1985 Putnam Question A-2: Solution Part 7
Overview
You made it! Here we are, finally at the conclusion of the series. In Section 6 we introduced the Cauchy-Schwarz inequality. This is an important inequality with applications in various fields of Mathematics, and an application to the problem at hand. In this section, we build upon our progress in Part 5, using the Cauchy-Schwarz inequality to arrive at the final solution for our problem.
Beginning of the End
At the end of Putnam 1985 A-2 Solution Part 5, we reduced our simplification of the given expression to be:
`\frac{A(R)+A(S)}{A(T)}=1-\sum_{i=1}^{n+1}(b_i)^2`
where `b_i` represents the ratio of the altitude of the `i^{"th"}` triangle to the Altitude of the acute triangle `T`, and there are `n` rectangles inscribed in `T`.
Implementing the Cauchy-Schwarz Inequality
In the previous section, Putnam 1985 A-2 Solution Part 6, we discussed the Cauchy-Schwarz inequality:
`[\sum_{i=1}^{n}(q_i)^2]\cdot [\sum_{i=1}^{n}(r_i)^2]` `\geq [\sum_{i=1}^{n}(q_i\cdot r_i)]^2`
Now we can use that equality to maximize the given expression.
Looking at the expression, we can see that the term `\sum_{i=1}^{n+1}(b_n)^2` is subtracted from 1. This term will always be a positive value, so the expression cannot possibly be more than 1. As such, we can find the maximum value of the expression by finding the minimum value of the term `\sum_{i=1}^{n+1}(b_n)^2`.
First let the `b_i` terms be the `q_i` terms given in the inequality:
`q_i = b_i`
furthermore, we can let the series of `r_i` terms be whatever we want. So lets define that series as:
`r_1=r_2=r_3=...=r_n=1`
(recall this is the same method we used to solve the example in the previous section).
Now our inequality looks like:
`\sum_{i=1}^{n}(b_i)^2\cdot \sum_{i=1}^{n}(1^2)\geq (\sum_{i=1}^{n}(b_i\cdot 1))^2`
But our maximum term is `n+1`, not `n`, so we must adjust all of the upper limits accordingly. The inequality now becomes:
`\sum_{i=1}^{n+1}(b_i)^2\cdot \sum_{i=1}^{n+1}(1^2)\geq (\sum_{i=1}^{n+1}(b_i\cdot 1))^2 \quad[1]`
Simplifying the Right Side of the inequality
First let's look at the right side.
Since `r_1=r_2=r_3=...=r_{n+1}=1`, the term on the right becomes:
`(\sum_{i=1}^{n}b_i)^2 \quad[2]`
Recall that `b_i=\frac{a_i}{\alpha}`, where `\alpha` is the total altitude of `T`, and `a_i` is the altitude of the `i^{th}` triangle.
We know that the sum of the altitudes of triangles `J_1` through `J_{n+1}` is the total altitude of T.
`\sum_{i=1}^{n+1}{a_i}=\alpha \quad[3]`
We can write out Expression [2] in long form as:
`\frac{a_1}{\alpha}+\frac{a_2}{\alpha}+\frac{a_3}{\alpha}+\frac{a_4}{\alpha}+...\frac{a_n}{\alpha}+\frac{a_{n+1}}{\alpha} \quad[4]`
And since all terms have the same denominator, we can simplify this further to be:
`\frac{a_1+a_2+a_3+a_4+...+a_n+a_{n+1}}{\alpha}`
Using Expression [3] in Expression [4] , we get
`\frac{\alpha}{\alpha} \quad[5]`
which will, of course, simplify to 1.
Using Expression [5], we can rewrite our inequality from Equation [1] as:
`\sum_{i=1}^{n+1}(b_i)^2\cdot \sum_{i=1}^{n+1}(1^2)\geq 1 \quad[6]`
Simplifying the left side
Let's begin by looking at the 2nd term on the left side:
`\sum_{i=1}^{n}(1^2) \quad[7]`
This simplifies pretty easily. Since `1^2=1`, we can rewite Expression [7] as:
`\sum_{i=1}^{n}(1)`
Now what's the sum `1+1+1+1+1+1+...+1`, if there are exactly n terms? Well it should be n of course!
`\sum_{i=1}^{n}(1^2)=\sum_{i=1}^{n}{1}=n`
So we can now rewrite the inequality Expression [6] as:
`n\cdot\sum_{i=1}^{n}(b_i)^2\geq 1`
Now we need only to divide each side by `n` to get the inequality:
`\sum_{i=1}^{n}(b_i)^2\geq\frac{1}{n}`
Since the inequality shows that `\sum_{i=1}^{n}(b_i)^2` is greater than or equal to `\frac{1}{n}`, we know that the smallest possible value can be achieved when the two sides are equal
So we can say that the minimum of `\sum_{i=1}^{n}(b_i)^2` is
`\frac{1}{n}`
Likewise, the maximum of the expression `1-\sum_{i=1}^{n+1}(b_i)^2` is
`1-\frac{1}{n+1}`.
Remember in Part 2, we generalized the problem to account for `n` rectangles inscribed within the acute triangle. Now, since we are dealing with 2 inscribed rectangles, we can let `n=2`.
So the maximum of `\frac{A(R)+A(S)}{A(T)}` is
`1-\frac{1}{2+1}`
`=1-\frac{1}{3}`
`=\frac{2}{3}`
That's it! We solved the problem!, We've found the maximum value of the given expression `\frac{A(R)+A(S)}{A(T)}` to be `\frac{2}{3}`
Conclusion
Congratulations! You've finally done it. An answer has been produced.
Actually, you've done far more than simply come up with an answer. You've produced what is known as a Mathematical Proof. We took small, incremental steps along the way, using facts that we know to be true. The result of this process is that we have arrived at an answer so rock-solid in its nature that it absolutely must be true.
We began by analyzing the problem A-2 from the 1985 Putnam Competition. We were given a geometric situation, and an expression that we had to maximize. Our first step towards a solution was determining that the expression could in fact be minimized. But the expression had too many variables: `A(S)`, `A(R)` and `A(T)` could be almost anything it seemed. So we used generalization as a tool to simplify the problem. We were able to do this by generalizing the situation to apply to `n` rectangles. The next step was to find some sort of relationship between the areas of the rectangles `R_i` and the acute triangle `T`. We were able to do this using some geometric properties of triangles. Using these properties in conjunction with some algebra allowed the expression to be simplified to a single variable. From here, we simply needed the Cauchy-Schwarz Inequality to arrive at the final solution.
At this point I'd like to take the time to thank you, the reader, for reading this series. I came up with this idea as a way to share my love of math with others in what I hope has been an non intimidating fashion. I think it is tragic how much of my personal school experience failed to foster a true love of mathematics. I am one of the lucky few who, by a matter of bizarre circumstance and dumb luck, has had the privilege of being exposed to the beauty of mathematics. I truly hope that I've been able to share a piece of my enthusiasm for the subject with you.
Thank you.
Acknowledgements
Special thanks to Murray Bourne, purveyor of IntMath.com for being open to this idea and for his help along the way.
Additionally I'd like to acknowledge the following people and organizations for providing references used in the solving of this problem:
- MAA, What is the Putnam Competition?
- Kiran Kedlaya, The Putnam Archive
- Po-Shen Loh, 2016 Putnam Seminar
- Razvan Gelca and Titu Andreescu, Putnam and Beyond
- Loren C. Larson, Problem Solving Through Problems
- D.J. Newman, A Problem Seminar
- Wolfram MathWorld
- Wikipedia
- Bjorn Poonen
- Ravi Vakil
