OF PRIME IMPORTANCE
How many primes are there? The famous Greek mathematician Euclid provides the following argument. Suppose that there are only a finite number of primes, say \(\{p_1,p_2, ... ,p_{k-1}, p_k\}\). Consider the number \(P = p_1 p_2 ... p_{k-1} p_k +1\). Note that \(P\) is another prime because it is not divisible by any of our primes yet it is not from our finite set of primes. Since we have shown that it is possible to keep constructing primes of this form, then it must be true that there are an infinite number of them (even if we cannot come up with nice formulas such as that of \(P\)).
These mysterious numbers have not escaped the cleverness (and funny bones) of some mathematicians. In fact, some primes have been given curious names such as being twins, cousins, good, happy, sexy, and even illegal!
A pair of primes are said to be twins (or twin primes) if they are of the form \(p\) and \(p+2\). Examples of such pairs are \(3\) and \(5\), and \(5\) and \(7\). Meanwhile, a pair of primes are cousins if they are of the form \(p\) and \(p+4\). Examples include \(3\) and \(7\), \(7\) and \(11\), and \(13\) and \(17\). However, notice that as we go further up the number line, the occurrences of such pairs become rarer. This leads one to ask whether or not there are an infinite number of cousin or twin primes. One unproven conjecture - known as de Polignac's conjecture - states that for every even number \(k\), there are infinitely many primes \(p_0\) and \(p_1\) such that \(p_1 - p_0 = k\). Once proven, then we know that there will be an infinite number of twin and cousin primes if \(k = 2\) or \(4\). The conjecture remains both unproven and unrefuted, so feel free to add your own solution!
While some prime pairs are considered members of numerical family trees, others are given human characteristics. A prime \(p_n\) is good if \((p_n)^2 > p_{n-1} p_{n+1}\); that is, a prime is good if it is greater than the geometric mean of its adjacent members. An example of a good prime is 5 because \(5^2 > (3)(7)\). It has been proven that there are infinitely many good primes. Meanwhile, a prime is happy if the sum of the squares of its digits, and the sums of the squares of the digits of these sums, will eventually reach 1. An example of a happy prime is \(19\). We have \[ 1^2 +9^2=82 \rightarrow 8^2 + 2^2 = 68 \rightarrow 6^2+8^2 = 100 \rightarrow 1^2 + 0^2 + 0^2 = 1. \]Lastly, a pair of primes is sexy if they are of the form \(p\) and \(p+6\), and they are named as such because the Latin word for "six" is "sex". Examples of sexy primes include \(5\) and \(11\), and \(7\) and \(13\). As you can see, being a sexy number has nothing to do with what that number wears or what it looks like.
How about illegal primes? Note that all the content we see online or in virtual form are actually codes that are acted upon by certain coding software owned by companies. This includes sensitive data such as copyrighted content, credit card passwords and confidential files that would be illegal to distribute in the public. Although originally written in binary form, these numbers can be represented in decimal form, and if it were illegal to possess the code, then its decimal form is an illegal number. An illegal number that is also a prime is known as an illegal prime.
We see that the prime numbers are not a trivial subset of the number system we use. There is definitely much to be known about them, and their applications can be found even in the most unexpected aspects of life.
ABOUT THE AUTHOR:
Arvin Lawrence N. Quinones is a graduate assistant at the Ateneo de Manila University. He obtained his B.S. in Applied Mathematics major in Mathematical Finance at the Ateneo de Manila University in 2015 and is currently taking his Master's degree at the same university.
REFERENCES:
- Weisstein, Eric W. "Prime Number." From MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/PrimeNumber.html.
- --. "Illegal prime numbers: What is it exactly?" Retrieved from http://stackoverflow.com/questions/4855972/illegal-prime-numbers-what-is-it-exactly.
OLYMPIAD CORNER
from the China Mathematical Olympiad, 2005 (posed by Ye Zhonghao)
Problem: A circle intersect sides \(BC\), \(CA\), \(AB\) of triangle \(ABC\) at two points for each side in the following order: \(\{D_1,D_2 \}\), \(\{E_1,E_2 \}\) and \(\{F_1,F_2 \}\). Line segments \(D_1E_1\) and \(D_2F_2\) at point \(L\); \(E_1F_1\) and \(E_2D_2\) intersect at point \(M\); \(F_1D_1\) and \(F_2E_2\) intersect at point \(N\). Prove that \(AL\), \(BM\) and \(CN\) are concurrent.
Note that \[ \frac{\sin \alpha _{1}}{\sin \alpha _{2}}=\frac{\frac{LL^{\prime }}{AL}}{\frac{LL^{\prime \prime }}{AL}}=\frac{LL^{\prime }}{LL^{\prime \prime }}=\frac{LF_{2}\sin \alpha _{3}}{LE_{1}\sin \alpha _{4}} \]
Draw line segments \(D_{1}F_{2}\) and \(D_{2}E_{1}\). By the intersecting chord theorem, we have \(\frac{LD_{1}}{LD_{2}}=\frac{LF_{2}}{LE_{1}}\). And since \(\angle D_{1}LF_{2}=\angle D_{2}LE_{1}\), it follows that \(\bigtriangleup LD_{1}F_{2}\sim \bigtriangleup LD_{2}E_{1}\). Therefore \[ \frac{LF_{2}}{LE_{1}}=\frac{D_{1}F_{2}}{D_{2}E_{1}} \]
Next we draw the line segments \(D_{2}F_{1}\) and \(D_{1}E_{2}\). By law of sines, we have \(\frac{\sin \alpha _{3}}{D_{2}F_{1}}=\frac{\sin \alpha _{4}}{D_{1}E_{2}}=2R\), where\(R\) is the circumradius. Hence \[\frac{\sin \alpha _{3}}{\sin \alpha _{4}}=\frac{D_{2}F_{1}}{D_{1}E_{2}}. \]
By substituting the second and third equations in the first equation, we have \[ \frac{\sin \alpha _{1}}{\sin \alpha _{2}}=\frac{LF_{2}\sin \alpha _{3}}{LE_{1}\sin \alpha _{4}}=\frac{D_{1}F_{2}}{D_{2}E_{1}}\cdot \frac{D_{2}F_{1}}{D_{1}E_{2}}. \]
Similarly, we write \(\angle BMC=\beta _{1},\angle MBA=\beta _{2},\angle NCA=\gamma _{1},\angle NCB=\gamma _{2}\), and we get \[ \begin{align*}
\frac{\sin \beta _{1}}{\sin \beta _{2}} &= \frac{D_{2}E_{1}}{E_{2}F_{1}}
\cdot \frac{D_{1}E_{2}}{E_{1}F_{2}} \\
\frac{\sin \gamma _{1}}{\sin \gamma _{2}} &= \frac{E_{2}F_{1}}{D_{1}F_{2}}
\cdot \frac{E_{1}F_{2}}{D_{2}F_{1}}
\end{align*} \]
Combining these equations, we obtain \[ \frac{\sin \alpha _{1}}{\sin \alpha _{2}}\cdot \frac{\sin \beta _{1}}{\sin\beta _{2}}\cdot \frac{\sin \gamma _{1}}{\sin \gamma _{2}}=1. \]
Extend \(AL\) and let it meet \(BC\) at \(E\). We now have the cevian \(AE\) in triangle \(ABC\) (similarly, we have the other cevians\(BF\) and \(CG\). By the inverse of (trigonometric) Ceva's theorem, the cevians \(AE\), \(BF\) and \(CG\) all intersect at the same point, which proves that\(AL\), \(BM\) and \(CN\) are concurrent.
Solution:
From point \(L\), draw two segments perpendicular to \(AB\) and \(AC\). Label the feet \(L^{\prime}\) and \(L^{\prime \prime }\) respectively. Let \(\angle LAB = \alpha_1\), \(\angle LAC = \alpha_2\), \(\angle LF_2A = \alpha_3\), and \(\angle LE_1A = \alpha_4\).
Note that \[ \frac{\sin \alpha _{1}}{\sin \alpha _{2}}=\frac{\frac{LL^{\prime }}{AL}}{\frac{LL^{\prime \prime }}{AL}}=\frac{LL^{\prime }}{LL^{\prime \prime }}=\frac{LF_{2}\sin \alpha _{3}}{LE_{1}\sin \alpha _{4}} \]
Draw line segments \(D_{1}F_{2}\) and \(D_{2}E_{1}\). By the intersecting chord theorem, we have \(\frac{LD_{1}}{LD_{2}}=\frac{LF_{2}}{LE_{1}}\). And since \(\angle D_{1}LF_{2}=\angle D_{2}LE_{1}\), it follows that \(\bigtriangleup LD_{1}F_{2}\sim \bigtriangleup LD_{2}E_{1}\). Therefore \[ \frac{LF_{2}}{LE_{1}}=\frac{D_{1}F_{2}}{D_{2}E_{1}} \]
Next we draw the line segments \(D_{2}F_{1}\) and \(D_{1}E_{2}\). By law of sines, we have \(\frac{\sin \alpha _{3}}{D_{2}F_{1}}=\frac{\sin \alpha _{4}}{D_{1}E_{2}}=2R\), where\(R\) is the circumradius. Hence \[\frac{\sin \alpha _{3}}{\sin \alpha _{4}}=\frac{D_{2}F_{1}}{D_{1}E_{2}}. \]
By substituting the second and third equations in the first equation, we have \[ \frac{\sin \alpha _{1}}{\sin \alpha _{2}}=\frac{LF_{2}\sin \alpha _{3}}{LE_{1}\sin \alpha _{4}}=\frac{D_{1}F_{2}}{D_{2}E_{1}}\cdot \frac{D_{2}F_{1}}{D_{1}E_{2}}. \]
Similarly, we write \(\angle BMC=\beta _{1},\angle MBA=\beta _{2},\angle NCA=\gamma _{1},\angle NCB=\gamma _{2}\), and we get \[ \begin{align*}
\frac{\sin \beta _{1}}{\sin \beta _{2}} &= \frac{D_{2}E_{1}}{E_{2}F_{1}}
\cdot \frac{D_{1}E_{2}}{E_{1}F_{2}} \\
\frac{\sin \gamma _{1}}{\sin \gamma _{2}} &= \frac{E_{2}F_{1}}{D_{1}F_{2}}
\cdot \frac{E_{1}F_{2}}{D_{2}F_{1}}
\end{align*} \]
Combining these equations, we obtain \[ \frac{\sin \alpha _{1}}{\sin \alpha _{2}}\cdot \frac{\sin \beta _{1}}{\sin\beta _{2}}\cdot \frac{\sin \gamma _{1}}{\sin \gamma _{2}}=1. \]
Extend \(AL\) and let it meet \(BC\) at \(E\). We now have the cevian \(AE\) in triangle \(ABC\) (similarly, we have the other cevians\(BF\) and \(CG\). By the inverse of (trigonometric) Ceva's theorem, the cevians \(AE\), \(BF\) and \(CG\) all intersect at the same point, which proves that\(AL\), \(BM\) and \(CN\) are concurrent.
PROBLEMS
- The sequence \(a_{0},a_{1},a_{2},\dots\) satisfies \[
a_{m+n}+a_{m-n}=\frac{1}{2}\left(a_{2m}+a_{2n}\right)
\]for all nonnegative integers \(m\) and \(n\) with \(m\ge n\). If \(a_{1}=1\), determine \(a_{2015}\).
- Show that if \(x\), \(y\), and \(z\) are nonnegative real numbers for which \(x+y+z=1\), then\[ x^{2}y+y^{2}z+z^{2}x\leq\frac{4}{27}. \]
- Determine the last four digits of \(2015^{2015}\).
- For \(\triangle ABC\), let \(a\), \(b\), and \(c\) represent the sides opposite \(\angle A\), \(\angle B\), and \(\angle C\), respectively. If \(b<\dfrac{1}{2}\left(a+c\right)\), prove that \(m\angle B<\dfrac{1}{2}\left(A+C\right)\).
- Consider a circle with center \(C\) and radius \(\sqrt{5}\), and let \(M\) and \(N\) be points on a diameter of the circle such that \(MO=NO\). Let \(AB\) and \(AC\) be chords passing through \(M\) and \(N\), respectively, such that \[ \frac{1}{MB^{2}}+\frac{1}{NC^{2}}=\frac{3}{MN^{2}}. \]Determine the length of \(MO\).
- Let \(A\) be the largest subset of \(\left\{ 1,2\dots,2015\right\}\) such that \(A\) does not contain two elements and their product. How many elements does \(A\) have?
We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may ONLY be submitted online via vantonio@ateneo.edu. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 12:30 PM September 26, 2015.
No comments:
Post a Comment