Two Olympiad Problems
 

1) Show that 1! + 2! + 3! + ...+ n! is not the square of any integer for n>3 .

2)Prove that of all the triangles with a given area K , the equilateral triangle has the shortest perimeter .

Neither of these problems should be too problematic but their solutions are simple . If you have an elegant solution to these problems then please post them .

Good luck !

Re: Two Olympiad Problems
 

Problem 2.

The equalateral is closer to a circle, I did the proof here-

proof

Re: Two Olympiad Problems
 

2) There are several methods, but maybe I am too old to
think of what is considered the simplest proof.

1) Too easy:
<font color="white">
For n&gt;3, the last digit of the sum ends in 3, so obviously
can't be a square.
</font>
Better problem IMHO is to show that 1! + 2! +...+ n! is not
a kth power for n&gt;3, k&gt;=2.
Hint for this:
<font color="white">
note the sum is divisible by 9 for n&gt;=5;
for n&gt;=8, look at (mod 27)
and simply check for n&lt;8
</font>

Re: Two Olympiad Problems
 

straightforward solution to 2)

write

K = \sqrt{s(s-a)(s-b)(s-c)}

by heron formula,

substitute a=x+y,b=x+z,c=y+z for positive x,y,z (always possible in triangle). heron's formula becomes k = \sqrt{(x+y+z)(xyz)}

now it is straightforward to verify that (x+y+z)^2 \geq 3^{3/2} \sqrt{(x+y+z)(xyz)} = 3^{3/2}*K, with equality iff x=y=z.

in particular, we see that in an arbitrary triangle, 1/4 * P^2 \geq 3^{3/2} * K, with equality iff P is equilateral.

Re: Two Olympiad Problems
 

Good answer Blah , and thx for the problem BigPooch .

For question 2 , I originally solved it using Herons Formula , but quickly realized that it wasn't even necessary .

Can you think of another solution ?

Re: Two Olympiad Problems
 

Given a triangle ABC, area = 1/2AB^2(Sin A)(Sin B)/(Sin C). For any triangle, holding AB and Sin C (the opposite angle) constant, and performing an infinitesimal change in angles A and B has the following property:

Because the second derivative of Sin is negative throughout the legal range of angles, bringing the angles infinitesimally closer together while keeping their sum the same increases the product (area), moving them further apart decreases the product (area).

So the maximum area is when the angles are equal, so any non-equilateral triangle does not maximize area for a given perimeter, which is equivalent to stating that perimeter is not minimized for a given area.

Re: Two Olympiad Problems
 

Good answer Tom .

Here is another solution :

Fix two points A and B and a variable point C on the line parallel to the line AB . Clearly the area of ABC is fixed but we wish to minimize the distance AC + CB . Let A' be the reflection of A on the parallel line and so A'C + CB is minimized when we have a straight line which happens when C is on the perpendicular bisector of AB . Now if we fix B and C and let A be the variable point , then it's immediately clear that the shortest perimeter occurs when we have an equilateral triangle .

Re: Two Olympiad Problems
 

Here is the way that requires neither geometrical ingenuity nor taking second derivatives of trigonemetric functions. (If Tom Cowley does that again his reign here will be short lived.)

Take a string six inches long and attach the ends. Thumbtack a potion of it on a horizontal line.Pull the rest up to make a triangle and notice that it is highest, and thus has greatest area, when it is isosceles.

Draw your altitude to form two right trianges. The hyptenuse (H) and the right triangle's base (b) add up to 3, half the perimeter. The area of the original triangle is the altitude times the right triangle's base.

The Pythagorean Theorem tells us that the altitude is the square root of [(3-b)squared - b squared]. Which is the square root of (9-6b).

So the area of the isosceles triangle is the square root of (9b squared -6b cubed).

The derivative of that is (18b-18b squared)/2 blah blah blah. Setting that derivative equal to zero we reduce to 18b = 18. b=1. Area is at a maximum when the triangle is 2 by 2 by 2.

Re: Two Olympiad Problems
 

[ QUOTE ]

Take a string six inches long and attach the ends. Thumbtack a potion of it on a horizontal line.Pull the rest up to make a triangle and notice that it is highest, and thus has greatest area, when it is isosceles.

[/ QUOTE ]
That applies to each possible base, so any two sides are equal. Nothing more is needed.

Re: Two Olympiad Problems
 

[ QUOTE ]
[ QUOTE ]

Take a string six inches long and attach the ends. Thumbtack a potion of it on a horizontal line.Pull the rest up to make a triangle and notice that it is highest, and thus has greatest area, when it is isosceles.

[/ QUOTE ]
That applies to each possible base, so any two sides are equal. Nothing more is needed.

[/ QUOTE ]

I don't think an average math student would get what you are saying with those few words.

Re: Two Olympiad Problems
 

When are you going to write that book Mr Sklansky? You've lived a life of sin and debauchery and if there is a God he won't be pleased.

Maybe a brilliant textbook will get you over the line.

Re: Two Olympiad Problems
 

The "elliptical idea" seems the most elegant IMO (but I'm
biased since this was the idea that struck immediately).

Re: Two Olympiad Problems
 

[ QUOTE ]
Here is the way that requires neither geometrical ingenuity nor taking second derivatives of trigonemetric functions. (If Tom Cowley does that again his reign here will be short lived.) Take a string six inches long and attach the ends. Thumbtack a potion of it on a horizontal line.Pull the rest up to make a triangle and notice that it is highest, and thus has greatest area, when it is isosceles.

[/ QUOTE ]

Bah. That's exactly the observation I made, except I gave a proof that it was true instead of just stating that it was true.

Re: Two Olympiad Problems
 

You don't have to use trigonometry to prove that the height is maximized when the triangle is isoceles. You can use the triangle inequality after adding a reflected copy of the triangle above. So, this whole problem can be done without calculus.

Re: Two Olympiad Problems
 

[ QUOTE ]
You don't have to use trigonometry to prove that the height is maximized when the triangle is isoceles. You can use the triangle inequality after adding a reflected copy of the triangle above. So, this whole problem can be done without calculus.

[/ QUOTE ]

All calculus problems can be done without calculus. When my father was teaching mathematical logic at City College he would sometimes ask for a calculus problem and demonstrate.

Re: Two Olympiad Problems
 

[ QUOTE ]
[ QUOTE ]
You don't have to use trigonometry to prove that the height is maximized when the triangle is isoceles. You can use the triangle inequality after adding a reflected copy of the triangle above. So, this whole problem can be done without calculus.

[/ QUOTE ]

All calculus problems can be done without calculus. When my father was teaching mathematical logic at City College he would sometimes ask for a calculus problem and demonstrate.

[/ QUOTE ]

Two comments:

1) I've never understood the resistance to using calculus on these types of problems, as if it is some kind of failure. Finding different roads to a solution is often very interesting, but what's wrong with calculus?

2) All calculus problems? How would one frame a generic differential equation in language that does not involve calculus? Is there some kind of geometric trickery that will give me Bessel functions?

Re: Two Olympiad Problems
 

[ QUOTE ]
I've never understood the resistance to using calculus on these types of problems, as if it is some kind of failure. Finding different roads to a solution is often very interesting, but what's wrong with calculus?

[/ QUOTE ]
First, it makes the solution more opaque. I'm not so interested in solving any one problem. I'm interested in acquiring powerful techniques which will let me attack additional problems. Since I know calculus, seeing a calculus solution generally does not help me with future problems, but seeing something like my reflection argument very well might.

Second, the non-calculus arguments were much simpler and easier to remember, both to show A=B, and then A=B=C.

[ QUOTE ]
All calculus problems? How would one frame a generic differential equation in language that does not involve calculus? Is there some kind of geometric trickery that will give me Bessel functions?

[/ QUOTE ]
Right, I don't buy that. Parts of calculus may be formally removable, but only by greatly increasing the complexity of calculations in general. I don't want to see someone try to remove calculus from a simple optimization with Lagrange multipliers, or a curvature calculation, or a deconvolution to sharpen a blurry image.

Re: Two Olympiad Problems
 

[ QUOTE ]
All calculus problems can be done without calculus.

[/ QUOTE ]
Yes, and all computer programs don't need computers to run them. I file this sort of thing under "true but irrelevant."

However, I do generally agree that a simple "stick in the sand" solution is preferred over one that mindlessly uses a technique. BUT, given that one understands the proof of why the technique works, the general method becomes more attractive. Lagrange multipliers (from another thread) is a good example, since the proof can be motivated very easily using geometry of graphs, tangent planes, etc. Of course, this is mostly speaking for advanced math students.

But this brings me to my question: DS, why are you so bent on explaining things in a way so that the average math student can understand them? I can understand the motivation to sell your publications, so if that is all, then so be it.

Beyond that, do you really think these simpler explanations really improve the average student's day to day decision making? If they weren't smart enough to solve a problem to begin with, what makes you think that they will be smart enough to recognize when they should be clever and actually employ what they have learned? I've found (in my own teaching) that the latter is really the most difficult hurdle. You can make people understand things, but you can't make them recognize when it is appropriate to apply what they have understood. That is, unless they are smart enough to understand more difficult and general techniques, and then we are back where we started.

Re: Two Olympiad Problems
 

[ QUOTE ]
[ QUOTE ]
All calculus problems can be done without calculus.

[/ QUOTE ]
Yes, and all computer programs don't need computers to run them. I file this sort of thing under "true but irrelevant."

However, I do generally agree that a simple "stick in the sand" solution is preferred over one that mindlessly uses a technique. BUT, given that one understands the proof of why the technique works, the general method becomes more attractive. Lagrange multipliers (from another thread) is a good example, since the proof can be motivated very easily using geometry of graphs, tangent planes, etc. Of course, this is mostly speaking for advanced math students.

But this brings me to my question: DS, why are you so bent on explaining things in a way so that the average math student can understand them? I can understand the motivation to sell your publications, so if that is all, then so be it.

Beyond that, do you really think these simpler explanations really improve the average student's day to day decision making? If they weren't smart enough to solve a problem to begin with, what makes you think that they will be smart enough to recognize when they should be clever and actually employ what they have learned? I've found (in my own teaching) that the latter is really the most difficult hurdle. You can make people understand things, but you can't make them recognize when it is appropriate to apply what they have understood. That is, unless they are smart enough to understand more difficult and general techniques, and then we are back where we started.

[/ QUOTE ]

Its strange that you used this thread as impetus for this question. Because my answer was not the one that used the least math, though I do think it was the easiest to understand. If you know basic calculus.

Anyway it isn't people who aren't smart enough to understand more difficult and general techniques that I try to reach. Its only the twenty percent or so of the population who could, with difficulty, understand them, but choose not to learn them.