Random Math Problems (3)

School finished yesterday. After one exam next week (for math), I’m done grade 10. I’ll probably have more time to write blog posts in the summer.

Problem 1

(from Geometry Revisited)

Prove that in this triangle (a being m+n), that

a(p^2+mn) = b^2m + c^2n

This seems like a really weird thing to prove, but apparently it’s known as Stewart’s theorem and is useful in some places.

Solution

Recall the Law of Cosines:

\cos C = \frac{a^2+b^2-c^2}{2ab}

We can apply the Law of Cosines twice to \angle AXB and \angle AXC:

\cos \angle AXB = \frac{p^2+m^2-c^2}{2mp}

\cos \angle AXC = \frac{p^2+n^2-b^2}{2np}

Adding the two together,

\cos \angle AXB + \cos \angle AXC = \frac{p^2+m^2-c^2}{2mp} + \frac{p^2+n^2-b^2}{2np}

Notice that because \cos(\theta) = \cos(180-\theta), the sum of cosines of supplementary angles is 0. Therefore,

\begin{array}{rcl} 0 &=& \frac{p^2+m^2-c^2}{2mp} + \frac{p^2 + n^2 - b^2}{2np} \\ &=& n(p^2+m^2-c^2) + m(p^2 + n^2 - b^2) \\ &=& np^2 + nm^2 - nc^2 + mp^2 + mn^2 - mb^2 \end{array}

Moving the two negative terms to the left hand side and factoring,

\begin{array}{rcl} b^2m + c^2n &=& np^2 + nm^2 + mp^2 + mn^2 \\ &=& n(p^2) + m(p^2) + n(mn) + m(mn) \\ &=& (m+n)(p^2+mn) \end{array}

The result follows:

b^2m + c^2n = a(p^2+mn)

Problem 2

(from the 2010 UVM contest)

How many ways are there to arrange the letters of the word NOODLES, so that the consonants (NDLS) appear in alphabetical order (although not necessarily consecutively)?

Solution

The four consonants must appear in the order of D, L, N, S. The three vowels, OOE, must appear somewhere in the five spaces between the consonants:

Thus the answer to the problem is the same as the number of ways to pack OOE into five boxes, a box being able to hold an arbitrary amount of letters, and permutations of letters within a box being counted as distinct.

This is half of the number of ways to pack ABC into five boxes, since there are two O’s in OOE. We will now calculate how many ways there are to pack ABC into five boxes.

We can split up this problem into several cases:

Case 1: All three letters ABC are in the same box. For any box there are 3! = 6 ways to permute ABC, while there are five boxes. The total for this case is 6*5 = 30.

Case 2: The three letters all go in three different boxes. This is simply P(5,3) which is 60.

Case 3: Two of the letters go in a box, while the third letter goes in a different box. This is a little bit tricky, so we break it in a few parts.

First of all we choose the two boxes of the five we use. The number of ways to do this is \binom{5}{2} which is 10. Secondly, if the two boxes are labelled Box 1 and Box 2, we can either put two letters in Box 1 and one in Box 2, or the other way around, giving an additional two ways.

Finally there are 3! or 6 ways to permute the three letters. The final count for case 3 is 10 * 2 * 6 or 120.

The number of ways to pack ABC into five boxes is the sum of the three cases, or 210. The answer to the problem is half of that, 105.

Problem 3

(again from the same UVM contest)

Point P is inside rectangle ABCD. If the distance from P to three corners of the rectangle are 5, 10, and 14, find the distance from P to the forth corner.

Solution

We first draw perpendicular lines from P to the four sides of the rectangle and label them a, b, c, and d, like this:

From pythagorean, we find that

b^2 + c^2 = 5^2

b^2 + d^2 = 10^2

a^2 + d^2 = 14^2

As we are looking for the length of DP, we can manipulate the three equations to get a value for a^2 + c^2:

(b^2 + c^2) + (a^2 + d^2)-(b^2+d^2) = a^2+c^2

Finally we can now substitute in real values to find a^2 + c^2:

a^2+c^2 = 5^2 + 14^2 - 10^2 = 121

The line DP can be found by taking a square root, it is 11.

General formula

When three of the four diagonals of a rectangle are known, the forth one can be calculated. Suppose we know m, n, and o. The formula for x is given by:

x = \sqrt{m^2 + n^2 - o^2}

Problem 4

This problem is from the Number Theory chapter of The Art and Craft of Problem Solving.

Prove that

\frac{n^3 + 2n}{n^4 + 3n^2 + 1}

is in lowest terms for all positive integers n.

Solution

Another way of saying this problem is that

\textrm{GCD}(n^3 + 2n,n^4 + 3n^2 + 1) = 1

for all positive integers n.

Then there does not exist any integer c, where c > 1, c | n^3+2n and c | n^4+3n^2+1. Supposing c exists, and it divides n^3+2n, we can prove by contradiction that c cannot simultaneously divide n^4+3n^2+1.

If c divides n^3+2n or n(n^2+2), then either c | n or c | n^2+2.

We can rewrite n^4 + 3n^2 + 1:

n^3 + 3n^2 + 1 = n^2(n^2+3) + 1

This way it can be seen that n \perp n^2(n^2+3) , as two consecutive integers are relatively prime.

To show that n^2+2 is also relatively prime, we rewrite it this way:

n^3 + 3n^2 + 1 = (n^2+1)(n^2+2)-1

This shows that, n^2+2 \perp (n^2+1)(n^2+2)-1.

As both n and n^2+2 are relatively prime to n^4 + 3n^2 + 1, it follows that c \perp n^4 + 3n^2 + 1, which is what needs to be shown.

One Response to Random Math Problems (3)

  1. Claudia says:

    I take pleasure in, result

    in I found just what I used to be having a look for.

    You have ended my four day long hunt! God Bless you man. Have
    a nice day. Bye

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