# 1985 Putnam Question A-2: Solution Part 4

## Overview

Congratulations on making it to Part 4, we're now over half way through!

In the previous section, 1985 A2 Solution Pt 3, we discussed various properties of triangles that will prove useful in our solving of the problem. This time, we will apply what we have learned about generalization and geometry to take the next step in our solving of the problem. There is a fair amount of algebra in this section, so I will be taking time to explain each step somewhat laboriously. If you feel the explanation to be too in depth for your ability, just skimming the more detailed sections would suit you well.

## Continuing our Generalization

The primary purpose of generalizing a problem (as mentioned in section 2) is to simplify the problem to as few variables as possible.

Now lets look again at our generalization of the problem from part 2.

*R*

_{1}

*R*

_{2}

*R*

_{3}

*R*

_{4}

*R*

_{5}

*R*

_{n}

Notice how next to each rectangle, there are two little triangles, one to the left of the rectangle and one to the right. These triangles continue to the top where they meet **above** the *n*^{th} rectangle, therefore creating *n* + 1 triangles on each side.

Remember our expression?

`\frac{A(R)+A(S)}{A(T)} \quad[1]`

Since we have already generalized the visual representation to apply to *n* rectangles, we can now generalize the given Expression [1] so that it too is in terms of **n** rectangles.

We have already replaced the rectangles *R* and *S* with rectangles `R_n` according to our generalization, so we can rewrite [1] as:

`\frac{A(R_1)+A(R_2)+A(R_3)+...+A(R_n)}{A(T)} \quad[2]`

Now we will use **summation notation** to rewrite this expression.

### Summation Notation

This section is a brief overview of summation notation, feel free to skip past this if you are already familiar with this notation.

Summation notation allows mathematicians to write long sums (like the one from Expression [2] ) in a neat and abbreviated fashion. It utilizes the Greek letter **sigma**, `\sum`, appended with **lower** and **upper** limits at the bottom and top of the **sigma** accordingly. Lets look at an example:

#### Example 1

Suppose `a_1, a_2, ... a_{100}` is a series of 100 positive integers. Writing the sum of this series using traditional notation (`a_1+a_2+a_3+a_4+a_5+...+a_{100}`) would become quite tiresome rather quickly.

But we can write this using summation notation as:

`\sum_{n=1}^{100}a_n`

Here `n=1` gives the **lower** limit of the function, and serves to identify **n** as the changing variable, while **100** is the upper limit of the sum.

You can also write this as the **sum** of separate sums, like below:

`\sum_{n=1}^{48}a_n+\sum_{n=49}^{100}a_n`

The upper and lower limits of the sum do **not** have to be the upper and lower limits of the function. For instance, you could write the sum of the `10^{"th"}` to the `37^{"th"}` term of the series as:

`\sum_{n=10}^{37}a_n`

#### Example 2

What if you wanted to write the sum of **all** positive even numbers?

Without Summation notation, this would look like:

`2+4+6+8+10+12+14+ . . .`

We can pull a 2 out of every factor using the distributive property:

`2\cdot(1+2+3+4+5+6+7+...)`

Now the term **inside** the parentheses just looks like the sum of all positive integers, `\sum_{k=1}^{\infty}k`. So we can write the sum of all positive even numbers as:

`2\cdot\sum_{k=1}^{\infty}k`

### Applying Summation notation to our Generalization

Now we can rewrite Expression [2] using summation notation as:

`\frac{\sum_{i=1}^{n}A(R_i)}{A(T)} \quad[3]`

### Reanalyzing Geometry

We have successfully simplified our problem to **two** variables as opposed to three, but now we will have to find some relation between `\sum_{i=1}^{n}A(R_i)` and `A(T)`.

Look again at the generalized version of the diagram for this problem.

*R*

_{1}

*R*

_{2}

*R*

_{3}

*R*

_{4}

*R*

_{5}

*R*

_{n}

We can see that subtracting the area in blue from the total area of the triangle `A(T)`, gives the total area of the inscribed rectangles, `\sum_{i=1}^{n}A(R_i)`, in green.

Now, if we can define the area in blue, `\sum_{i=1}^{n}A(R_i)` and `A(T)` in terms of a single variable, we will have successfully generalized the problem at hand to a point where we can solve it.

### Separating the Triangle

Consider our **acute** triangle, `T` below, inscribed with an altitude:

Now let's separate `T` into the two right triangles resulting from the altitude, which we will call `J` and `K`.

Now we can rewrite `A(T)` using `J` and `K`:

`A(T)=A(J)+A(K)`

As we said just before the start of this subsection, we need to relate the three areas.

Note how the colored areas are really just two series of triangles, one on each side of the rectangles.

*R*

_{1}

*R*

_{2}

*R*

_{3}

*R*

_{4}

*R*

_{5}

*R*

_{n}

Also note how there are *n* + 1 triangles on each the left and right side (since there are *n* rectangles.)

The triangles on the **left** side will be similar to triangle *J*, and the triangles on the **right** side to *K*, accordingly.

But why is this?

#### Example 3

Let's look at the following example:

Consider the right Triangle `Q` with acute angles `30^"o"` and `60^"o"`:

^{o}

^{o}

Now let's draw a vertical line that is exactly perpendicular to the **base** of `Q`

_{1}

^{o}

^{o}

Drawing this line creates a smaller right triangle, which we can call `Q_1.`

Now `Q_1` has the same angle of **30**° that `Q` has in the bottom right corner. These two triangles also both have right angles in the bottom left corner. As we learned in section 3, the sum of all angles of a triangle is **180°**. As such, we know that the remaining angle must be 180° − 90° − 30° = 60°. Therefore, all angles of `Q_1` are the same as the angles of `Q`, meeting the conditions of **similar** triangles.

### Back to the problem

The vertical line of each `R_i` acts in the same way, creating triangles similar to `J` on the left and `K` on the right. Since the triangles are similar, we can call each small triangle `J_i` and `K_i` accordingly.

*R*

_{1}

*J*

_{1}

*K*

_{1}

*R*

_{2}

*J*

_{2}

*K*

_{2}

*R*

_{3}

*J*

_{3}

*K*

_{3}

*R*

_{4}

*J*

_{4}

*K*

_{4}

*R*

_{5}

*J*

_{5}

*K*

_{5}

*R*

_{n}

*J*

_{n}

*K*

_{n}

Now we can rewrite `\sum_{i=1}^{n}R_i` as:

`A(T)- \sum_{i=1}^{n+1}[A(J_i)+A(K_i)]`

(Remember how we said there are `n+1` triangles on each side? This gives the upper limit.)

We can now rewrite the entire expression [3]** **as:

`\frac{A(T)- \sum_{i=1}^{n+1}[A(J_i)+A(K_i)]}{A(T)}`

Simplifying:

`\frac{A(T)}{A(T)} - \frac{\sum_{i=1}^{n+1}[A(J_i)+A(K_i)]}{A(T)}`

But since `\frac{A(T)}{A(T)}=1`, we can simplify this further to be:

`1 - \frac{\sum_{i=1}^{n+1}[A(J_i)+A(K_i)]}{A(T)}`

As we said before,

`A(T)=A(J)+A(K)`, so `\frac{A(R)+A(S)}{A(T)}=1 - \frac{\sum_{i=1}^{n+1}[A(J_i)+A(K_i)]}{A(J)+A(K)}`

This concludes the simplification of our generalization for this section of the problem.

## Summary

In this section, we used some of the properties of triangles discussed in section 3 to continue our generalization of the problem. We also found a way to generalize by analyzing the geometry of the acute triangle. We then used these generalizations to rewrite the given expression `\frac{A(R)+A(S)}{A(T)}` as:

`1 - \frac{\sum_{i=1}^{n+1}[A(J_i)+A(K_i)]}{A(J)+A(K)}`

In the next section, we will continue our simplification of the problem as we explore how to relate `A(K_i)` and `A(J_i)` to `A(K)` and `A(J)`.

As always, thanks for reading and I hope you enjoyed this section!