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?
The polynomial \(P(x)=x^3+3x^2-7x+1\) has three distinct roots: \(a,b,\) and \(c\). What is the value of \(a^2+b^2+c^2\)?
So far we have discussed polynomials in one variable, i.e: with only an \(x\) as our variable. We can however, include as many as we want. For example, we can talk of a polynomial such as \[P(x,y)=x^2-y^2,\] where both \(x\) and \(y\) are variables. This is an example of an antisymmetric polynomial, which means that \(P(x,y)=-P(y,x)\) (i.e: switching \(x\) for \(y\) gives the original polynomial with a minus sign). Conversely, a polynomial \(Q(x,y)\) such that \(Q(x,y)=Q(y,x)\) is called symmetric. Show that every antisymmetric polynomial \(P(x,y)\) can be factored as \[P(x,y)=(x-y)Q(x,y),\] where \(Q(x,y)\) is a symmetric polynomial.
Find the remainder of dividing \(x^{100}-2x^{51}+1\) by \(x^2-1\). Try not to do a long calculation.