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Page by Pat Lachapelle. Last updated: 03 Dec 2019

1985 Putnam Question A-2: Solution Part 5

Overview

In section 4, we worked to further simplify our problem. We successfully broke it down into terms of A(Ji), A(Ki), A(J), and A(K).

We also established in the previous installment of this series that `J_i` & `J` are similar triangles, as are `K_i` & `K`. We will use this fact to further simplify this problem so that it is in terms of a single variable.

Relating A(Ki) to A(K)

So how does A(Ki) relate to A(K)?

For reference, use the diagram below comparing the similar triangles.

In the above diagram, `a_i` is the altitude of the ith triangle `K_i`, while `\alpha` is the total altitude of the triangle `T` (as before).

Also, `p_0` and `p_i` are corresponding sides (in proportion), as are `s_0` and `s_i`.

Using the formula for the area of a right triangle (given in section 3), we can write:

`A(K)=\frac{1}{2}\cdot\alpha\cdot p_0 \quad[1]`

Likewise, we can rewrite A(Ki) as:

`A(K_n)=\frac{1}{2}\cdot a_i \cdot p_i \quad[2]`

According to the properties of similar triangles:

`\frac{\alpha}{a_i}=\frac{p_0}{p_i}=\frac{s_0}{s_i}`

Now using the equality:

`\frac{\alpha}{a_i}=\frac{p_0}{p_i}`

we can solve for `p_i`:

1. Multiply each side by `p_i` to get:

`\frac{\alpha\cdot p_i}{a_i}=\frac{p_0\cdot p_i}{p_i}`

and since the `p_i` on the top and bottom will cancel on the right side,

`\frac{\alpha\cdot p_i}{a_i}=p_0`

2. Now multiply each side by `a_i` to get:

`\alpha\cdot p_i=a_i\cdot p_0`

3. Finally, we can divide each side by `\alpha` to get the equation:

`p_i=\frac{p_0\cdot a_i}{\alpha} \quad[3]`

Now we can substitute Equation [3] into Equation [2] to write A(Ki) in terms of `\alpha`, `a_i` and `p_0`

`A(K_i)=\frac{1}{2}\cdot a_i \cdot \frac{p_0\cdot a_i}{\alpha}`

Simplifying this further, we get:

`A(K_i)=\frac{a_i^2}{\alpha}\cdot[\frac{1}{2}\cdot p_0] \quad[4]

Now we can use Eq. [1] to find a definition for the second term in Eq. [4], `[\frac{1}{2}\cdot p_0]`:

1. `A(K)=\frac{1}{2}\cdot \alpha \cdot p_0`

2. Rearranging the left side, we get:

`A(K)=\alpha\cdot[\frac{1}{2}\cdot p_0]`

3. Now, isolating for `[\frac{1}{2}\cdot p_0]`(by dividing each side by `\alpha`), we get:

`[\frac{1}{2}\cdot p_0]=\frac{A(K)}{\alpha}\quad[5]`

We can substitute Eq. [5] into Eq. [4] to get:

`A(K_i)=\frac{a_i^2}{\alpha}\cdot\frac{A(K)}{\alpha}`

Simplifying further, we can write A(Ki) as:

`A(K_i)=[\frac{a_i}{\alpha}]^2\cdot A(K) \quad[6]`

Now we have successfully written A(Ki) in terms of the altitude of each ith triangle (`a_i`), the altitude of `T` (`\alpha`), and the area A(K).

We can simplify this even further by assigning a new variable, `b_i`

`b_i=\frac{a_i}{\alpha} \quad[7]`

We are able to do this because the value of `\alpha` will remain the same regardless of which ith triangle is in question. So now `b_i` is simply the ratio of the ith altitude to the total altitude of `T`

Using Equations [7] and [6], we can write:

`A(K_i)=(b_i)^2\cdot A(K) quad[8]`

Simplifying the rest of the problem

Since `J_i` relates to `J` in the same manner that `K_i` relates to `K`, we can write the relation between A(Ji) and A(J) as:

`A(J_i)=(b_i)^2\cdot A(J) \quad[9]`

Since we are dealing with rectangles that are inscribed, the `b_i` term we used for the `K_i` to `K` relation, can also be applied to the relationship between `J_i` and `J`.

At the end of section 4, we rewrote the given expression as:

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

Now we can use Equations [8] and [9] to rewrite this again. Since A(Ji) and A(Ki) are in the numerator of the second term only, we will begin by just looking at that:

`\sum_{i=1}^{n+1}A(J_i)+A(K_i)` `=\sum_{i=1}^{n+1}[(b_i)^2\cdot A(K) + (b_i)^2\cdot A(J)]`

As we said in the last section (under the explanation of summation notation), the addition of two terms in the sum can be written as two separate sums. Applying this here, we get:

`\sum_{i=1}^{n+1}[(b_i)^2\cdot A(K) + (b_i)^2\cdot A(J)]` `=\sum_{i=1}^{n+1}{A(K)\cdot (b_i)^2}+\sum_{i=1}^{n+1}{A(J)\cdot (b_i)^2}`

Going further, we can write:

`\sum_{i=1}^{n+1}{A(K)\cdot (b_i)^2}+\sum_{i=1}^{n+1}{A(J)\cdot (b_i)^2}` `=A(K)\cdot\sum_{i=1}^{n+1}(b_i)^2+A(J)\cdot\sum_{i=1}^{n+1}(b_i)^2`

Now recognize that this can be further simplified using the distribution property of multiplication (`k\cdot a+k\cdot b=k(a+b)`). So we can rewrite this again as:

`\sum_{i=1}^{n+1}[A(J_i)+A(K_i)]` `=[\sum_{i=1}^{n+1}(b_i)^2]\cdot[A(J)+A(K)]`

Remember, this is only the numerator of the second term of the given expression. So, rewriting the entire expression:

`1 - \frac{\sum_{i=1}^{n+1}[A(J_i)+A(K_i)]}{A(J)+A(K)}` `=1-\frac{[\sum_{i=1}^{n+1}(b_i)^2]\cdot [A(J)+A(K)]}{A(J)+A(K)} \quad[10]`

Looking at the second term of this expression, we can see that both the numerator and denominator contain the term `[A(J)+A(K)]`.

Since `\frac{A(J)+A(K)}{A(J)+A(K)}=1`,

we can rewrite Equation [10] as:

`1-\sum_{i=1}^{n+1}(b_i)^2`

This is the full simplification of the expression. We can now write:

`\frac{A(R)+A(S)}{A(T)}=1-\sum_{i=1}^{n+1}(b_i)^2`

Where `b_i` is the ratio of the height of the `i^{"th"}` inscribed rectangle, `R_i` to the overall height of the acute triangle `T` in which the rectangles are inscribed.

Summary

In this section we continued our simplification of the expression, picking up where we left off in Part 4 of this series. We finally simplified the given expression `\frac{A(R)+A(S)}{A(T)}` so that it is in terms of a single variable, `b_i`.
While there are likely a variety of methods to find the maximum value of our reduced form of the given function,

`1-\sum_{i=1}^{n+1}(b_i)^2`

we will do so using what is known as the Cauchy-Schwarz Inequality.

Since this series is intended for people without an extensive background in Mathematics, this will be a rather complicated idea. It will most likely take some time for you to wrap your head around it. (at least it did for me!) As such, I will dedicate the entirety of section 6 to the thorough explanation of the inequality. Once the inequality has been sufficiently explained, we will pick up where we leave off here in Section 7, as we apply the Cauchy-Schwarz Inequality to our problem.

As for now, give yourself some credit. We've dealt with some challenging ideas and concepts. This section was very Math intensive and I hope you have learned something new. As always, thank you for taking the time to read and attempt to understand. We're on the home stretch!

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