The Rewards of Risk (Tuklas Vol. 14, No. 3 - Dec. 1, 2012)

THE REWARDS OF RISK

In our capitalist society, money is power. Money drives the economy on a large scale. On a basic level, money allows us to buy the things we want and/or need. However, some people have lots of money and some don't. This now creates an interesting phenomenon wherein some people have more money than what they need and some need more money than what they have. Since there is a discrepancy, those who are in need borrow money from those who have a lot. We distinguish between these two types of people by calling them borrowers and lenders (respectively). 

Borrowers need money for reasons as varied as starting up a business, buying a car or a house, or paying for tution. Lenders, on the other hand, lend money primarily because they have a lot of it. Nevertheless, lending is not the only option they have. They may opt to spend the money, keep it in their wallets, invest it in their existing or start-up business, or lend it. The question now is, what do they do with the money?

Since we're assuming that the money in question is in excess of what they spend (a surplus, if you will) then we can assume that they will not choose to spend the money. Choosing to keep the money in their wallets, inside their safes or under their mattresses would not be a very wise decision. When you keep P1000 in your wallet, it will still be P1000 by the end of the day. Keep it in your wallet for a year, and it will still be P1000 after a year.

What may be better options? Bank deposits return 0.375% annually. That means, if you deposit P1000 in your bank, your money will be P1003.75 after 1 year. On the other hand, starting up a business may give a higher return for your investment. I emphasize the word may since there is a chance that won't happen. Consider the likes of Mark Zuckerberg, the founder of Facebook. He invested his money to start up a business and it is now worth $50 billion. This business can and will earn so much more than choosing to deposit your money in a bank. Of course, that's one out of the hundreds and thousands of people struggling to establish their own business. How many other Zuckerbergs have you heard of? Furthermore, by starting up a business, you will need to invest a lot of time, effort and courage to take risks for you to end up with that return... and that return might not even materialize.

Entrepreneurs like Mark Zuckerberg might not have much money to start the business but need to borrow to be able to provide the resources. That is where lenders come to the picture. To simplify things, let's say these lenders, also called investors, will help finance the project for an exact return. For example, Zuckerberg borrows $40,000 from Investor A for an interest of 30% after a year. Investor A will receive $52,000 after a year. If \(r\) is the interest rate, \(P\) is the principal or initial amount and \(A\) is the final amount, the equation of interest is \[ \begin{align*} A &=P+Pr \\ A &=P(1+r) \\ 52,000&=40,000*(1+0.3), \end{align*} \] which means Mark Zuckerberg pays $12,000 extra at the end of the year to borrow $40,000.

Aside from start-up businesses, governments can also borrow money from the public to build roads and other projects. The Philippine government issues what is called Treasury bills and Treasury bonds. Interest will be around 4% per annum. That means, if the government borrowed $40,000 from you, you will only expect $41,600 at the end of the year earning only $1,600. Why is there a discrepancy?

First, we have to understand that there is much larger risk in investing in a start-up business compared to investing in the government. Start-up businesses are more likely to be unable to pay back the debt due to bad management, uncontrollable market conditions, operational risks, etc. The Philippine Government, on the other hand, will never run out of cash because it can always print more money. If both of them have an interest of 4%, of course you would prefer the one with the higher chance to pay back. That is why riskier borrowers increase their interest rate to attract lenders.

There are many things that one can do with excess money to produce even more money. It all  just depends on the amount of money that one wishes to make versus the level of risk that one is willing to take.

ABOUT THE AUTHOR:
Allen Dominique Torres is an Instructor at the Ateneo de Manila University. He obtained his B.S. and Master's degree in Applied Mathematics, major in Mathematical Finance at the Ateneo de Manila University in 2010 and 2011, respectively.

OLYMPIAD CORNER

from the Singapore National Team Selection Test for IMO, 2001/2002

Problem: In the acute \(\Delta ABC\), let \(D\) be the foot of the perpendicular from \(A\) to \(BC\), let \(E\) be the foot of the perpendicular from \(D\) to \(AC\), and let \(F\) be a point on the line segment \(DE\). Prove that \(AF\) is perpendicular to \(BE\) if and only if \(FE/FD=BD/DC\).

Solution: 

Let \(G\) be the point on \(CE\) such that \(DG\) is parallel to \(BE\). Hence \(\Delta EBC\sim \Delta GDC\), which implies that \(\angle EBC=\angle GDC\) and \(EG/GC=BD/DC\). 

Note also that \(\Delta ADE\sim \Delta DCE\), since \(\Delta ADC\) is a right triangle and \(DE\perp AC\). 

Let \(H\) be the point of intersection between \(AF\) and \(BE\). We now make the following equivalent statements. \[ \begin{align*} FE/FD &= BD/DC \\ &\Leftrightarrow FE/FD=EG/GC \\ &\Leftrightarrow \Delta ADF\text{ }\sim \Delta DCG \\ &\Leftrightarrow \angle DAF=\angle GDC \\ &\Leftrightarrow \angle DAF=\angle EBD \\ &\Leftrightarrow \text{quadrilateral }ABDH\text{ is cyclic} \\ &\Leftrightarrow \angle ADB=\angle AHB=90^{\circ } \\ &\Leftrightarrow AF\perp BE \end{align*} \]

Remark: The statement \(FE/FD=EG/GC\Leftrightarrow \Delta ADF \sim \Delta DGC\) is based on the similarity of \(\Delta ADE\) and \(\Delta DCE\), where the cevians \(AF\) and \(DG\) divide \(DE\) and \(EC\), respectively, in the same proportion (\(FE/FD=EG/GC\)).

PROBLEMS

7. Let \(S(x)\) be the sum of the digits of the positive integer \(x\) in its decimal representation. Show that \(\frac{S(3x)}{S(x)}\) is bounded, or provide a counterexample that says otherwise.
   
8. From the \(xy\)-plane, select five distinct points that have integer coordinates. Find the probability that there is a pair of points among the five whose midpoint has integer coordinates.
9. Let \(ABCD\) be a convex quadrilateral. Prove that \(AC \perp BD\) if and only if \(AB^2+CD^2=AD^2+BC^2\).

We welcome readers to submit solutions to the problems posed above for publication consideration. Solutions may be submitted to the PEM facilitators on the deadline date or online via mactolentino@math.admu.edu.ph. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:00 PM December 8, 2012. 

SOLUTIONS

(for November 24, 2012)
NOTE: The answers sent via email have not yet been checked. Those who provided correct and complete solutions prior to this problem set's deadline (12:00PM December 1, 2012) will be credited within the week.

4. Prove that the sum of squares of 3, 4, 5, or 6 consecutive integers is not a perfect square.  (Mathematical Olympiad Summer Program, 1998)
   (solved by Clyde Wesley Ang [Chiang Kai Shek College], Jecel Manabat [Valenzuela City Sci HS], Jazzrine Tagle [Valenzuela City HS], Renzo Tan [Chiang Kai Shek College], Terence Tsai [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy]; partial credit for Kimberly Co [St. Stephen's])
   
   
   SOLUTION: 
   Define \(S(n,k)\) to be \[S(n,k)=n^2+(n+1)^2+\ldots+(n+k-1)^2,\] or the sum of squares of \(k\) consecutive integers starting with \(n\). We now consider each case separately.
        
   CASE 1: \(S(n-1,3)=3n^2+2\equiv 2\text{ mod }3\) hence the sum of 3 consecutive integers is not a perfect square.
        
   CASE 2: \(S(n,4)=4(n^2+3n+3)+2\equiv 2\text{ mod }4\) hence the sum of 4 consecutive integers is not a perfect square.
        
   CASE 3: \(S(n-2,5)=5(n^2+2)\equiv 2\text{ mod }4\) if \(n\) is even or \(3\text{ mod }4\) if \(n\) is odd, hence the sum of 5 consecutive integers is not a perfect square.
   
   CASE 4: \(S(n-2,6)=6n(n+1)+19\equiv 3\text{ mod }4\) since \(6n(n+1)\) is always even, hence the sum of 6 consecutive integers is not a perfect square.
   
   Note that \(16 = 4^2 = 3^2 + 5 + 2\) is an example indicating that a perfect square may not necessarily be expressed as a perfect square trinomial.
5. If it is possible to construct a triangle with side length \(a<b<c\), prove that it is possible to construct a triangle with side lengths \(\sqrt{a}<\sqrt{b}<\sqrt{c}\); also, show that the converse is false. (Taken from Inequalities by Manfrino, Ortega, and Delgado)
   (solved by Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy]; partial credit for Gabrielle Bruce [AC] and Terence Tsai [Chiang Kai Shek College])
   
   
   SOLUTION: 
   Note that \[ c<a+b \Rightarrow c<a+b+2\sqrt{ab} = \left(\sqrt{a} + \sqrt{b}\right)^2 \Rightarrow \sqrt{c}<\sqrt{a}+\sqrt{b}. \] To disprove the converse, consider the example with \(a=4\), \(b=9\) and \(c=16\).
6. Prove that among any seven distinct integers, there must be two such that their sum or difference is divisible by 10. (China Mathematical Competitions for Secondary Schools, 2002)
   (solved by Clyde Wesley Ang [Chiang Kai Shek College], Milet Aquino [Saint Pedro Poveda College], Jecel Manabat [Valenzuela City Sci HS], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College], Renzo Tan [Chiang Kai Shek College], Terence Tsai [Chiang Kai Shek College], Farrell Eldrian Wu [MGC New Life Christian Academy])
   
   
   SOLUTION: 
   We replace the integers by their remainder modulo 10. Recall that a number is divisible by 10 if its ones digit is a zero.
   
   Any pair of numbers whose sum or difference is divisible by 10 would fall under one of the following sets: \[ \{0,0\}, \{5,5\}, \{1,9\}, \{2,8\}, \{3,7\}, \{4,6\}. \]Since there are 7 remainders to choose from, we are assured that at least one set pair will appear, and thus would have a sum or difference divisible by 10.
   
   

ERRATA

There was a typographical error in the solution to Problem #1 that read \(\sin^\theta(2)\), which has been corrected to read \(\sin^2\theta(2)\).

There have been typographical errors in the past issues of Tuklas regarding the directions for the Problems section; "posed below" has been changed to "posed above."