I remember this problem amazed me. I was surprised by its beautiful solution that I will show you later not to spoil some fun to those of you who wants to try to solve it.
Four robots are initially at the vertices of a unit square. They start moving simultaneously with the same speed, each one keeping its direction to the right closest neighbor. Where do they meet and what distance will each one cover to the meeting point?
Sunday, March 31, 2013
Saturday, March 30, 2013
Fun with toothpicks
There are many fun problems that involve constructions with toothpicks. One nice thing about them is that they can be enjoyed by adults and kids alike, starting at a fairly young age. Here is by far my favorite problem of this type.
It is easy to make 4 equilateral triangles using 9 toothpicks (see below), but can you make 4 equilateral (and equal) triangles using just 6 toothpicks?
Here are a few easier problems with toothpicks that I got from this book by Martin Gardner.
Warning: the solution to the problem above is completely different from the ones below.
1) Move four toothpicks to make three identical squares (with no toothpicks left over).
I apologize for the image. I had some trouble aligning the toothpicks!
2) Move one toothpick and make the house face east instead of west.
3) Move two toothpicks to make four identical squares (with no toothpicks left over).
4) Move three toothpick so that the triangular pattern points down instead of up.
Please share your favorite toothpick problems!
It is easy to make 4 equilateral triangles using 9 toothpicks (see below), but can you make 4 equilateral (and equal) triangles using just 6 toothpicks?
Warning: the solution to the problem above is completely different from the ones below.
1) Move four toothpicks to make three identical squares (with no toothpicks left over).
I apologize for the image. I had some trouble aligning the toothpicks!
2) Move one toothpick and make the house face east instead of west.
Please share your favorite toothpick problems!
A disclaimer and a plea
I came to the realization that I am not very good at judging the difficulty of certain problems that I am very familiar with. In particular, in my previous post I said that the problem that I discussed can be understood without any prior knowledge of the subject. This is true, but I did not realize that it might not be so easy to follow for someone not familiar with the notation and terminology. This was pointed out to me by several people and in a future post I would like to approach the same subject but taking a step backwards. I will present some specific examples and explain the concepts a bit more.
One of the goals of this blog is to collect fun math problems that could get and keep kids (and potentially some adults :-) interested in the subject. But perhaps an equally important goal is to find good ways to present problems/concepts/solutions and to become aware of what sort of prerequisites are needed to understand them. To that end, it would be great to receive comments from people with different levels of mathematical familiarity, saying what is understandable and what needs more explanation, what are better/different ways of presenting the same material, and finally just which problems they like and which ones they find uninteresting and uninspiring. All comments are welcome as long as they are in some way relevant (and this should be interpreted in a very broad sense). We are also always very happy when people share their fun problems with us, whether they are related to something we post about or not.
One of the goals of this blog is to collect fun math problems that could get and keep kids (and potentially some adults :-) interested in the subject. But perhaps an equally important goal is to find good ways to present problems/concepts/solutions and to become aware of what sort of prerequisites are needed to understand them. To that end, it would be great to receive comments from people with different levels of mathematical familiarity, saying what is understandable and what needs more explanation, what are better/different ways of presenting the same material, and finally just which problems they like and which ones they find uninteresting and uninspiring. All comments are welcome as long as they are in some way relevant (and this should be interpreted in a very broad sense). We are also always very happy when people share their fun problems with us, whether they are related to something we post about or not.
Another favorite problem
One more from the list of my favorites.
Prove that there are 100 consecutive composite (nonprime) natural numbers.
I like it because everyone can understand and solve it, but the idea of the proof is relatively deep.
Prove that there are 100 consecutive composite (nonprime) natural numbers.
I like it because everyone can understand and solve it, but the idea of the proof is relatively deep.
Friday, March 29, 2013
My favorite problem
I should have started with my favorite problem. I gave it to different audiences, ages 10 to whatever. Usual results are always the same: two or three people do it correct. The problem is many times older than I am. Here it is (more modern variation).
The distance between A and B is 100 km. Two bicyclists start moving simultaneously towards each other from A and B respectively with speeds 10 km/h and 15 km/h. A dog starts running from A at the same moment as bicyclists do with the speed 20 km/h. When it reaches the second bicyclist, it turns back, runs until meeting the first one, turns again etc. The dog runs back and forth until bicyclists meet. What distance does it run?
As I mentioned above, the number of fifth graders who solved this problem is approximately the same as the number of high school seniors. BTW the results were the same in USA, Russia, and Ukraine :-)
The distance between A and B is 100 km. Two bicyclists start moving simultaneously towards each other from A and B respectively with speeds 10 km/h and 15 km/h. A dog starts running from A at the same moment as bicyclists do with the speed 20 km/h. When it reaches the second bicyclist, it turns back, runs until meeting the first one, turns again etc. The dog runs back and forth until bicyclists meet. What distance does it run?
As I mentioned above, the number of fifth graders who solved this problem is approximately the same as the number of high school seniors. BTW the results were the same in USA, Russia, and Ukraine :-)
Another good book
The only part of a school math course I have never taught is geometry. My favorite part of a standard school course :-) I come back to it from time to time in our Math club. But most of my students do not share my passion. Maybe one day I will ask to try to teach it. Not now, I am not ready yet.
BTW I should mention a book that I like. It connects geometry and nature (physics in particular) according to the ideas of a late V. Arnold. This book is Measurement by Paul Lockhart. I do enjoy it, there are many good geometry problems on each page. If you like geometry, open this book, and you will have a lot of fun.
BTW I should mention a book that I like. It connects geometry and nature (physics in particular) according to the ideas of a late V. Arnold. This book is Measurement by Paul Lockhart. I do enjoy it, there are many good geometry problems on each page. If you like geometry, open this book, and you will have a lot of fun.
Thursday, March 28, 2013
Construction problems in geometry II
First of all I would like to share with you a book that I got yesterday (it was a real surprise from Amazon, the book was delivered at 8:00 PM, and I expected it only today or tomorrow). The secrets of triangles is a great book. It includes, among others, a big section on constructing triangles using a compass and straight edge given three different elements of a triangle. I started to read it and I am really enjoying this reading.
Now back to problems. A separate group of construction problems consists of ones that should be solved by a compass only. My favorite one was introduced by Napoleon (yes, the emperor).
Given a segment of the length one, construct a segment of the length $\sqrt{2}$ units.
Obviously, you cannot draw a segment without a straight edge, just find the endpoints.
The solution to the problem from yesterday's post is given by two figures below.
Now back to problems. A separate group of construction problems consists of ones that should be solved by a compass only. My favorite one was introduced by Napoleon (yes, the emperor).
Given a segment of the length one, construct a segment of the length $\sqrt{2}$ units.
Obviously, you cannot draw a segment without a straight edge, just find the endpoints.
The solution to the problem from yesterday's post is given by two figures below.
AB is a given diameter and C is a given point. AC intercepts given circle at D, and BC intercepts it at E. Angles ADB and AEB are right angles because AB is a diameter. AE and BD are altitudes of BC and AC respectively. The line from C through the point their intersection is an altitude to AB.
Counting in two ways
One of the many reasons why I like the subject of combinatorics is that many problems in this area can be understood by someone with very little (or sometimes even zero) knowledge of the field. The solutions to these problems can sometimes be likewise simple (but often beautiful and insightful) or extremely difficult (a classic example is the Four Color Theorem which until very recently was a conjecture that remained unsolved for several hundred years). Here I will stick with problems of the first kind.
In this post I will present some formulas (or combinatorial identities) that can be explained by counting the same set of objects or quantities in two different ways. Here is a simple example:
Let \( {n \choose k}\) be the number of ways that k objects can be chosen from a set of n objects (this is the familiar binomial coefficient). We want to show that \[ {n\choose k} = {n-1 \choose k} + {n-1 \choose k-1}.\] Of course one can write out the formula for \({n \choose k}\) using factorials and then algebraically manipulate the right hand side so that it matches the left hand side. However, that will provide you with no intuition about what is actually going on. Here is the combinatorial, and much more insightful, explanation. As already mentioned, the left hand side counts the number of ways we can select k objects from a set of n objects. We want to show that the right hand side counts the same things. For a specific object (out of the n), every subset of k objects either contains it or does not (there is no third possibility!). Now it is easy to see that there are \({n-1 \choose k}\) ways to choose k objects without picking our specific one and there are \({n-1 \choose k-1}\) to choose k objects if we want to include our specific one. This explains the formula!
Here are a few more combinatorial identities that have similarly simple combinatorial interpretations:
\[\sum_{k=0}^n {n\choose k} = 2^n\]
\[ \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} {n \choose 2k} = 2^{n-1} \]
\[ \sum_{k=0}^m {n-k \choose m-k} = {n+1 \choose m} \]
\[ \sum_{k=0}^m {u \choose k}{v \choose m-k} = {u+v \choose m}\]
I will write explanations for these in a future post. All of these identities, as well as many others can be found in the book Combinatorial Problems and Exercises.
In this post I will present some formulas (or combinatorial identities) that can be explained by counting the same set of objects or quantities in two different ways. Here is a simple example:
Let \( {n \choose k}\) be the number of ways that k objects can be chosen from a set of n objects (this is the familiar binomial coefficient). We want to show that \[ {n\choose k} = {n-1 \choose k} + {n-1 \choose k-1}.\] Of course one can write out the formula for \({n \choose k}\) using factorials and then algebraically manipulate the right hand side so that it matches the left hand side. However, that will provide you with no intuition about what is actually going on. Here is the combinatorial, and much more insightful, explanation. As already mentioned, the left hand side counts the number of ways we can select k objects from a set of n objects. We want to show that the right hand side counts the same things. For a specific object (out of the n), every subset of k objects either contains it or does not (there is no third possibility!). Now it is easy to see that there are \({n-1 \choose k}\) ways to choose k objects without picking our specific one and there are \({n-1 \choose k-1}\) to choose k objects if we want to include our specific one. This explains the formula!
Here are a few more combinatorial identities that have similarly simple combinatorial interpretations:
\[\sum_{k=0}^n {n\choose k} = 2^n\]
\[ \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} {n \choose 2k} = 2^{n-1} \]
\[ \sum_{k=0}^m {n-k \choose m-k} = {n+1 \choose m} \]
\[ \sum_{k=0}^m {u \choose k}{v \choose m-k} = {u+v \choose m}\]
I will write explanations for these in a future post. All of these identities, as well as many others can be found in the book Combinatorial Problems and Exercises.
Wednesday, March 27, 2013
I am a rare guest on Google +, but today I went there and found two posts that I really liked - a picture of Paul Erdos and Terence Tao and a trailer of a new movie "Travelling salesman" about a group of four mathematicians who solved P vs. NP problem.
A picture is a great symbol of the continuity of math - the major figure of the 20th century explains something to a kid who is a face of 21st century math.
I am not so sure about movie, but I'll try to see it when it comes to theaters or DVDs.
A picture is a great symbol of the continuity of math - the major figure of the 20th century explains something to a kid who is a face of 21st century math.
I am not so sure about movie, but I'll try to see it when it comes to theaters or DVDs.
Construction problems in geometry I
I always loved construction problems. I will write a lot about them. Today I talked to my high school math teacher and among other things I mentioned constructions with a compass only. He replied that constructions with a straight edge only are the interesting kind as well. I have never seen one before, so he immediately gave me one to solve. It is relatively simple, but I like it.
You are given an arbitrary circle. A diameter is drawn in this circle. You are also given a point outside the circle. Using straight edge only, construct a perpendicular line from a given point to a given diameter.
When solving, do not forget that a perpendicular line can intercept a diameter itself or its continuation.
I will post a solution in the next post in order not to spoil someone a pleasure to find a solution by oneself.
You are given an arbitrary circle. A diameter is drawn in this circle. You are also given a point outside the circle. Using straight edge only, construct a perpendicular line from a given point to a given diameter.
When solving, do not forget that a perpendicular line can intercept a diameter itself or its continuation.
I will post a solution in the next post in order not to spoil someone a pleasure to find a solution by oneself.
Off we go! Why did we decide to start this blog? Because we love math! There are a lot of great problems, models, facts etc. that upon discovery we share with each other, or sometimes friends and family, but then forget or lose forever. Similarly, we lose track of fun/interesting websites, books, magazine articles, so that when we want to refer to them again, we cannot find the link or bookmark. One might ask why not to use Evernote, Diigo, or other such social bookmarking services? Mainly because we want to share and discuss our findings with others, test our ways of presenting stuff, and get new "gems" from our potential readers :-) In short, we would like to have some fun with math.
Not last on our list of priorities is to be able to share our love of math with our kids and grandchildren. By gathering problems of various levels now, we can share them with the kids when they become age appropriate. Of course we are also interested in cool ideas that we can already present to them, so we are always on the lookout for fun facts that can be shared with 3-4 year olds.
Not last on our list of priorities is to be able to share our love of math with our kids and grandchildren. By gathering problems of various levels now, we can share them with the kids when they become age appropriate. Of course we are also interested in cool ideas that we can already present to them, so we are always on the lookout for fun facts that can be shared with 3-4 year olds.
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