# 1985 Putnam Question A-2: Solution Part 3

## Overview

In Part 2 of this series we reviewed the concept of **generalization** and applied it to our problem.

In this section, we will not actually *solve* any more of the problem but we will lay the groundwork necessary for the succeeding sections. We will begin by discussing and reviewing different types of triangles, then go on to explain some properties of triangles. These properties will help us solve the problem.

As usual, there will be numerous graphics and visual aids to assist along the way.

## Triangles

### Types of Triangles

In Part 1 of this series we explained **acute triangles**, or triangles where each angle is **less** than 90 degrees.

It is worth noting that there are also **obtuse triangles** (where one of the three angles is **more** than 90 degrees).

### Right Triangles

For our purposes, it is important to discuss right triangles in particular.

As shown in the diagram, a **right triangle** has one angle that measures **exactly** 90 degrees, as denoted by the small square in the corner of that angle.

It is worth noting that only *one* of the three angles in any triangle can be 90 degrees or larger. This is because the sum of all angles in a triangle is 180^{o}.

In *Euclidian Geometry* (the geometric concepts discussed and explained in ancient Greek mathematician Euclid's timeless classic, *The Elements*), this is known as the **Triangle Postulate**.

### Proof

If one angle is a right angle (*exactly* 90^{o}), the other two angles must be 180^{o} − 90^{o}, or 90^{o}. Then if a *second* angle is 90^{o}, the remaining angle must be 0^{o}. This, however is impossible, as a 0^{o} angle, is actually no angle at all, rather a straight line.

### Properties of right triangles

Listed below are some properties of right triangles (refer to the diagram of the right triangle below)

1. Side *C* (also known as the hypotenuse) is the longest side of the triangle (since it is opposite the right angle)

2. Angles `\alpha` and `\beta` add up to 90 degrees. This is an extension of Euclid's **Triangle Postulate** which states that the sum of all angles within a triangle is 180 degrees.

3. The **area** of a right triangle can be found by multiplying `\frac{1}{2}xx "base"xx"height"`.
For the above triangle, this is: `\frac{1}{2}AB.`

### Similar Triangles

Similar triangles are triangles that have corresponding angles that are equal to each other. Look at the triangles below:

For clarification:

- Each side is labeled with a dark brown, lowercase letter;
- Each triangle is labeled in the middle with a large, black, uppercase letter. This letter serves as the "name" of the triangle;
- Each angle that is
**not**a right angle is denoted with an arc and the measure of the angle in*degrees*.

Triangles *A* and *B* are similar, since they share the angles `27^"o"`, `63^"o"` and `90^"o".`

However, triangle *C* is similar to neither because it doesn't share **any** of the same angles with *A* or *B*.

Similar triangles do not need to be the same size or the same area, although they can (in which case they are known as **identical** triangles).
Similar triangles do however, have sides that are proportional to each other. For instance, on the above diagram,

`\frac{d}{m}=\frac{e}{j}=\frac{f}{k}.`

This property of similar triangles proves to be extremely useful in the field of Geometry and likewise will be implemented in our solving of the problem at hand.

These proportions also apply to the ratio of the **areas** of the two triangles.

### Altitude of a triangle

It should also be noted that **any** triangle can be separated into two right triangles by drawing what is known as an **altitude**.

The two triangles created by the altitude are **similar** to each other in this case, because the original triangle is isosceles (two equal sides).

## Summary

While we did not actually go further in *solving* the problem in this section, we made great progress in laying the foundation for the next section. In Mathematics, the speed with which one solves a problem is often trivial, whereas an accurate answer with a rock-solid foundation is highly coveted. As such, we are content to build on our foundation for the problem. We first discussed various topics pertaining to triangles, including types of triangles, right triangles, similar triangles and altitudes of triangles.

In the next installment of the series, we will use what we learned in the section (along with some new techniques) to further reduce the given expression into a more generalized form.

Thanks for reading!