Let’s find the angles of the triclinic crystal! A triclinic crystal has all sides and angles different. The triclinic crystal we’re dealing with is a prism. All of its sides are parallelograms. With faces with areas 30cm2, 35cm2, and 42cm2, it could be orthorhombic, with all right angles. That would give it sides 5cm, 6cm, and 7cm, but we know that those aren’t the side lengths. It could also have 1 right angle, and then it could have sides 3cm, 11cm, and 14cm. 3*10 = 33, and 3*14 = 42. And 11*14 = 154. 90 degree angles give the biggest area, so 2 bigger areas are fine. If you didn’t know about triclinic crystals, you’d probably guess that. It’s the obvious solution. But for no angles to be 90, all the products have to be higher than necessary. The 33 must go to 30. The 42 can’t go to 42, the 154 must. And so the 42 goes to 35. 33>30, 42>35, and 154>42. No problem. The portion of the area is the sin of the angle. Simple trig. So our answers are arcsin(30/33), arcsin(35/42), and arcsin(42/154), which gives us 65.38°, 56.44°, and 15.83°.
Author: zachary
Exquisite Equation
Advanced algebra is required to solve this challenge. You should read Perplexing Protons before beginning this challenge.
We left an unsolved equation after solving Perplexing Protons. In this challenge, we will solve it. Good luck!
x=5\,000\,000+E^+\frac{2-e^ {\frac{-1}{N^+}}-e^\frac{-3}{N^+}}{N^+} +E_{t=3.25}-3E^+-1\\[8pt] \frac{E^+-E^+e^\frac{-3}{N^+}} {N^+e^\frac1{N^+}}=1\,000\,000Confounding Crystals
Chemistry knowledge is needed to attempt this challenge.
Crystals are very interesting and beautiful. There are seven types of crystal shapes. Plagioclase feldspar makes triclinic crystals. You have one that’s a prism whose faces have areas 30 cm2, 35 cm2, and 42 cm2. The sides have lengths 3cm, 11cm, and 14cm. The challenge is to use what you know about triclinic crystals to determine what the angles of the crystal must be. This challenge may seem to be about math, but really it’s just a little bit of trig. The important part is that you know the defining features of triclinic crystals.
Solution to Bouncy Ball
The new instantaneous velocity of the ball becomes the negative square root. This bouncy ball defies physics, but we can still calculate it’s velocity after 100 jumps. Most importantly, The ball’s velocity at the end of a bounce is -1 times it’s velocity at the beginning of it. So the velocity is square-rooted every bounce. If a non bouncy object is dropped from a height of A metres, and it’s terminal velocity is B metres per second, then B = √(2gA)/2. g is the force of gravity, 9.807 metres per second squared. The initial height, A, is 1 metre. 2gA is 19.614 m2/s2. Thus B is 4.429 m/s. Square-rooting 99 times is the equivalent of raising it to the power of 2-99. We get approximately 1 metre per second upwards. It’s slightly more, but very little.
Solution to Strange Sums
Let’s solve an equation! The sum of a decreasing geometric series is described by the second equation. If you would like to know why, see Peculiar Pets. Knowing this, the full equation is clearly the harmonic series, 1/1 + 1/2 + 1/3 + … . The harmonic series adds to infinity, our answer.
\sum_{k=2}^{\infin}\Bigg(\sum_{l=0}^{\infin} \Big(\frac1{k^l}\Big)\Bigg)\\[8pt] \sum_{l=0}^{\infin}\Big(\frac1{k^l}\Big)= \frac1{k-1}\\[8pt] \sum_{k=1}^{\infin}\Big(\frac1{k}\Big)= \infin\\[8pt]Bouncy Ball
Physics simulations are highly reccomended to solve this science challenge.
You release a bouncy ball from 1 metre above the Earth. Neglect decrease from surface gravity. It will fall and bounce, then fall again, and so on.With each bounce, the ball’s velocity after the bounce is the negative root of it’s speed before it. For example, if it’s velocity was 4 m/s down, it’s velocity would become -2 m/s down, or 2 m/s up. Clearly, this is not a normal bouncy ball. In fact, it breaks several laws of physics. But still, calculate the velocity of the ball after 100 jumps.
Strange Sums
Algebra and geometry are needed to solve this challenge.
Sigma notation is the topic of today’s challenge. At the bottom, we set a variable, usually ‘K’. K typically starts at 0 or 1. We would write ‘k=0’ or ‘k=1’ below the Sigma. Sigma notation is often used to denote the partial or full sum of an infinite series. At the top, we say when the series will end, often infinity. The value of the expression is the sum of the stuff to the right of the sigma for every value of K. The challenge for today is to find the sum of the expression. Look out for the sequel to this challenge, Peculiar Products, in which we will discuss Pi notation.
\sum_{k=2}^{\infin}\Bigg(\sum_{l=0}^{\infin} \Big(\frac1{k^l}\Big)\Bigg)Solution to Perplexing Protons
We want to find the number of protons in the particle collider. We know that 1 Kj of energy was released. 1 electron capture releases 511 electron volts. There are ≈6.242 * 1021 electron volts in a kilojoule. Thus there were 1.221 * 1019 reactions. The rate of reactions is proportional to the number of electrons. Let the number of neutrons be N and the electrons E. The increase in electrons is E+, and that of neutrons, N+. E is increased linearly and decreased proportionally to itself. The distance from equilibrium will decrease by a factor of e in every length of time over which it would become zero if it were only being decreased, and at a constant rate equivalent to the instantaneous force of decrease at t=0. We know this from Battle of the Blobs and Spinning Stick. This duration of time equals 1/n. The equilibrium is at nE=E+, so the number of electrons will approach E+/N+. E, at any time, equals E+/N+ * (1-e-t/N+). The number of neutrons is E+t minus all that stuff. We make the equations below for t=1, t=3, t=3.25 and t=3.5 to find our answer. The final equation to get E+ and N+ is rather difficult, and it is necessary to find them to solve the problem. This equation will be a challenge of it’s own, Exquisite Equation, coming next week.
P_{t=1}=5\,000\,000+\frac{E^+}{N^+} (1-e^{\frac{-1}{N^+}})\\[8pt] E_{t=3}=\frac{E^+}{N^+}1-e^{\frac{-3}{N^+}} \\[8pt] E_{t=3.25}=\frac{E_{t=3}}{e^\frac1{N^+}} =1\,000\,000\\[8pt] P_{t=3.5}=P_{t=1}+\frac {E^+}{N^+}(1-e^{\frac {-3}{N^+}})+E_{t=3.25}-3E^+-1\\[8pt] \frac{E^+-E^+e^\frac{-3}{N^+}} {N^+e^\frac1{N^+}}=1\,000\,000Solution to Complex Coins
We wanted to find the probability that you beat John at your 3 sided coin game. John has 3 points and you have only one. In 2 more flips, John has a 4/9 probability of winning. In any given roll, either round or tails gives him a point. There is a 1/9 probability you will tie it up. The chances are then fifty fifty. The remaining 4/9 have John ending at 4 points. 2 of them give you 1 point, 2 give you two. When the score is 4-2, there is a 1/9 chance of getting 2 heads in a row to tie is up. With 4-3, it’s 1/3. When it’s tied up, you have a 50% chance of winning. 50% of a ninth of 2/9 is 1/81 or 2/162. 50% of a third of 2/9 is 1/27 or 6/162. And 50% of 1/9 is 1/18 or 9/162. (2+6+9) / 162 = 17/162. Not very likely at all.
Table \space of \space Probabilities\\[16pt] \begin{matrix} Points&0&1&2&3&4&5\\[8pt] 0&1\\[4pt] 1&\frac12&\frac12\\[4pt] 2&\frac16&\frac23&\frac16\\[4pt] 3&\frac1{18}&\frac49&\frac49&\frac1{18}\\[4pt] 4&\frac1{54}&\frac29&\frac{14}{27}&\frac29& \frac1{54}\\[4pt] &&&… \end{matrix}Perplexing Protons
You will need to use chemistry principles to solve this challenge.
You have some protons in a particle collider, but you don’t know how many. Your computers said there were 5 million at 12 noon based on their magnetic field. The problem is that you just found out that some electrons were leaking into the collider. You discovered the leak and patched it at 2pm. You know the electrons were leaking since 11 o’clock, and that they leaked at a constant rate. There are now 1 million electrons at 2:15, and they disrupted the magnetic field from which you counted the protons. However, some electrons collided with protons to make neutrons. You need to know how many protons there are for your experiment at 2:30. You don’t have time to count all the neutrons or recount the protons. You must figure it out based on the amount of energy released, which was 1 kilojoule. The challenge is to use the information above to determine the number of protons that will be in the collider at 2:30 if the rate of protons colliding with electrons is linearly proportional to the number of electrons.