## Leon Vaserstein’s Problems

I met Leon Vaserstein at a party. What do you think I do at parties? I bug people for their favorite problems, of course. The first riddle Leon gave me is a variation on a famous problem I had already written about. Here’s his version:

The hypotenuse of a right triangle is 10 inches, and one of the altitudes is 6 inches. What is the area?

When Leon told me that he had designed some problems for the Soviet Olympiads, naturally I wanted to hear his favorite:

Share:A closed polygonal chain has its vertices on the vertices of a square grid and all the segments are the same length. Prove that the number of segments is even.

## Simon:

Yay, problems I could solve for once. Here are my solutions.

The triangle in the first problem is a scaling of the 3-4-5 right triangle (a 6-8-10 triangle), so the area is 6*8/2 = 24.

In the second problem, place one vertex of the polygon at (0,0) and an adjacent one in the first quadrant (a,b). Also impose the restriction that a >= b. Starting from the origin, as you move to an adjacent vertex of the polygon, the Manhattan distance or L-1 norm to the origin either increases or decreases by a + b or a – b. Since a + b and a – b have the same parity, if you follow a path made up of an odd number of edges, then you cannot end up a distance 0 from the origin, so the polygon must have an even number of sides.

14 June 2011, 2:08 pm## Brenda:

1. the area is 24 square inches? what am I missing?

2. all the shapes matching the problem description are either squares, or polygons which vertices can be formed by superimposing rotated squares upon each other, such as an octagon. Therefore the number of sides must be some multiple of four, hence even.

14 June 2011, 2:30 pm## Ambarish:

I guess the only thing left in 1 is proving that the altitude cannot be from the vertex with the right-angle.

Let the triangle be ABC, with angle B = 90 degrees. Let BD be the altitude with D on AC. So |AC| = 10, }BD| = 6.

Let |AD| = x, so |DC| = 10-x. Now, triangles BDA and CDB are similar, so

BD/DA = CD/DB, or

6/x = (10-x)/6, or

36 = 10x-x^2, or

x^2-10x+36 = 0, or

(x-5)^2 = -11,

clearly impossible for real x.

In other words, given |AC| = 10, and BD perpendicular to AC, with |BD| = 6, and D on AC, ABC can never be a right-angle.

14 June 2011, 5:11 pm## Cristi:

Probably the catch of the first question is that you need to exclude the case when the altitude corresponds to the right angle, because 6>10/2.

For the second problem, in addition to Simon’s solution, we need to consider also the case when the length is the hypotenuse of a Pythagorean triangle (or of more such triangles). One way is to use alternate colorings of the grid’s vertices. If the length is odd, any two adjacent vertices have opposite color. If the length is even, the allowed points form a subgrid, and we use alternate coloring for that subgrid.

14 June 2011, 6:04 pm## Andrew MW:

Surely in the first problem, the altitude *could* be a perpendicular from the hypotenuse, meaning the area would be 10*6/2 = 30.

So the result is indeterminate?

14 June 2011, 6:11 pm## M:

1. Can also be 30 (10*6/2). Is the trick that this is not a well-posed problem?

14 June 2011, 6:52 pm## Andrew MW:

Actually that’s a dumb answer – the altitude from the hypotenuse can’t be more than 5 – so the area must be 24 I think.

14 June 2011, 7:55 pm## Paul Owen:

I think I disagree with both Simon and Brenda about the second problem, although I don’t have a proof of my own.

Suppose my polygon has the following vertices:

(0,0)

(4,3)

(4,8)

(0,11)

(-4,8)

(-4,3)

These six points are the vertices of an equilateral hexagon, each of whose segments has length five. This hexagon will serve as a counterexample for both Simon’s and Brenda’s proofs, though it will not disprove the hypothesis (since it has an even number of sides).

Simon claims that as you move from one vertex to the next on the polygon, the Manhattan distance to the origin increases or decreases by a+b or a-b. In my hexagon, the Manhattan distances from the origin to the second and third vertices are 7 and 12. Their difference is 5, which is neither 4+3 nor 4-3. So my polygon meets the conditions of Simon’s construct but contradicts his claim regarding Manhattan distances.

Brenda claims to have proven that a polygon satisfying the problem description must have a number of sides that is a multiple of four. My hexagon has six sides, which is not a multiple of four but satisfies the conditions of the problem.

So neither proof is sufficient.

14 June 2011, 8:08 pm## K:

There’s more than one solution to the first problem. The area can also be 30 inches^2.

14 June 2011, 8:11 pm## Marcial Fonseca:

Hi there; as always, I am enjoying this blog. Now, about the first problem, it’s clear it should be 24 or 30. Anyway, I have just one comment to that problem: the wording “of a right triangle” is redundant.

14 June 2011, 8:30 pmMarcial Fonseca

## Student:

K:

no, thats what the trick is to the problem

14 June 2011, 8:57 pmif you draw the triangle in a circle, the hypotenuse will be a diameter, and the largest the altitude can be for the hypotenuse of length 10 is 5 (a radius)

so that means that the altitude of 6 isn’t for the hypotenuse, but is the length of one of the legs, (right triangle), so the other length must be 8 (B^2 = C^2- A^2), so the area is 6*8/2 = 24

## K:

@Student you’re right, I didn’t even think about that. I need more caffeine.

15 June 2011, 3:05 am## Paul Owen:

I think I agree with Cristi’s proof of the second problem if she can prove that no Pythagorean triangle exists with two odd legs and an even hypotenuse. Otherwise, the segment length would be even but there would not be a subgrid to use to make the colorings of the vertices alternate. For some reason I’m having a hard time convincing myself that no such Pythagorean triangle exists.

15 June 2011, 9:49 pm## Bill:

Paul, I think I can convince you that no Pythagorean triangle exists with two odd legs and an even hypotenuse.

All odd squares are 1 (mod 4). So any two odd squares sum to a number that is 2 (mod 4). But all even squares are 0 (mod 4). Therefore, two odd squares can never sum to an even square.

16 June 2011, 3:34 am## Austin:

Re: the second problem:

I think Cristi’s solution is closest, but needs a little more. Here is my attempt. Let the square grid be the usual integer grid, generated by (1,0) and (0,1). Let the common length of the segments in the polygonal chain be sqrt(N). Then N is either 0, 1, or 2 (mod 4). Let us consider these cases in reverse order.

If N is 2 (mod 4), then each segment has odd x- and y-components, so both coordinates alternate in parity as we travel around the chain. Thus the number of segments in the chain is even.

If N is 1 (mod 4), then each segment must have one odd and one even component, so the sum of coordinates alternates in parity. Again, the chain must consist of an even number of segments. (The N=2 and N=1 cases are similar, but the checkerboards colorings apply to different grids!)

If N is 0 (mod 4), then each segment has even x- and y-components, so all points visited lie on a subgrid of the original integer grid. We may divide a common power of 2 out of all lengths so as to pass to one of the cases already considered.

17 June 2011, 12:11 pm## Animesh:

Considering a triangle as a polygon. Can’t we make an equilateral triangle inside a square.

28 June 2011, 5:22 amSince triangle has 3 sides it’s odd….so not an even number of segments.

What am I missing….?

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6 July 2011, 3:13 pm