Tag: solution

Solution to Galvanizing Growth

We’re trying to solve a rather insane system of equations. Due to the size of this problem, It’s publishing was delayed. Also, the solution is not typed, that would take a very long time, but shown is a scan of my rough work.

The initial equation system problem was:

a_{n1}=125\ e^\frac{-c_n}{100}\\[8pt] a_{n2}=5\ e^\frac{-c}{5}\\[16pt] g_n=\left\lfloor a_{n1}s_n{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_n}{p_n}\right)\right\rfloor\\[16pt] g_{a_b}=\left\lfloor a_{b2}s_b{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_a}{p_a}\right)\right\rfloor\\[16pt] i_n=s_n+10\left(s_n-\frac{p_n}5+|s_n-\frac{p_n}5|\right)\\[16pt] x=5-3\sqrt2\operatorname{erfc^{-1}}{(2r)}\\[8pt] y=11-5\sqrt2\operatorname{erfc^{-1}}{(2r)}

The solution is:

solution_to_galvanizing_growthDownload

Solution to Seriocomic Set

a,\,b,\,c,\,d\in\mathbb Z\\[8pt] a\leqslant b\leqslant c\leqslant d\\[8pt] a+b+3=a+d=\cos'{\pi}\\[8pt] \int \frac{d-c}a+1\;\;\delta a\bigg\lt\cos'{(a+b)\pi}\bigg\lt\int\frac{1-a}b-2\;\;\delta b\\[8pt] 0\lt|bc|+1\leqslant a+b+c+d\lt5\\[8pt] bc\notin\{a^2|a\in\mathbb N\}

Begin with the equation above and the assumption that any leftover constants after integrating always equal c. We can simplify to the 7 simpler expressions below. Consider only up to before the statement regarding their sum, that the sum is positive. There are at this point four possible tuples (a, b, c, d). They are -2, -1, 0, and 2; -3, 0, 1, and 3; -4, 1, 2, and 4; and -5, 2, 3, and 5. Now, the first and last of these have sums that are out of bounds. The first of the remaining options is eliminated by the final clue. This leaves us with an answer of (a, b, c, d) = (-4, 1, 2, 4).

a,\ b,\ c,\ d\in\Z\\[8pt] a\leqslant b\leqslant c\leqslant d\\[8pt] a+b=-3\\[8pt] 2c-d-b\lt3\\[8pt] 2b-c-a\gt0\\[8pt] 0\lt a+b+c+d\lt5\\[8pt] \sqrt{bc}\notin\Z\\[16pt] \Rightarrow\therefore(a,b,c,d)=(-4,1,2,4)

Solution to Radical Rectangles

To begin, the total of the perimeters of rectangles X Y and Z is 46 cm. This is equal to 2(A+B+C+2D+E). So A+B+C+2D+E is 23 centimetres. If we can find the value of D, we can solve the problem. Knowing that rectangles X and Y have equal perimeters and X and Z have equal areas, we can form a system of equations below. There are only 2 possible values of D, 3 cm and 13 cm. 13 cm obviously doesn’t work. D=3, and so A+B+C+D+E = 20 cm.

A+B=C+D\\AB=DE\\A+B+C+2D+E=23\\AB+CD+DE=39\\[8pt] \\D(C+2E)=39\\C,D,E\in\Z\\\Rightarrow D=3

Solution to Esoteric Elastics

In this challenge, the objective is to find the distance between 2 rods pulling elastics given that those elastics have equal heights. Now, because the elastics have different nominal lenghts and spring constants, they will only have equal lengths at a specific distance. This is because they must both be pulling with equal force. The force exerted by the two elastics is given by the first 2 equations below. They are equal, so we can solve them like a system of equations. We get that both are stretched to a length of 9 cm. The elastics are folded over twice, so their total length is thus only 4.5 centimetres. The combined force of the 2 elastics is 24 cmkg/s2. The weight resting upon the elastics will exert a downward force of 1 g * ≈9.807 m/s2 ≈ 9.807 mg/s2. In order for the elasitics to support it’s weight, the elastics must have an elasics must have this upwards force component. Let the angle below horizontal of the elastics be theta. θ ≈ asin(≈9.807/240), and the vertical distance between the rods is given by 4.5 * cos(asin(≈0.04086)). Using trigonometric identities, we simplify to the equation below. We get a distance of ≈8.992 centimetres.

\vec F_A=-{2kg\over s^2}(D-3cm)\\[8pt] \vec F_B=-{3kg\over s^2}(D-5cm)\\[16pt] \vec F_A=\vec F_B\\[8pt] {Dkg\over s^2}={9cmkg\over s^2}\\[8pt] D=9cm\\[32pt]F_A+F_B=24{cmkg\over s^2}\\[8pt] F_G\approx9.807{mg\over s^2}\\[16pt] x\approx4.5\cdot2\cdot\cos{\Bigg(\arcsin{\bigg({\approx9.807\over240}\bigg)}\Bigg)}\\[8pt] \approx\sqrt{1-\approx{0.04086}^2}\\[8pt] \approx8.992cm

Solution to Curious Colonies

We wanted to find a way to position the colonies so that they all have the same area. It’s very interesting and fun to make cool pictures, models, and simulations of bacterial colony growth. But even just theorizing will do for this problem. The slowest colony, the Dotty bacteria colony, only grows at 1 centimetre a minute, but must still occupy 20% of the area. If the Dotty colony is with only the Linear bacteria, and opposite them, it will still fall short of 20%. Thus it is clearly impossible. You must tell Joana that the colonies cannot all start at the border and occupy equal areas. She must choose one.

Solution to Centrifuge System

Free body diagram of setup for challenge.

We are trying to find the rotations per minute of this system after a force of 1 newton from stationary for 1 second. It is important to note that, since the small weight is being pulled by the second pivot, it will be at a right angle to the wire connecting the two pivots. The Pythagorean theorem says that it’s distance from the centre will be 5 centimetres. This is the same as that to the first weight and force, so we do not need to consider torque. The total weight is 8 kg. A newton is just a mkg/s2. Multiply by time and divide by weight to get the total change in velocity. The angle is of no importance because we are dealing with angular momentum. 1 times 1 divided by 8 gives 0.125 metres per second. The system began stationary, so this is the new velocity. The circumference is 2πr=31.42 centimetres or ≈0.3142 metres. Divide speed (m/s) by distance (m) to get rate (r/s). Multiply by 60 to get rpm. 0.125/≈0.3142*60≈23.87 rpm. The answer is 23.87 rotations per minute.

Solution to Recondite Raking

Illustration of you raking leaves in the optimal method.

We’re trying to find the most efficient way to rake leaves over an infinite plane. The simple answer is to rake in a straight line. But how frequently should you bag the leaves? Well, the amount of seconds per leaf unit picked up is proportional to the number of leaves you already have. Let the number of leaf units per bag be l, and the average seconds per leaf unit, a. Your rake has a length of 0.5 metres, so you must walk 2 metres. That means the a is 2, when l is 0. This is when we bag every zero seconds. But at that point, the time required to bag them is undefined. The limit, however, as leaf units per bag decreases, is infinite. The average time spent pushing leaves per leaf unit increases by one second per 20 leaves, so the number of seconds spent pushing the leaves is dictated by the first equation below. Now, 1 bag takes 2 seconds plus 1 second per five leaf units. Divide this by the number of leaf units to get the seconds per leaf unit. Multiply the top and bottom by 5 to simplify. Separate the l to get the second equation below. The first equation is linear and increasing. The second is hyperbolic and decreasing. The total is concave up. What is it’s minimum? The minimum is the point at which the slope is zero. The derivative of the first equation is a constant twentieth. What is the derivative of 2/l? Remember, 2/l = 2l-1. The exponent rule says it’s derivative is -2l-2. At the minimum, this plus a twentieth equals zero. Solve for l to get √10. Divide by 2 and get the optimal strategy. Rake in a straight line and bag the leaves every ≈1.581 metres. It takes only ≈0.7906 seconds to bag a square metre of leaves!

a_p=2+{l\over20}\\[8pt] a_b=\frac2l+0.2\\[8pt] a=2.2+{l\over20}+\frac2l\\[16pt] {\delta\over\delta l}\bigg(\frac2l\bigg)=-2l^{-2}\\[8pt] 2l^{-2}=\frac1{20}\\[8pt] l^2=10\\[8pt] l=\sqrt{10}\approx 3.162\\[8pt] a={\sqrt{10}\over4}\approx 0.7906

Solution to Captivating Circles BONUS

Free body diagram of cylinders illustrating why they do not roll.

We’re trying to find the rate at which a cylinder rolls. This is actually rather simpler than you would think. For every metre the small cylinder rolls, the big one rolls three. The small one has internal friction with the big one. But, this is more than that of the big one and ground! The push of this friction is not only exerted by the big cylinder onto the small, but vice versa! Now, the cylinder would roll because of the counterclockwise torque applied by the friction with the ground. But the internal friction is so much stronger that it overpowers that torque with an even stronger clockwise torque threshold. Remember, it only happens when the counterclockwise torque is applied, so it is only a threshold. It will not cause the cylinder to roll backwards or anything crazy like that. This principle also applies to the normal force exerted by the large cylinder unto the small. Because of this, it is as if the small cylinder did not exist! Well, not really. But in a purely mathematical concept, it will be once equilibrium is reached. Knowing this, we need only look at the big cylinder. The acceleration gravity applies downward, combined with the normal force from the ground, exerts a total force dictated by the equation below. For 9.807 m/s2 and 15°, it gives 9.473 m/s2. Per metre, the kinetic friction is 1 kg/s. To equal the acceleration downwards, we divide, and get 9.473 metres per second.

\vec F_g-\vec F_n=F_g\cos\theta\\[4pt] =9.807\frac m{s^2}\times0.9659\\[4pt] =9.473\frac m{s^2}

Solution to Loopy Lines

Graph of Loopy lines for this challenge.

We’re trying to find the centre of a circle given the radius, a point on its circumference, and a point on a line that intersects 2 lines on its circumference. We already know one of the intersections. Since it takes 2 points to define a line, we can define 2 of the lines and find their intersection at (4.5, -0.75). Now, to find the centre of a circle, we draw the perpendiculars of 2 of the lines between the 3 points we are defining it with. Their intersection is the centre. We can use the 2 lines we know! That way, we know the slopes. Finding the intersection of 2 lines in y=mx+b form is easy peasy! x=(b1-b2)/(m2-m1), and y = m1x+b1. No problem. For the perpendiculars, we need the midpoints of the lines between intersections. No problem either, x = (x1+x2)/2, and y = (y1+y2)/2. To get a line given it’s slope and a point it passes through, we use b=y-mx. The slope is -1/m, where m was the old slope. The distance between 2 points is √((x1-x2)2+(y1+y2)2). Put all these simple things together and the answer should be easy. Let the slope of the line be x. Plugging x in through these equations gives the big equation below. A little bigger than expected. It equals 5 when x≈-2.954 and x≈-0.2364. These give the centres (5.503, 4.419), and (9.185, 0.9967).

\sqrt{\bigg(-{33x+166\over8(2x+3)}-2.053125\bigg)^2+\bigg(-{51x+214\over4(2x+3)}-{2x+3\over2x-3}-5.98125\bigg)^2}

Solution to Unbalanced Balance

Diagram of the now balanced balancing scale.

We’re trying to determine how much water must be poured into the bucket on the left to balance the balance. On the right is 2.5 metres of plank and a 10 kilogram weight. The weight is partially held up by a rope from a pulley. The weight opposite the pulley has a weight of 7 kilograms, exerting 7 kg * ≈9.807 m/s2 ≈ 68.65 mkg/s2 of tension on the rope. This lifts the 10kg weight, and so the force it exerts is equivalent to that of a 10-7 = 3 kilogram weight. It exerts a downward force of 3 kg * ≈9.807 m/s2 ≈ 29.42 mkg/s2. It is 2 metres away from the pivot, so it exerts ≈29.42 * 2 ≈ 58.84 m2kg/s2 of clockwise torque on the plank. Now, the plank on the right has a weight of 2.5 kilograms, so it exerts a downward force of 2.5 kg * ≈9.807 m/s2 ≈ 24.52 mkg/s2. It’s centre of mass is 1.25 metres away from the pivot, so it exerts a clockwise torque of ≈24.52 * 1.25 ≈ 30.65 m2kg/s2. The total clockwise torque is ≈58.84 + ≈30.65 ≈ 89.49 m2kg/s2. On the left, there is 1.5 metres of plank, weighing 1.5 kilograms. It exerts a downward force of 1.5 kg * ≈9.807 m/s2 ≈ 14.71 mkg/s2. It’s centre of mass is 0.75 metres to the left of the pivot, so it exerts a counterclockwise torque of ≈14.71 * 0.75 ≈ 11.03 m2kg/s2. Finally, water at 20° celsius has a density of 998.2 kg/m3. The bucket is 2 metres to the left of the pivot. The counterclockwise torque the water must create is ≈89.49 – ≈11.03 ≈ 78.46 m2kg/s2. This requires a downward force of ≈78.46 / 2 ≈ 39.23 mkg/s2. The mass required to exert this force is ≈39.23 / ≈9.807 ≈ 4 kg. The volume of water that has this mass is 4 / 998.2 = 4.007 × 10-3 m3. We have our answer. There is an easier way to do this problem, where we do not calculate gravity. This works because all the pieces that we start and end with do not use gravity. It makes sense because a scale would still work on the moon.