Gadfly

Let \(z=a+bi\). Then

\(z^5=(a+bi)^5=a^5+{5\choose 1}a^4bi+{5\choose 2}a^3(bi)^2+{5\choose3}a^2(bi)^3+{5\choose4}a(bi)^4+(bi)5\)

                           =\(a^5+5a^4bi-10a^3b^2-10a^2b^3i+5ab^4+b^5i\)\(=(a^5-10a^3b^2+5ab^4)+(5a^4b-10a^2b^3+b^5)i\)

S0\(Im(z^5)=5a^4b-10a^2b^3+b^5\) and

\(\frac{Im(z^5)} {[Im(z)]^5}=\frac{5a^4b-10a^2b^3+b^5}{b^5}\)\(=\frac{5a^4b}{b^5}-\frac{10a^2b^3}{b^5}+\frac{b^5}{b^5}\)=\(5 ( \frac{a}{b})^4-10(\frac{a}{b})^2 + 1\).

This is a quadratic-type expression that can be converted to a quadratic expression by replacing \((\frac{a}{b})^2\)with x. So we need to find the minimum value of \(5x^2-10x+1\), which can be found at the vertex of \(y=5x^2-10x+1\). Since \( \frac{-b}{2a}=1\), the minimum value is \(5(1)^2-10(1)+1 =-4\)

You are very welcome and I am glad I could be of help.

There are 15 splits if we do not count the order in which the 4 numbers are arranged, This comes from the observation that picking two distinct numbers out of the 6 possible, not counting order, is \({ 6\choose 2}\)=15. Each of the splits can be ordered in 6 different ways. For instance, two ones and two twos can be arranged as

(1, 1, 2, 2)

(1, 2, 1, 2)

(1, 2, 2, 1)

(2, 1, 2, 1)

(2, 2, 1, 1)

(2, 1, 1, 2)

so there are \({6 \times15}\)=90 ordered splits. There are a total of \(6^4\)outcomes when 4 six-sided dies are cast. So the probability of an ordered split is \(\frac {90} {6^4}=\frac{5}{72}\)

If "equally strong "means that each has probability \(\frac {1}{2}\) of winning at any match , then

P(game over  after the first 3 matches) = P( B wins the first 3 matches) +P( I wins the first 3 matches)= \((\frac{1}{2})^3 + (\frac{1}{2})^3 = \frac {1}{4}\)

Our guest is asolutely right; the number 286 is way too large. The problem is that the statement I made, that "each partition would be uniquely associated with a lattice point", is not correct since there are duplications. For instance, (0, 4, 1, 5) and (2, 2, 3, 3) both take you to the terminal point (-1, -1) on the plane. I am trying to figure out how many duplications there are among the 286 partitions so that I can subtract thatnumber from the total.

Also geometrically, an approach that the first answer hinted at, the terminal points, I think, may lie on the following lattice :

If so , the number of terminal points would be \(2(1+3+5+...+19) +21=2(100)+21=221\). That would mean that among the 286 partitions I counted there were 65 duplications.

The answer is 286. It comes from the following sum:\(1+(1+2)+(1+2+3)+(1+2+3+4)+...+(1+2+3+...+11)=\)

1+3+6+10+15+21+28+35+45+55+66 = 286.

It is the number of ordered partitions of 10 as the sum of four numbers taken from the set \({{0, 1, 2, ..., 10}}\). If we write these partitions as 4-tuples such as (3, 2, 0, 5), and take the coordinates to mean, say,

(3 units to the right, 2 units up, no movement to the left, 5 units down), then each partition would be uniquely associated with a lattice point in the plain (or is that plane?!). I have carefully counted the number of such ordered partitions and have a list of them, but am too tired at the moment to list them. The sum can also be written as

\({2 \choose 0}+{3 \choose 1}+{4 \choose 2}+{5 \choose 3}+ {6 \choose4}+{7 \choose 5}+{8 \choose 6}+{9 \choose 7}+{10 \choose 8}+{11 \choose 9}+{12 \choose 10}\)

Sorry,I posted an EXCEL chart without explanation. For the \(2 \times 50\) grid, the number of arrangements is

\({50 \choose 0}+{49 \choose 1} + {48 \choose 2} + ...+{26 \choose 24} +{25 \choose 25}= 20,365,011,074\) arrangements. EXCEL was the only tool I could use to calculate such an enormous sum.

I am looking for a more elegant way to count this giant number of arrangement. I am certain there is one somewhere on the web!

The sum of the measures of the interior angles of the quadrilateral in the middle is 180(4-2) =360. This makes the sum of the four angles inside the four triangles that are opposite to the quadrilateral's interior angles, the ones that are not orange,  also 360 degrees. The sum of the interior angles of the four triangles is 4(180)= 720. So the orange angles must add up to 720-360= 360 as well.

Let's call a \(2 \times 2\) square a 'block' and a \(2 \times 1\) rectangle consisting of a column of two \(1 \times 1\) squares a 'blank'. So a \(2\times 10\) grid, for instance, consists of 10 consecutive blanks, numbered 1 to 10 from left to right, and a block occupies two consecutive blanks, say blanks 1 and 2, which I will denote by (1, 2). The max number of blocks that can be arranged on our \(2 \times 10\)grid is, of course 5, and the places they would occupy  would be denoted as (1, 2), (3, 4), (5, 6), (7,8), and (9,10). We will first look at the number of arrangements on the \(2 \times 10\) grid and then extend the results to a \(2 \times 50\) grid.

First note that a block can occupy a \(2 \times k\) blan grid in k - 1 different ways. For instance, on a \(2 \times 10\) blank grid, the block can be placed on (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9), or (9, 10).

The following table shows that the number of arrangements for a \(2 \times 10\) grid is

\({10 \choose 0 } + {9\choose 1} +{8\choose 2}+{7\choose3}+{6\choose 4}+{5\choose5}=89\)

Thiscan be generalized to a \(2 \times 50 \) Grid. see below

The triangle ABC is made up of three smaller triangle, namely AIC, AIB, and BIC. These triangles have the same height, namely r, the inradius of the triangle, and their bases are \(a, b\ and\ c.\) Adding the areas of these triangle we get the area of the triangle ABC, and in terms of r, that area is

\(\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr=\frac{a+b+c}{2} \cdot r\) \(=\frac{6+8+10}{2} \cdot r=12r\).

So if we can find the area of triangle ABC , then by dividing that area by 12, we would have the inradius. Fortunately Heron, or Hero of Alexandria, some Greek dude who lived around 10 AD, left us a formula by which we can calculate thee area we need using only the known sides of the triangle ABC. If we let \(s=\frac{a+b+c}{2}\), a quantity we calculated to be 12, the the area of ABC would be\(\sqrt {s(s-a)(s-b)(s-c)}\)\(=\sqrt{12(12-10)(12-8)(12-6)} = \sqrt{12\cdot48} =\sqrt{36\cdot16}=24\). So the measure of the inradius is 24 divided by 12 or 2 units of length.