1985 Putnam Question A-2: Solution Part 1

1. The Problem (1985 Ques A-2)

Let T be an acute triangle. Inscribe a rectangle R in T with one side along a side of T. Then inscribe a rectangle S in the triangle formed by the side of R opposite the side on the boundary of T, and on the other two sides of T, with one side along the side of R. For any polygon X let A⁢(X) denote the area of X. Find the maximum value, or show that no maximum exists, of

`(A(R)+A(S))/(A(T))`

where T ranges over triangles and R and S over all rectangles as above.

2. Discussion

Since this is the first installment of the series, we will spend most of this section breaking down the problem and addressing any parts and terminology that may be confusing. We will also begin to solve the problem. But don’t worry! We will only be taking the smallest of steps towards our final solution.

a. Acute Triangle

An acute triangle is a triangle where each of the three angles is less than 90 degrees. For example:

For comparison, below is a right triangle (where one of the angles is exactly 90 degrees), and an obtuse triangle, where one of the angles is greater than 90 degrees.

b. Inscribe

"Inscribe" means to draw, specifically inside. In the following diagram, rectangle R is inscribed inside the acute triangle T, as required by the first part of the question, which said "Inscribe a rectangle R in T with one side along a side of T".

The next sentence in the question asks us to inscribe another rectangle in "the triangle formed by the side of R opposite the side on the boundary of T, and on the other two sides of T".

Let's take this part first: "the side of R opposite the side on the boundary of T". This refers to the (top) side of R in magenta (pink) in this diagram:

Then, the "triangle" bounded by the magenta segment and the other two sides of T refers to the magenta one outlined in this diagram:

We need to inscribe a second rectangle, S, in this magenta triangle (it may, or may not be the same height as rectangle R):

c. Polygon

A "polygon" is any two-dimensional shape that has at least three straight sides and angles. The lines must be straight or else it is not a polygon.

Our acute triangle, T and rectangles R and S all qualify as polygons. More examples of polygons are:

If you think the second polygon here looks kind of wonky, you’re right. The only requirements are more than three straight sides or angles, meaning there are infinitely many polygons we can draw.

d. Areas written as a function

Consider the diagram below, this time shaded with colors.

A⁢(R) is the area in magenta (pink)

A⁢(S) is the area in yellow

A⁢(T) is the area of the entire triangle (5 small green triangles + y⁢e⁢l⁢l⁢o⁢w + magenta)

e. Maximum value

The problem asks us to determine the maximum value of the given expression.

There can only be two cases:

(i) there is a maximum value

(ii) there is no maximum value

For example, consider the expression: 3⁢x

To easier visualize this, we can write this expression in equation form as y = 3⁢x.

As x increases, so will y. A graph of this equation will look like a straight line. If you follow the straight line forever, you will never see y stop increasing.

So y = 3⁢x does not have a maximum value

Now consider `-(x-1)^2+2`

Writing this in equation for as we did above gives `y=-(x-1)^2+2`

We can clearly see that there is a maximum value for this expression/equation. As x increases, the expression reaches a value of 2 when x is 1. This is the maximum value of the expression, since no other values of x will yield a value this high for the expression.

So some expressions will have a maximum and some will not.

f. Expression

The expression in the question is given as:

`(A(R)+A(S))/(A(T))`

The problem asks us to find the maximum value or show that no maximum exists.

Look at the shaded figure from part d.

As we said before, A⁢(R) is everything in magenta, A⁢(S) is everything in yellow and A⁢(T) is everything in magenta, yellow, and green.

For the sake of simplicity, lets call the green area A⁢(G).

We can see then, that

`A(T)=A(R)+A(S)+A(G)`

so we can rewrite the given expression as:

`(A(R)+A(S))/(A(R)+A(S)+A(G))`

Thinking about the above expression, we can see that the denominator (bottom part of the fraction) must be larger than the numerator (top part of the fraction).

Look at other fractions with this characteristic:

`1/2, 3/4, 19/86, 4/5, 99/100`

It is clear that all of these must be less than 1.

So we can conclude that our expression must be less than 1.

When discussing the maximum (or lack thereof) of a function, we said that expressions with a maximum have a point at which they stop decreasing. Our expression cannot be 1, or anything greater than 1, so we can conclude that our expression does have a maximum.

It might be tempting to say that the equation has a maximum value of 1, but this too would be incorrect. Here’s why:

If the maximum value is 1, then

`A(R)+A(S)=A(T)`

So there would be no green space on the diagram, and it would look like this:

Which of course does not involve a triangle, a key part of the question. So there must be some amount of green space to satisfy the conditions of the problem.

3. Summary

The problem we will be solving in this series is problem A2 from the 1985 Putnam Exam. This problem falls under the larger category of Geometry. We were given an expression:

`(A(R)+A(S))/(A(T))`

We defined terms used in the problem (acute triangle, inscribe, polygon, etc.) and explained their importance. The problem requires us to ”find the maximum value, or show that no maximum exists” of the given expression. In the Discussion section, we showed that a maximum value does indeed exist. For the next installment of the series, we will begin an attempt to find that maximum.

Thanks for reading.