2 Olympiad Geometry problems I like

[TomCowley]

I think you're forgetting that no distance can ever be greater than r, since all points are at most r from P_1. The hexagon gives 7r^2.

[jay_shark]

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I think you're forgetting that no distance can ever be greater than r, since all points are at most r from P_1.

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P1 is center of the circle or circumference ?
I've heard of a point being a center or a hemisphere , but I've never heard of a point being the center of a circumference .

[TomCowley]

I assumed it was the center of the circle. I don't know what center of the circumference would even mean (it's on the circumference?). OP can clarify.

[TomCowley]

Ahh, the problem is a pain in the ass because it's actually <=7 (hexagon), but I guess it's really hard to prove that's optimal. Assuming that points are evenly spaced (tedious but not hard to show that non-evenly spaced points can never be optimal), we can treat this as a circle packing problem. Each circle will have radius R/n. Now, since the minicircles must be centered in the big circle, they can fill a maximum area of pi*(R+R/n)^2. Dividing areas of the circles means that there can be at most pi*(R+R/n)^2/(pi*(R/n)^2) = n^2 + 2n + 1 minicircles of radius R/n. Now the sum-of-squares metric that we're trying to compute states that we have at most n^2 + 2n +1 points at a distance of 2R/n apart, so the sum is at most (n^2 + 2n + 1)*4R^2/n^2 = 4R^2 (1 + 2/n + 1/n^2). Obviously the bigger n is, the smaller this expression is, so we want the minimum n. Since n<2 means there can only be one point (since we have a point at the middle of the circle, the center of the next minicircle would be outside the circle), which is trivially true, the smallest meaningful n is 2. Plugging that in gives 4R^2 (1 + 1 + 1/4) = 9r^2, which is an upper bound.

And, amazingly, the n=2 solution is the hexagon.

[thylacine]

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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.

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Good start. Actually AE, AC, CE (and BD, DF, FB) are upper bounds for hexagon width so you just need one being at most sqrt(2).

Also if ACE (or BDF) has no obtuse angle, I believe your argument does this case.

But you still have to the case where ACE and BDF both have an obtuse angle.

[jay_shark]

Tom , the sum of the squares of the sides of a hexagon inscribed in a circle of radius r is 6*r^2 .

An equilateral triangle works .

Let a be the side length of the equilateral inscribed in a circle of radius r .


a^2 = r^2+r^2-2*r^2cos120
a^2 = 3^r^2
a = r*sqrt3

So the sum of the squares is 9*r^2

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The sum of the squares of the sides of a hexagon inscribed in a circle is 6*r^2 .

In general , the sum of the squares of the sides of a polygon should decrease as n increases .

[sirio11]

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I think you're forgetting that no distance can ever be greater than r, since all points are at most r from P_1. The hexagon gives 7r^2.

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This is correct. P_1 is the center of the circle

[TomCowley]

You're forgetting the center point, which has distance r, so 6r^2 for the hexagon sides, + r^2 for the center point, =7r^2.

[jay_shark]

Right .

Good thing he cleared up that P1 is on the center of the circle .

[TomCowley]

Good catch. If ACE is obtuse, then it's tougher to cover ACF and/or BCF, where C/F are the obtuse angles, than it is to cover ACE, and the shortest distance across the hexagon is CF, not AC/CE/AE. Then the problem reduces to the parallelogram ACEF, and by similar analysis as for ACE in the other post, the CF distance is maximized to sqrt(2) as EFA reduces to 90 degrees from above.