Math Trivia 5 Trivia Quiz

Answer: 101

10 to the 100th power when written out is a 1 followed by 100 zeros, so the number has 101 digits.

Answer: square

A square has the greater area. Both are regular polygons. It can be shown that a regular polygon of m sides inscribed in a circle of radius 1 will always have greater area than a regular polygon of n sides inscribed in a circle of radius 1 if m > n.

In this case, the area of the equilateral triangle is 3/4 times the square root of 3, which is 1.299 to three decimal places and the area of the square is exactly 2.

Answer: 3

A quadratic equation with roots a and b is given by

(x - a)(x - b) = 0.

Multiplying this out gives x^2 - (a + b)x + ab = 0.

Now x^2 - (a + b)x + ab = x^2 - 3x - 7, so we have a + b = 3.

Note also that the product of the roots is -7. You can also solve this problem by finding the roots by using the quadratic formula, then adding them.

Answer: 8

The coordinates of the midpoint of a line segment is found by averaging the coordinates of the endpoints. So the x-coordinate is

(1 + 15)/2 = 16/2 = 8.

Answer: true

Both sine and cosine are periodic with period 2*pi. In fact, one is just a phase shift of the other:

sin(x) = cos(x - pi/2)

This means that the graph of y = sin(x) can be obtained from the graph of y = cos(x) by shifting it to the right pi/2 units.

Answer: no

There is more than one way to see the answer has to be no. If you compute the determinant of the matrix, you will get 0, so it is not invertible. Also, since the second row is twice the first row, the matrix has rank less than 2 (it has rank 1, meaning the row space has dimension 1), thus it cannot be invertible for that reason.

Here's yet another reason: If you solve the system AX = 0, you will get more solutions than just the 0 solution (x_1 = 2 and x_2 = -1 gives a solution). Therefore A cannot be invertible.

Answer: 9

Notice I gave hints: The three midpoints of each side, the three feet of the altitudes, and the three other points, which total 9 points, and this is indeed the famous "nine point circle".

Interestingly, wikipedia says that this circle is also called the 6 point circle and the 12 point circle, so I accepted these answers as well. Check out the article by searching for "nine point circle" in wikipedia - it has a nice picture. Any three noncollinear points in the plane lie on a circle, but it is surprising that 9 interesting points can lie on the same circle (although I've been told there are many more than 9 interesting points on this circle!)

Answer: true

Yes, you can have a field with 4 elements! Here's how you constuct one:

Let Z2[x] denote the ring of polynomial with coefficients in the ring (field) Z/2Z. The polynomial x^2 - x - 1 is irreducible in Z2[x] since neither x nor x - 1 is a factor (check by putting x = 0 or x = 1). Therefore the ideal generated by x^2 - x - 1 in the ring Z2[x] is maximal, thus the quotient

Z2[x]/(x^2 - x - 1) is a field.

There are four distinct equivalence classes - I'll list their representatives:

0, 1, x, 1 + x

We immediately have the relationship x^2 = 1 + x by construction. You can also verify that x(1 + x) = 1. So the multiplicative inverse of x is 1 + x and vice-versa.

In general, if p is a prime, then there exists a unique field with p^n elements.

Answer: true

This follows from the following interesting fact: The conjugate of a product is equal to the product of the conjugates. Algebraically:

The conjugate of (a + bi)(c + di) is equal to (a - bi)(c - di).

So in this case, the conjugate of (2 + 3i)^4 is the same as

(the conjugate of (2 + 3i))^4 = (2 - 3i)^4.

Of course you can do this by multiplying each out:

(2 + 3i)^2 = 2^2 + 2*2*3i + (3i)^2 = 4 + 12i - 9 = -5 + 12i

So (2 + 3i)^4 = (-5 + 12i)^2 = (-5)^2 + 2*(-5)*12i + (12i)^2 = 25 - 120i - 144 = -119 - 120i.

Similarly you can show (2 - 3i)^4 = -119 + 120i.

Answer: true

This is true. Here is a nice mathematical proof: To show two sets are equal, all you need to show is each is a subset of the other. I want to show that (A union B) equals B in this case. First of all, B is always a subset of (A union B), so I must show that (A union B) is a subset of B. So let x be an element of (A union B). Then x is an element of A or x is an element of B. If x is an element of B we're done, so assume x is an element of A. Since A is a subset of B, we have x is an element of B, so we're done.

I hope you enjoyed this quiz! Thanks for playing!