Two Plus Two Newer Archives  

Go Back   Two Plus Two Newer Archives > Other Topics > Science, Math, and Philosophy

Reply
 
Thread Tools Display Modes
  #1  
Old 11-22-2007, 04:50 PM
jay_shark jay_shark is offline
Senior Member
 
Join Date: Sep 2006
Posts: 2,277
Default 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 !
Reply With Quote
  #2  
Old 11-22-2007, 06:05 PM
hitch1978 hitch1978 is offline
Senior Member
 
Join Date: Aug 2007
Posts: 466
Default Re: Two Olympiad Problems

Problem 2.

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

proof
Reply With Quote
  #3  
Old 11-22-2007, 06:49 PM
bigpooch bigpooch is offline
Senior Member
 
Join Date: Sep 2003
Location: Hong Kong
Posts: 1,330
Default 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>
Reply With Quote
  #4  
Old 11-22-2007, 10:00 PM
blah_blah blah_blah is offline
Senior Member
 
Join Date: Feb 2007
Posts: 378
Default 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.
Reply With Quote
  #5  
Old 11-22-2007, 10:35 PM
jay_shark jay_shark is offline
Senior Member
 
Join Date: Sep 2006
Posts: 2,277
Default 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 ?
Reply With Quote
  #6  
Old 11-23-2007, 03:38 AM
TomCowley TomCowley is offline
Senior Member
 
Join Date: Sep 2004
Posts: 354
Default 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.
Reply With Quote
  #7  
Old 11-24-2007, 02:49 PM
jay_shark jay_shark is offline
Senior Member
 
Join Date: Sep 2006
Posts: 2,277
Default 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 .
Reply With Quote
  #8  
Old 11-25-2007, 03:55 AM
David Sklansky David Sklansky is offline
Senior Member
 
Join Date: Aug 2002
Posts: 5,092
Default 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.
Reply With Quote
  #9  
Old 11-25-2007, 04:35 AM
pzhon pzhon is offline
Senior Member
 
Join Date: Mar 2004
Posts: 4,515
Default 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.
Reply With Quote
  #10  
Old 11-25-2007, 05:03 AM
David Sklansky David Sklansky is offline
Senior Member
 
Join Date: Aug 2002
Posts: 5,092
Default 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.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 10:24 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.