2 Olympiad Geometry problems I like

[sirio11]

Problem #1:

Let P_1,P_2,P_3,.......,P_n points contained in a circle with radius = r with P_1 being the center of the circumference.
Let d_k = min{//P_j - P_k// j <> k } (This is the minimum of the distances from the other points to P_k)

Prove that: (d_1)^2+(d_2)^2+(d_3)^2+.......+(d_n)^2 <= 9(r^2)

Problem # 2 (Hard):

Let abcdef an hexagon with every pair of sides parallel. Any 3 vertices can be covered by a wide stripe of length = 1 (This is the length between the lines of the infinite strip).

Prove that the hexagon can be covered by a wide stripe of length = sqrt(2)

[furyshade]

here is another one that one of my brothers friends got when he did it, i liked it a lot and the solution is cool

what is the remainded of (1!+2!+3!...100!)/12

[blah_blah]

[ QUOTE ]
here is another one that one of my brothers friends got when he did it, i liked it a lot and the solution is cool

what is the remainded of (1!+2!+3!...100!)/12

[/ QUOTE ]

these problems have absolutely nothing in common at all. also, n! is divisible by 12 for n>3, so your problem is trivial (the answer is 1!+2!+3! = 9)

[furyshade]

[ QUOTE ]
[ QUOTE ]
here is another one that one of my brothers friends got when he did it, i liked it a lot and the solution is cool

what is the remainded of (1!+2!+3!...100!)/12

[/ QUOTE ]

these problems have absolutely nothing in common at all. also, n! is divisible by 12 for n>3, so your problem is trivial (the answer is 1!+2!+3! = 9)

[/ QUOTE ]

i never said they had anything to do with it, other than that they both were olympiad questions

[blah_blah]

pray tell which olympiad that was on?

[thylacine]

[ QUOTE ]
pray tell which olympiad that was on?

[/ QUOTE ]

The special olympiad.

[sirio11]

Meh, I don't have a problem with calling your question Olympiad, there all all kind of levels in the math Olympiad.
Probably for this forum easy problems are better, since we don't use too much time to post answers (Look at my problems, no answers yet).

A little variation I thought for your problem is:

For which k, 1+2+3+....+k divides 1!+2!+3!+....+k!

[Phil153]

#2 is tilting me. I can see it's true by drawing a diagram (since equal sides form the maximum ratio of hexagon width:strip length) but the proof is eluding me. It's something to do with the angles of the largest width internal triangle and how the points of that triangle relates to the lengths of ab,bc,and cd (and therefore the angles and therefore extra width required).

I'm missing something simple.

[jay_shark]

1) Are all the points on the circumference of the circle ?

The lhs of the inequality can be shown to be maximized when the n points form a regular polygon . Equality happens when there are 3 points equally spaced from each other .

Proof: Suppose we have n-1 points and we've determined d1,d2,d3,...,d(n-1). Now let the variable point Pn be between 2 random points . It's easy to see that if we select pn to be bisect the arc formed from the 2 points, then we've maximized dn . Now that dn has been chosen , we can continue this argument to prove that all the points must form a regular pentagon to maximize the lhs of the inequality .
Now use cosine law to prove that the sum of the squares of the sides cannot exceed 9r^2 .

Equality occurs when we have an equilateral triangle with sides of length sqrt3r .

[TomCowley]

Let ACE be the hardest set of vertices to cover. Now, the easiest way to cover these is to have two points on a line opposite the side of the triangle ACE with the largest angle. So the distance between the vertex with the largest angle and the opposite side will be 1. Now the shortest distance across the hexagon will be either AE, AC, or CE (since the sides are parallel). The shortest distance will be opposite the smallest angle in ACE, and by the definition of sine will be 1/sin(180-largest angle-angle).

Since the largest possible angle in ACE is approaching 90 degrees (can't be bigger in a standard no angle>180 hexagon), that leaves 90 degrees to divide among the smaller angles. So the largest possible smallest angle is approaching 45 degrees. So the value of 180-largest angle-angle will be approaching 45 from above. So the value of sin(180-largest angle-angle) will be approaching 1/sqrt(2) from above. So the value of 1/sin(180-largest angle-angle), which is our final answer, will be approaching sqrt(2) from below, so the upper bound of sqrt(2) is proven.

Edit: took the wrong complement, fixed.