What? Day 3 (Beijing)

Note: Day 3, Beijing implies the third day I’ve been in Beijing.

Another note: A regular and hopefully non-boring blog post is coming up. As a note to self, things on the back-burner are (in order of priority) the following:

  1. Hard Work – How people’s idea of “You need to work hard to get really good at something” is completely wrong, and you’ll only ever get good if you don’t consider your investments “hard work.” Also talks about motivations to get top-level at certain activities and how “nice benefits from doing something” has been frequently misconstrued as “reasons to do something” (mathematics and sports are the most common victims of this).
  2. An Attempt at a Public School – Given enough money, time, and influence, it would be very easy for me to construct a good private school (or a series of outside classes or something of the sort). Most of the time when we complain about public schools all we get is an echo chamber, and I think we’ve spent so much time complaining we haven’t really been bothering to find a solution.
  3. Exploring Euclidean Geometry – A progress report and why I am so behind schedule. Estimated release times are completely inaccurate and I really hope I can get it released before I finish my freshman year of high school.

Also, remind me of this in 5 years: Mosquitoborne viruses such as malaria might be able to be solved by getting the mosquitoes to die whenever they bite a human. Whenever I take the time to learn Chem/Bio better, I might want to look into stuff that won’t affect humans but will debilitate/kill mosquitoes. If that isn’t possible, we could engineer a virus that doesn’t do anything against humans but debilitates mosquitoes (sort of the reverse of malaria).

Short explanation: Mosquitoes only get malaria from biting infected humans. This way the amount of mosquitoes with malaria can be limited and we only have to deal with the existing ones. This is probably very far off, but if in 5 years I see this and realize it is actually feasible when I have more knowledge then this is a lead worth pursuing.

Problems

Since I did 2017 AIME I #6-10 for homework a few hours before writing this post, solution outlines, thought process, and comments on these are fresh from the mind.

(Problem 6)

A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$. Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$. Find the difference between the largest and smallest possible values of $x$.

Solution:

Let y = 2x. Then we have arcs of lengths y, y, 360-2y. Let’s complementary count. The probability of no intersection is the probability both point are on the same arc, namely (\frac{y}{360})^2+(\frac{y}{360})^2+(\frac{360-2y}{360})^2. If you solve this quadratic, you see that the difference between the two values of y is 24. But we want double that so the answer is 48.

(Problem 7)
For nonnegative integers $a$ and $b$ with $a + b \leq 6$, let T(a,b)=\binom{6}{a}\binom{6}{b}\binom{6}{a+b}. Let $S$ denote the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a + b \leq 6$. Find the remainder when $S$ is divided by $1000$.

Solution: Let c=6-(a+b). Then T(a,b)=\binom{6}{a}\binom{6}{b}\binom{6}{c}. Combinatorially, this is basically the same as saying “take 18 objects and choose a any amount of them from each group so you choose 6 in total,” which is the same as saying “18 choose 6.” Evaluating it gives a remainder of 564.

(Problem 8)

Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution: These are arcs, arcs, arcs of a circle with diameter OP. The measures of the arcs 2a and 2b are from (0,150). If QR is less than 100, then |2a-2b|\leq 60 as the radius is $latex $100.$ This implies |a-b|\leq 30, and finding the probability that this happens is very standard. Using a graph, you get \frac{16}{25}, so the answer is 41.

(Problem 9)

Let $a_{10} = 10$, and for each integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least $n > 10$ such that $a_n$ is a multiple of $99$.

Solution: Essentially you can let a_n=a_{n-1}+n, meaning a_n=n+(n-1)+\dots+11+10. This evaluates to \frac{1}{2}(n+10)(n-9). Now it is not hard to see that the answer is 45.

(Problem 10)

Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$.

Solution: Terrible problem.

The condition just means z must be on the circumcircle of \triangle z_1z_2z_3. So if we translate this to Cartesian Coordinates, we just want to find the x value of the highest point on the circle; this is the x value of the circumcenter, which is easy to find. The answer is 56.

Pointlessness

I believe what we did this day can be described as “hilariously pointless.”

We decide to take a taxi to the National Aviation Museum. It’s closed, and five minutes later we desperately call the same taxi driver to come pick us up as literally nobody else is there. (Fortunately, they come back and deliver us home.)

We then eat food and do more boring stuff. When we decide to go home, we walk to the further metro station. Basically, if the closer station is Station A and the further one is Station B, we walked to Station B and took the metro to Station A… when we could’ve just walked to Station A.

Our bus ride is even more idiotic; we were supposed to take Line 430 (which we end up taking), but we sit for 7 or 8 stations before realizing we are going the wrong way, which ends up taking us about 30 minutes further away from home. Oops. I would like to say we get home safely that day despite mishaps, but that wouldn’t be true; we didn’t even get home that day. When we got back, it was (technically) already the next day. Fun!

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