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Notice a bit of symmetry here. The angles \(36^\circ,324^\circ\) are centered around \(0^\circ\) at \(0^\circ\pm36^\circ\). The angles \(108^\circ,252^\circ\) are centered around \(180^\circ\) at \(180^\circ\pm72^\circ\). We can use sums and differences: \[\cos(0^\circ-36^\circ)+\cos(0^\circ+36^\circ) +\cos(180^\circ-72^\circ)+\cos(180^\circ+72^\circ)\] Using the sum/difference identity for cosine, we can show \[\cos(a-b)+\cos(a+b)=2\cos(a)\cos(b)\] So if we apply it to the problem, we have \[2\cos(0^\circ)\cos(36^\circ)+2\cos(180^\circ)\cos(72^\circ) =2(\cos(36^\circ)-\cos(72^\circ))\] Now we need some way to handle these angles. Let \(\theta=36^\circ=\pi/5\). We can solve by using \(\sin(2\theta)=\sin(\pi-2\theta)=\sin(3\theta)\), starting with the double and triple angle formulas. \[2\sin(\theta)\cos(\theta)=3\sin(\theta)-4\sin^3(\theta)\] Divide by \(\sin(\theta)\neq0\) and apply \(\sin^2(\theta)=1-\cos^2(\theta)\), which leads us to the following quadratic equation. \[4\cos^2(\theta)-2\cos(\theta)-1=0\Rightarrow\cos(\theta) ={1\pm\sqrt{5}\over4}\] Take the positive value since \(\theta\) is in quadrant 1. Then we can solve the problem with this quantity. For \(72^\circ\), use the double angle identity \(\cos(2\theta)=2\cos^2(\theta)-1\). \[2\left({1+\sqrt{5}\over4}-2\left({1+\sqrt{5}\over4}\right)^2+1\right) =2\left({1+\sqrt{5}\over4}-{2\left(6+2\sqrt{5}\right)\over16}+1\right)\] \[=2\left({1+\sqrt{5}\over4}-{3+\sqrt{5}\over4}+{4\over4}\right) =2\left({1-3+4\over4}\right)=2\cdot{1\over2}=1\]
Another way to do this is with the formula from roots of unity / a regular polygon (\(n\) sides) in the complex plane. \[\sum_{k=0}^{n-1}\cos\left(\theta+k{2\pi\over n}\right)=0\] This formula can be thought of as the center of mass of the vertices of a regular polygon. We can prove this by using Euler's formula and properties of geometric series. We are looking for the real part of the following expression. \[\sum_{k=0}^{n-1}e^{i\left(\theta+k{2\pi\over n}\right)} =e^{i\theta}\sum_{k=0}^{n-1}e^{ik{2\pi\over n}} =e^{i\theta}\left(e^0+e^{i{2\pi\over n}}+e^{2i{2\pi\over n}}+\ldots+ e^{(n-1)i{2\pi\over n}}\right)\] \[=e^{i\theta}{1-e^{ni{2\pi\over n}}\over1-e^{i{2\pi\over n}}} =e^{i\theta}{1-e^{2\pi i}\over1-e^{i{2\pi\over n}}} =e^{i\theta}{1-1\over1-e^{i{2\pi\over n}}}=0\] Note that if \(n=1\) then this proof would divide by \(0\), so the proof is valid for \(n\geq2\). If \(n=1\) then we actually get \(\cos(\theta)\) instead of \(0\).
Going back to the problem, let \(\theta=36^\circ\) and \(n=5\). Then \[\cos(36^\circ)+\cos(108^\circ)+\cos(180^\circ) +\cos(252^\circ)+\cos(324^\circ)=0\] But \(\cos(180^\circ)=-1\) so if we add \(1\) to both sides, we have the answer. \[\cos(36^\circ)+\cos(108^\circ)+\cos(252^\circ)+\cos(324^\circ)=1\]
Let \(r\) be the radius of the large circle. Then using some arc length formulas \[c=2\pi r\left({x\over2\pi}\right)=rx, \quad b=2\pi(r-a)\left({x\over2\pi}\right)=rx-ax\] First substitute \(c=rx\) into the formula we were given so \(2a+b=rx\). Now substitute \(b=rx-ax\) so \(2a+rx-ax=rx\). We can subtract away the \(rx\) terms relealing \(2a=ax\Rightarrow x=2\).
First use the identity \(\cos^2(\theta)=1-\sin^2(\theta)\) to find a quadratic equation. \[{\sin^4(\theta)\over8}+(1-\sin^2(\theta))^2={1\over9} \Rightarrow {9\over8}\sin^4(\theta)-2\sin(\theta)+{8\over9}=0\] Its solution is (turns out to only have 1) \[\sin^2(\theta)={2\pm\sqrt{2^2-4\cdot{9\over8}\cdot{8\over9}}\over9/4} ={8\over9}\] We don't actually have to compute \(\sin(\theta)\). Focus on the expression for \(x\). Use the positive root since we are given \(x>0\). \[x=\sec(\theta)={1\over\cos(\theta)}={1\over\sqrt{1-\sin^2(\theta)}} ={1\over\sqrt{1-8/9}}={1\over\sqrt{1/9}}=\sqrt{9}=3\]
Find the only \(x\) such that there exists a differentiable manifold that is homeomorphic but not diffeomorphic to \(\mathbb{R}^x\).
Homeomorphic is a topology term that means one space can be transformed into another space by smooth deformation. Mathematically, it means there exists a continuous bijection between the spaces whose inverse is also continuous.
Diffeomorphic is similar, but also requires that the bijection is smooth in addition to continuous. This means that the bijection and its inverse are infinitely differentiable.
People showed that in dimensions \(1,2,3\) and \(\geq5\), any smooth manifold homeomerphic to the space is also diffeomorphic to it. The 4th dimension is an anomaly and exotic smooth structures exist. Here, exotic means homeomorphic to the space but not diffeomorphic to it.
The solution is \(x=4\), the only dimension with this different property.
We can recognize that \(3125=5^5\), a power of \(5\). This makes it noticeable that \(x=5\) is the solution. If we do not know this, we could try increasing \(x\) to look for a positive solution, observing that \(3125^{1\over x}\) decreases as \(x\) increases.
To see that this is the only solution in positive \(x\), notice how \(x\) is increasing. Then find the derivative to see that \(3125^{1\over x}\) decreases. \[{d\over dx}\left(3125^{1\over x}\right) =3125^{1\over x}\left({-1\over x^2}\ln(3125)\right)\] So for positive \(x\), there is only 1 solution. If \(x\) is negative, then there can be no solution because \(3125^{1\over x}\) would be positive.
Find the sum of the lengths of all sides and diagonals of a regular hexagon with side length \(2-\sqrt{3}\).
First we have \(6\left(2-\sqrt{3}\right)=12-6\sqrt{3}\) from the lengths of the sides. Knowing that a regular hexagon can be divided into 6 equilateral triangles, we can see that the 3 "big" diagonals count 6 of the same segments doubling our current total to \(24-12\sqrt{3}\).
Now focus on the smaller diagonals. If we draw one of them, and a line to the vertex in the middle, we create two 30-60-90 triangles. The leg we care about, which is half of the diagonal, has length \[\left(2-\sqrt{3}\right){\sqrt{3}\over2}=\sqrt{3}-{3\over2}\] One of these diagonals is twice this length so \(2\sqrt{3}-3\). There are 6 of them so they contribute \(12\sqrt{3}-18\) to our sum. Add this to what we had from the sides and big diagonals to find the solution. \[24-12\sqrt{3}+12\sqrt{3}-18=6\]
Square both sides first. \[9+\sqrt{80}=a^2+b+2a\sqrt{b}\] So this gives us equations \[a^2+b=9,\quad2a\sqrt{b}=\sqrt{80}\] Square the second one to find \(4a^2b=80\Rightarrow a^2b=20\). Multiply the first equation by \(b\) on each side to be able to substitute this. \[a^2b+b^2=9b\Rightarrow20+b^2=9b\Rightarrow b^2-9b+20=0 \Rightarrow(b-4)(b-5)=0\] So \(b=4,5\). If we try \(b=4\), then \(a=\pm\sqrt{5}\not\in\mathbb{Z}\). So we find \(b=5\) and \(a=\pm2\). The one that works is \(a=2\). \(a=-2\) is extraneous. Finally the solution is \(x=2+5=7\).
Find the number of regions in a Venn diagram with three intersecting sets.
Venn diagrams divide a plane into regions by all possible set containment. For each set, either the item is inside it or not. The outer region (not in any set) also counts as a region. In total, there would be \(2^3=8\) regions for 3 intersecting sets.
Find the integer that is a perfect square and also the product of four consecutive odd integers.
To make a nice equation for this, we search for an even integer \(n\) right in the middle so for some integer \(k\geq0\), \[(n-3)(n-1)(n+1)(n+3)=k^2\] We can multiply together pairs of terms to get differences of squares \[(n^2-1)(n^2-9)=k^2\] Now let \(m=n^2-5\) which is right in the middle, so we can get more differences of squares. \[(m+4)(m-4)=k^2\Rightarrow m^2-16=k^2\Rightarrow m^2-k^2=16 \Rightarrow(m-k)(m+k)=16\] The difference between \(m-k\) and \(m+k\) is \(2k\) so they must both be even to multiply to \(16\). Also since \(n\) is even, \(m\) is odd, which also shows \(k\) is also odd. By searching \(0\leq m\leq16\), we find \(m=5,k=3\). We also need to check \(-16\leq m\leq0\) since we have not restricted things to just positive numbers. We also find the corresponding negative number solution with \(m=-5,k=3\). If \(m=5\), then \(n^2=10\) which does not result in an integer. If \(m=-5\) then \(n=0\). From this, we find \((-3)(-1)(1)(3)=9\) is the perfect square solution.
Noticing that \(4^x=(2^x)^2\), the numerator is a difference of squares and can be factored. Also notice the powers of 2 in the fraction. \[{(2^x+2^{-x})(2^x-2^{-x})\over2^x+2^{-x}}=2^x-2^{-x}={2^{20}-1\over2^{10}}\] Multiply each side by \(2^{10}\) \[2^{10+x}-2^{10-x}=2^{20}-1\] Here we can see that \(x=10\) makes the 2 sides match.
We can also do this by arranging the right side to look like the left. \[{2^{20}-1\over2^{10}}\cdot{2^{20}+1\over2^{20}+1} ={2^{40}-1\over2^{30}+2^{10}}\cdot{2^{-20}\over2^{-20}} ={2^{20}-2^{-20}\over2^{10}+2^{-10}}={4^{10}-4^{-10}\over2^{10}+2^{-10}}\] This also makes it clear that \(x=10\) is the solution.
The last thing we might want to do is check that it is the only solution. If we let \(u=2^x\), then we differentiate to find \[{d\over dx}{u^2-u^{-2}\over u+u^{-1}}={d\over dx}{u^4-1\over u^3+u} ={4u^3(u^3+u)-(u^4-1)(3u^2+1)\over(u^3+u)^2}\] \[={u^6+3u^4+3u^2+1\over u^2(u^2+1)^2}={(u^2+1)^3\over u^2(u^2+1)^2} ={u^2+1\over u^2}\] This is clearly positive on the entire interval which means the function is always increasing so the \(x=10\) we found is the only solution.
First the derivative \[f'(t)={d\over dt}t^{1/3}={1\over3}t^{-2/3}\] Then notice that \({1\over216}=6^{-3}\) and evaluate \[x=f'(6^{-3})={1\over3}\left((6^{-3})^{-2}\right)^{1/3}={1\over3}(6^6)^{1/3} ={1\over3}6^2=12\]
Find the sum of the coefficients of the odd powers of \((1+x+x^2)^3\).
Let \(f(x)=(1+x+x^2)^3\). Then \(f(1)\) is the sum of all coefficients and \(f(-1)\) is the alternating sum of coefficients. If we compute \(f(1)-f(-1)\) then we get \(3^3-1^3=26\). This has the even exponents subtract away since \[c\cdot(1)^{2n}-c\cdot(-1)^{2n}=c-c=0\] But the odd exponents add to double \[c\cdot(1)^{2n+1}-c\cdot(-1)^{2n+1}=c+c=2c\] So \(26\) is twice the sum of the odd coefficients and \(13\) is the sum of the odd coefficients.
First multiply the 2nd equation by \(2\) so we have only integers \[18x+22y+z=980\] Then subtract the first to eliminate \(z\) \[17x+21y=490\] We need positive integer solutions to this. Consider operations modulo \(21\). Since \(490=21\cdot23+7\), \(17x\equiv7\). Now we would like to isolate \(x\). The inverse of \(17\) modulo \(21\) is \(5\). So \(x\equiv7\cdot5\equiv14\). We have that \(x=14+21k\) for some integer \(k\geq0\) since \(x>0\). Now for \(y\), \(21y=490-17x=490-17(14+21k)\). This simplifies to \(y=12-17k\). Since \(y>0\), we have \(k\leq0\). The only possible value is \(k=0\) so we find \(x=14\) and \(y=12\). Using the first equation, \(z=490-x-y=464\). These can all be verified to solve both equations.
Since there is a square root, and we will square things, be wary of extraneous solutions and that a square root needs a nonnegative quantity. Eliminate the square root by squaring both sides of the first equation. \[{(x-y)^2\over xy-y^2}=4 \Rightarrow x^2-2xy+y^2=4xy-4y^2 \Rightarrow x^2-6xy+5y^2=0\] This factors nicely to \[(x-y)(x-5y)=0\] If \(x-y=0\) then the left side of the first equation is \(0\), so that would not equal \(2\). So we must have \(x-5y=0\). Substituting \(x=5y\) into the second equation find us \(5y+3y=24\Rightarrow8y=24\Rightarrow y=3\). Then we can find \(x=24-3y=24-9=15\). This solution satisfies both equations.
First observe \(2048=2^{11}=\sqrt{2}^{22}\). So by taking \(\log_{\sqrt{2}}\) of each side, we find \[\sqrt{500-x}=22\Rightarrow500-x=484\Rightarrow x=16\]
First rearrange the equations into a standard form \[x-3y=-4,\quad x-6y=-25\] Now subtract the second from the first to eliminate \(x\) \[3y=21\Rightarrow y=7\] Finally we can find \[x=-4+3y=-4+21=17\]
First focus on line \(AB\). Let \(D\) be the other point where it intersects the radius \(2\) circle, so \(D\) lies somewhere on segment \(OA\). Also let \(C_2\) and \(C_3\) be the centers of the circles of radius \(2\) and \(3\) respectively.
By the power of a point and related theorems, we have \(|OD|\cdot|OA|=\Pi(O)\) where \(\Pi(O)\) is the power of a point for \(O\) and the radius \(2\) circle. We can see that \(|OC_3|=6\) and by similar triangles, \(|OC_2|=4\). So \[|OD|\cdot|OA|=|OC_2|^2-2^2=4^2=2^2=12\] Next we use homothety scaling. Since \(OD\) points to the radius \(2\) circle, a scaling by \(-3/2\) maps it to \(OB\) pointing to the radius \(3\) circle. The negative is because it switches to the other side of the homothety center. So \[|OB|={3\over2}|OD|\] Now substitute this into what we found with power of a point \[{2\over3}|OB|\cdot|OA|=12\Rightarrow|OA|\cdot|OB|=18\]
This problem does not really appear to be suitable for pencil and paper. Using a calculator, \[\sqrt{{40\over3}-2\sqrt{3}}\approx3.1415333387050945\] \[y\approx0.00005931488469856916\] \[321000y\approx19.0400779882407\] So \(x=19\).
Find the length of the interval defined by \(\left|{2x+1\over5}\right|<4\).
First multiply each side by \(5\) so we have \(|2x+1|<20\). Then for the positive case \(2x+1<20\Rightarrow x<19/2\). For the negative case \(-2x-1<20\Rightarrow x>-21/2\). The interval length is \[{19\over2}-{-21\over2}={40\over2}=20\]
If 4 students take 3 hours to solve 20 problems, how many hours will it take 3 students to solve 105 problems (at the same rate)?
The rate is \[{20\ \text{problems}\over4\ \text{students}\cdot3\ \text{hours}} ={5\over3}\ \text{problems/student/hour}\] If there are \(3\) students, then the rate is \(5\ \text{problems/hour}\). So divide the 105 problems by this \[105\ \text{problems}\cdot{1\ \text{hour}\over5\ \text{problems}} =21\ \text{hours}\]
We have an isosceles triangle so both of the sides adjacent to the \(45^\circ\) angle have length \(\sqrt[4]{3872}\). We can use the side-angle-side area formula. Also use \(3872=2^5\cdot11^2\) \[{1\over2}ab\sin(45^\circ)={1\over2}\sqrt{3872}{\sqrt{2}\over2} ={1\over4}\sqrt{2^6\cdot11^2}={1\over4}\cdot2^3\cdot11=2\cdot11=22\]
First subtract the 2 equations, so we can factor a difference of squares. Division by \(r-s\) is justified since \(r\neq s\). \[r^2-s^2=s-r \Rightarrow (r+s)(r-s)=-(r-s) \Rightarrow r+s=-1\] Now make \(rs\) appear by multiplying the first equation by \(r\) and the second by \(s\) on each side. \[r^3=24r+rs,\quad s^3=24s+rs\] However, we will not use this \(rs\) to find the solution directly. Instead it appears from the difference of cubes obtained by subtracting these equations. \[r^3-s^2=24(r-s) \Rightarrow (r-s)(r^2+rs+s^2)=24(r-s) \Rightarrow r^2+s^2+rs=24\] Now from adding the 2 given equations, \(r^2+s^2=48+r+s\). Substitute this in \[48+r+s+rs=24\Rightarrow rs+(r+s)=-24\] But we had \(r+s=-1\) from before so we can substitute that too. \[rs-1=-24\Rightarrow rs=-23\] And finally \(x=-rs=-(-23)=23\). If we were to try to solve for \(r,s\), we would find 4 solutions at the intersections of the 2 parabolas described by the two given equations.
Notice how the bottom left triangle has side lengths in ratios \(1,2,\sqrt{5}\) and its side lengths are \(4\sqrt{3},8\sqrt{3},4\sqrt{15}\). The little triangle at the top left can be shown to be similar to it scaled down by \(2\), so its side lengths are \(2\sqrt{3},4\sqrt{3},2\sqrt{15}\). Their hypotenuses have ratio \(1:2\) and form the legs for the next triangle, so it is similar with the same \(1,2,\sqrt{5}\) ratio and has side lengths \(2\sqrt{15},4\sqrt{15},10\sqrt{3}\). Now the angle markings show us the final right triangle, bordering the \(x\) region, is similar, and we know its \(10\sqrt{3}\) longer leg, so its side lengths are \(5\sqrt{3},10\sqrt{3},5\sqrt{15}\).
From this, we have shown that 2 sides of the \(x\) region have lengths \(5\sqrt{15}\) and \(8\sqrt{3}\). Now we can find the angle at the bottom to be able to solve with the \({1\over2}ab\sin(\theta)\) area formula. Since we showed the 3 right triangles stacked up rotating each are similar, the smaller angle of each is \(\theta=\arctan(1/2)\). So this means the angle we need to solve for \(x\) is \(\pi/2-3\arctan(1/2)\). We don't actually need to find this angle directly, but to find its sine. Start with cofunction identity. \[\sin\left({\pi\over2}-3\arctan\left({1\over2}\right)\right) =\cos\left(3\arctan\left({1\over2}\right)\right)=\cos(3\theta)\] If we draw a triangle for \(\theta=\arctan(1/2)\), we will find \(\cos(\theta)=2/\sqrt{5}\). Now apply the triple angle identity. \[\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta) =4\cdot{8\over5\sqrt{5}}-3{2\over\sqrt{5}} ={32\over5\sqrt{5}}-{6\cdot5\over5\sqrt{5}}={2\over5\sqrt{5}}\] Now we can put in the 2 side lengths and \(\sin(\pi/2-3\theta)\). \[x={1\over2}\left(5\sqrt{15}\right)\left(8\sqrt{3}\right){2\over5\sqrt{5}} =20\sqrt{45}{2\over5\sqrt{5}}={2\cdot20\cdot3\sqrt{5}\over5\sqrt{5}}=24\]
What is the least number of coins needed (among pennies, nickels, dimes, and quarters) that would add up to $4.99?
In general, a greedy algorithm does not work for least number of coins, but it does happen to work for american coin amounts. So we make $4.75 with 19 quarters first. Then for the rest, we use 2 dimes and 4 pennies. The total number of coins is \(19+2+4=25\).
\(11010_2\) in base \(10\)
We can convert this with place value of the ones: \[2^4+2^3+2^1=16+8+2=26\]
Find the maximum postage value that cannot be made with 5 cent and 8 cent stamps.
We can do this by brute force. Once there are 5 consecutive values that can be made, we can make any larger value. Below is the table. Make this by starting with zero 8 cent coins and adding 5 cent coins to it repeatedly, then one 8 cent coin, and so on.
| Cost | Coins | Cost | Coins |
|---|---|---|---|
| 1 | 17 | ||
| 2 | 18 | 8,5,5 | |
| 3 | 19 | ||
| 4 | 20 | 5,5,5,5 | |
| 5 | 5 | 21 | 8,8,5 |
| 6 | 22 | ||
| 7 | 23 | 8,5,5,5 | |
| 8 | 8 | 24 | 8,8,8 |
| 9 | 25 | 5,5,5,5,5 | |
| 10 | 5,5 | 26 | 8,8,5,5 |
| 11 | 27 | ||
| 12 | 28 | 8,5,5,5,5 | |
| 13 | 8,5 | 29 | 8,8,8,5 |
| 14 | 30 | 5,5,5,5,5,5 | |
| 15 | 5,5,5 | 31 | 8,8,5,5,5 |
| 16 | 8,8 | 32 | 8,8,8,8 |
After doing this, we can see that 27 is the largest we cannot make with these coin values. For anything larger, we have a way for 5 consecutive values (28-32) so beyond that we can add as many 5 cent coins as necessary after choosing one.
This is special case of the Frobenius coin problem. For 2 coprime integers \(a,b\), it has solution \(ab-a-b\). Here, we would have \(5\cdot8-5-8=27\).
There is actually a problem with the way it is stated. We can see this from the british flag theorem. Imagine we project the top vertex of the pyramid into its plane. Then we satisfy \[a^2+b^2=c^2+d^2\] Now add \(2h^2\) to each side where \(h\) is the height of the pyramid. \[(a^2+h^2)+(b^2+h^2)=(c^2+h^2)+(d^2+h^2)\] Now we can see each of these new quantities with the \(h^2\) added form the length of one of the 4 diagonals going up the pyramid. This means our pyramid should satisfy the british flag theorem with its diagonal lengths. But we see \(8^4+4^2\neq7^2+1^2\). However, if we swap the \(7\) and \(8\), then we have \(7^2+4^2=8^2+1^2\). I believe this is how the problem was meant to be.
So with that correction out of the way, consider the 2 triangles with the side length \(1\). If \(a,b\) are side lengths of a triangle, \(a>b\), then the remaining side \(c\) must satisfy \(b+c>a\) and \(c<a+b\). For the \(7,1,l\) triangle, we have (remembering that \(l\) must be an integer) \[1+l>7,\ l<7+1\ \Rightarrow\ 6<l<8\ \Rightarrow\ l=7\] Similarly for the \(4,1,w\) triangle \[1+w>4,\ w<4+1\ \Rightarrow\ 3<w<5\ \Rightarrow\ w=4\] So the area of the base is \(x=lw=7\cdot4=28\).
Find the sum of the denominators of the Egyptian fraction expansion of \({41\over55}\).
The systematic way to do this is to subtract away the largest possible unit fraction until we reach zero. This can be shown to always terminate. In general, we would look for \[{1\over n}\leq{a\over b}\Rightarrow n\geq{b\over a}\] So we would choose \(n=\lceil b/a\rceil\). For the first step, we find \[\left\lceil{55\over41}\right\rceil=2,\ {41\over55}={1\over2}+{27\over110}\] Next, \[\left\lceil{110\over27}\right\rceil=5,\ {27\over110}={1\over5}+{1\over22}\] So the Egyptian fraction expansion is \[{}={1\over2}+{1\over5}+{1\over22}\] And the denominator sum is \(2+5+22=29\).