Two Olympiad Problems

[jay_shark]

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 !

[hitch1978]

Problem 2.

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

proof

[bigpooch]

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>

[blah_blah]

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.

[jay_shark]

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 ?

[TomCowley]

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.

[jay_shark]

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 .