Pub Trivia
1 day ago
Monday, July 31, 2006
one-third base times height
Today I had to teach that the volume of a cone or pyramid is one-third base times height. This is not very exciting, and not very intuitive since we weren't proving it, only stating it, and it isn't even intuitive when you prove it (with calculus or with the formula for the volume of a sphere). So I made a model out of cardboard of the three congruent right pyramids that, when put together, make a cube. It was great. I ran out of white cardboard, so I made the inside faces colored, which means that when you assemble the cube, all you can see is white, which kind of helps with putting the cube together if you can't figure out how to do it. The students liked it, as did my master teacher. The pieces don't fit together exactly, but they are close enough to get the point across. I am very proud of my model to illustrate this principle, and I will probably keep my three right pyramids for a long time, if they don't get smushed.
Tuesday, July 25, 2006
Actually, maybe I don't deserve rights
... where "rights" is a clever allusion to the clever pun I made in my last title. I actually do deserve rights.
An astute mathematician pointed out to me that there is a fatal error in my reasoning in my last post, so ridiculously terrible that I am thinking about deleting the whole post and pretending it never happened. But I'm honest, so I won't delete it; I'll leave it for historical interest. The key is this: Triangles actually have to be on the plane x+y+z=180. Inside the pyramid, there are points like (1,2,3) which clearly are not the angles of any triangle. So, welcome to the plane.
The plane intersects the first octant in an equilateral triangle with vertices at (180,0,0), (0,180,0), and (0,0,180). An equilateral triangle appears at the centroid (60,60,60), and isosceles triangles appear along the medians, which I am not going to bother expressing in parametric form. I think that right triangles appear along the midlines (connect the midpoints to make a small equilateral triangle).
Here is what I propose: Draw in all the medians. This creates six 30-60-90 triangles. I assert that each one carries exactly the same triangles, just with the angle coordinates in a different order. There are six triangles, which corresponds to the 3!=6 ways to rearrange three different numbers. There are three different hypotenuses and three different short legs, which correspond to the isosceles triangles. The 3 corresponds to the 3!/2! = 3 ways to rearrange three numbers when two of them are the same. And there is one center point, because there is only one way to "rearrange" 60-60-60.
Thus, we need only inspect one of these small 30-60-90 triangles. The coordinates of one of them, for example, are (60,60,60), (0,0,180), and (0,90,90). In my picture, this is the one on the top right.
Now look at the midline that cuts across each small triangle (from (45,45,90) to (0,90,90) in the example). The midline acts as an altitude drawn from the right angle to the hypotenuse. This cuts the 30-60-90 triangle into two smaller, similar triangles, one of which is 1/4 the area and one of which is 3/4 of the area of the original. The smaller one, whose vertices are (60,60,60), (45,45,90), and (0,90,90) in the example, contains all of the acute triangles. The larger triangle contains the obtuse triangles, and the right triangles are on the dividing line.
So, in conclusion, 1/4 of triangles are acute, 3/4 are obtuse, and none of them are right. QED.
An astute mathematician pointed out to me that there is a fatal error in my reasoning in my last post, so ridiculously terrible that I am thinking about deleting the whole post and pretending it never happened. But I'm honest, so I won't delete it; I'll leave it for historical interest. The key is this: Triangles actually have to be on the plane x+y+z=180. Inside the pyramid, there are points like (1,2,3) which clearly are not the angles of any triangle. So, welcome to the plane.
The plane intersects the first octant in an equilateral triangle with vertices at (180,0,0), (0,180,0), and (0,0,180). An equilateral triangle appears at the centroid (60,60,60), and isosceles triangles appear along the medians, which I am not going to bother expressing in parametric form. I think that right triangles appear along the midlines (connect the midpoints to make a small equilateral triangle).
Here is what I propose: Draw in all the medians. This creates six 30-60-90 triangles. I assert that each one carries exactly the same triangles, just with the angle coordinates in a different order. There are six triangles, which corresponds to the 3!=6 ways to rearrange three different numbers. There are three different hypotenuses and three different short legs, which correspond to the isosceles triangles. The 3 corresponds to the 3!/2! = 3 ways to rearrange three numbers when two of them are the same. And there is one center point, because there is only one way to "rearrange" 60-60-60.
Thus, we need only inspect one of these small 30-60-90 triangles. The coordinates of one of them, for example, are (60,60,60), (0,0,180), and (0,90,90). In my picture, this is the one on the top right.
Now look at the midline that cuts across each small triangle (from (45,45,90) to (0,90,90) in the example). The midline acts as an altitude drawn from the right angle to the hypotenuse. This cuts the 30-60-90 triangle into two smaller, similar triangles, one of which is 1/4 the area and one of which is 3/4 of the area of the original. The smaller one, whose vertices are (60,60,60), (45,45,90), and (0,90,90) in the example, contains all of the acute triangles. The larger triangle contains the obtuse triangles, and the right triangles are on the dividing line.
So, in conclusion, 1/4 of triangles are acute, 3/4 are obtuse, and none of them are right. QED.
Monday, July 24, 2006
Acutes or obtuses? And give back my rights!
We had a problem in geometry a few days ago that involved saying whether something was an acute triangle or an obtuse triangle. This is an easy problem, because if one of the angles is greater than 90 degrees, you say "obtuse," and if none of the angles is greater than 90 degrees, you say "acute," and if two of the angles are greater than 90 degrees, you say "get back to plane geometry."
This problem got me to thinking: what are there more of, acute triangles or obtuse triangles? Which naturally led me to wonder: what is the ratio of acute triangles to obtuse triangles?
Now, one way you could do the problem would be to make a "simplification" like say that the angles have to be whole numbers, and then count. But then you'd have to do a lot of counting, and you'd get all confused with the right triangles. So this would be a bad way to solve the problem.
The way I decided to solve the problem was to imagine that the first angle is plotted on the x-axis, the second angle is plotted on the y-axis, and the third angle is plotted on the z-axis. Then your x, y, and z values can all range between 0 and 180, but with the constraint that x+y+z=180. This is the equation of a plane that intersects the axes at (180,0,0), (0,180,0), and (0,0,180), respectively.
All the triangles are contained between this plane and the x-y, y-z, and x-z planes, not inclusive (because the planes contain the "triangles" with angles of 0 and 180, which are not really triangles at all). This region is a triangular pyramid, whose volume is 1/3*(180*180/2*180) because it's 1/3 times base (180 x 180 / 2) times height (180). This is better known as 1/6*(1803).
Then you can think about just acute triangles. Plot the angles as above, and notice that they are constrained both by x+y+z=180 and by x, y, z < 90. x, y, z < 90 is a cube with a side length of 90, with corner coordinates like (0,0,0), (90,90,90), (0,90,90), etc. Note, then, that the solid that includes all the acute triangles is the intersection of this cube with the pyramid described above. Note also that the cube is not contained within the pyramid, because one of its corners is (90,90,90) and the point (60,60,60) is on the surface of the pyramid, the plane x+y+z=180.
The key to knowing the volume of the acute triangles is, then, the intersection of these two solids -- or, more helpfully, the volume of the cube that is inside the pyramid. To find this, we will figure out how much of the cube is outside of the pyramid. Notice that three vertices of the cube -- (0,90,90), (90,0,90), and (90,90,0) -- are on the plane x+y+z=180. Thus, the part of the cube that is outside the pyramid is a small pyramid, similar to the one described above, except with edges of 90 instead of 180. So the volume of this is 1/3*(90*90/2*90), which is 1/6 the volume of the cube, so the volume of acute triangles is 5/6*(903).
Thus, the ratio of acute triangles to all triangles is (5/6*903)/(1/6*1803), which is
(5/6*903)/(1/6*(2x90)3), which is
(5/6*903)/(1/6*23x903), and we can cancel out the 903 to get
(5/6)/(1/6*23), and 23 is 8, so we get
(5/6)/(8/6), which is
5/8.
So, 5/8 of triangles are acute triangles, and so 3/8 of triangles are obtuse triangles. QED.
For example, if you asked the question: If you randomly chose three angles that added up to 180, and made a triangle with those angles, what is the probability that it would be an acute triangle? The answer would be 5/8. That sort of thing.
This may make you wonder: What about right triangles? There are no eighths left to be right triangles!
That is because a 90 degree angle is a very precise number. If you could pick any three angles that added up to 180 -- not just positive integers, not just rational numbers, but any positive angles -- the probability of your picking exactly 90 degrees for one of them would be exactly 0. Sorry, Pythagoras.
My question is, clearly there is some overcounting here. For example, the math above counts (40, 60, 80) and (80, 60, 40) -- six different points like this -- as different pairs of angles, but clearly they all describe the same triangle. So clearly the math above is wrong (you better have caught that). How to fix it?
This problem got me to thinking: what are there more of, acute triangles or obtuse triangles? Which naturally led me to wonder: what is the ratio of acute triangles to obtuse triangles?
Now, one way you could do the problem would be to make a "simplification" like say that the angles have to be whole numbers, and then count. But then you'd have to do a lot of counting, and you'd get all confused with the right triangles. So this would be a bad way to solve the problem.
The way I decided to solve the problem was to imagine that the first angle is plotted on the x-axis, the second angle is plotted on the y-axis, and the third angle is plotted on the z-axis. Then your x, y, and z values can all range between 0 and 180, but with the constraint that x+y+z=180. This is the equation of a plane that intersects the axes at (180,0,0), (0,180,0), and (0,0,180), respectively.
All the triangles are contained between this plane and the x-y, y-z, and x-z planes, not inclusive (because the planes contain the "triangles" with angles of 0 and 180, which are not really triangles at all). This region is a triangular pyramid, whose volume is 1/3*(180*180/2*180) because it's 1/3 times base (180 x 180 / 2) times height (180). This is better known as 1/6*(1803).
Then you can think about just acute triangles. Plot the angles as above, and notice that they are constrained both by x+y+z=180 and by x, y, z < 90. x, y, z < 90 is a cube with a side length of 90, with corner coordinates like (0,0,0), (90,90,90), (0,90,90), etc. Note, then, that the solid that includes all the acute triangles is the intersection of this cube with the pyramid described above. Note also that the cube is not contained within the pyramid, because one of its corners is (90,90,90) and the point (60,60,60) is on the surface of the pyramid, the plane x+y+z=180.
The key to knowing the volume of the acute triangles is, then, the intersection of these two solids -- or, more helpfully, the volume of the cube that is inside the pyramid. To find this, we will figure out how much of the cube is outside of the pyramid. Notice that three vertices of the cube -- (0,90,90), (90,0,90), and (90,90,0) -- are on the plane x+y+z=180. Thus, the part of the cube that is outside the pyramid is a small pyramid, similar to the one described above, except with edges of 90 instead of 180. So the volume of this is 1/3*(90*90/2*90), which is 1/6 the volume of the cube, so the volume of acute triangles is 5/6*(903).
Thus, the ratio of acute triangles to all triangles is (5/6*903)/(1/6*1803), which is
(5/6*903)/(1/6*(2x90)3), which is
(5/6*903)/(1/6*23x903), and we can cancel out the 903 to get
(5/6)/(1/6*23), and 23 is 8, so we get
(5/6)/(8/6), which is
5/8.
So, 5/8 of triangles are acute triangles, and so 3/8 of triangles are obtuse triangles. QED.
For example, if you asked the question: If you randomly chose three angles that added up to 180, and made a triangle with those angles, what is the probability that it would be an acute triangle? The answer would be 5/8. That sort of thing.
This may make you wonder: What about right triangles? There are no eighths left to be right triangles!
That is because a 90 degree angle is a very precise number. If you could pick any three angles that added up to 180 -- not just positive integers, not just rational numbers, but any positive angles -- the probability of your picking exactly 90 degrees for one of them would be exactly 0. Sorry, Pythagoras.
My question is, clearly there is some overcounting here. For example, the math above counts (40, 60, 80) and (80, 60, 40) -- six different points like this -- as different pairs of angles, but clearly they all describe the same triangle. So clearly the math above is wrong (you better have caught that). How to fix it?
Tuesday, July 04, 2006
Not enough homework
Some of the students finished their homework before the end of the afternoon period today, so I think we need to give them more homework. They are just so advanced. Well, sort of.
Today I got to do more teaching, because I decided to actually organize the class by presenting solutions on the board at one time, and having the students work quietly on their homework at a different time, because it didn't work very well when I was trying to have one student show how to solve a problem on the board while the other students were working on their homework and not paying attention. It worked better today. I have decided that at this point I will have students come up to the board and work through problems just with me, and then later I will have them present their solutions to the class. It was a bit too early to try to get them to present to their classmates, and to get their classmates to listen, on the first day.
I realized recently how much free time we have here -- unless we have duty, we have all of Wednesday afternoon free, and Saturday afternoon and almost all of Sunday. So I could go somewhere even if it wasn't my weekend off. For example, I could up and decide to go to Williamstown, you know, tomorrow. So that's what I did, actually. The current plan is for me and another Williams student to go to Willytown tomorrow, just because we can, and also because we have both run out of cash and need free ATM. Clearly. And also because they have The Night Cafe' at the Clark right now, and really, how could I not see it when it's right there? So I had better see it when it is right there. Probably everyone I know at Williams will be busy watching the World Cup when I am there and won't want to talk to me, but I will just go over to the Clark and look at the excellent paintings and that will be okay.
Today I got to do more teaching, because I decided to actually organize the class by presenting solutions on the board at one time, and having the students work quietly on their homework at a different time, because it didn't work very well when I was trying to have one student show how to solve a problem on the board while the other students were working on their homework and not paying attention. It worked better today. I have decided that at this point I will have students come up to the board and work through problems just with me, and then later I will have them present their solutions to the class. It was a bit too early to try to get them to present to their classmates, and to get their classmates to listen, on the first day.
I realized recently how much free time we have here -- unless we have duty, we have all of Wednesday afternoon free, and Saturday afternoon and almost all of Sunday. So I could go somewhere even if it wasn't my weekend off. For example, I could up and decide to go to Williamstown, you know, tomorrow. So that's what I did, actually. The current plan is for me and another Williams student to go to Willytown tomorrow, just because we can, and also because we have both run out of cash and need free ATM. Clearly. And also because they have The Night Cafe' at the Clark right now, and really, how could I not see it when it's right there? So I had better see it when it is right there. Probably everyone I know at Williams will be busy watching the World Cup when I am there and won't want to talk to me, but I will just go over to the Clark and look at the excellent paintings and that will be okay.
Sunday, July 02, 2006
Once again, saving the day
Yesterday I saved the day in several ways.
Why does this happen? You may say it is because I am relatively short, but others are the same height and do not have this problem. And I haven't had bangs for four years, so it can't be that. When adults mistake me for a student -- the bookstore worker hesitated to give me the employee discount because I apparently don't look old enough to be a student -- they tell me it's a good thing, because I will be glad to look young when I am older. And perhaps it is. At least, it's very fortunate that I don't drink, because I am sure I would be heavily carded everwhere I went.
- I fixed the locks on two different doorknobs: one that wouldn't lock, and one that wouldn't unlock. Now they both do both. I accomplished this by hitting them really hard and twisting the doorknobs a lot.
- I provided a Phillips-head screwdriver to someone who needed it to fix her lamp. Impressively, she returned my multitool to me with the screwdriver put away, which means that she was able to figure out how to get past the safety feature.
- I unstuck a stuck suitcase zipper on the first try, thereby ending 10 minutes of frustrating effort by my advisee.
Why does this happen? You may say it is because I am relatively short, but others are the same height and do not have this problem. And I haven't had bangs for four years, so it can't be that. When adults mistake me for a student -- the bookstore worker hesitated to give me the employee discount because I apparently don't look old enough to be a student -- they tell me it's a good thing, because I will be glad to look young when I am older. And perhaps it is. At least, it's very fortunate that I don't drink, because I am sure I would be heavily carded everwhere I went.
Saturday, July 01, 2006
I got to sleep in until 8:00 today: delicious. I had a good workout, doing a loop that I learned about yesterday in the woods and on the roads. There were five pieces, each supposedly a little slower than 5k pace, and my HR at the end of each was 178, 181, 184, 183, and 188, respectively. So it was a good workout.
In other sporting news, Williams sports will be on CBS today:
I wonder if they will mention that Williams is once again, for the third year in a row, #1 in both academics (US News) and #1 in athletics (Sears Cup). Probably CBS sports doesn't care about academics. Too bad, really.
In other sporting news, Williams sports will be on CBS today:
Senior crew captain Meaghan Rathvon will be one of the featured athletes on the CBS NCAA Spring Championships Highlights Show airing this Saturday (7/1) at 3:00 p.m. EST. The CBS highlight show chronicles the NCAA championship events that were not televised live.I'll be busy with registration during that time, but I'm glad Williams sports are getting publicity on actual normal television (not television that you pay for, like cable), and getting publicity on television at all. That's great.
Rathvon overcame injury to help lead the Ephs to their second NCAA Division 3 women's rowing title in the past five years. Footage of the championship weekend and a brief feature on Rathvon will be included.
Other Eph spring teams may also appear on the show as all six Williams teams that entered a spring NCAA tourney finished in the top seven -- women's track & field (2nd), men's tennis and men's track & field (4th), women's tennis (5th) and softball (7th).
I wonder if they will mention that Williams is once again, for the third year in a row, #1 in both academics (US News) and #1 in athletics (Sears Cup). Probably CBS sports doesn't care about academics. Too bad, really.
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