Planet System Superposition Chart, Part 1

This chart is a to-scale representation of the near parts of ring and moon systems of the Solar system, superimposing the centers of all the planets at the top line of the chart. This chart goes up to 300000 kilometers from a planet’s center. Each line represents one megameter (1000 kilometers) of distance.

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Charitability versus Theism: Western Europe

The following chart graphs, for the countries of Western Europe (excepting Iceland and microcountries, as statistics for them are unavailable; availability is also the reason this is only a “Western Europe” graph) the charitability of a country by Official Development Assistance as a percentage of Gross National Income (2009) with the percentage of people who answered “I believe in God” in a Eurobarometer poll (2005).

Density Comparison of Major Solar System Bodies

(So I tried making these graphs with a newer Microsoft Excel and a newer Microsoft Paint. Boy was that a pain. Overdone antialiasing made it so that I couldn’t even reasonably fill in the points with a particular color on Paint. I guess from now on I’ll just go back to the good ol’ XP versions.)

Anyway, in the following graph, objects with the same mass are on the same shallowly negatively-sloped lines and objects with the same volume are on the same shallowly positively-sloped lines. Mass is measured in grams and volume is measured in cubic centimeters (and density in grams per cubic centimeter) for this graph. Moving directly right means greater mass and lesser volume. Moving directly left means lesser mass and greater volume. Moving up means both greater volume and greater mass. Note, for example, that Uranus has a greater volume than Neptune, but Neptune has a greater mass than Uranus, both of which are both much smaller and much less massive than Jupiter. Note that along a vertical, the difference between the two scales is equal, and since both scales are logarithmic, the difference preserves ratio; specifically, the ratio of mass to volume is density, and thus the resultant horizontal measure is ln Density, and objects along the same vertical have the same density, and the further toward the right the object is, the more dense it is.

In the following graph, I also placed a few reference densities.

These data points are plotted from mass and volume information gathered from Wikipedia.

A Guide to Pronouncing Chinese Names (Pinyin Transliteration)

Very frequently, non-Chinese people incorrectly pronounce Chinese names. Of course, it’s completely not their fault. One reason of course is that there are always those new nuances in pronunciation that are pretty much inevitable when one gets to a new language. (I still don’t think I roll r’s properly, let alone differentiate all those h’s in Arabic.) But another reason, one that actually bothers me, is that Chinese people outside China do not usually pronounce them correctly either.

I could partially understand why, because the sudden shift in sound in mentioning a Chinese name with the correct pronunciation in the middle of an English language sometimes is very linguistically awkward. I in fact do it too, although to a much lesser extent than others, for example frequently with the names “Lin” and “Sun” (although the surge of people that were talking about “Linsanity” has made me question that a bit more). (For those of you who currently want an advanced touch, Lin is pronounced “leen” and Sun is pronounced “swen.”) But in any case, Chinese people pronouncing more English-friendly versions of Chinese names started bothering me when I heard someone talk about the “many ways” to spell Chang, when actually each of those “different ways” that that someone was mentioning was actually a different Chinese name, pronounced a different way. I figured that at least those that wish to take the effort to correctly pronounce Chinese names should have the right to have a fairly easy guide to the actual way to pronounce them, and thus I’m writing this. Note that in no way am I requiring or asking that the general public perfect their Chinese pronunciation, and that this article exists for the purpose of giving more to the people who wish to know more.

A few notes before I start:

1. In actual Chinese, there are various accents (or tones) on the way any phonetic can be pronounced to produce different words, and that sounds even more out-of-place in a sentence of another language. For the purposes of this article, the accents are left out.

2. This guide is a guide to pronouncing Pinyin. Note that that is not the only method of Romanization of Chinese characters, but it will be the only one thoroughly discussed here, although I would occasionally insert a sidenote about various other translations. If you use this guide for pronunciation, make sure you’re looking at Pinyin.

3. This article references Chinese sounds with respect to English approximations. If you don’t know English, 1) you’re either a liar or an amazing guesser as you’re reading this and 2) this article won’t help you much.

Consonants I

Let’s start with some consonants pronounced just like in English.

They are b, d, f, g, h, k, l, m, n, p, s*, and t. We’ll get to the other consonants later.

*Nuance later. Also, as an aside, this is notated “sz” is other Romanizing systems.

Vowels I

In any Chinese syllable, one never finds consonants after vowels, so you can identify vowels by their location within a syllable. Note that we are slightly changing the definition of a vowel here in a way that will be clearly apparent when examples are given. There are many vowels in Chinese.

“a” is pronounced “ah” as in the a in “wasp.”

“e” is pronounced “uh” as in the e in “the.”**

“ai” is pronounced long i as in the i in “find.”

“ei” is pronounced long a as in a in “mane.”

“ao” is pronounced the not-the-long-o “ow” as in the ow in “cow.”

“ou” is pronounced long o as in o in “tone.”

**Exception coming later.

Exercise: How would you pronounce Chinese “la”?

Answer: Just like as if you were singing. Or, you know, the note that’s after sol.

Exercise: How would you pronounce Chinese “he”?

Answer: Like the interjection “huh.”

Exercise: How would you pronounce Chinese “tai”?

Answer: Like the word “tie.”

Exercise: How would you pronounce Chinese “bei”?

Answer: Like the word “bay.”

Exercise: How would you pronounce Chinese “nao”?

Answer: Like the word “now.”

Exercise: How would you pronounce Chinese “hou”?

Answer: Like the word “hoe.”

Now, for what is in my opinion a harder one. For “o,” pretend that you’re saying the short “o” as in “not” but that you are getting there from saying a long “oo.” The effect sounds somewhat like “uo” where u represents the long oo, although usually the vowel u will come in front anyway.

We are classifying the following combinations of vowels with what would in English be considered a consonant for purposes here, as warned above.

“an” is pronounced like the “an” in “ran.”

“en” is pronounced like the “en” in “hen.”

“ang” is pronounced like “ah” as in “wasp” followed by English “ng.”

“eng” is pronounced like “uh” as in “the” followed by English “ng.”

“er” is pronounced like “ur” as in the “ir” in “stir.”

Exercise: How would you pronounce Chinese “tan”?

Answer: Like the English word “tan.”

Exercise: How would you pronounce Chinese “dang”?

Answer: Not like the English “dang”! Look above.

Exercise: How would you pronounce Chinese “deng”?

Answer: Like the English word “dung.” Yeah, that’s unfortunate. It’s like how one definition of “ben” in Chinese is “stupid.”

“i” is pronounced long “e” as in the ee in “see.”

“u” is pronounced long “oo” as in the oo in “noon.”

“y” before i is silent.

“w” before u is silent.

Exercise: How would you pronounce Chinese “yi”?

Answer: just long e: “ee”

Exercise: How would you pronounce Chinese “wu”?

Answer: just long oo: “oo”

Note that “i” and “u” is sometimes followed consecutively by another vowel above in a syllable. In some of these cases, some letters are left out, and will be discussed later. There is also one more vowel to introduce later.

Consonants II

“z” is pronounced like “dz” in English. Note that “zh” is a totally different consonant, mentioned later in this section.

“c” is pronounced like “tz” in English.

Exercise: How would you pronounce Chinese “zu”?

Answer: “dzoo”

Exercise: How would you pronounce Chinese “ceng”?

Answer: “tzuhng”

“j” is pronounced very close to “j” in English. Smile more than when pronouncing the English “j.”

“q” is pronounced very close to “ch” in English. Smile more than when pronouncing the English “ch.”

“x” is pronounced very close to “sh” in English. Smile more than when pronouncing the English “sh.”

Notes: In some methods of Romanizing, the last of the above-mentioned three sounds is Romanized as “hs.” Also, in the exercises, I will sometimes leave out the smiling remark in pronunciation for “j,” “ch,” and “sh,” but remember the subtlety.

Exercise: How would you pronounce Chinese “ji”?

Answer: Like the English interjection “gee” with a smile.

“zh” is a sound that does not exist in English. In my opinion, it is closest to “dr,” and can be approximated closer by mashing the two sounds temporally closer.

“ch” is a sound that does not exist in English. In my opinion, it is closest to “tr,” and can be approximated closer by mashing the two sounds temporally closer.

“sh” is a sound that does not exist in English. In my opinion, it is closest to “shr,” and can be approximated closer by mashing the two sounds temporally closer.

“r” is very much like the English “r,” but roll the tongue even less than in English.

Exercise: How would you pronounce the Chinese name “Chan”?

Answer: approximately “tran”

Exercise: How would you pronounce the Chinese name “Chen”?

Answer: approximately “tren”

Exercise: How would you pronounce the Chinese name “Chang”?

Answer: approximately “trahng”

Exercise: How would you pronounce the Chinese name “Cheng”?

Answer: approximately “truhng”

Exercise: How would you pronounce the Chinese name “Zhang”?

Answer: approximately “drahng”

Exercise: How would you pronounce the Chinese name “Zheng”?

Answer: approximately “druhng”

Exercise: How would you pronounce the Chinese name “Zhu”?

Answer: approximately “droo”

Exercise: How would you pronounce the Chinese name “Zu”?

Answer: approximately “dzoo”

Exercise: How would you pronounce the Chinese name “Ren”?

Answer: approximately “ren”

Vowels II

Sometimes parts of sounds are replaced in the spelling of a word. For example, when indicating “i” followed by “ou,” the “o” is truncated and “iu” is used instead. Note that in some systems of Romanizing, the truncation is left out. Additionally, “eng” is sometimes written “ong.”

Exercise: How would you pronounce the Chinese name “Liu”?

Answer: like “Leo” (the lion), except quickly, as one syllable

Exercise: How would you pronounce the Chinese name “Huang”?

Answer: “hooahng” (one syllable) (most people actually bother to pronounce this last name correctly, just listen to what they say)

Exercise: How would you pronounce the Chinese name “Zhuang”?

Answer: approximately “drooahng” (one syllable)

Exercise: How would you pronounce the Chinese name “Min”?

Answer: “meen”

Exercise: How would you pronounce the Chinese name “Sun”?

Answer: “swen”

Exercise: How would you pronounce the Chinese name “Jin”?

Answer: approximately “jeen”

Exercise: How would you pronounce the Chinese name “Jing”?

Answer: approximately “jeeng”

In Chinese, some words consist of only a consonant, pronounced with some sound. In these cases, a silent “i” is attached at the end.

Exercise: How would you pronounce the Chinese name “Shi”?

Answer: approximately “shr”

Exercise: How would you pronounce the Chinese name “Ci”?

Answer: approximately “tz”

One last vowel. This is sometimes notated “ü,” sometimes “uu,” sometimes “eu,” and sometimes “v.” (Yes, “v” being a vowel. Imagine that. It’s used because the English “v” sound does not exist in Chinese and is close to “u.”)  I will use the “uu” notation in words and the “ü” notation in pronunciations from here forward.

The vowel “uu” is a sound that does not exist in the English language, and is not even close to any sound in the English language. To produce this sound, pucker your lips and make a vowel sound. One only finds this vowel specifically written in its form after l or n; after j, q, x, or y, any “u” following is implied to be this u, not the long-oo u.

Exercise: How would you pronounce the Chinese name “Lu”?

Answer: “loo”

Exercise: How would you pronounce the Chinese name “Luu” or “Leu”?

Answer: “lü”

Exercise: How would you pronounce the Chinese name “Fu”?

Answer: “foo”

Exercise: How would you pronounce the Chinese name “Ju”?

Answer: “jü”

Exercise: How would you pronounce the Chinese name “Mu”?

Answer: “moo”

Exercise: How would you pronounce the Chinese name “Qu”?

Answer: “chü”

Exercise: How would you pronounce the Chinese name “Wu”?

Answer: “oo”

Exercise: How would you pronounce the Chinese name “Xu”?

Answer: “shü”

Exercise: How would you pronounce the Chinese name “Yu”?

Answer: “ü”

Exercise: How would you pronounce the Chinese name “Zu”?

Answer: “dzoo”

Exercise: How would you pronounce the Chinese name “Yuan”?

Answer: “üen”

Exercise: How would you pronounce the Chinese name “Yun”?

Answer: “üin”

Exercise: How would you pronounce the Chinese name “Jun”?

Answer: “jüin”

Apostrophes

In Pinyin, apostrophes indicate that the two parts split by an apostrophe are to be pronounced as separate syllabes. For example, “Xi’an” would be “shee an” whereas “Xian” would “sheeen” (that may look confusing; that’s a long e followed by “en”). Note that in other Romanizing systems, apostrophes have other roles, sometimes representing a different sound. Also note that when it is clear that the word has multiple syllables, apostrophes usually aren’t used.

Ending Exercises

Try pronouncing the following names.

1. Sun Yixian
2. Jiang Jieshi
3. Mao Zedong
4. Li Xiaolong
5. Hua Mulan
6. He Xiaoyu
7. Wang Weiwei
8. Lin Ying
9. Hu Jintao
10. Bei Yuming
11. Deng Xiaoping
12. Xu Yushi

Answers:

1. swen ee sheeen
2. jeeahng jeeeh shr
3. mow dzuh dohng
4. lee sheeow lohng
5. hooah moo lan
6. huh sheeow ü
7. wahng wayway
8. leen eeng
9. hoo jeen tow
10. bay ü meeng
11. duhng sheeow peeng
12. shü ü shr

Mission Math Tournament: Regular Round Part 2 Solutions

11. Find the sum of all real values of x that satisfy the equation x(x-1)=1.

Solution: x(x-1)=1 \implies x^2-x=1 \implies x^2-x-1=0. For any quadratic equation ax^2+bx+c=0, the two roots can be found via the quadratic formula to be \frac{-b+\sqrt{b^2-4ac}}{2a} and \frac{-b-\sqrt{b^2-4ac}}{2a}. Adding these two together gives that the sum of the roots of any quadratic equation is -\frac{b}{a}. This is one of a set of useful equations regarding sums and products of roots of polynomials known as Vieta’s Formulae. Applying this sum equation here, we find the sum of the values of x that satisfy the given equation to be -\frac{-1}{1}=\boxed{1}.

12. Find the area of the smallest square that can contain a line segment of length 2.

Solution: The longest line segment that can be contained in any square is clearly the diagonal. By properties of a 45º-45º-90º right triangle formed by dividing a square along a diagonal, the side length of the square is \sqrt{2}. The area is therefore (\sqrt{2})^2=\boxed{2}.

13. Find all real solutions to (x^{10}+30)^{(x^9+27)^{(x^8+24)^{.^{.^{. ^{(x^3+9)^{(x^2+6)^{(x+3)}}}}}}}}=1.

Solution: When a real number a is raised to the power of a real number b, there are three ways a result of 1 could occur: the base is 1, the base is -1 and the exponent is even, or the base is nonzero and the exponent is zero. As x^{10}+30\geq30 for any real x, only the third case mentioned above could occur. Thus, we seek values of x that make the exponent 0. The only way a real number raised to a real number could equal 0 is if the base is zero and the exponent is nonzero. Thus, we set the base of this quantity, x^9+27, to be 0. Solving, we find that the only real solution to x^9+27=0 is \boxed{-\sqrt[3]{3}}. Now, all we need to do is to make sure this term’s exponent is nonzero. Again, we use the fact that a real number raised to a real number is 0 if and only if the base is zero and the exponent is nonzero. For any real x, x^8+24\geq24, so the  exponent of x^9+27 cannot be zero, and thus we have found the one and only solution.

14. What is the smallest possible number of sides a polygon could have if its centroid is outside of itself?

Solution: This would not be a triangle, as a triangle can’t even be concave. It is possible, however, for a \boxed{4}-sided polygon to have a centroid outside of it, and an example is depicted below.

15. If \frac{x^2}{x^3+1}=\frac{2}{7} and \frac{x^4}{x^6+1}=\frac{4}{65}, find all possible values of x.

Solution: Squaring the first equation, we find that \frac{x^4}{x^6+2x^3+1}=\frac{4}{49}. Cross multiplying, we find that 49x^4=4x^6+8x^3+4. Cross multiplying the other equation, we find that 65x^4=4x^6+4. Subtracting the second equation from the squared first equation gives -16x^4=8x^3. This has two solutions, 0 and \boxed{-\frac{1}{2}}. The first solution is extraneous.

Problem 16 was dropped.

17. An ant starts at (1,1,1,1,1,1) in a six-dimensional world consisting of the 729 points (u,v,w,x,y,z) such that u,v,w,x,y,z,\in \{-1,0,1\}. Each move the ant takes a stroll from its current location to another point in its world exactly n units of distance away. Find the product of all n such that given unlimited moves the ant can eventually visit all 729 points in its world.

Solution: For a move to take the ant from a lattice point to a different lattice point, it must be of distance \sqrt{n} for some positive integer n. For any n greater than 6, the ant will never be able to reach the point in the center of the six-dimensional cube, so we now only have six cases for n to check. Note that whenever n is even, The parity of the sum of the coordinates of the location of the ant is conserved through any legal move, so no even n works. If we prove that the ant can move to an adjacent point given an arbitrary location and we can prove that the ant can get to the center, we have proven that the value of n works. Proving this for n=1 is trivial, as the ant is simply moving to any adjacent square. For n=3, we find (1,1,1,1,1,1)\rightarrow(1,1,1,0,0,0)\rightarrow(0,0,0,0,0,0) brings the ant to the center and (1,1,1,1,1,1)\rightarrow(1,1,1,0,0,0)\rightarrow(1,1,0,-1,1,0)\rightarrow(0,0,0,-1,0,0)\rightarrow(0,1,1,0,0,0)\rightarrow(1,1,1,1,1,0) brings the ant to an adjacent square. For n=5, we take the complement of any move in n=1. Thus, all odd values of n under 6 are valid, and multiplying the square roots gives \sqrt{1}\sqrt{3}\sqrt{5}=\boxed{\sqrt{15}}.

18. Participants in the regional lottery pick two of the numbers 1, 2, 3, 4, and 5. A ticket wins if neither of the two numbers matches the two numbers randomly drawn by the lottery official. What is the smallest number of tickets a participant should have in order to guarantee that one of the tickets wins?

Solution: The minimum number of tickets is \boxed{4}, for example if one chooses the tickets 1 and 2, 1 and 3, 2 and 3, and 4 and 5. Since any ticket wins for only \binom{5-2}{2}=3 outcomes, and there are \binom{5}{2}=10>3(3) possible outcomes, 3 tickets cannot cover all possibilities.

19. How many complex numbers are roots of f(z)=\frac{(z-1)(z^2-1)(z^3-1)...(z^{14}-1)(z^{15}-1)(z^{16}-1)}{(z+1)(z^2+1)(z^3+1)...(z^{14}+1)(z^{15}+1)(z^{16}+1)}?

Solution: The numerator consists of the nth roots of unity, up to n=16, and the denominator consists of the nth roots of minus unity, up to n=16. Thus, a number is a root if it is a root of unity while not being a root of minus unity, when considering all degrees of roots up to 16. Thinking in polar coordinates using De Moivre’s Theorem, one can form a Farey Sequence representing the roots of unity in terms of the fraction of the away around the ring of unity at which they are located; in this case, we’re looking for the 16th Farey Sequence, with the last term omitted due to being a doubling of the first case: \{0,\frac{1}{16},\frac{1}{15},\frac{1}{14},\frac{1}{13}...\frac{1}{8},\frac{2}{15},\frac{1}{7}...\frac{14}{15},\frac{15}{16}\}. In the case of the denominator, we double denominators and take only odd numerators of terms in the sequence.

We now perform parity casework on the fraction of a circle at which each potential root is located: specifically, we note the parity of both the numerator and the denominator. When the parity of both is even, the fraction will eventually simplify into one of the other cases, and thus all of these cases can be discounted. Whenever the denominator is odd, the fraction is not equivalent to any with an even denominator and an odd numerator. Whenever the numerator is odd and the denominator is even, the corresponding complex number is not a root. Thus, we count, for each odd number less than 16, the number of simplified fractions that have that number as the denominator. There are are 49 such fractions, and thus the equation has \boxed{49} roots.

20. In a certain game, you pick two relatively prime numbers less than or equal to 1000. Then, you subtract the smaller number from the larger number. You earn one point if the resulting difference is less than the original smaller number, and repeat the process using these two numbers. The game ends when the difference of 1 appears. Find the two numbers you should pick at the start to achieve the highest possible score. Here’s a sample run of this game’s process using starting values of 77 and 111:

111-77=34, 34<77, score 1 point

77-34=43, 43>34, score 0 points

43-34=9, 9<34, score 1 point

34-9=25, 25>9, score 0 points

25-9=16, 16>9, score 0 points

16-9=7, 7<9, score 1 point

9-7=2, 2<7, score 1 point

7-2=5, 5>2, score 0 points

5-2=3, 3>2 score 0 points

3-2=1, 1<2, score 1 point

Difference of 1 appears, game ends, score=5

Solution: Think from the bottom up. Whenever one adds a term to another, one gets the previous term in one of these sequences. Thus, we try to maximize the number of addings while staying under a certain value, and thus we add starting from 1 and 2. By adding only once, we ensure that no moves coming down are “wasted,” or in other words, producing 0 points. By adding starting with 1 and 2, we move up the Fibonacci sequence, stopping at the given value of 1000, to find the two numbers we want to be \boxed{610 \text{ and } 987}.

Mission Math Tournament: Regular Round Part 1 Solutions

(With the exception of Regular Round #4, written by Jonathan Chen, all of these problems are written by me.)

1. What is the probability that a randomly selected letter in “RATIONAGE” is a vowel?

Solution: There are 5 vowels (A,I,O,A,E) out of 9 letters (R,A,T,I,O,N,A,G,E), and \boxed{\frac{5}{9}} is already a simplified fraction.

2. A triangle with side lengths 5, 12, and 13 is inscribed in a circle. What is the circumference of this circle?

Solution: Since 5^2+12^2=13^2, the triangle is a right triangle, with right angle opposite the side of length 13. By Thales’ Theorem, a right angle is subtended by a semicircle, so the edge of length 13 is a diameter of the circle the triangle is inscribed in. Since a circle’s circumference is π times its diameter, this circle’s circumference is \boxed{13\pi}.

3. Let a be the arithmetic mean of 24\sqrt{7} and 25\sqrt{7} and h be the harmonic mean of 24\sqrt{7} and 25\sqrt{7}. Find the geometric mean of a and h.

Solution: The arithmetic mean of a and h is \frac{a+h}{2}. The harmonic mean of a and h is \frac{2}{\frac{1}{a}+\frac{1}{h}}=\frac{2}{\frac{a+h}{ah}}=\frac{2ah}{a+h}. The geometric mean of two positive numbers is the square root of their product. Thus, the geometric mean of the arithmetic and harmonic means of a and h is \sqrt{\frac{a+h}{2}\left(\frac{2ah}{a+h}\right)}=\sqrt{ah}. Using this result, the sought geometric mean is equal to \sqrt{24\sqrt{7}*25\sqrt{7}}=\boxed{10\sqrt{42}}.

4. There exists two numbers a and b such that a+b=10 and \frac{1}{a}+\frac{1}{b}=\frac{1}{5}. Find a^3b+2a^2b^2+ab^3.

Solution: a^3b+2a^2b^2+ab^3=ab(a^2+2ab+b^2)=ab(a+b)^2. We could thus compute the value of the original expression if we can find the values of a+b and ab, without needing to find what a and b are. We were already given the fact that a+b=10. Since \frac{1}{a}+\frac{1}{b}=\frac{b}{ab}+\frac{a}{ab}=\frac{a+b}{ab}, we now have \frac{a+b}{ab}=\frac{1}{5}, giving ab in terms of a+b, which we already know. Specifically, ab=\frac{a+b}{\frac{1}{5}}=\frac{10}{\frac{1}{5}}=50. Plugging the values of a+b and ab into the formula we found above, ab(a+b)^2, we find the sought value to equal 50(10)^2=\boxed{5000}.

5. After removing face cards and jokers, how many of the remaining 40 standard poker cards have a front side with 180-degree rotational symmetry? Neglect imperfections in printing.

Solution: Note one important fact about suit pips: the diamond is 180-degree rotationally symmetric, whereas the spade, club, and heart are not! The 2, 3, 4, 5, 6, 7, 8, 9, and 10 of diamonds are 180-degree rotationally symmetric, whereas only the 2, 4, and 10 of spades, clubs, and hearts are, as in poker cards all pips along the middle of the card face in one direction. In total, there are 9+3+3+3=\boxed{18} of these cards.

6. Find all solutions (x,y), with x and y both rational, to the equation 40x+ 72y=16x^2+79+24y^2.

Solution: First, move all terms to one side to arrive at 16x^2-40x+24y^2-72y+79=0. Moving the constant to the right side, we find that 16x^2-40x+24y^2-72y=-79. Completing the square in both x and y, we arrive at 16x^2-40x+25+24y^2-72y+54=-79+25+54, or 16x^2-40x+25+24y^2-72y+54=0. Dividing the square trinomials of x and y by the respective variables’ coefficients gives \frac{(x-5/4)^2}{16}+\frac{(y-3/2)^2}{24}=0. Here, it can be identified that the conic relation is that of a point, but this equation can also be analyzed algebraically. Any real value squared is nonnegative, so both terms on the left hand side must equal 0 for this equation to be true. The only real solution, and also the only rational solution, is \boxed{(5/4,3/2)}.

7. Equilateral triangle ABC has side length \sqrt{3}. D is on \overline{AC}, and E and F are both on \overline{BC}, between B and C. Square DEFG has side length 1. Circle O is the smallest circle that passes through G and is tangent to \overline{AB}. This point of tangency is H. Find GH.

Solution: Draw altitude \overline{AJ}, and call its point of intersection with \overline{BG} K. Label the intersection of \overline{AB} and \overline{BG} L. See the below diagram.

\angle A\cong\angle B\cong\angle C, because equilateral triangles are also equiangular, and m\angle A=m\angle B=m\angle C=60^\circ. Since opposite sides of a square are parallel, by theorems regarding parallel lines m\angle A=m\angle{ADK}=m\angle{ALK}=60^\circ. By vertical angles, m\angle{ALK}=m\angle{GLH}=60^\circ. The smallest circle through G and H is drawn when \overline{GH} is the diameter. The diameter is perpendicular to a tangent at the point of tangency, so m\angle{LHG}=90^\circ, and \triangle{GHL} is a 30º-60º-90º triangle, similar to \triangle{AJB}, \triangle{AJC}, \triangle{AKD}, and \triangle{AKL}.

Now that many 30º-60º-90º right triangles are established, repeatedly apply side-length ratios of 30º-60º-90º right triangles (1:\sqrt{3}:2) from here on. Since AB=\sqrt{3}, AJ=\frac{3}{2}. Since JK=DE=1, AK=\frac{3}{2}-1=\frac{1}{2}, and thus DK=KL=\frac{\sqrt{3}}{6}, and DL=\frac{\sqrt{3}}{3}. Since DG=1, LG=1-\frac{\sqrt{3}}{3}, and finally, GH=\frac{\sqrt{3}}{2}(1-\frac{\sqrt{3}}{3})=\frac{\sqrt{3}}{2}-\frac{3}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\boxed{\frac{\sqrt{3}-1}{2}}.

8. What is the 18th digit after the decimal point in the decimal representation of (9-4\sqrt{5})^{9+4\sqrt{5}}?

Solution: 9 and 4\sqrt{5} are very close to each other in value. As the exponent in the given expression is significantly greater than 1, the expression itself has an even smaller value. Let’s try to see how small this value is.

Representing both 9 and 4\sqrt{5} as the square root of a positive integer, we find the base to equal \sqrt{81}-\sqrt{80}. \sqrt{81}-\sqrt{64}=1, and the difference between square roots of consecutive integers decreases as the integers themselves increase, so \sqrt{81}-\sqrt{80}<\frac{1}{17}<\frac{1}{16}. The exponent, 9+4\sqrt{5}, can be similarly written as \sqrt{81}+\sqrt{80}>17=\sqrt{81}+\sqrt{64}. Thus, (9-4\sqrt{5})^{9+4\sqrt{5}}<\left(\frac{1}{16}\right)^{17}=\frac{1}{2^{68}}.

Since 5^2<2^5, \frac{1}{2^{68}}=\frac{1}{2^{18}*2^{50}}<\frac{1}{2^{18}*5^{20}}=\frac{1}{10^{18}*5^2}<\frac{1}{10^{18}}. Thus, (9-4\sqrt{5})^{9+4\sqrt{5}}<\frac{1}{10^{18}}, and the first nonzero digit of the expression has not even occurred by the 18th digit, and the 18th digit, like all digits between it and the decimal point (and a few after), is \boxed{0}.

9. Equilateral triangle XYZ has side lengths 60. Two line segments are drawn from X to \overline{YZ}, trisecting \overline{YZ}; two line segments are drawn from Y to \overline{XZ}, trisecting \overline{XZ}; and two line segments are drawn from Z to \overline{XY}, trisecting \overline{XY}. What is the perimeter of the smallest polygon described by these line segments that contains the center of the triangle?

Solution: We first draw a diagram, like the one below.

We see, after considering symmetry, that the smallest polygon described by these line segments that contains the center of the triangle is an equilateral, but not equiangular, hexagon. Thus, we can find the perimeter of this hexagon by finding the length of one of its sides and multiplying that length by six.

Label the center of the triangle O. Label the midpoint of \overline{YZ} A. Draw \overline{XA}, which is a perpendicular bisector, because a line segment from a vertex to the midpoint of the opposite side of an isosceles (including equilateral) triangle is an altitude and thus perpendicular to the side. By the properties of a 30º-60º-90º right triangle, XA=30\sqrt{3}. Label the points of trisection on \overline{XZ} as B and C, with B closer to X. Since Drop a perpendicular from B to \overline{YZ}, labeling the point of contact W. Draw \overline{WX}. Label the point of trisection on \overline{XY} closer to X as V. Drop a perpendicular from V to \overline{AX}, labeling the point of contact S. Label the point of intersection of \overline{ZV} and \overline{AX} M. Label the point of intersection of \overline{ZV} and \overline{BW} P. Label the point of intersection of \overline{ZV} and \overline{WX} N, which is a vertex of the inner hexagon and between M and P. Draw \overline{CO}. Draw \overline{BS}. See the diagram below.

\angle AXZ is shared by \triangle COX and \triangle ZAX. The centroid of any triangle is two-thirds of the way from a vertex to the midpoint of the opposite edge, so OA=10\sqrt{3} and \frac{XO}{OA}=2. Since C is a point of trisection, \frac{XC}{CZ}=2. Thus, by SAS, \triangle COX \sim \triangle ZAX. Thus, since $\angle ZAX$ is a right angle, \angle COX is a right angle. Additionally, since AZ=30, OC=20.

\angle XZY is shared by \triangle BZW and \triangle XZA. Both \angle XAZ and \angle BWZ are right angles, and are thus congruent. Thus, by AA, \triangle BZW \sim \triangle XZA. Since XZ=60, BZ=40, and XA=30\sqrt{3}. BW=20\sqrt{3}.

Since OPWA is a rectangle and OA=10\sqrt{3}, PW=10\sqrt{3}, and since BW=20\sqrt{3}, BP=10\sqrt{3}. Since B, S, and V are collinear because of symmetry, SO=10\sqrt{3}, and AS=20\sqrt{3}.

\angle BYZ is shared by \triangle MAY and \triangle BWY, and \angle MAY and \angle BWY are both right angles and thus congruent, so \triangle MAY \sim \triangle BWY. Hence, since \frac{YA}{YW}=\frac{3}{4}, \frac{AM}{WB}=\frac{3}{4}. Thus, since BW=20\sqrt{3}, AM=15\sqrt{3}, and since AO=10\sqrt{3} and AS=20\sqrt{3}, OM=MS=5\sqrt{3}. By vertical angles, \angle OMP \cong \angle SMV, and \angle MOP \cong \angle MSV because they’re both right angles. Thus, by ASA, \triangle{MOP}\cong\triangle MSV.

Since AW=10 and AX=30\sqrt{3}, by the Pythagorean Theorem, WX=20\sqrt{7}. By symmetry, BY=20\sqrt{7}, and by the similarity of triangles  MAY and BWY, BM=5\sqrt{7}. By symmetry, VM=5\sqrt{7}. By the congruence of triangles MOP and MSV, MP=5\sqrt{7}.

Since XA=30\sqrt{3} and AM=15\sqrt{3}, MX=15\sqrt{3}. Since XA||BW, \angle XMP \cong \angle WPM, and \angle MXW \cong \angle PWX. Thus, by AA, $\triangle MXN \sim \triangle PWN$. Hence, because \frac{MX}{PW}=\frac{15\sqrt{3}}{10\sqrt{3}}=\frac{3}{2}, \frac{MN}{NP}=\frac{3}{2}. Thus, \frac{MN}{MP}=\frac{3}{5}. Since MP=5\sqrt{7}, this means MN=\frac{3}{5}(5\sqrt{7})=3\sqrt{7}. Now that we have found a side length of the equilateral hexagon, we sextoople it to find that this hexagon’s perimeter is 6(3\sqrt{7})=\boxed{18\sqrt{7}}.

Problem 10 was dropped.