Problems
Prove the general version of Sperner’s lemma: Consider an \(n\)-dimensional simplex \(\mathcal{A} = A_1A_2...A_{n+1}\). Strictly speaking a simplex is a convex linear combination of \(n+1\) points in general position (when \(k\) points are never in one subspace of dimension \(k-1\)). One can view it as an \(n\)-dimensional tetrahedron or a body spanned over vertices \((0,0,...,0), (1,0,0,...,0), (0,1,0,0...,0), ... (0,0,...0,1)\). \[\mathcal{A} = \{\sum_{i=1}^{n+1}a_i(0,0,...,1,...,0), \,\,\, a_i \geq 0, \,\,\,\, \sum_{i=1}^{n+1}a_i = 1\}.\]
A simplicial subdivision of an \(n\)-dimensional simplex \(\mathcal{A}\) is a partition of \(\mathcal{A}\) into small simplices (cells)
of the same dimension, such that any two cells are either disjoint, or
they share a full face of a certain dimension.
Define a Sperner’s coloring of a simplicial subdivision as an assignment
of \(n+1\) colors to the vertices of
the subdivision, so that the vertices of \(\mathcal{A}\) receive all different colors,
and points on each face of \(\mathcal{A}\) use only the colors of the
vertices defining the respective face of \(\mathcal{A}\).
Prove that every Sperner’s coloring of any subdivision of an \(n\)-dimensional simplex contains an odd
number of rainbow simplexes, namely whose vertices are colored using all
\(n+1\) colors.
Draw Sperner’s coloring for the following triangulation. Try to avoid rainbow triangles at all costs.
Consider an \(n\)-dimensional simplex \(\mathcal{A} = A_1A_2...A_{n+1}\), namely a body spanned over vertices \((0,0,...,0), (1,0,0,...,0), (0,1,0,0...,0), ... (0,0,...0,1)\). \[\mathcal{A} = \{\sum_{i=0}^{n}a_i(0,0,...,1,...,0), \,\,\, a_i \geq 0, \,\,\,\, \sum_{i=1}^{n+1}a_i = 1\}.\] Where next to \(a_i\) there is a point with coordinate where \(1\) is in \(i\)-th place. The point \((0,0,...,0)\) belongs to the simplex as well.
A simplicial subdivision of an \(n\)-dimensional simplex \(\mathcal{A}\) is a partition of \(\mathcal{A}\) into small simplices (cells)
of the same dimension, such that any two cells are either disjoint, or
they share a full face of a certain dimension.
Define a Sperner’s coloring of a simplicial subdivision as an assignment
of \(n+1\) colors to the vertices of
the subdivision, so that the vertices of \(\mathcal{A}\) receive all different colors,
and points on each face of \(\mathcal{A}\) use only the colors of the
vertices defining the respective face of \(\mathcal{A}\).
Consider a simplicial subdivision given by pairwise connected middles of
all the segments in the original simplex. Assign the numbers \(0,1,2...,n\) to the subdivision vertices in
such a way as to conduct a Sperner’s coloring in such a way that you
will have only one rainbow simplex.
Consider an equilateral triangle \(ABC\). Parallel to each side, five equally spaced segments are drawn across the triangle so that \(ABC\) is subdivided into \(36\) smaller equilateral triangles. Vertices \(A,B\) and \(C\) are painted red, blue and green, respectively in counterclockwise order. Alice and Bob take turns painting each vertex of the smaller triangles either red, green, or blue with the following restrictions: the segment \(AB\) can only have red or green, the segment \(BC\) can only have green and blue, and the segment \(AC\) can only have blue and red. The interior vertices can be drawn freely. Once all vertices have been painted, Alice gets a point for every smaller triangle whose vertices have the three colors (red, green, and blue) appearing in the counterclockwise direction. Bob gets a point for every smaller triangle whose vertices have the three colors but in the clockwise direction. Who wins?
There are \(100\) people in a room, and each person has at least one friend in the room. Prove that amongst them there are two people with the same number of friends in the room (we don’t count being friends with oneself).
Can we obtain the polynomial \(h(x)=x\) by adding, subtracting, or multiplying the polynomials \(f(x)=x^2+x\) and \(g(x)=x^2+2\)?
In this example we will discuss division with remainder. For polynomials \(f(x)\) and \(g(x)\) with \(\deg(f)\geq \deg(g)\) there always exists polynomials \(q(x)\) and \(r(x)\) such that \[f(x)=q(x)g(x)+r(x)\] and \(\deg(r)<\deg(g)\) or \(r(x)=0\). This should look very much like usual division of numbers, and just like in that case, we call \(f(x)\) the dividend, \(g(x)\) the divisor, \(q(x)\) the quotient, and \(r(x)\) the remainder. If \(r(x)=0\), we say that \(g(x)\) divides \(f(x)\), and we may write \(g(x)\mid f(x)\). Let \(f(x)=x^7-1\) and \(g(x)=x^3+x+1\). Is \(f(x)\) divisible by \(g(x)\)?
In this example we will discuss division with remainder. For polynomials \(f(x)\) and \(g(x)\) with \(\deg(f)\geq \deg(g)\) there always exists polynomials \(q(x)\) and \(r(x)\) such that \[f(x)=q(x)g(x)+r(x)\] and \(\deg(r)<\deg(g)\) or \(r(x)=0\). This should look very much like usual division of numbers, and just like in that case, we call \(f(x)\) the dividend, \(g(x)\) the divisor, \(q(x)\) the quotient, and \(r(x)\) the remainder. If \(r(x)=0\), we say that \(g(x)\) divides \(f(x)\), and we may write \(g(x)\mid f(x)\). Let \(f(x)=x^7-1\) and \(g(x)=x^3+x+1\). Is \(f(x)\) divisible by \(g(x)\)?
In this example we discuss the Factor Theorem. First, let us recall the following concept: if \(P(x)\) is a polynomial, then a number \(z\), is a root of \(P(x)\) if \(P(z)=0\). For example, \(x=1\) is a root of the polynomial \(Q(x)=x-1\). Show:
If \(0\) is a root of \(P(x)\), i.e: \(P(0)=0\), then \(P(x)=xQ(x)\) for some polynomial \(Q(x)\).
Use part 1 to show the Factor Theorem: if \(z\) is a root of \(P(x)\), then \(P(x)=(x-z)K(x)\) for some polynomial \(K(x)\).
In this example we discuss one of Vieta’s formulae. Consider the polynomial \(P(x)=x^2+5x-7\). You can take it as a fact that this polynomial has exactly two distinct roots. What is the sum of its roots? What about their product?