The Viewable Sphere (Tuklas Vol. 17, No. 2 - September 26, 2015)

THE VIEWABLE SPHERE

Imagine yourself standing in the middle of a room. Keep your body still, while turning your head in all directions. Try to observe everything around you. Your perception of your surroundings is roughly called a viewable sphere. You can imagine yourself as being on the center of a sphere with images of your surroundings pasted across the inner surface of the sphere. We perceive the entire world in terms of viewable spheres, and this sphere changes as we move across space.

To make this description more precise, imagine a sphere in space with a radius that is sufficiently small so that it does not touch any object in its surroundings. Consider another point in space. The line segment joining this point and the center of the sphere intersects the sphere at a single point. Using this association between the surroundings and the sphere, one can essentially paint the world on the surface of the sphere. This is called a viewable sphere.

What is usually of interest is how these viewable spheres are represented or projected on a plane or a flat surface. Whenever you take a picture of your surroundings, you are essentially capturing a portion of the camera's viewable sphere and flattening it to produce a photo. This gets more interesting when you are taking a larger portion of a viewable sphere, such as when you are capturing a panorama. Recently, cameras are capable of applying different ways of mapping these viewable spheres to a flat surface. A kind of mapping that is gaining popularity is the stereographic projection, where it transforms a panorama, as in Figure 1, to a photo that resembles a small world, as in Figure 2.

Figure 1: A panorama of Neumünster Abbey, Luxembourg. 1



Figure 2: A small world photo of Neumünster Abbey, Luxembourg. 2

Imagine the viewable sphere sitting on top of a plane. Then, consider a light beam that is being emitted from the upper tip of the viewable sphere towards some region of the sphere. Let us assume that the viewable sphere is translucent, thus allowing the light beam to pass through. Then the light beam will create a projection of a region of the sphere on the plane surface. This type of mapping is called a stereographic projection, as illustrated in Figure 3. This type of projection produces a photo that resembles a small world because it projects lines on the sphere as circles on the plane. As such, a horizon on a viewable sphere (created using a panoramic photo) projects as a circle on the plane surface and creates an illusion of a small world. Furthermore, objects located below the horizon in the viewable sphere will fall inside the small world, while objects located above the horizon will fall outside of the small world.

Figure 3: A stereographic projection from a sphere to a plane. 3

After reading this article, you probably won't look at a photograph or a map the same way again!

ABOUT THE AUTHOR:
Jayhan Regner is a part-time faculty member 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.

ENDNOTES:

[1] Photo taken by Sébastian Pérez-Duarte. Retrieved from http://flickr.com/photos/sbprzd.

[2] Photo taken by Sebastian Perez-Duarte. Retrieved from http://flickr.com/photos/sbprzd.
[3] Retrieved from http://dimensions-math.org.

REFERENCES:

- Swart, David and Bruce Torrence. "Mathematics Meets Photography." Math Horizons, 19(1), 14-17 (2011).

OLYMPIAD CORNER

from the 22nd Irish Mathematical Olympiad, 2009

Problem: For any positive integer $n$, define $$E\left( n\right) =n\left( n+1\right) \left( 2n+1\right) \left( 3n+1\right) \cdots \left( 10n+1\right)$$ Find the greatest common divisor of $E\left( 1\right) ,E\left( 2\right) ,\ldots, E\left( 2009\right)$.

Solution (Proposed by Marius Ghergu): 

Let $m$ be the GCD of $E\left( 1\right) ,E\left( 2\right) ,...,E\left(2009\right)$.

Since $m|E\left( 1\right) =11!$, then $m$ does not have any prime divisor greater than $13$, i.e. if $p|m$ and $p$ is prime, then $p\in \{2,3,5,7,11\}$.

Moreover, $p$ is less than $2009$, hence $m|E\left( p\right) =p\left(p+1\right) \left( 2p+1\right) \cdots \left( 10p+1\right)$. Note that $\gcd \left( p,kp+1\right) =1$ for $k=1,2,...,10$, hence $E\left( p\right)$ is divisible by $p$ but not by $p^{2}$. It also follows that $m$ is not divisible by $p^{2}$, which establishes that $m$ is not divisible by the square of any prime number.

Since $m|E\left( 1\right) =11!$, then $m$ must divide the product of all primes less than or equal to $11$, that is, $m|2310$. If $m$ divides a larger number (product) and does not divide $2310$, then $m$ is divisible by the square of some prime number, which isn't suppose to be. Hence $m$ must necessarily divide $2310$.

We also make the following observation. For any $n\geq 1$,

(a) one of the numbers $n$ or $n+1$ is even, so $2|E\left( n\right)$;

(b) one of the numbers $n,n+1,2n+1$ is divisible by $3$, hence $3|E\left(n\right)$;

(c) one of the numbers $n,n+1,2n+1,3n+1,4n+1$ is divisible by $5$, hence $5|E\left( n\right)$;

(d) one of the numbers $n,n+1,2n+1,3n+1,4n+1,5n+1,6n+1$ is divisible by $7$, hence $7|E\left( n\right)$;

(e) one of the numbers $n,n+1,2n+1,3n+1,4n+1,...,9n+1,10n+1$ is divisible by $11$, hence $11|E\left( n\right)$.

This shows that $E\left( n\right)$ is a multiple of $2\cdot 3\cdot 5\cdot 7\cdot 11=2310$ for $n\geq 1$, hence $m\geq 2310$. But as shown earlier, $m$ divides $2310$, which means $m\leq 2310$. Thus, $m=2310$.

Remark:

To establish the above statements, consider the following solution for (c):

The number $n$ can take one of these five forms, $$5k,5k+1,5k+2,5k+3,5k+4.$$ If $n=5k$, then $5|n$ hence $5|E\left( n\right)$.

If $n=5k+1$, then $5|4n+1=20k+5$, hence $5|E\left( n\right)$.

If $n=5k+2,$ then $5|2n+1=10k+5$, hence $5|E\left( n\right)$.

If $n=5k+3$, then $5|3n+1=15k+10$, hence $5|E\left( n\right)$.

If $n=5k+4$, then $5|n+1=5k+5$, hence $5|E\left( n\right)$.

We have thus shown that one of the numbers $n,n+1,2n+1,3n+1,4n+1$ is divisible by $5$.

Do the same thing for the other statements.

PROBLEMS

7. Let $ABCD$ be a rhombus with $\angle A=120^{\circ}$ and $P$ is any point on the plane of the rhombus. Prove that $$PA+PC>\frac{BD}{2}.$$
8. Determine, with proof, the values of $a$, $b$, and $c$, with $a<b<c$ such that $2015=a^{2}+b^{2}-c^{2}$.
9. Consider $2015$ nonnegative real numbers $a_{1},a_{2},\dots,a_{2015}$ whose sum is $1$. Define $M$ to be $$M=\max_{1\leq k\leq}a_{k}+a_{k+1}+a_{k+2}.$$ As $a_{k}$ varies, $M$ varies as well. Find the minimum value of $M$.

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 October 3, 2015.

SOLUTIONS

(for September 12, 2015)

1. 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 $a_{2015}$. (Russia, 1995)
   (solved by Patricia Faith E. Capito [Philippine Science High School], Jarrett Ian G. Lim [Philippine Academy of Sakya], Ryan Jericho Sy [Chiang Kai Shek College]; partial credit for Maria Monica Manlises [St. Stephen's High School] and Madeline L. Tee [Chang Kai Shek College])
   
   
   SOLUTION: 
   We first note that if $m=n=0$, then we have $$a_{0}+a_{0}=\frac{1}{2}\left(a_{0}+a_{0}\right),$$ and solving for $a_{0}$, we get $a_{0}=0$.
   
   Since it is given that $a_{1}=1$, we can obtain $a_{2}$ as follows: $$\begin{align} a_{1+0}+a_{1-0} & = \frac{1}{2}\left(a_{2\left(1\right)}+a_{2\left(0\right)}\right)\\ 1+1 & = \frac{1}{2}a_{2}\\ a_{2} & = 4. \end{align}$$ In general, if we use any $m$ and $n=0$, we get $$\begin{align} a_{m+0}+a_{m-0} & = \frac{1}{2}\left(a_{2\left(m\right)}+a_{2\left(0\right)}\right)\\ 4a_{m} & = a_{2m}. \end{align}$$ And this is true for all $m$.
   
   However, we still don't have a pattern for when the goal term is odd. Checking for $m=2$ and $n=1$, we get $a_{3}=9$. The pattern that we are seeing is that $a_{m}=m^{2}$.
   
   We use strong induction on $a_{m}$.
   
   Note that if $m=0$, then $a_{0}=0^{2}=0$. Now, let our conjecture be true for all values $m=k-1$ and $m=k$. This means that we will assume, for the purposes of induction, that for those two values of $m$, we have $a_{m}=m^{2}$, and our goal is to show that this conjecture is also true for $m=k+1$. Now, note that if we have $n=1$, then $$a_{k+1}+a_{k-1}=\frac{1}{2}\left(a_{2k}+a_{2}\right),$$ from the result above, we know that $a_{2k}=4a_{k}$. So, $$a_{k+1}+a_{k-1}=2a_{k}+2,$$ but from our inductive hypothesis, we have $a_{k-1}=\left(k-1\right)^{2}$ and $a_{k}=k^{2}$, so we have $$\begin{align} a_{k+1}+\left(k-1\right)^{2} & = 2k^{2}+2\\ a_{k+1} & = k^{2}+2k+1\\ & = \left(k+1\right)^{2}. \end{align}$$ Thus, our conjecture holds. This means that $a_{m}=m^{2}$. So $a_{2015}=2015^{2}$.
2. Show that if $x$, $y$, and $z$ are nonnegative real numbers for which $x+y+z=1$, then $$x^{2}y+y^{2}z+z^{2}x\leq\frac{4}{27}.$$ (Canadian Mathematical Olympiad, 1999)
   (solved by Jarrett Ian G. Lim [Philippine Academy of Sakya]; partial credit for Patricia Faith E. Capito [Philippine Science High School], Maria Monica Manlises [St. Stephen's High School], Joyce Heidi Ong [Chiang Kai Shek College], Ryan Jericho Sy [Chiang Kai Shek College] and Madeline L. Tee [Chang Kai Shek College])
   
   
   SOLUTION: 
   Without loss of generality (and because of the cyclical property of the inequality), we can assume that $x\geq y\geq z$.
   
   Now, note that, if $x$, $y$, and $z$ are nonnegative, then $$\begin{align} 0 & \leq y^{2}\left(x-z\right)+z^{2}y\\ & = xy^{2}-zy^{2}+z^{2}y+\left(xyz-xyz\right)\\ & = z^{2}y+yz\left(x-y\right)+xy\left(y-z\right) \end{align}$$ is an expression that's also nonnegative. Since we are trying to get an upper bound for the cyclic sum, again, without loss of generality, we can assume that $z=0$. So we have $x+y=1$, and we want to show that $$x^{2}y\leq\frac{4}{27}.$$ We use the generalized AM-GM inequality on $x$, $x$, and $2y$ to get $$\sqrt[3]{2x^{2}y}\leq\frac{2x+2y}{3},$$ but since $y$ is nonnegative, we have $$\begin{align} \sqrt[3]{2x^{2}y} & \leq \frac{2x+2y}{3}\\ 2x^{2}y & \leq \left(\frac{2}{3}\right)^{3}\\ x^{2}y & \leq \frac{4}{27}, \end{align}$$ which is what we want.
3. Determine the last four digits of $2015^{2015}$. (Modified from the Rice Math Tournament 2005 Advanced Test)
   (solved by Patricia Faith E. Capito [Philippine Science High School], Bianca Guerrero [Assumption Antipolo], Jarrett Ian G. Lim [Philippine Academy of Sakya], Joyce Heidi Ong [Chiang Kai Shek College], Madeline L. Tee [Chang Kai Shek College], and Carissa Elaine F. Varon [Assumption Antipolo]; partial credit for Maria Monica Manlises [St. Stephen's High School])
   
   
   SOLUTION: 
   Note that $$\begin{align} 2015^{2015} & = \left(2000+15\right)^{2015}\\ & = \sum_{i=0}^{2015}\binom{2015}{i}2000^{2015-i}15^{i}\\ & \equiv 15^{2015}\left(\mod10000\right). \end{align}$$ This means that we just have to consider the remainder when powers of 15 are divided by 10000. $$\begin{align} 15^{1} & = 0015\\ 15^{2} & = 0225\\ 15^{3} & = 3375\\ 15^{4} & = 0625\\ 15^{5} & = 9375\\ 15^{6} & = 0625\\ 15^{7} & = 9375\\ & \vdots & \end{align}$$ $9375$ and $625$ will already alternate as the last four digits, the former for odd exponents, while the latter for even exponents. This means that the last four digits of $2015^{2015}$ is $9375$.
4. 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 $m\angle B<\dfrac{1}{2}\left(A+C\right)$. (China, 1990)
   (partial credit for Joyce Heidi Ong [Chiang Kai Shek College] and Madeline L. Tee [Chang Kai Shek College])
   
   
   SOLUTION: 
   We first extend sides $c$ and $a$ (the two longer sides) of the triangle to an isosceles triangle with two of the sides with length $a+c$. So define $D$ and $E$ such that $BD=a+c$ and $BE=a+c$. Locate $F$ on $\triangle BDE$ such that $ACFD$ is a parallelogram. (Similarly, we can look for $F$ such that $ACEF$ is a parallelogram. Note that $F$, in that case, will be located outside $\triangle BDE$, but the result will still follow.)
   
   

   

   
   Note that, by SSS congruence, $\triangle CBA\cong\triangle ECF$.
   
   The triangle inequality says that $DE< DF+FE=2b$. From the given, we have $$\frac{DE}{2}<b<\frac{1}{2}\left(a+c\right).$$ This means that $DE<a+c$, but $BE=BD=a+c$. This means that $DE<BE$, and by Hinge Theorem, we have $\angle B<\angle BDE$. Note that, because $\triangle BDE$ is isosceles, we have $$\begin{align} 2\angle BDE+\angle B & = 180^{\circ}\\ \angle BDE & = \frac{180^{\circ}-\angle B}{2}. \end{align}$$ Combining this equation with $\angle B<\angle BDE$, we have $$\begin{align} \angle B & < \frac{180^{\circ}-\angle B}{2}\\ 3\angle B & < 180^{\circ}\\ 3\angle B & < \angle A+\angle B+\angle C\\ \angle B & < \frac{1}{2}\left(\angle A+\angle C\right). \end{align}$$
5. 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 $MO$. (Bulgarian Winter Mathematical Competition 2006 43 (177))
   (partial credit for Madeline L. Tee [Chang Kai Shek College])
   
   
   SOLUTION: 
   Let segment $M_{1}N_{1}$ be the diameter on which $M$ and $N$ are both located. Let $x$ be the length of $MO$ and $NO$. Then we have the following figure.
   
   

   

   
   If $MO=NO=x$, then $M_{1}M=N_{1}N=\sqrt{5}-x$, while $MN_{1}=NM_{1}=\sqrt{5}+x$. Using Power of a Point Theorem on $M$ and $N$, respectively, we have $$\begin{align} MA\cdot MB & = MM_{1}\cdot MN_{1}\\ & = \left(\sqrt{5}-x\right)\left(\sqrt{5}+x\right)\\ & = 5-x^{2}\\ MB & = \frac{5-x^{2}}{MA} \end{align}$$ We can also use the same argument for $N$ and see that $NA\cdot NC=5-x^{2}$ as well. Moreover, $NC=\dfrac{5-x^{2}}{NA}$. From the given condition on $MB$ and $NC$, we have $$\begin{align} \frac{1}{MB^{2}}+\frac{1}{NC^{2}} & = \frac{MA^{2}}{\left(5-x^{2}\right)^{2}}+\frac{NA^{2}}{\left(5-x^{2}\right)^{2}}.\\ \frac{3}{MN^{2}} & = \frac{MA^{2}+NA^{2}}{\left(5-x^{2}\right)^{2}}\\ \frac{3}{4x^{2}} & = \frac{MA^{2}+NA^{2}}{\left(5-x^{2}\right)^{2}}. \end{align}$$ Consider now $\triangle AMN$, where $O$ is the midpoint of $MN$. This means that $AO$ is a median of $\triangle AMN$. By the median formula, we have $$\begin{align} \left(\sqrt{5}\right)^{2} & = \frac{1}{4}\left[2\left(MA^{2}+NA^{2}\right)-MN^{2}\right]\\ 5 & = \frac{1}{4}\left[2\left(MA^{2}+NA^{2}\right)-4x^{2}\right]\\ 10+2x^{2} & = MA^{2}+NA^{2}. \end{align}$$ Substituting this into $$\frac{3}{4x^{2}}=\frac{MA^{2}+NA^{2}}{\left(5-x^{2}\right)^{2}},$$ we have $$\begin{align} \frac{3}{4x^{2}} & = \frac{10+2x^{2}}{\left(5-x^{2}\right)^{2}}\\ 3\left(25-10x^{2}+x^{4}\right) & = 40x^{2}+8x^{4}\\ 0 & = 5x^{4}+70x^{2}-75\\ & = 5\left(x^{2}+15\right)\left(x^{2}-1\right)\\ x & = \pm1. \end{align}$$ This just means that $x=1$, that is, $MO=NO=1$ unit.
6. 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?
   (solved by Jarrett Ian G. Lim [Philippine Academy of Sakya])
   
   
   SOLUTION: 
   First, we take out $1$, since if $1$ is in $A$, then any element multiplied with $1$ will given the product. In addition, note that $44^{2}<2015$ and $45^{2}>2015$, so we also need to remove all numbers less than or equal to $44$, because their squares will also be in $A$. Moreover, if $x$ and $y$ are both greater than or equal to 45, then we are sure that their product will not be inside $\left\{ 1,2,\dots,2015\right\}$. This means that we just have to remove $1,2,\dots,44$. This means that we have 1971 elements in $A$.

Of Prime Importance (Tuklas Vol. 17, No. 1 - September 12, 2015)

OF PRIME IMPORTANCE

A number $n$ is said to be prime if it has no other whole number factor other than $1$ and $n$. Prime numbers, usually referred to as "primes", are known as the building blocks of arithmetic because they form all the other natural numbers when multiplied among themselves. You have most likely encountered these numbers when you were doing mental maths or factorization. These numbers have a lot of interesting properties, and they have been regarded as one of the most elusive mysteries in number theory since there is no known closed or efficient formula to compute any of them.

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.

- Caldwell, Chris K. "Illegal Prime." Retrieved from http://primes.utm.edu/glossary/xpage/Illegal.html.

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.

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

1. 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 $a_{2015}$.
   
2. Show that if $x$, $y$, and $z$ are nonnegative real numbers for which $x+y+z=1$, then $$x^{2}y+y^{2}z+z^{2}x\leq\frac{4}{27}.$$
3. Determine the last four digits of $2015^{2015}$.
4. 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 $m\angle B<\dfrac{1}{2}\left(A+C\right)$.
5. 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 $MO$.
6. 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.