Do the other two mean anything? Google Translate says Fugori is falsehood and Orokana just comes back as Orokana. I know it works pretty well, but I’d trust someone with personal knowledge over the program.
@@sabereagle9277 Originally, it was very easy... you just have Fugori follow the first plane and give all of its fuel once it has the same amount remaining as the first plane has empty capacity. Of course this means Fugori crashes and dies. Orokana does similar but the opposite way around the world and with a pause to wait for the right time. This doesn't work when Fugori and Orokana have to survive :) Though, the answer is somewhat similar.
I just realized. In order for the professor to travel around the globe at 1 degree longitude per minute, he'd be travelling roughly 4,150 miles per hour!
probably faster, as that is assuming he is literally dragging his plane on the water the whole time. any elevation above sea level increases the circumference of his journey, thus he must increase his speed to maintain the 1 degree per minute.
Flat earthers are people who don't go to space even once so they still believe about what the 16th's believes (They do have claims, but they don't know the world is too big to their claims).
watch these puzzles, think about them for like one min (so you don't bad about yourself like at least you tried) watch the solutions , laugh at the funny comments ..... Next puzzle please
"now all we need to do is make the lunch not fly everywhere" what you need to do is make it so that passengers don't instantly die from the g force of turning around
@@soso3792 fun fact: if the plane weighs 0.5 kg or higher the force exerted to turn the plane that fast is enough force to break the human femur which is the bodies strongest bone, so all of your bones would instantly break and you would almost definitely immediately die
There are 2 optimizations that can be made to the solution. 1. Fugori can actually travel past 90 degrees with the professor, if Orokana, after refueling, heads back out to help Fugori on the way back to base. 2. Orokana can meet the professor with more fuel at 270 degrees if Fugori accompanies her when moving clockwise, and refills her tank to full after 60 degrees. With this optimization, a fuel tank with capacity 162 kiloliters is sufficient - the professor can be left with a full fuel tank 99 degrees out, and meet up with an assistant as he runs out of fuel, 99 degrees before the end of his trip.
But Orokana and Fugori would take time to reach back to the airport and then take even more time to fly east to meet the prof. By that time prof would be dead
@@thesmartnerd732 Well, technically the prof could go on another 9 degrees to the 279 degree point, wich is exactly what they need to meet up with him since on of the planes left at 99 degrees and 180 degrees to the other direction is 279, the problem with that is that it doesn't offer a solution in wich "meet the professor with more " as the comment said would be possible
@@josephc.9520 Because he's wrong. As TheSmartNerd says, if Orokana helps Fugori return to base, they can't get to the professor in time. Stern Huang doesn't justify his figures of 162kl tank size and 99°-261° for the professor's solo flight, but we don't need to look at those specific numbers to see the problem. Let's assume just a 1°/kl optimization. Professor Fukano begins flying solo at 91° with 179kl fuel. As in the original solution, he will reach 270° and run out of fuel at the 270 minute mark. That's 90 degrees from base; therefore, to meet him at that time, someone must take off 90 minutes prior, at minute 180. But wait! Fugori was with the professor at 91°. A round trip takes 91 minutes for each leg, so Fugori cannot reach base until 182 minutes have elapsed. If Orokana helps Fugori, neither of them can be at the base to take off at minute 180, so the professor will crash. If she's en route to the professor at minute 180, she cannot have met and refueled Fugori, so he will crash. The solution is already optimized.
@@noodle_fc Ah I hadn't thought through this. You know the 3am urge to binge but not think through anything right? Really sorry for taking up your time, I had figured out why a while ago
You don't have to go through security more than once if you stay within the airport lounges, so once they all go through security before taking off at the beginning there's no issue.
But plot twist, Fugori and Orokana were always jealous of the professor, having stolen their design for the plane and giving no credit where credit was due. So when the time to give the Professor his fuel came, the two left with no transfer of fuel. The Professor plummeted to the ground thinking the tubes had malfunctioned, and died. Fugori and Orokana flew back to the airport and found a small party had been thrown together (planned by the professor) with streamers and balloons in honour of the two assistant's help with the project, as well as an offer at a full time job alongside the professor. As the fellow scientists and plane technicians on the ground saw Fugori and Orokana leave their planes, the only question on their minds was "Where is the professor?". The two lied and said that they had not see him where they agreed to meet. Overwhelmed with guilt, Fugori and Orokana took off once more after refueling, claiming they would go back to look for the professor at the spot where he had fallen. They were never heard from again.
No, they were heard from again. They landed back in Tokyo and were met with heavy fines after losing a case after Fukano's family sued them for deliberating getting rid of a loved one's life and so the assistants were fined heavily. They were charged with illegally landing aircraft and neglecting a comrade's life. They were very controversial.
Jacob Griffin But nobody ever saw them again. Did they die? Did they go into hiding? Did they start anew as farmers, cut off from society, to repent for their sins?
I came up with another solution to the riddle. 12:00 → all three planes depart 12:00 - 12:45 → assistant C and B and the Professor fly 45 longitudes from 0°l to 45°l 12:45 → assistant C transfers 2x 45kl, the other two pilots now have a full tank again 12:45 - 13:30 → assistant B and the Professor fly 45 longitudes from 45°l to 90°l 13:30 → assistant B transfers 45kl to the Professors plane, the Professor has now a full tank again 13:30 → assistant B has 90kl fuel left, which is enough to fly the 90 longitudes back home 13:30 → assistant C arrives back home 13:30 - 16:30 → the Professor flies 180 longitudes from 90°l to 270°l 15:00 → assistant B arrives back home 15:00 - 16:00→ assistant C and B fly 60 longitudes from 360°l to 300°l 16:00→ assistant C transfers 60kl to assistant B, assistant B has now a full tank again 16:00→ assistant C has 60kl fuel left, which is enough to fly the 60 longitudes back home 16:00 - 16:30→ assistant B flies 30 longitudes from 300°l to 270°l 16:30 → assistant B transfers half of his tank to the Professor. Both have 75kl fuel 16:30 - 17:45 → assistant B and the professor fly 75 longitudes from 270°l to 345°l 17:00 → assistant C arrives back home 17:30 - 17:45 → assistant C flies 15 longitudes from 360°l to 345°l 17:45 → assistant C transfers a third to assistant B and the Professor, everyone has 55kl fuel 17:45 - 18:00 → assistant C and B and the Professor fly 15 longitudes from 345°l to 360°l 18:00 → all three planes arrive safely back home with 40kl fuel in each tank.
I figured it out but in a differnt way. Let me explain, also I will use the amount of fuel they have in % instead of using the exact amount like shown in the video to make it easier. We know that all of your fuel will take you to just the other side of the planet and you can only land on one spot which is the airport you started from. To accomplish this task ive decided to devide the whole trip around the globe into 10 sections. Every section is exactly 36 degree of longitude (One tenth of the whole trip). Due to our understanding that half the trip will take up 100% of our fuel we can safely assume that every section will take up 20% of our fuel. So basically. 36 deegres of longitude = 1 section = 20% fuel. Now to the actual flight we will (just like in the video) fly every plane west. But instead of stopping at 1/8 (or 1.25 sections) of the total flight you will now stop at 1/5 or exactly 2 section. All the planes should now have 60% of their fuel left and has all used 40% of their fuel to get there. Plane 2 and 3 will now transfer 20% of their fuel to plane 1 (each) which will put plane 1 at 100% fuel while the other planes will have 40% each. The second and third plane will now have exactly enough to fly back to the airport to refuel for their next mission. Meanwhile plane 1 will have just enough fuel after the refuel to go 180 degrees of longitude plus the 72 it has already travelled which puts the total at 252 degrees of longitude or in more simple terms, 7 sections. Meanwhile, plane 2 and 3 has refueled at the airport and plane 2 has gone 3 sections or 108 defrees longitude and has now reached plane 1. With the remaing 40% they split it even and can now go one more section. Meanwhile plane nr 3 has also gone east and meets up with plane 1 and 2 and they all split the fuel to get 20% each. With this fuel they all aim for the airport and they all crash in the pacific, 18 degrees of longitude from their destination and Fukanōs dreams were crushed and yes I just made you read all that and yes I thought it would work while writing it.
Lmao , I have a better solution . Why can't the planes stack on top of each other ?Assistant #1 will carry both Assistant #2 and the professor . At the 180 degree mark , they'll switch places and then Assistant #2 will carry Assistant #1 and the professor to the airport to complete the trip and the professor still has his fuel tank completely full. Don't worry , I'm only kidding.
I found a solution where all three planes return to the airport with 30kL of fuel each to spare, assuming it doesn't take time to refuel at the airport (if it doesn't take time to transfer fuel in midair, why would it?). At the halfway point, as soon as plane 2 lands, he refuels and both of the support planes take off. After 60 minutes, plane 2 transfers 60 kL of fuel to plane 3 and turns around, having just enough fuel to get back to the airport. 30 minutes later, planes 1 and 3 meet each other as plane 1 is running out of fuel and plane 3 has 150 kL, which it gives half of to plane 1. In another 30 mintues, plane 2 arrives back at the airport to refuel and the other two planes are 60 degrees from the end, each having 45 kL of fuel. Plane 2 goes back and meets the other two in 30 minutes, planes 1 and 3 have 15 kL each and plane 2 has 150. Collectively, the have 180 kL and they only need 90 to get all three planes back to the airport. Now, the real riddle: how do you reconstruct this problem so that it's only possible to solve it without any fuel left over. PLEASE LIKE THIS COMMENT SO I CAN FEEL VALIDATED AND SMART.
** if plane 1 starts before the others how is plane 3 supposed to catch up to the 1st plane in time before the 1st plane runs out of fuel mid way? It travels at a constant speed , can't accelerate remember?
I came up with a similar solution, albeit it does assume that the airport can refuel the planes in zero time (same as transferring fuel between planes). The 'official' solution is more elegant in that (1) this assumption is not required, and (2) both assistants get to enjoy a break before relaunching from the airport.
well i got all 3 of them crashing together 20 degrees short of the goal. *scene: planes running out of fuel with all the turbulence, panic lights, and beeping sounds* "it will be an honor to die with two of my most dedicated students." "you were the best professor i ever had, dr. fukano." "i.. i've always loved you, orokana!" "i feel the same way! oh fugori, if only you hadn't waited 'til now, our love could have blossomed into beautiful math!" "then our love shall burn in our hearts as brightly as our planes hurdle into the earth!" "but wait, jet fuEL CAN'T MELT STEEL BEA--***C R A S H***" such a tragedy ;__;
I got the answer but would like to add that the parameters allow for slight variations and extra fuel in various spots. The important part in the parameters is this: Refueling doesn't have a time cost. This means that Orokana could have only given fuel to the professor, returned back to refuel and then met Fugori at the 45 mark to refuel him. They both get back with a bit of fuel to spare. Then they can both leave instantly (refueled of course), one of them refueling the other at the 45 mark again, returning again to refuel and then going back with more fuel. The other continues to the 90 mark to meet the professor as normal - now having fuel to spare for the return trip. Although that spare fuel isn't enough for the full way back with both planes, the first plane going back to the 45 mark will compensate. What I'm wondering (and is a much harder question than the riddle) is whether this strategy would allow for a smaller tank of fuel. Maybe the support planes could get the professor to the 95 mark and meet him at the same on the other side? Then a tank of 170 kL would be enough. The main difficulty here is the timing. Both support planes can't get to the airport together, or they won't have time to meet the professor. My educated guess (I made a few preliminary attempts) is that it's impossible... but I'm not entirely sure.
"You can instantly transfer fuel at any time provided..." We've forgotten how logic works. My solution: land at another airport even though it hasn't allowed you to. None of the rules say you can't break the law.
Technically, if you run out of fuel and try to land it counts as an emergency landing which I’m pretty sure is allowed at all airports so you could refuel at a airport halfway.
I did it differently. I didn't try to minimize the amount of fuel used since that wasn't part of the riddle. My method uses 1020kL whereas theirs use 900: 60 min: Plane A. 120+60=180 kL Plane B. 120-60=60 kL Plane C. 120 kL Plane B flies back. 80 min: Plane A. 160+20=180 kL Plane C. 100-20=80 kL Plane C flies back. 120 min: Plane B lands and refills. 160 min: Plane C lands and refills. Both take off in the opposite direction. 220 min: Plane B. 120+60=180 kL Plane C. 120-60=60 kL Plane C flies back. 260 min: Plane A. 0+70=70 kL Plane B. 140-70=70 kL 280 min: Plane C lands and refills. 300 min: Plane C takes off. 330 min: Plane A. 0+30=30 kL Plane B. 0+30=30 kL Plane C. 150-60=90kL 360 min: Plane A. 0kL Plane B. 0kL Plane C. 60kL
A few seconds into the explanation, my first thought was "Oh, they're allowed to land and take off again?" Kinda gets easy then, so I didn't consider the possibility.
I found a different solution that works assuming that refueling is instantaneous. Instead of just one plane (Orokana) going in the opposite direction to meet the Fukano at the 3/4 mark and then both being refueled by the third plane (Fugri) on the way back, you can have two planes setting out together in the other direction and one giving fuel to the other on the way with enough fuel to make it back. At the 3/4 mark when the two remaining planes meet, they both just have enough fuel to make it back. It's really the same solution but different order. By the way, I love the Japanese names 不可能 (impossible), 不合理 (unreasonable), and 愚かな (foolish)
Your method cannot be done. To get home from 3/4 point, two planes both need half a tank. Meaning you need one full tank and disperse it between the two. You cannot get 1 plane to 3/4 with a full tank.
Simgen x This alternative solution works if the plane that got home takes off a third time to give 45L to the plane that needs it on the final way home. Therefor this solution is less optimal than the one in the video
Simgen x Sorry, it's a little hard to explain without diagrams. While Fukano was going the long way around, I had the two other planes going in the opposite direction with one giving excess fuel to the other and then returning. When they meet at the 3/4 mark they have enough fuel to get back because there is further support from the other plane that did not go all the way to the 3/4 mark. This solution has more steps that the one in the video. It also assumes that refueling on the ground is instantaneous which was not really specified in the rules (although implied). The video solution is much better.
+blabby102 Your solution is incorrect. Assuming two planes go the other way, at the 3/4 mark, both would have to start to come back for them to survive. They would not have any excess to supply to each other for one to meet Fukano and survive. If you really want another answer to this, Fukano could literally just bring the two tanks in the other planes with him and refuel midair. He doesn't want to redesign, but he can just let the tanks dangle along some rope or sit with him.
I have a different solution: All three Planes start simultaniously, heading west. After 30 Minutes, Assistent 2 gives each other 30 kl and heads home again. At 60 Minutes, A2 reaches home, refuels (has to be as fast as transferring fuel between planes, else this doesn't work) and heads off west again. At 90 Minutes, A1 transfers 60 kl to the professor and heads home. At 120 Minutes, A1 and A2 meat, A2 transfers 30 kl to A2, they both head home. At 180 Minutes, they reach home and refuel, heading east afterwards. At 210 Minutes, A2 transfers 30 kl to A1 and heads home. At 240, A2 reaches home, refuels and heads east. At 270, A1 reaches the Professor and transfers 60 kl. At 300, all meet, A2 transfers 30 kl each to everyone, so everyone has 60 kl left to head home
I had to think long and hard about it. I did not think it was going to work because of how it looked to me like you were transferring fuel too early in the beginning but it did. Nice job.
See mine - (calling the 2 side planes A and B) 1) first part same as explained in video 2) as soon as B returns to airport, A and B takeoff 3) After 60 mins A gives 60 kl fuel to B and returns 4) Professor and B meet 90 degrees from airport, B gives half the fuel (75 kl) to prof. 5) After 75 mins, A is back with (180-15) kl of fuel, give B 15 kl and keeps 15 giving prof. 135 kl Since fuel is left it means they can do the job with max capacity less than 180 kl
Also found a different and much less elegant (more complicated) solution: All planes leave west. After 72 minutes both assistants transfer 36 kl to the professor (have not seen any rule that says they cannot transfer fuel simultaneously), and head back home. At 144 minutes they reach home, refuel and take off instantly, flying east. At 216 minutes, assistant 1 transfer 72 kl to assistant 2 and returns home. Assitant 2 continues to meet the professor at 252 minutes, transferring half his tank (72 kl) to the professor. Assitant 1 gets back home at 288, refuels, and takes off instantly, reaching the other two at 324. It transfers 36 kl to each of them and they all return home. In the end, assistant 1 still has 72 kl left in the tank.
@@shotguneu I believe assistant 1 would crash. If assistant 1 and 2 leave headed east at 144 minutes and travel together until 216 minutes (72 miles traveled together) and then assistant 1 transfers 72 kL to assistant 2 (180-72-72=36) that would leave assistant 1 with 36 kL left while still being 72 kL from the airport.
There are solutions that don't require two planes to be refueled at once. A, B and C fly to 60 degree mark. C transfers 60 units of fuel to A and turns back. A and B fly to 90 degree mark. B transfers 30 units to A. B flies back to 30 degree mark where C meets him to refuel him. Both return to airport, refuel and take off again in the opposite direction. At 330 degrees, C transfers 30 units to B and returns to the airport. B flies on to 270 degrees to meet A. B transfers -half of his- 90 units to A, keeping 30 for himself. A flies the remaining 90 degrees to the airport. C has meanwhile refueled and meets B at 300 degrees, transfers 60 of his 120 units and they fly to the airport.
I'm sure others have discussed this but I just found this video. I came up with a different solution but when I did the math, it turns out the solution in the video uses less fuel. My solution was to have P3 refuel P1 and P2 at 45 and fly back. P2 refuels P1 at 105, and flies back. P2 will hit 0kl at the 45 degree mark. P3 leaves at 120 to go to the 45 mark and refuels P2. They fly back to the base together. P2 refuels back up to 180kl, but P3 only needs 135kl to refuel. They do this instantly and fly in the opposite direction. At 255, P3 refuels P2. P3 flies back and lands at the base with 45kl left in his tank. P2 flies on to 285 where he meets the professor. P2 refuels P1 so they both have 75kl left. They fly back to the base and land with 0kl left in their tanks. My solution uses 1035kl for fuel and ends with 45kl left in the tank for a net of 990kl used. Their solution uses only 900kl of fuel.
There is a safer solution that leaves some room for error or technical difficulties, as the pilots have some fuel left in the end. The short version is that one assistant can support the other in the second half of the professor's journey, when they fly to refuel him. The long version (spoilers!): Professor (P), Fugori (F) and Orokana (O) fly 1/8 of the way west around the equator (45 degrees of longitude). Each has 135 kl of fuel left. O refuels P and F for 45 kl each (to 180 kl), leaving her with 45 kl. O returns to the airport and refuels. Meanwhile P and F travel another 1/8 of the equator (45 degrees), for a total of 1/4 (90 degrees) from the airport. Each has 135 kl left. F refuels P for 45 kl (to 180 kl), leaving him with 90 kl. F returns to the airport and refuels. P continues, using all of his fuel to travel around 1/2 of the equator (180 degrees), for a total of 3/4 (270 degrees) travelled. He now has 0 kl. In the middle of the last step, when P is halfway around the equator, F and O both take off and fly east (in the direction opposite to the professor's) for 1/6 of the equator (60 degrees). Each has 120 kl left. O refuels F for 60 kl, leaving her with 60 kl. O returns to the airport and refuels. F continues for 1/12 of the equator (30 degrees) toward the professor. He now has 150 kl of fuel. He refuels the professor for 75 kl (to 75 kl), leaving himself with the same amount. P and F continue west around 5/24 of the equator (75 degrees), for a total of 23/24 (345 degrees) travelled. They both have 0 kl. Meanwhile, O has flown 1/24 of the way around the equator (15 degrees) east to meet them. She has 165 kl of fuel left. She refuels each of them for 55 kl (to 55 kl), leaving herself with the same amount. P, F and O finish the journey, travelling 1/24 of the equator (15 degrees) west to the airport. Each of them has 40 kl of fuel left in the tank.
Factoring the earth's rotation doesn't change much. It only increases the starting velocity of the planes. For example, if the earth "spins" at 100 mph at the equator, then while standing on the equator I also have a velocity of 100mph. Side note this example is VERY simplified
But can the air's density change it? www.braeunig.us/space/atmos.htm . There is a wide different between air's density on the ground and at the height of 20km.
I like how they helped us in a sneaky way on that 30 second mark, when they actually displayed what fuel meter looks like, so if you are paying attention you realize that you will be transferring 45 kiloliters around.
My Solution: Never CREATE the problem in the first place. The best way to solve any problem is to avoid it. It is better to avoid a war than to win it. It is better to avoid running out of fuel by building a better plane or more airports.
You can actually complete it with fuel to spare if B "overextends" on the initial westward trip (past 90) with A, refuels A to propel A more than 3/4 of the way around, and then C makes *another* westward trip to rescue B.
At first i didnt believe this method pass pilot time constraint but it pass. It seem for a degree you push, a degree less time is required on rescue mission of the second half. It can extend 20 degree more, since third plan can safely rescue one plane at max of 60 degree distance
Just wait a few hundred million years until the earth's rotation slows by half. Then fly in the direction opposite that the Earth spins. That was my guess.
EchoL0C0 that would not work. the relative air on the athmosphere will go relative to the ground thus putting you in the same spot in the air (given no wind condition).
Honestly I'm so bad at even small numbers maths that it hurts my brain to even try, but I adore riddles so I watch them anyway without trying to solve these ones :') I prefer logic, language, verbal reasoning, riddles with context clues, where are they at!
I got a really good one, the answer is also very logical (don't use dumb reasoning to answer this question). Please try it out, I think you'll really enjoy it and get a great sense of accomplishment if you get it. In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
CaptainRay The treasurer found a weak poison to drink before he was summoned, so that he would then drink a stronger one from the pharmacist...he took a regular potion not a poison so the pharmacist wouldn't drink anything stronger, but the pharmacist worked this out and did the same, so the treasurer wasn't neutralised, and the pharmacist never drank any poison at all. The king wanted the strongest poison but he got a useless potion so he didn't get what he wanted??
CaptainRay Instead of bringing poison, treasurer brings Pepsi. He knows pharma will die because his 'poison' which is actually Pepsi, won't neutralise the pharma. Also, treasurer drinks weak poison before he is summoned so the pharma's poison will neutralise him, but the Pharma works out this plan and also brings Pepsi. The treasurer dies because he didn't get his poison neutralised, pharma stays alive because he never drank any poison, but king doesn't get what he wants because he's just left with useless Pepsi instead of the strongest poison
There's another solution: EDIT: Everything below is incorrect 1. They leave West at noon 2. At 1:00 Fugori gives 60 kl of fuel to professor Fukano 3. At 1:20 Okokama gives 20 kl of fuel to professor Fukano 4. At 2:00 Fugori returns and refuels 5. At 2:40 Okokama returns and refuels 6. At 2:40 Fugori leaves East 7. At 4:00 Okokama leaves East 8. At 4:20 (lol) Fugori gives 40 kl of fuel to professor Fukano 9. At 5:00 Okokama gives 60 kl of fuel to both Fugori and professor Fukano 10. At 6:00 they all return to the airport
Sorry mate, but you are wrong. If Fugori leaves at 2:40 to reach the professor at 4:20 (when the professor will be out of fuel according to your refuels), he will have traveled 1 hour(60 minutes) and 40 minutes, meaning a total of 100 minutes, which means 100 kl. So even if he didn't give any fuel to the professor(which he must for your plan to work), he wouldn't be able to return because he would need another 100 kl.
@@Georgetsiki13 Yea, my mistake was that I knew Okokama could save either one of then, but I accidentally thought "both" instead of "either" (but idk for sure cus I made this awhile ago).
Ok here's how I did it before watching the solution*** I did this by dividing a circle into eighths then labeling each line 45, 90, and so on.... and I'm typing this by times and the Professor is going counterclockwise: *45 mins* _Prof_: Goes to 45° and receives 90 from _A1_ Remainder- 225 _A1_: Goes to 45° and gives 90 to _Prof._ Remainder- 45 _A2_: Stays put *90 mins* _Prof_: Goes to 90° Remainder- 180 _A1_: Goes back to airport _A2_: Goes to 45°(C) Remainder - 135 *135 mins* _Prof_: Goes to 135° Remainder- 135 _A1_: Goes to 45° (C) Remainder- 135 _A2_: Goes to 90° (C) Remainder- 90 *180 mins* _Prof_: Goes to 180° Remainder- 90 fuel _A1_: Gives 90 fuel to _A2_ Remainder- 45 _A2_: Goes to 45°(C) and receives 90 fuel from _A1_ *225 mins* _Prof_: Goes to 225° Remainder- 45 _A1_: Goes to airport _A2_: Goes to 90° (C) Remainder- 90 *270 mins* _Prof_: Goes to 270°/90°(C) and receives 45 fuel from _A2_ Remainder- 45 _A1_: Goes to 45°(C) Remainder- 135 _A2_: Stays at 90°(C) and gives 45 fuel to _Prof_ Remainder- 45 *315 mins* _Prof_: Goes to 315°/45°(C) and receives 45 from _A1_ Remainder- 45 _A1_: Stays at 45°(C) and gives 45 fuel to both _Prof_ and _A2_ Remainder- 45 _A2_: Goes to 45°(C) and receives 45 fuel from _A1_ Remainder- 45 *360 mins* _They all return with 0 fuel left :D_ *NOTE*: I just decided that standing there wouldn't be fuel taken away for just standing in one area bc they never said so, fuel would be instantly refueled, and being at 0 was okay if someone was there to help ;-; C= Clockwise
I think that I have another solution but it's late at night so please correct me if I go wrong somewhere. I cut this in 6 instead of 8. Professor and Assistant 1 go West together for 60 minutes (120 left for each). Assistant 1 gives Professor 60 kL (Assistant has 60 and Professor has 180) Assistant 1 turns back. (60 minutes pass) Professor is now at 120 kL (2/6 completed) and Assistant 1 lands, refuels immediately and leaves with Assistant 2 for the East (All in 1 second) Assistants 1 & 2 travel 60 minutes East and Assistant 1 gives Assistant 2 60 kL then turns back while Assistant 2 continues forward (Assistant 1 at 60 kL ; Assistant 2 at 180 kL). All of this happens while the Professor reaches the halfway mark with 60 kL left. (60 minutes pass) Assistant 1 lands while Assistant 2 and Professor meet at the 4/6 completed mark, where Assistant 2 gives Professor 60 kL (Assistant 2 at 60 kL ; Professor at 60 kL) Assistant lands refuels and takes off in 1 second again, then meets Assistant 2 and Professor at 5/6 completed mark. Assistant 1 gives each 60 kL (All at 60 kL) Which is the perfect amount to get back to the airport safely!
There's a problem: at the very last exchange of fuel, the assistant has already burned 60 kL of fuel, so they only have 120 left. If they give 60 to the other assistant and the professor, they won't have any left for their own plane to get back.
A different solution, which someone may have already mentioned in the comment. Time 0: Planes 1 and 2 take off with full tanks Time 60: Plane 2 transfers 60KL to Plane 1, and then turns around to airport. Plane 1 now has full tank again with 180KL and plane 2 has 60KL remaining. Time 120: Plane 2 lands at airport, refuels and takes off along with Plane 3. On the other hand, plane 1 has 120KL remaining Time 180: Plane 2 transfers 60KL to Plane 3, and then turns around to airport. Now, plane 1 has 60KL remaining, while plane 3 has a full tank with 180KL Time 240: At this point, plane 1 meets plane 3. Plane 1 has no fuel, so plane 3 transfers 60KL to plane 1, and turns back to the airport. Now, plane 1 and 3 are travelling together. After transferring fuel, plane 3 has 60KL remaining. Back at the airport, plane 2 takes off again with full tank. Time 300: Plane 2 meets both Planes 1 and 3 at this point and transfers 60KL each to both of them. Now all three have 60KL in their tanks, and all head to the airport. Time 360: All three planes land at airport, with zero fuel in their tanks.
Interestingly, for some reason all the solution has zero redundancy. P.s. oh wait.. seem everyone dies in your case. At the last fuel transfer. Plane 2 has used 60 fuel unit to go there at the position 60 unit far from airport, so it has 120 left... not enough for everyone to survive?
I'm an engineer. Here's the more simple solution: bring the gas tanks of the other planes with the professor. 1.5 times the amount of gas, able to refuel in the air, nobody dies, continuous trip: it follows all the rules stated, even the professor's intention of not wanting to redesign. He simply brings more that's not designed into the plane which may add to the mass, but 1.5 times the amount needed is well over enough.
If you devide the globe into 8 parts as in the video you can also use another strategy in Fukanos last 3 hours. With Plane 2 and 3 loaded up with 180KL fuel each they both fly east for 45 minutes ending up with 135KL each, but 45KL from Plane 3 is transfered to Plane 2. Plane 3 goes back to the airport to refuel meanwhile Plane 2 and Fukano meet at 270Longitude. Fukano has 0KL Fuel, meanwhile Plane 2 has 135KL. Plane 2 transfer 90KL to Fukano which is enough for Fukano to go back home. Plane 2 on the other hand doesn't have enough and will run out of fuel at 315 Longitude with 45 minutes left but Plane 3 is there to meet up and Transfer 45KL extra to Plane 2 and both planes makes it back with Plane 3 having 90KL to spare.
My solution was less efficient and I still made it: I used in total 1060kl of fuel, while their solution used only 900 kl. That means there is room for error here :) Wasn't that amazing after all
i think I have a solution using 1000kl, so also less efficient. In my solution no plane rests or fuels the other two simultaneously, which is also kind of nice.
There are some typos in the video, the the solution still works. At 2:36, where is says "45 minutes later", it is adding 45 minutes to the the time when plane 3 lands back at the airport (13:30). In reality, it should be 45 minutes later from when she starts heading back. So at 13:30, when #3 lands back at the airport, the other two are 1/4 of the way through their journey.
Dan K a plane with 81 people run out of fuel when they were reaching the airport, there is there only 6 survivors. Was a charter airplane were a football/soccer Brazilian club, they were going to play the first row of the SouthAmerica Cup Final in Colombia.
I assumed the planes would refuel instantly with my solution, since the video didn't mention how long it takes to refuel. So here's what I got: all 3 planes fly out. At 60 minutes west, assistant 1 gives the professor 60kL of fuel and heads back. At 80 minutes west, assistant 2 gives the professor 20 kL of fuel and heads back. As soon as assistant 2 arrives to refuel, both assistants fly east. At 60 minutes east, assistant 1 gives assistant 2 60kL of fuel and heads back. At 100 minutes east, assistant 2 gives the professor 70kL of fuel and begins to head back with the professor. And finally at 30 minutes east, assistant 1 gives the professor and assistant 2 30kL of fuel each and they all head back.
KeanuIsAwesome refueling time is literally zero because they can transfer fuel in absolutely no time in mid air.. So on land it's a piece of cake...that is wat i think
It’s not possible because when you switch directions you would probably snap your neck due to the sudden change in direction combined with the g forces.
More safe solution by me is, (before that, lets name professor’s airplane #1 and his students #2 & #3) : Now, all three of them will start at the EA (equator airport) with full capacity fuel, 4/4. At the 45* all three of them will have 3/4 fuel. At this point, airplane #2 will give 1/4 both #1 & #3 (in total 2/4) and so on it (#2) will head back to EA when #1 & #3 will go forward from 45* with max fuel, to 90* with 3/4 fuel. At this point #3 will give 1/4 to #1, which will let #3 with 2/4 and #1 with 4/4. Now they both will arrive to 135* with #1 3/4 and #3 with 1/4. Now #3 will give the #1 1/8 which will leave #3 to goes back to EA with 1/8 fuel and #1 with 7/8 fuel. Now #1 goes to 225* alone, and in there (in 225*), #1 will have 5/8 fuel remaining. #1 moves forward to 270* with 3/8 remaining. As soon as #1 arrives to 270*, #2 start the journey the opposite way with max capacity of fuel. Now they both (#1 & #2) will meet at 315* with #1 1/8 fuel remaining and #2 with 6/8 fuel remaining. At that point (315*), #2 give 4/8 to #1 and head back with 2/8 remaining and #1 head back with 5/8 fuel remaining which is more than enough fuel to head back safely bc there is only 45* remanding from the EA… ( 45* = 2/8 fuel). So no one will have to reach 0 or to gamble their lifes…
For anyone wondering, Fukanō (Fukanou) means "impossible" in Japanese.
What I was saying
No one was wondering
I was about to comment that 😂😂😂
😮
Do the other two mean anything? Google Translate says Fugori is falsehood and Orokana just comes back as Orokana. I know it works pretty well, but I’d trust someone with personal knowledge over the program.
“The professor probably could have designed the plane to hold more fuel, but where’s the fun in that?”
Me: NOT DYING
lmfao this made me chuckle
THIS MADE ME DIE ON THE FLOORRRRRRRRRR💀
And where's the fun in that?
Your dying
@@CCABPSacsach so?
I GOT IT A FIGURED IT OUT!!!! Wait all three have to survive? Dammit
IKR!
Why can't the assistants just have parachutes or something?
they ll land in the sea and drown.
#AssistantLivesMatter
ur funny
Me: Hears first part of puzzle.
Me: This is easy
Narrator: Fugori and Orokana cannot crash
Me: damn, there goes my strategy
What was your strategy?
Just curious
@@sabereagle9277 Originally, it was very easy... you just have Fugori follow the first plane and give all of its fuel once it has the same amount remaining as the first plane has empty capacity. Of course this means Fugori crashes and dies. Orokana does similar but the opposite way around the world and with a pause to wait for the right time.
This doesn't work when Fugori and Orokana have to survive :) Though, the answer is somewhat similar.
@@ericfleet9602 oh okay
"some of you are probably going to die, but that's a sacrifice i'm willing to make" lol
@@Iggybart05 idk man, i still think that Fugori and Orokana are disposable...
he doesn’t even need to fly around the earth nonstop, being able to make a plane that goes over 5 times the speed of sound is impressive enough
this comment is severely underrated
Fukano. *builds a revolutionary supersonic airplane*
Also Fukano: M O R E !
And being able to dock a fuel nozzle midflight at that speed instantaneously too!
actually, the real world X-15 goes at mach 5.93 making it almost 1 speed of sound faster
@@crow2464how much fuel does that consume?
The names are brilliant. "Impossible "(Fukano), "unlogical" Fugori, and Silly/stupid/obsurd "Orokana"
Alice Beck lmao
"Absurd "
Lol
@@unico17324 lol
LOOOOOL
不可能 = Fukano = not possible
不合理 = Fugori = absurd
愚かな = Orokana = foolish
Terence Wang you’re right!!!
불가능 불합리 어리석음....
교수의 불가능해보이는 작업에 불합리하게 차출되어 바보같은 작업을 강제로 하는 대학원생들이였다ㅠㅜ
曰本人👋🏻
@@GMNYU 오 한국인.....
Terence Wang if this is true amazing
I just realized. In order for the professor to travel around the globe at 1 degree longitude per minute, he'd be travelling roughly 4,150 miles per hour!
Well at exactly 3600 nm/h
probably faster, as that is assuming he is literally dragging his plane on the water the whole time. any elevation above sea level increases the circumference of his journey, thus he must increase his speed to maintain the 1 degree per minute.
*sniff sniff * smell that? A NERD
@Joseph ONI it's a joke chill
@Joseph ONI well dont take offense. Dont let it bug you. And if you are one, then embrace it. Live your life, and love every minute of it. God bless
I just had orokani give all her fuel to the prof. RIP orokani she took one for the team
lamo
Zachary Hughes 😂😂
😂😂😂😂😂😂
Helani The sacrifice is 3/4 of the way back
Zachary Hughes Lmao
Alt Solution 1: beg and bribe another airport for permission.
Alt Solution 2: stop being so ambitious and make a larger fuel tank.
AndyHappyGuy maybe it’s a riddle? Hmm idk man tho
-- I would want more fuel
because I dont want to die
@@skelapi7723 and where's the fun in that?
Alt solution 3: use the solution BUT say fukano put on his SECOND favorite album instead of his first
@@Post_Stall_Maneuver And where's the fun in that?
when professor realizes his favourate album takes fuel
You died!
😂
I thought of that too
oh no
To Be Continued
*Insert Roundabout*
Viewers: ok
Flat earthers: REEEEEE
Flat earthers are people who don't go to space even once so they still believe about what the 16th's believes (They do have claims, but they don't know the world is too big to their claims).
@@Ripurlife they don't have proofs, they have claims
@BG - 11ZZ - Port Credit SS (2272) ouch! Wonder how the landing went though
365 likes.
BG - 11KM - Port Credit SS (2272) he was building his own rockets? As in rocket ships? Was he alone building these first of all?
Too much fuel was wasted on this experiment.
yea he only needs to design to increase the fuel capacity.
@@Ripurlife yeah *WhErEs ThE FuN In ThAt*
As long as he didn’t use oil, natural gasses, or anything like that, I’m fine with it
I ruijed the 555 👺
Hwat
watch these puzzles, think about them for like one min (so you don't bad about yourself like at least you tried) watch the solutions , laugh at the funny comments ..... Next puzzle please
Saaaaaameeeee
Absolutely me
Bet you didn't play KSP. There's these riddles all over
Next puzzle please 😂😂 feeling accomplished and shiz
That’s me doing the same
"now all we need to do is make the lunch not fly everywhere"
what you need to do is make it so that passengers don't instantly die from the g force of turning around
Imagine going from a velocity of 4150mph to -4150mph in a split second
@@soso3792 fun fact: if the plane weighs 0.5 kg or higher the force exerted to turn the plane that fast is enough force to break the human femur which is the bodies strongest bone, so all of your bones would instantly break and you would almost definitely immediately die
@@c0smic_eve "fun" fact indeed lmao
@@c0smic_eve And, this is how you see if someone is crazy, have so much free time, a genious or them together.
@@c0smic_eve This makes no sense, the mass of the plane has nothing to do with the force applied to your body during such a turn
There are 2 optimizations that can be made to the solution.
1. Fugori can actually travel past 90 degrees with the professor, if Orokana, after refueling, heads back out to help Fugori on the way back to base.
2. Orokana can meet the professor with more fuel at 270 degrees if Fugori accompanies her when moving clockwise, and refills her tank to full after 60 degrees.
With this optimization, a fuel tank with capacity 162 kiloliters is sufficient - the professor can be left with a full fuel tank 99 degrees out, and meet up with an assistant as he runs out of fuel, 99 degrees before the end of his trip.
Why hasn't Ted Ed pinned this comment lmao
But Orokana and Fugori would take time to reach back to the airport and then take even more time to fly east to meet the prof. By that time prof would be dead
@@thesmartnerd732 Well, technically the prof could go on another 9 degrees to the 279 degree point, wich is exactly what they need to meet up with him since on of the planes left at 99 degrees and 180 degrees to the other direction is 279, the problem with that is that it doesn't offer a solution in wich "meet the professor with more " as the comment said would be possible
@@josephc.9520 Because he's wrong. As TheSmartNerd says, if Orokana helps Fugori return to base, they can't get to the professor in time. Stern Huang doesn't justify his figures of 162kl tank size and 99°-261° for the professor's solo flight, but we don't need to look at those specific numbers to see the problem.
Let's assume just a 1°/kl optimization. Professor Fukano begins flying solo at 91° with 179kl fuel. As in the original solution, he will reach 270° and run out of fuel at the 270 minute mark. That's 90 degrees from base; therefore, to meet him at that time, someone must take off 90 minutes prior, at minute 180.
But wait! Fugori was with the professor at 91°. A round trip takes 91 minutes for each leg, so Fugori cannot reach base until 182 minutes have elapsed. If Orokana helps Fugori, neither of them can be at the base to take off at minute 180, so the professor will crash. If she's en route to the professor at minute 180, she cannot have met and refueled Fugori, so he will crash.
The solution is already optimized.
@@noodle_fc Ah I hadn't thought through this. You know the 3am urge to binge but not think through anything right? Really sorry for taking up your time, I had figured out why a while ago
Wait what ? I'm a professor now ? xD
Fukano YO FUKANO LOL
haha when i searched "fukano" in google a pokemon popped up. i thopught 'professor fukano' was a real person
Fukano HAHAHAHAHAHA
Good thing the Gyrobowl was invented to solve the meal spilling problem.
Go ask Marshmallow.
FAIL: This doesn't take into consideration time spent going through airport security. :/
Professor Fukanō's airport doesn't require them to go through airport security.
Yeah his airport is fancy
You don't have to go through security more than once if you stay within the airport lounges, so once they all go through security before taking off at the beginning there's no issue.
YOU HAD ONE MIND
AND THEN YOU JUST WITHERED IT AWAY INTO 1 BRAIN CELL
stop taking ted ed's fun riddles so literally 😡
Fuck dude I didn't think they were allowed to refuel
Ikr
M224 SAME
SAME
The puzzle did say you could go all the way back and refuel, but I forgot about it
***** I heard but I thought it meant they can refuel when they all reach after the end of the puzzle xD
3:05 women when I come near
That's nice
Im So~
LOooOOOOoooooooooOoOOoOoooOOOnnNeLyyy
LMaOOOOoO me too
underrated comment
I feel that bro
But what's the fun in that?
THE FUN IS YOU GUARENTEED SURVIVE
Yeah, but who wants that?
@@BappO-is-me not fukano atleast
Isabelita Alon the professor crashed when he tryed he’s dead now
It’s not guaranteed to survive jim
Srry that jim part is prob not nessesarly
He's Livin Like Larry
Yeah, but its 2020, so who cares anymore?
I know this one. The horse's name is Friday!
Wait what
xD
Wrong! The horse was white!
the octopus was on top of the house so the bear is 573 pounds. your wrong.
I cant get doune off friday
But plot twist, Fugori and Orokana were always jealous of the professor, having stolen their design for the plane and giving no credit where credit was due. So when the time to give the Professor his fuel came, the two left with no transfer of fuel. The Professor plummeted to the ground thinking the tubes had malfunctioned, and died.
Fugori and Orokana flew back to the airport and found a small party had been thrown together (planned by the professor) with streamers and balloons in honour of the two assistant's help with the project, as well as an offer at a full time job alongside the professor. As the fellow scientists and plane technicians on the ground saw Fugori and Orokana leave their planes, the only question on their minds was "Where is the professor?". The two lied and said that they had not see him where they agreed to meet.
Overwhelmed with guilt, Fugori and Orokana took off once more after refueling, claiming they would go back to look for the professor at the spot where he had fallen. They were never heard from again.
Nicolas K (slow clap) well done
Swigity Swooty I'm coming for that booty
No, they were heard from again. They landed back in Tokyo and were met with heavy fines after losing a case after Fukano's family sued them for deliberating getting rid of a loved one's life and so the assistants were fined heavily.
They were charged with illegally landing aircraft and neglecting a comrade's life. They were very controversial.
Jacob Griffin But nobody ever saw them again. Did they die? Did they go into hiding? Did they start anew as farmers, cut off from society, to repent for their sins?
Nicolas K
Did they fly into the Bermuda Triangle?
Why didnt he just confirm he had green eyes and then ask to leave?
BUt YoU caN'T LeAvE yOu neED to gO aRouNd tHE wORld!!1!111!!
This is my favorite joke.
This is not the 100 green eye prisoners riddle
@@Dmanrique6 r/woosh
Because he had 2 fuses that each burned for a minute, and he still had to catch all his robo-ants
I live for ted ed riddles
InkJelly ME TOO
InkJelly same
InkJelly me too
InkJelly that's a sad life
ikr
Yeeee!!! Apart from the Kilometers calculations I figured out the way how he can do it!
"But where's the fun in that?" I love the professor's smirk there. XD
I came up with another solution to the riddle.
12:00 → all three planes depart
12:00 - 12:45 → assistant C and B and the Professor fly 45 longitudes from 0°l to 45°l
12:45 → assistant C transfers 2x 45kl, the other two pilots now have a full tank again
12:45 - 13:30 → assistant B and the Professor fly 45 longitudes from 45°l to 90°l
13:30 → assistant B transfers 45kl to the Professors plane, the Professor has now a full tank again
13:30 → assistant B has 90kl fuel left, which is enough to fly the 90 longitudes back home
13:30 → assistant C arrives back home
13:30 - 16:30 → the Professor flies 180 longitudes from 90°l to 270°l
15:00 → assistant B arrives back home
15:00 - 16:00→ assistant C and B fly 60 longitudes from 360°l to 300°l
16:00→ assistant C transfers 60kl to assistant B, assistant B has now a full tank again
16:00→ assistant C has 60kl fuel left, which is enough to fly the 60 longitudes back home
16:00 - 16:30→ assistant B flies 30 longitudes from 300°l to 270°l
16:30 → assistant B transfers half of his tank to the Professor. Both have 75kl fuel
16:30 - 17:45 → assistant B and the professor fly 75 longitudes from 270°l to 345°l
17:00 → assistant C arrives back home
17:30 - 17:45 → assistant C flies 15 longitudes from 360°l to 345°l
17:45 → assistant C transfers a third to assistant B and the Professor, everyone has 55kl fuel
17:45 - 18:00 → assistant C and B and the Professor fly 15 longitudes from 345°l to 360°l
18:00 → all three planes arrive safely back home with 40kl fuel in each tank.
Sir you are truly the smartest out of all of us
Nice
DIDN'T PAUSE THE VIDEO AND WAITED FOR THREE SECONDS
Just push the right button on your computer to skip time :P
themantsang app user :P
김재곤 well then :| ignore me then :D
김재곤 Is there by any chance that you are a kpop fan?
EXO-L and Xingstan I'm a big fan of some old Korean singers, but I'm not interested in K-pop.
I figured it out but in a differnt way. Let me explain, also I will use the amount of fuel they have in % instead of using the exact amount like shown in the video to make it easier.
We know that all of your fuel will take you to just the other side of the planet and you can only land on one spot which is the airport you started from.
To accomplish this task ive decided to devide the whole trip around the globe into 10 sections. Every section is exactly 36 degree of longitude (One tenth of the whole trip).
Due to our understanding that half the trip will take up 100% of our fuel we can safely assume that every section will take up 20% of our fuel.
So basically. 36 deegres of longitude = 1 section = 20% fuel.
Now to the actual flight we will (just like in the video) fly every plane west. But instead of stopping at 1/8 (or 1.25 sections) of the total flight you will now stop at 1/5 or exactly 2 section. All the planes should now have 60% of their fuel left and has all used 40% of their fuel to get there. Plane 2 and 3 will now transfer 20% of their fuel to plane 1 (each) which will put plane 1 at 100% fuel while the other planes will have 40% each. The second and third plane will now have exactly enough to fly back to the airport to refuel for their next mission. Meanwhile plane 1 will have just enough fuel after the refuel to go 180 degrees of longitude plus the 72 it has already travelled which puts the total at 252 degrees of longitude or in more simple terms, 7 sections.
Meanwhile, plane 2 and 3 has refueled at the airport and plane 2 has gone 3 sections or 108 defrees longitude and has now reached plane 1. With the remaing 40% they split it even and can now go one more section. Meanwhile plane nr 3 has also gone east and meets up with plane 1 and 2 and they all split the fuel to get 20% each. With this fuel they all aim for the airport and they all crash in the pacific, 18 degrees of longitude from their destination and Fukanōs dreams were crushed and yes I just made you read all that and yes I thought it would work while writing it.
he can glide if hes high enough :D
So god darn close
Not only his dreams getting crushed there
Wow I read that whole thing that was hilarious!!! Also I’m pissed.
Lmao , I have a better solution . Why can't the planes stack on top of each other ?Assistant #1 will carry both Assistant #2 and the professor . At the 180 degree mark , they'll switch places and then Assistant #2 will carry Assistant #1 and the professor to the airport to complete the trip and the professor still has his fuel tank completely full.
Don't worry , I'm only kidding.
I found a solution where all three planes return to the airport with 30kL of fuel each to spare, assuming it doesn't take time to refuel at the airport (if it doesn't take time to transfer fuel in midair, why would it?). At the halfway point, as soon as plane 2 lands, he refuels and both of the support planes take off. After 60 minutes, plane 2 transfers 60 kL of fuel to plane 3 and turns around, having just enough fuel to get back to the airport. 30 minutes later, planes 1 and 3 meet each other as plane 1 is running out of fuel and plane 3 has 150 kL, which it gives half of to plane 1. In another 30 mintues, plane 2 arrives back at the airport to refuel and the other two planes are 60 degrees from the end, each having 45 kL of fuel. Plane 2 goes back and meets the other two in 30 minutes, planes 1 and 3 have 15 kL each and plane 2 has 150. Collectively, the have 180 kL and they only need 90 to get all three planes back to the airport. Now, the real riddle: how do you reconstruct this problem so that it's only possible to solve it without any fuel left over.
PLEASE LIKE THIS COMMENT SO I CAN FEEL VALIDATED AND SMART.
I shall validate this comment so you can feel smart
if plane 1 starts before the others how is plane 3 supposed to catch upto the 1st plane ? It travels at a constant speed , can't accelerate remember?
** if plane 1 starts before the others how is plane 3 supposed to catch up to the 1st plane in time before the 1st plane runs out of fuel mid way? It travels at a constant speed , can't accelerate remember?
I came up with a similar solution, albeit it does assume that the airport can refuel the planes in zero time (same as transferring fuel between planes). The 'official' solution is more elegant in that (1) this assumption is not required, and (2) both assistants get to enjoy a break before relaunching from the airport.
I came up with a very similar solution if you don’t take time to refuel
0:41 This is the most badass smirk I have ever seen
Well, every airport must let a plane that ran out fuel land, otherwise that is a crime, but this is a riddle .
Well they CLEARLY wan't him dead, so i'm assuming that's a risk they are willing to take
Simple, professor Funkand uses his shrink ray to make the earth 50% smaller, then all three planes fly around.
lol
Lol
It will have 360 degrees at any size
then say goodbye to the following
> the three pilots
> their planes
> gravity
> all 7.6 billion people
but there will be less distance from the airport and to the airport :D
I thought it was okay if the other assistants crashed, as long as the professor made the round trip. :/
Untapped Happiness lmao just constantly refill him xD
You heartless monster! >:(
Untapped Happiness lmao XD
Untapped Happiness I
Untapped Happiness I
Drinking challenge: take a shot everytime he says 45
*vomiting everywhere*
mY LiVer StOpPed wOrKInG
I played this with my dad. He passed out halfway. For some reason he doesn't have a pulse tho, should I be worried?
Of water and also maybe for every time he says 5
*UH OOOOOOOOOW*
Me counting the number of times the narrator said “45”
I don't think professor Fukano knows exactly how air travel works
I dont think he knows how logic works, deadass be making a small fuel tank for fun inb4 he crashes and 9/11 starts again, smh
Peachy Keen well because you said that I have nothing to say but this, oh Fuk a no
Peachy Keen I
Peachy Keen true
Snowishy lol
well i got all 3 of them crashing together 20 degrees short of the goal.
*scene: planes running out of fuel with all the turbulence, panic lights, and beeping sounds*
"it will be an honor to die with two of my most dedicated students."
"you were the best professor i ever had, dr. fukano."
"i.. i've always loved you, orokana!"
"i feel the same way! oh fugori, if only you hadn't waited 'til now, our love could have blossomed into beautiful math!"
"then our love shall burn in our hearts as brightly as our planes hurdle into the earth!"
"but wait, jet fuEL CAN'T MELT STEEL BEA--***C R A S H***"
such a tragedy ;__;
I love this comment.
Creative af
you are my new favorite person
Pepe Bawagan Top 10 saddest anime deaths
Shippers: OMG TEDEDS BEEN HOLDING THE SHIPPING
READ THIS COMMENT NOW! FUGOROKANA COMFIRMED!
Me:🤔
All shippers say #fugorokana
I got the answer but would like to add that the parameters allow for slight variations and extra fuel in various spots.
The important part in the parameters is this: Refueling doesn't have a time cost.
This means that Orokana could have only given fuel to the professor, returned back to refuel and then met Fugori at the 45 mark to refuel him. They both get back with a bit of fuel to spare.
Then they can both leave instantly (refueled of course), one of them refueling the other at the 45 mark again, returning again to refuel and then going back with more fuel. The other continues to the 90 mark to meet the professor as normal - now having fuel to spare for the return trip. Although that spare fuel isn't enough for the full way back with both planes, the first plane going back to the 45 mark will compensate.
What I'm wondering (and is a much harder question than the riddle) is whether this strategy would allow for a smaller tank of fuel. Maybe the support planes could get the professor to the 95 mark and meet him at the same on the other side? Then a tank of 170 kL would be enough. The main difficulty here is the timing. Both support planes can't get to the airport together, or they won't have time to meet the professor.
My educated guess (I made a few preliminary attempts) is that it's impossible... but I'm not entirely sure.
Me: Does a few barrel rolls and trick shots, therefore missing the runway and crashing to my death.
At least you got the bird with a 360 no scope
4:10 I totally thought it would be "...just as soon as they put bathrooms on the planes"
TheMadGopher JY
TheMadGopher JY lol
XD
"You can instantly transfer fuel at any time provided..."
We've forgotten how logic works. My solution: land at another airport even though it hasn't allowed you to. None of the rules say you can't break the law.
Rain Storm das true
"We've forgotten how logic works"
Someone's forgotten how hypothetical thought experiment situations work.
There is actual a real plane that the people who flies planes at the government
Step 1) Declare emergency.
Step 2) Land at the closest airport.
Step 3) Refuel and continue.
Technically, if you run out of fuel and try to land it counts as an emergency landing which I’m pretty sure is allowed at all airports so you could refuel at a airport halfway.
I like how the planes (and professors) are named "Impossible" (Fukanou), "Absurd" (Fugouri) and "Foolish" (Orokana)
I did it differently. I didn't try to minimize the amount of fuel used since that wasn't part of the riddle. My method uses 1020kL whereas theirs use 900:
60 min:
Plane A. 120+60=180 kL
Plane B. 120-60=60 kL
Plane C. 120 kL
Plane B flies back.
80 min:
Plane A. 160+20=180 kL
Plane C. 100-20=80 kL
Plane C flies back.
120 min:
Plane B lands and refills.
160 min:
Plane C lands and refills.
Both take off in the opposite direction.
220 min:
Plane B. 120+60=180 kL
Plane C. 120-60=60 kL
Plane C flies back.
260 min:
Plane A. 0+70=70 kL
Plane B. 140-70=70 kL
280 min:
Plane C lands and refills.
300 min:
Plane C takes off.
330 min:
Plane A. 0+30=30 kL
Plane B. 0+30=30 kL
Plane C. 150-60=90kL
360 min:
Plane A. 0kL
Plane B. 0kL
Plane C. 60kL
MrSlothJunior i like how u organized ur answer gj btw
y that's exactly how I solved it, too. I knew I'll find this answer in comments when I saw their solution.
I got the same and I think it's actually better than the one shown since there is a margin for not dying! (Although it's less eco-friendly haha)
For all of you who say "Make a larger tank" Know that that's not the point, the planes where just physical representations of the problem
Ali Hakam that's what I'm thinking as I read pretty much every single comment.
Ali Hakam that's what I said! Some people just don't get it...
And to make it worse, the narrator even said it at 0:37.
Exactly!
Exactly, they could just say the planes have weight limits
“Then they get arrested because they aren’t pilots and what they just did was illegal “
A few seconds into the explanation, my first thought was "Oh, they're allowed to land and take off again?"
Kinda gets easy then, so I didn't consider the possibility.
3:29 top 10 scenes that science cannot explain
Who just listens to the answer and dosent even think on the question part lol
spicy meme me
spicy meme me
I think, but fail.
ME
me
I found a different solution that works assuming that refueling is instantaneous. Instead of just one plane (Orokana) going in the opposite direction to meet the Fukano at the 3/4 mark and then both being refueled by the third plane (Fugri) on the way back, you can have two planes setting out together in the other direction and one giving fuel to the other on the way with enough fuel to make it back. At the 3/4 mark when the two remaining planes meet, they both just have enough fuel to make it back. It's really the same solution but different order.
By the way, I love the Japanese names 不可能 (impossible), 不合理 (unreasonable), and 愚かな (foolish)
Your method cannot be done. To get home from 3/4 point, two planes both need half a tank. Meaning you need one full tank and disperse it between the two. You cannot get 1 plane to 3/4 with a full tank.
+Simgen x Yahtzee
Simgen x This alternative solution works if the plane that got home takes off a third time to give 45L to the plane that needs it on the final way home. Therefor this solution is less optimal than the one in the video
Simgen x
Sorry, it's a little hard to explain without diagrams. While Fukano was going the long way around, I had the two other planes going in the opposite direction with one giving excess fuel to the other and then returning. When they meet at the 3/4 mark they have enough fuel to get back because there is further support from the other plane that did not go all the way to the 3/4 mark. This solution has more steps that the one in the video. It also assumes that refueling on the ground is instantaneous which was not really specified in the rules (although implied). The video solution is much better.
+blabby102 Your solution is incorrect. Assuming two planes go the other way, at the 3/4 mark, both would have to start to come back for them to survive. They would not have any excess to supply to each other for one to meet Fukano and survive. If you really want another answer to this, Fukano could literally just bring the two tanks in the other planes with him and refuel midair. He doesn't want to redesign, but he can just let the tanks dangle along some rope or sit with him.
i had another solution that worked but two of them died so...
space kitten One of the rules is that they all survive
he was joking
I have a different solution: All three Planes start simultaniously, heading west. After 30 Minutes, Assistent 2 gives each other 30 kl and heads home again. At 60 Minutes, A2 reaches home, refuels (has to be as fast as transferring fuel between planes, else this doesn't work) and heads off west again. At 90 Minutes, A1 transfers 60 kl to the professor and heads home. At 120 Minutes, A1 and A2 meat, A2 transfers 30 kl to A2, they both head home. At 180 Minutes, they reach home and refuel, heading east afterwards. At 210 Minutes, A2 transfers 30 kl to A1 and heads home. At 240, A2 reaches home, refuels and heads east. At 270, A1 reaches the Professor and transfers 60 kl. At 300, all meet, A2 transfers 30 kl each to everyone, so everyone has 60 kl left to head home
I had to think long and hard about it. I did not think it was going to work because of how it looked to me like you were transferring fuel too early in the beginning but it did. Nice job.
See mine - (calling the 2 side planes A and B)
1) first part same as explained in video
2) as soon as B returns to airport, A and B takeoff
3) After 60 mins A gives 60 kl fuel to B and returns
4) Professor and B meet 90 degrees from airport, B gives half the fuel (75 kl) to prof.
5) After 75 mins, A is back with (180-15) kl of fuel, give B 15 kl and keeps 15 giving prof. 135 kl
Since fuel is left it means they can do the job with max capacity less than 180 kl
Johanniklas Lp BRUH
Also found a different and much less elegant (more complicated) solution:
All planes leave west. After 72 minutes both assistants transfer 36 kl to the professor (have not seen any rule that says they cannot transfer fuel simultaneously), and head back home. At 144 minutes they reach home, refuel and take off instantly, flying east. At 216 minutes, assistant 1 transfer 72 kl to assistant 2 and returns home. Assitant 2 continues to meet the professor at 252 minutes, transferring half his tank (72 kl) to the professor. Assitant 1 gets back home at 288, refuels, and takes off instantly, reaching the other two at 324. It transfers 36 kl to each of them and they all return home. In the end, assistant 1 still has 72 kl left in the tank.
@@shotguneu I believe assistant 1 would crash. If assistant 1 and 2 leave headed east at 144 minutes and travel together until 216 minutes (72 miles traveled together) and then assistant 1 transfers 72 kL to assistant 2 (180-72-72=36) that would leave assistant 1 with 36 kL left while still being 72 kL from the airport.
Yes... This one is fairly easy
Quinn Gauder yes I am very smart I can figure out riddles in 3 seconds. :b
Yesman
yep.
There are solutions that don't require two planes to be refueled at once.
A, B and C fly to 60 degree mark. C transfers 60 units of fuel to A and turns back.
A and B fly to 90 degree mark. B transfers 30 units to A.
B flies back to 30 degree mark where C meets him to refuel him. Both return to airport, refuel and take off again in the opposite direction.
At 330 degrees, C transfers 30 units to B and returns to the airport.
B flies on to 270 degrees to meet A. B transfers -half of his- 90 units to A, keeping 30 for himself.
A flies the remaining 90 degrees to the airport.
C has meanwhile refueled and meets B at 300 degrees, transfers 60 of his 120 units and they fly to the airport.
DrGerbils plane c wouldn't be able to make it back to plane b in time to refuel it
A riddle, or a math calculation?
math calculation, it's tough
Is it? It seemed to be basic adding and subtracting.
Zox Sonic It's not a riddle. Just math.
Hans Smirnov Thats Maths isnt it?
Zox Sonic it's physics.
This is really good but I hate how nobody understands that planes can glide around for hours without fuel
Mika Stojkovich that's not the point. The planes are just a way of relating the riddle to the real world. It's the math that matters
Mika Stojkovich tell that to the plane that just crashed and killed 70 people
LOL what math simple subtraction and addition? This aint rocket science...
0zfer It's rocket fuel science :D
No they can't, lol.
I'm sure others have discussed this but I just found this video. I came up with a different solution but when I did the math, it turns out the solution in the video uses less fuel.
My solution was to have P3 refuel P1 and P2 at 45 and fly back. P2 refuels P1 at 105, and flies back. P2 will hit 0kl at the 45 degree mark. P3 leaves at 120 to go to the 45 mark and refuels P2. They fly back to the base together. P2 refuels back up to 180kl, but P3 only needs 135kl to refuel. They do this instantly and fly in the opposite direction.
At 255, P3 refuels P2. P3 flies back and lands at the base with 45kl left in his tank. P2 flies on to 285 where he meets the professor. P2 refuels P1 so they both have 75kl left. They fly back to the base and land with 0kl left in their tanks.
My solution uses 1035kl for fuel and ends with 45kl left in the tank for a net of 990kl used. Their solution uses only 900kl of fuel.
nope.
Синий дракон eeyup
Ok,yep.
Синий дракон
Синий дракон how did you get +30? .-.
dunno.
How I feel about this puzzle: Fuk-a-no.
lmfao
lul
Lololol
lol
CamNM89 *FUCK-a-no
nah, both assistants are super dead. no way they survive the massive g-forces involved in "turning on a dime" at those speeds
Nah, you super pretentious with your "facts" go cite your sources.
LMAO
@@aidanwilson4574 Bruh moment
@@joepat1279 ｐｅｒｉｏｔ
k
There is a safer solution that leaves some room for error or technical difficulties, as the pilots have some fuel left in the end.
The short version is that one assistant can support the other in the second half of the professor's journey, when they fly to refuel him.
The long version (spoilers!):
Professor (P), Fugori (F) and Orokana (O) fly 1/8 of the way west around the equator (45 degrees of longitude). Each has 135 kl of fuel left. O refuels P and F for 45 kl each (to 180 kl), leaving her with 45 kl.
O returns to the airport and refuels.
Meanwhile P and F travel another 1/8 of the equator (45 degrees), for a total of 1/4 (90 degrees) from the airport. Each has 135 kl left. F refuels P for 45 kl (to 180 kl), leaving him with 90 kl.
F returns to the airport and refuels.
P continues, using all of his fuel to travel around 1/2 of the equator (180 degrees), for a total of 3/4 (270 degrees) travelled. He now has 0 kl.
In the middle of the last step, when P is halfway around the equator, F and O both take off and fly east (in the direction opposite to the professor's) for 1/6 of the equator (60 degrees). Each has 120 kl left. O refuels F for 60 kl, leaving her with 60 kl.
O returns to the airport and refuels.
F continues for 1/12 of the equator (30 degrees) toward the professor. He now has 150 kl of fuel. He refuels the professor for 75 kl (to 75 kl), leaving himself with the same amount.
P and F continue west around 5/24 of the equator (75 degrees), for a total of 23/24 (345 degrees) travelled. They both have 0 kl.
Meanwhile, O has flown 1/24 of the way around the equator (15 degrees) east to meet them. She has 165 kl of fuel left. She refuels each of them for 55 kl (to 55 kl), leaving herself with the same amount.
P, F and O finish the journey, travelling 1/24 of the equator (15 degrees) west to the airport. Each of them has 40 kl of fuel left in the tank.
I was wondering why this wasnt offered as another solution. I thought I had gotten it wrong!
1:29 He is already doing it!!
Actually in Japan, each pilot's name means...lol
FUKANO=impossible
FUGORI=absurd
OROKANA=stupid
Lolol
LOL how do you know that? 😂
Google translate i guess
Jayden Palmer he's Chinese/Japanese, at least his name is.
蟹 蟹 That's Orokana
My reaction when I saw a new riddle was out:
YESSSSSSSS!!!
Next Riddle :
Professor gonna make the world flying around his plane
That sounds like flying around the world with a plane, with extra steps
"But lets get real" then says they can turn on a dime and instant refuel each other
I solved it slightly different, assuming landing/take-off/refuelling takes no-time
same
The puzzle assumed that landing, take off and refueling took no time. A solution that required those things to take time would be interesting.
DrGerbils refuelling could take up time in this solution.
but yeah, acceleration = infinite.
Did the video mention refueling at airport take no time?
Sany Liew I don't think so, but it's a reasonable inference from the fact that refueling in the air takes no time.
the professor made a plane for fun and if he fails to get fuel thats his fault
all the 45's in the video:
45 (1): 2:16
45 (2): 2:23
45 (3): 2:25
45 (4): 2:30
45 (5): 2:36
45 (6): 2:45
45 (7): 3:32
45 (8): 3:35
45 (9): 3:46
45 (10): 3:54
45 (11): 3:57
So the rotation of the Earth never comes into play? I may have been over thinking this.
Leon Hager bruh the earth is flat dummy
Planes fly in the atmosphere, which is held into place with Earth's orbit by its gravity, I think. Anyway, it does not come into account.
But the earth itself is rotating below the planes.
Factoring the earth's rotation doesn't change much. It only increases the starting velocity of the planes.
For example, if the earth "spins" at 100 mph at the equator, then while standing on the equator I also have a velocity of 100mph. Side note this example is VERY simplified
But can the air's density change it? www.braeunig.us/space/atmos.htm . There is a wide different between air's density on the ground and at the height of 20km.
fukanou = impossible
fugouri = irrational
orokana = fool
he shouldve chosen a team with different names
Didn't notice that before your comment, LOL. I guess Ted-Ed didn't expect some of the TH-camrs here could understand Japanese.
Water Spray I believe it's intentional. kinda like a hidden joke
Professor Fukano makes sense
It's impossible, irrational fool!
TheNinjas lmaoo
Do the assistants need to live?
Yes.
#AssistantLivesMatter
John Roy haha
I like how they helped us in a sneaky way on that 30 second mark, when they actually displayed what fuel meter looks like, so if you are paying attention you realize that you will be transferring 45 kiloliters around.
My Solution: Never CREATE the problem in the first place.
The best way to solve any problem is to avoid it.
It is better to avoid a war than to win it.
It is better to avoid running out of fuel by building a better plane or more airports.
You can actually complete it with fuel to spare if B "overextends" on the initial westward trip (past 90) with A, refuels A to propel A more than 3/4 of the way around, and then C makes *another* westward trip to rescue B.
That's the results that I got too! I was able to get the professor full at 105 degrees. B can be rescued at 45 degrees by C.
At first i didnt believe this method pass pilot time constraint but it pass. It seem for a degree you push, a degree less time is required on rescue mission of the second half. It can extend 20 degree more, since third plan can safely rescue one plane at max of 60 degree distance
Just wait a few hundred million years until the earth's rotation slows by half. Then fly in the direction opposite that the Earth spins.
That was my guess.
EchoL0C0 that would not work. the relative air on the athmosphere will go relative to the ground thus putting you in the same spot in the air (given no wind condition).
I figured it out!
1st.go to the South Pole
2nd.from a few feet away from the South Pole rotate yourself
Then boom your finished!
How about the fuel?
@@ScreamMario then poor it at your shoe
@@danteungas958 I got r/woosh in a cool way right?
@@ScreamMario idk
Bro, I don't think you understand this riddle.
fukano means impossible in japanese
fugori means nonsense and orokana means stupid
oldcowbb oh shit
i thought it meant fuck no in english
stupid impossible nonsense
Putting aside practicality and almost running out of fuel everytime.. I was on point!!!!
Honestly I'm so bad at even small numbers maths that it hurts my brain to even try, but I adore riddles so I watch them anyway without trying to solve these ones :')
I prefer logic, language, verbal reasoning, riddles with context clues, where are they at!
I got a really good one, the answer is also very logical (don't use dumb reasoning to answer this question). Please try it out, I think you'll really enjoy it and get a great sense of accomplishment if you get it.
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison.
The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted.
What exactly happened there?
CaptainRay The treasurer found a weak poison to drink before he was summoned, so that he would then drink a stronger one from the pharmacist...he took a regular potion not a poison so the pharmacist wouldn't drink anything stronger, but the pharmacist worked this out and did the same, so the treasurer wasn't neutralised, and the pharmacist never drank any poison at all. The king wanted the strongest poison but he got a useless potion so he didn't get what he wanted??
CaptainRay Nooo I mean the treasurer figured out the pharma's plan and also brought in a useless potion, not a poison
Maybe not 'potion'. Imagine if they brought Pepsi lmao.
CaptainRay Instead of bringing poison, treasurer brings Pepsi. He knows pharma will die because his 'poison' which is actually Pepsi, won't neutralise the pharma. Also, treasurer drinks weak poison before he is summoned so the pharma's poison will neutralise him, but the Pharma works out this plan and also brings Pepsi. The treasurer dies because he didn't get his poison neutralised, pharma stays alive because he never drank any poison, but king doesn't get what he wants because he's just left with useless Pepsi instead of the strongest poison
There's another solution:
EDIT: Everything below is incorrect
1. They leave West at noon
2. At 1:00 Fugori gives 60 kl of fuel to professor Fukano
3. At 1:20 Okokama gives 20 kl of fuel to professor Fukano
4. At 2:00 Fugori returns and refuels
5. At 2:40 Okokama returns and refuels
6. At 2:40 Fugori leaves East
7. At 4:00 Okokama leaves East
8. At 4:20 (lol) Fugori gives 40 kl of fuel to professor Fukano
9. At 5:00 Okokama gives 60 kl of fuel to both Fugori and professor Fukano
10. At 6:00 they all return to the airport
Sorry mate, but you are wrong. If Fugori leaves at 2:40 to reach the professor at 4:20 (when the professor will be out of fuel according to your refuels), he will have traveled 1 hour(60 minutes) and 40 minutes, meaning a total of 100 minutes, which means 100 kl. So even if he didn't give any fuel to the professor(which he must for your plan to work), he wouldn't be able to return because he would need another 100 kl.
@@Georgetsiki13 Yea, my mistake was that I knew Okokama could save either one of then, but I accidentally thought "both" instead of "either" (but idk for sure cus I made this awhile ago).
Fukano: “What did Fugori leave me to listen to?”
James Hetfield: “GIMME FUEL, GIMME FIRE, GIMME THAT WHICH I DESIRE!”
For real this riddle had to be the only one that had me thinking real hard on the answer
Ok here's how I did it before watching the solution***
I did this by dividing a circle into eighths then labeling each line 45, 90, and so on.... and I'm typing this by times and the Professor is going counterclockwise:
*45 mins*
_Prof_: Goes to 45° and receives 90 from _A1_ Remainder- 225
_A1_: Goes to 45° and gives 90 to _Prof._ Remainder- 45
_A2_: Stays put
*90 mins*
_Prof_: Goes to 90° Remainder- 180
_A1_: Goes back to airport
_A2_: Goes to 45°(C) Remainder - 135
*135 mins*
_Prof_: Goes to 135° Remainder- 135
_A1_: Goes to 45° (C) Remainder- 135
_A2_: Goes to 90° (C) Remainder- 90
*180 mins*
_Prof_: Goes to 180° Remainder- 90 fuel
_A1_: Gives 90 fuel to _A2_ Remainder- 45
_A2_: Goes to 45°(C) and receives 90 fuel from _A1_
*225 mins*
_Prof_: Goes to 225° Remainder- 45
_A1_: Goes to airport
_A2_: Goes to 90° (C) Remainder- 90
*270 mins*
_Prof_: Goes to 270°/90°(C) and receives 45 fuel from _A2_ Remainder- 45
_A1_: Goes to 45°(C) Remainder- 135
_A2_: Stays at 90°(C) and gives 45 fuel to _Prof_ Remainder- 45
*315 mins*
_Prof_: Goes to 315°/45°(C) and receives 45 from _A1_ Remainder- 45
_A1_: Stays at 45°(C) and gives 45 fuel to both _Prof_ and _A2_ Remainder- 45
_A2_: Goes to 45°(C) and receives 45 fuel from _A1_ Remainder- 45
*360 mins*
_They all return with 0 fuel left :D_
*NOTE*: I just decided that standing there wouldn't be fuel taken away for just standing in one area bc they never said so, fuel would be instantly refueled, and being at 0 was okay if someone was there to help ;-;
C= Clockwise
.......That's so simple..... 😥
so is your mind
Harold Wang
Ok lol
I don't think you can have more than 180 kL in the tank though. In step 1 the professor gets 90 kL which brings his fuel up to 225 kL.
Manerkire
Gassssppp you're right.... Drat... oh well
Thanks for pointing it out
Ted-Ed: "Can you solv-
Me: "Oh, Fukano!"
FINALLY! whos been waiting for a riddle?!
I think that I have another solution but it's late at night so please correct me if I go wrong somewhere.
I cut this in 6 instead of 8.
Professor and Assistant 1 go West together for 60 minutes (120 left for each). Assistant 1 gives Professor 60 kL (Assistant has 60 and Professor has 180) Assistant 1 turns back.
(60 minutes pass) Professor is now at 120 kL (2/6 completed) and Assistant 1 lands, refuels immediately and leaves with Assistant 2 for the East (All in 1 second)
Assistants 1 & 2 travel 60 minutes East and Assistant 1 gives Assistant 2 60 kL then turns back while Assistant 2 continues forward (Assistant 1 at 60 kL ; Assistant 2 at 180 kL). All of this happens while the Professor reaches the halfway mark with 60 kL left.
(60 minutes pass) Assistant 1 lands while Assistant 2 and Professor meet at the 4/6 completed mark, where Assistant 2 gives Professor 60 kL (Assistant 2 at 60 kL ; Professor at 60 kL)
Assistant lands refuels and takes off in 1 second again, then meets Assistant 2 and Professor at 5/6 completed mark. Assistant 1 gives each 60 kL (All at 60 kL) Which is the perfect amount to get back to the airport safely!
Can't really tell if your right or wrong...
There's a problem: at the very last exchange of fuel, the assistant has already burned 60 kL of fuel, so they only have 120 left. If they give 60 to the other assistant and the professor, they won't have any left for their own plane to get back.
A different solution, which someone may have already mentioned in the comment.
Time 0: Planes 1 and 2 take off with full tanks
Time 60: Plane 2 transfers 60KL to Plane 1, and then turns
around to airport. Plane 1 now has full tank again with 180KL and plane 2 has
60KL remaining.
Time 120: Plane 2 lands at airport, refuels and takes off
along with Plane 3. On the other hand, plane 1 has 120KL remaining
Time 180: Plane 2 transfers 60KL to Plane 3, and then turns
around to airport. Now, plane 1 has 60KL remaining, while plane 3 has a full
tank with 180KL
Time 240: At this point, plane 1 meets plane 3. Plane 1 has
no fuel, so plane 3 transfers 60KL to plane 1, and turns back to the airport.
Now, plane 1 and 3 are travelling together. After transferring fuel, plane 3
has 60KL remaining. Back at the airport, plane 2 takes off again with full tank.
Time 300: Plane 2 meets both Planes 1 and 3 at this point
and transfers 60KL each to both of them. Now all three have 60KL in their
tanks, and all head to the airport.
Time 360: All three planes land at airport, with zero fuel
in their tanks.
exactly ! I found this one only.
Interestingly, for some reason all the solution has zero redundancy.
P.s. oh wait.. seem everyone dies in your case.
At the last fuel transfer.
Plane 2 has used 60 fuel unit to go there at the position 60 unit far from airport, so it has 120 left... not enough for everyone to survive?
“The professor could’ve designed the plane to hold more fuel, but where’s the fun in that?”
Goddamn it, I give up-
in my solution they all explode in mid air, and I'm a mathematician but can't find a way around it
Alex Raxach then your pretty stupid because this was one of the easy riddles
ImYourSenpai No need to call him stupid. Why are people so rude nowadays?
no kidding, and i imagine he was joking since there is no reason for anything to explode, only run out of fuel and crash if that what was meant..
Vatsyayana the reason for the planes to explode would be atmospheric friction, but I was kidding anyway.
I'm an engineer. Here's the more simple solution: bring the gas tanks of the other planes with the professor. 1.5 times the amount of gas, able to refuel in the air, nobody dies, continuous trip: it follows all the rules stated, even the professor's intention of not wanting to redesign. He simply brings more that's not designed into the plane which may add to the mass, but 1.5 times the amount needed is well over enough.
I found the solution, took me way more time than it should
smart one, I just watched the answer - I know I'm stupid, but I just don't have enough patient!!! ^-^
patience*
Case in point
we have all something to learn ^~^
yes you are really the smart one, I now I saw my mistake =) haha
Solution: JUST DESIGN THE PLANE WITH MORE FUEL INSTEAD OF RISKING YOU,AND YOUR ASSISTANTS LIVES!!!
TheGamingParadiseBlue - Roblox They said that was no fun
WHO CARES HES BASICALLY SUICIDING WHAT YOUR SAYING
ok sorry ik but still who cares professorlivesmater
there are things called boats, cars and parachutes
TheGamingParadiseBlue - Roblox The professor is OP and a daredevil. He doesn't need to play safe when he's got that awesome math skills😂
If you devide the globe into 8 parts as in the video you can also use another strategy in Fukanos last 3 hours.
With Plane 2 and 3 loaded up with 180KL fuel each they both fly east for 45 minutes ending up with 135KL each, but 45KL from Plane 3 is transfered to Plane 2.
Plane 3 goes back to the airport to refuel meanwhile Plane 2 and Fukano meet at 270Longitude. Fukano has 0KL Fuel, meanwhile Plane 2 has 135KL. Plane 2 transfer 90KL to Fukano which is enough for Fukano to go back home.
Plane 2 on the other hand doesn't have enough and will run out of fuel at 315 Longitude with 45 minutes left but Plane 3 is there to meet up and Transfer 45KL extra to Plane 2 and both planes makes it back with Plane 3 having 90KL to spare.
My solution was less efficient and I still made it: I used in total 1060kl of fuel, while their solution used only 900 kl.
That means there is room for error here :) Wasn't that amazing after all
Mine used a full tank and a half. Calculate that for me lmao
Netts Deli
in total? That is not possible as just the professor himself has to use at least 2 tanks :)
i think I have a solution using 1000kl, so also less efficient. In my solution no plane rests or fuels the other two simultaneously, which is also kind of nice.
+alexander reusens I realized I made a miscalculation, I was so annoyed lmao
Yeah, I got one that used 1040 kl.
This is really just a math problem, not a riddle.
what is the difference?
This was not originally a riddle, it was for war. Google "Operation Black Buck" for more details.
Lord Lordington of Awesomeshire ikr
Literally all of the riddles on this channel are math or logic problems.
Jonathan Sharman that's what a riddle is...
I SOLVED A TED ED RIDDLE FOR THE FIRST TIME
Evan James me too!!!
Jacklyn Yeh me three
Evan James Me too!
of all the riddles this is the one u solved!!!!
There are some typos in the video, the the solution still works. At 2:36, where is says "45 minutes later", it is adding 45 minutes to the the time when plane 3 lands back at the airport (13:30). In reality, it should be 45 minutes later from when she starts heading back. So at 13:30, when #3 lands back at the airport, the other two are 1/4 of the way through their journey.
TED-Ed You needed seriously to place an AIRPLANE riddle after what happened few days ago?
What happened?
Wobbufet Stylish Yes and you now what was the cause of the acident the facts that the plane got Out of fuel
Wobbufet Stylish what is wrong with that?
Dan K a plane with 81 people run out of fuel when they were reaching the airport, there is there only 6 survivors.
Was a charter airplane were a football/soccer Brazilian club, they were going to play the first row of the SouthAmerica Cup Final in Colombia.
Are you claiming that this logic puzzle shows a lack of respect for the victims of the plane crash? If so, how?
I assumed the planes would refuel instantly with my solution, since the video didn't mention how long it takes to refuel. So here's what I got: all 3 planes fly out. At 60 minutes west, assistant 1 gives the professor 60kL of fuel and heads back. At 80 minutes west, assistant 2 gives the professor 20 kL of fuel and heads back. As soon as assistant 2 arrives to refuel, both assistants fly east. At 60 minutes east, assistant 1 gives assistant 2 60kL of fuel and heads back. At 100 minutes east, assistant 2 gives the professor 70kL of fuel and begins to head back with the professor. And finally at 30 minutes east, assistant 1 gives the professor and assistant 2 30kL of fuel each and they all head back.
KeanuIsAwesome refueling time is literally zero because they can transfer fuel in absolutely no time in mid air.. So on land it's a piece of cake...that is wat i think
It’s not possible because when you switch directions you would probably snap your neck due to the sudden change in direction combined with the g forces.
More safe solution by me is, (before that, lets name professor’s airplane #1 and his students #2 & #3) :
Now, all three of them will start at the EA (equator airport) with full capacity fuel, 4/4. At the 45* all three of them will have 3/4 fuel. At this point, airplane #2 will give 1/4 both #1 & #3 (in total 2/4) and so on it (#2) will head back to EA when #1 & #3 will go forward from 45* with max fuel, to 90* with 3/4 fuel. At this point #3 will give 1/4 to #1, which will let #3 with 2/4 and #1 with 4/4. Now they both will arrive to 135* with #1 3/4 and #3 with 1/4. Now #3 will give the #1 1/8 which will leave #3 to goes back to EA with 1/8 fuel and #1 with 7/8 fuel. Now #1 goes to 225* alone, and in there (in 225*), #1 will have 5/8 fuel remaining. #1 moves forward to 270* with 3/8 remaining. As soon as #1 arrives to 270*, #2 start the journey the opposite way with max capacity of fuel. Now they both (#1 & #2) will meet at 315* with #1 1/8 fuel remaining and #2 with 6/8 fuel remaining. At that point (315*), #2 give 4/8 to #1 and head back with 2/8 remaining and #1 head back with 5/8 fuel remaining which is more than enough fuel to head back safely bc there is only 45* remanding from the EA… ( 45* = 2/8 fuel).
So no one will have to reach 0 or to gamble their lifes…