Posted September 8, 2020 18 minutes ago, Ozymandias said: The area required is the are of A-D-C-B-A (an area twice the size of the triangle ABC = ADC) less the area of the circle sector swept out by the angle DCA. Angle CAD = arctan(10/25) = 21.8deg >> angle ACD = angle ACB = 90deg - 21.8deg = 68.2deg Angle DCB = angle ACD + angle ACB = 2(68.2deg) = 136.4deg Area of Circle Sector swept out by the angle DCB = (136.4/360)(pi)(10^2) = 119 cm^2 Area of triangle ADC = (1/2)(base)(perp.height) = 1/2(25)(10) = 125cm^2 Area of A-D-C-B-A = 2(area triangle ADC) = 2(125cm^2) = 250cm^2 'Black' area = 250cm^2 - 119cm^2 = 131cm^2 Why give a hint if you reveal it all 11 minutes later? I was about to typewrite my solution when I realized you revealed it all 5 minutes earlier. Link to comment Share on other sites More sharing options...

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