Sides = Diagonal

[BMAD]

How many sides of the convex polygon can equal its longest diagonal?

[k-man]

at most 2. These sides must have a common vertex that will also be one of the vertices for the longest diagonal. A simplest example of such polygon is a kite formed by 2 identical tall isosceles triangles connected by a common side. The common side becomes the longest diagonal of the kite.

[harey]

Make the sides as small as possible and you approach a circle.

[plasmid]

I agree with k-man.

Find the longest diagonal (or one of the longest diagonals if there are multiple diagonals that are tied for longest) and call the two vertices involved A and B. Because AB is a longest diagonal, all other points of the polygon must lie within a circle of radius AB centered on A, and similarly within a circle of radius AB centered on B. Suppose there are two vertices C and D that share a side CD of length AB. They must fall within the intersection of the circles of radius AB centered on A and B, and they must both fall on the same side of line AB (or else the polygon could not be convex). All pairs of points within that area (if you exclude points A and B) will be less than distance AB apart. So no side CD can be as long as the longest diagonal unless it includes one of the vertices involved in the longest diagonal.

This should be easy to see if you consider "diagonal" to be any line between two vertices. If you consider "diagonal" to mean any line between two vertices but excluding any pairs of vertices that share a side, then consider what happens if you try to make a side CD of length AB where one (or both) of C or D shares a side with A or B so they don't need to be within the intersection of circles A and B. C and D must still both be on the same side of line AB in order for the polygon to be convex, and you can define point C as the closest to A (out of C and D) if you were to walk toward A along the perimeter of the polygon without crossing AB, and D as the closest to B. Then C must still lie within circle B, D must still lie within circle A, and they must form a convex polygon while C and D are both on the same side of AB. While it's difficult to explain in words why creating such a line CD of length AB is impossible, it's probably easy to convince yourself of the fact by looking at a picture. The pertinent conclusion is that, regardless of which definition of diagonal you choose, any side of length AB must include point A or B.

Suppose there are more than two sides with length AB and, as shown in the previous paragraph, they each must include at least one of points A and B. In that case, two of the three sides must both fall on the same side of line AB. The only way you could have two sides of the polygon of length AB that fall on the same side of AB and are within the intersection of the circles at A and B is if you pick the point C at the intersection of the circles centered on A and B to form sides AC and CB. If you have a third side that is a distance AB away from point B (or point A) on the opposite side of AB as point C, then it must have a vertex somewhere on the perimeter of circle B (or A). But if it has a vertex on the perimeter of B and is on the opposite side of line AB as point C, then that vertex's distance from point C will be larger than AB and you would violate the initial statement that AB is a longest diagonal.

Edit: added side note

Edited February 7, 2015 by plasmid