Archive for the ‘Brain Teaser’ Category

Q: If you could see through the Earth, how big would Australia look from the other side?

Sunday, June 27th, 2010

The original question was: Relative to the size my feet appear when I’m standing up and looking at the ground, how large would Australia appear if I could see all the way through the Earth and observe its shape?  Also, if we considered my location to be a new “north pole”, how large would the “northern” hemisphere I observe seem relative to the “southern” hemisphere? In other words, due to the direct inverse relation between apparent size and distance, how much smaller does one half of a sphere appear from a point directly centered on the surface of the other half?

Physicist: This is example of “party trick mathematics”, the kind of math that you can do in your head, but that looks really complicated.  There’s a seriously old theorem from the days when togas meant math (not frat parties) called the “inscribed angle theorem”.  It says that if something has an angle on a circle of 2ϕ when seen from the center of the circle, then when seen from a point on the edge it will have an angle of ϕ.  What’s really surprising is that it doesn’t matter where you are on the circle.  It always works.

The Inscribed Angle Theorem: Surprising, but true.

I estimate that Australia spans about 34°.  Which means that, if you could see it through the Earth, it would take up about 17° of your vision.  Also, it wouldn’t matter where you are on the planet, it would always be 17°.  Unless you’re in Australia.  The size of where ever you are is always 180°.  Unless you’re on the beach or something (90°?).

Lucky for us (people), we all scale about the same.  There are some (literal) rules of thumb that you can use to estimate angles.  From standing, your feet are about 10°.  With your arm outstretched, the width of your thumb is about 1.5°, and your fist is about 7°.

So if you could see Australia through the ground, it would span about two and a half fists-at-arms-length, or a little less than two of your-own-feet-while-standing.  If you could see the other hemisphere (pick one), then it would appear to be exactly 90° across.

What follows is answer gravy:

Finally, for those of you who want to find exact arcangles on the Earth’s surface: If you have two locations at latitudes \gamma and \phi, and the difference in longitudes is \theta, then the true arcangle between them is:

\cos^{-1}{\left(\cos{(\gamma)}\cos{(\phi)}\cos{(\theta)}+\sin{(\gamma)}\sin{(\phi)}\right)}

Also, if you multiply this number by 6365, then you’ve got the distance between those points in km (as the crow flies).

Posted in -- By the Physicist, Brain Teaser, Geometry | 1 Comment »

Q: What would happen if an unstoppable force met with an unmovable, impenetrable object?

Thursday, April 22nd, 2010

Mathematician: Sometimes, when we don’t use language carefully enough, we can get ourselves into philosophical trouble. For example, consider the following statement:

If a barber shaves all those men (and only those men) who do not shave themselves, does he shave himself?

If the barber shaves himself, then he is shaving a man who shaves himself, which is something that (by definition) he does not do. On the other hand, if the barber does not shave himself, then there is a man who doesn’t shave himself that the barber doesn’t shave, which again contradicts our definition of the barber.

So what is the answer? Well, the question has no answer, because the definition we use for our barber contains within it a logical contradiction. What’s more, it is impossible for such a barber to actually exist in the real world, since the razor burn associated with simultaneously shaving yourself and not shaving yourself is too much for any single person to withstand.

Now, let’s return to the original question:

What would happen if an unstoppable force met with an unmovable, impenetrable object?

Well, let’s suppose that we define an “unstoppable” force to be one that can move absolutely any matter. Furthermore, let’s define an “unmovable” object to be one that cannot be moved by any force. In that case, this question is unanswerable, because like the barber paradox above, it relies on contradictory information. By definition our force can move anything, but then, also by definition, there is an object that the force cannot move. This is a bit like saying “suppose X is true, and not X is true. Then is X true?”. Here  X is the idea that the force can move anything, and not X is the idea that there is at least one object that cannot be moved by the force (which in this case is our unmovable object). Hence, this question has no answer because it relies on assumptions which contradict each other.

Posted in -- By the Mathematician, Brain Teaser, Philosophical | No Comments »

Q: If you were to break down an average human body into its individual atoms, and then laid the atoms out in a single straight line, how far would it stretch?

Wednesday, March 3rd, 2010

Physicist: Atoms are a little “fuzzy”, so their exact size is a little tricky to define.  So taking their size in terms of bond length, and looking at the most common elements in the human body (by mass: 65% oxygen, 18% carbon, and 10% hydrogen), you’ll find that 1kg of person will stretch about 7 trillion km.  So an average (80kg) human would extend about 550 trillion km, or about 14 billion loops around the equator, or 1.4 billion trips to the moon, or about 58 light years.

So you can fit a rich man through the eye of a needle, but be sure to coil him up after you string him out.  Otherwise the process will take at least 58 years.

Posted in -- By the Physicist, Biology, Brain Teaser | No Comments »

Q: If two trains move towards each other at certain velocities, and a fly flies between them at a certain constant speed, how much distance will the fly cover before they crash?

Thursday, January 21st, 2010

The brain teaser comes in a many variations. For example:

Trains A and B, 700 miles apart, are heading toward each other on a straight piece of track. Train A is going 85 mph while train B is going 55 mph. At the same moment, a bee that flies 110 mph is sitting on the nose of train A and begins flying toward train B. When it reaches train B it makes an instantaneous reversal of direction and flies back toward train A. It continues to change direction every time it runs into a train until both trains and the bee meet in a spectacular crash. What total distance did the bee fly before the big collision?

Mathematician: The difficult way to solve this problem is to figure out how much distance the bee (or fly) traveled before turning around each time it approached a train, and then sum these distances together. The easy way to solve it is simply to figure out how long it took the trains to crash, and then calculate how far the bee, which travels at a constant speed, must have gone during this amount of time.

More specifically: The bee always travels at the same speed V. If we can figure out how much time, T, the bee flew before the trains collided with each other, then the total distance D it flew will just be V T, the product of the velocity and time. We know V, so all that remains is to figure out T. To do this, we just need to calculate how long it takes for the trains to crash. If the first train has velocity v1 and the second v2, and the distance between them initially is d, then the time T before the crash will just be d/(v1+v2), which is equivalent to the amount of time that it takes a train going velocity v1+v2 to travel the distance d. The total distance traveled by the bee is given by:

D = V d / (v1+v2)

= (700 miles)  * (110 mph)/((55 mph)+(85 mph))

= 550 miles

Posted in -- By the Mathematician, Brain Teaser, Math | 1 Comment »

Q: How does the Monty Hall Problem work?

Saturday, December 26th, 2009

For those of you who aren’t familiar with The Monty Hall Problem: You’re on a game show where there is a prize hidden behind one of three doors (A, B, or C), and the objective is to guess the correct door.  After you make a guess the host of the show opens another door (that is not the one you picked) that has no prize behind it.  You are now given the option to stay with your original guess or switch to the remaining unopened door.  The “paradox” is that if you stay you’ll have a 1 in 3 chance of getting the prize, but if you switch to the remaining door you’ll have a 2 in 3 chance.

Mathematician: In my opinion, the easiest way to understand the Monty Hall problem is this: Suppose there are three doors, A, B and C and you originally chose door A. If you stay with your original door, then the only way that you win is if originally the prize was behind A, which has a chance of 1 in 3. If the prize was originally behind door B on the other hand (which also has a chance of 1 in 3), then when you pick door A, door C will be removed. Hence, if you switch you will be switching to door B, and therefore you will win. Finally, if the prize was originally behind door C (which again has a chance of 1 in 3) then door B will be removed, and if you switch you will be switching to door C and therefore will win. Hence, if you stay with your original door, you win if and only if the prize was originally behind door A. If you switch though, you win if it was originally behind either door B or door C. Since the chance of the prize being behind door A from the get go is 1 in 3, whereas the chance of it being behind either B or C from the get go is 2 in 3, you are better off switching!

Posted in -- By the Mathematician, Brain Teaser, Math | No Comments »