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Let \(y=2^{-x}\). Then \(y^2+y^2+y=1\Rightarrow2y^2+y-1=0\) and this has solutions \(y={-1\pm3\over4}=1/2,-1\). Clearly \(-1\) is extraneous since \(y=2^{-x}>0\) so \(y=1/2\) which gives us \(x=1\).
Let \(s\) be the side length of the smaller square. The larger square has side length \(s\sqrt{2}\) so its area is \(2s^2\), twice the area of the smaller square which is \(s^2\). So the answer is 2.
For this problem, we can use the triangle area formula with angle \(\theta\) between the sides \(a\) and \(b\) \[{1\over2}ab\sin(\theta)\] In this case, \(a=3\) and \(b=4\), which can be used for both the area 3 triangle and the \(x\) triangle. Since the 2 squares (we assume they are squares) have right angles, the angle for the area 3 triangle and the area for the \(x\) triangle sum to \(180^\circ\). This means that \(\sin(\theta)\) will be the same in both cases since \[\sin(\theta)=\sin(180^\circ-\theta)\] so we must also have \(x=3\).
Note: we used the double angle identity: \(\sin(2x)=2\sin(x)\cos(x)\)
We have a sort of grid-aligned non convex polygon. Notice how it kind of nicely fits into a rectangle of size \(3\times3.5=10.5\). We can draw some lines to see how to exclude the parts inside the rectangle containing this weird shape. Then it becomes a matter of subtracting some rectangles and triangles from \(10.5\).
Subtracting top left first, \(10.5-1\times1.5-{1\over2}\times0.5\times1 =10.5-1.5-0.25=8.75\). Next subtract the top right, \(8.75-1\times1 -{1\over2}\times1\times1=8.75-1-0.5=7.25\). Then the bottom right, \(7.25-{1\over2}\times0.5\times1-0.5\times1.5=7.25-0.25-0.75=6.25\). Finally subtract the bottom triangle, \(6.25-{1\over2}\times2.5\times1 =6.25-1.25=5\), so the area is \(5\).
According to the Gauss-Wantzel theorem, a regular \(n\)-gon can be constructed with straightedge and compass iff \(n=2^{k}p_1p_2\ldots p_m\) for an integer \(k\geq0\) and distinct Fermat primes \(p_i\). The first 2 Fermat primes are \(3,5\). We can test that \(3,4,5,6\) are all divisible by only \(2,3,5\). But \(7\) is not, so a regular \(7\)-gon (heptagon) cannot be constructed with straightedge and compass.
This theorem involves constructible numbers obtained by repeatedly adjoining square roots to rational numbers. It also requires showing that we can construct an angle \({2\pi\over n}\).
The reasonable way to solve this is listing the primes and counting pairs. The primes below \(102\) are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101. By symmetry, we can go through the primes \(p\leq51\) and check if \(q=102-p\) is prime. The pairs we find are (5,97), (13,89), (19,83), (23,79), (29,73), (31,71), (41,61), (43,59). In total, there are 8 pairs.
Consider its binomial expansion and the conjugate binomial expansion.
\[\left(17+\sqrt{280}\right)^{17}=\sum_{k=0}^{17}{17\choose k} 17^k\left(+\sqrt{280}\right)^{17-k}\] \[\left(17-\sqrt{280}\right)^{17}=\sum_{k=0}^{17}{17\choose k} 17^k\left(-\sqrt{280}\right)^{17-k}\] When \(k\) is odd, the terms are integers, and identical in both summations. When \(k\) is even, they have opposite signs, positive in the first, negative in the second. Therefore, for some positive integers \(A,B\) \[\left(17+\sqrt{280}\right)^{17}=A+B\sqrt{280}\] \[\left(17-\sqrt{280}\right)^{17}=A-B\sqrt{280}\] \[\left(17+\sqrt{280}\right)^{17}+\left(17-\sqrt{280}\right)^{17}=2A\] Next, consider that \((16+1/2)^2=256+16+1/4=272+1/4<280\). So \(\sqrt{280}>16.5\) and \(0<17-\sqrt{280}<0.5\). From this, it clearly follows that \(0<\left(17-\sqrt{280}\right)^{17}<0.1\). Now we have \[\left(17+\sqrt{280}\right)^{17}=2A-\left(17-\sqrt{280}\right)^{17} >2A-0.1\] So we subtract a tiny number from an integer \(2A\), meaning its tenth digit must be 9.We need to know the factorization. This one can be factored with a little small number trial division and \(265837=11^2\times13^3\). Using the exponents, there are \((2+1)(3+1)=12\) divisors, but we exclude the number itself since that is not a proper divisor, so there are \(11\).
These solids are convex polyhedra which have regular polygon faces and are vertex transitive (isogonal). The solids are listed below.
There are 13 total. Sometimes it might be counted as 15 because the snub cube and snub dodecahedron are chiral.
Consider the triangle area formula \({1\over2}abc\sin(\theta)\). This means that if 2 triangles share an angle and direction of 2 sides, the area scales with the lengths of those 2 sides.
Label the midpoint of \(AC\) as \(M\), the point separating the \(m\) and \(2m\) segments as \(P\), and the point separating the \(n\) and \(3n\) segments as \(Q\).
The area of \(\triangle ABC\) is 48 and is \({1\over2}(AC)(AB)\sin(A)\) even though we do not know some of these quantities. But by replacing \(AC\) with \(AM\) and \(AB\) with \(AP\), we scale the area by \({1\over2}\) and \({1\over3}\). So the area of \(\triangle APM\) is \(8\).
Similarly the area of \(\triangle BPQ\) is \(48\cdot{2\over3}\cdot{1\over4}=8\) and the area of \(\triangle CMQ\) is \(48\cdot{1\over2}\cdot{3\over4}=18\).
Finally subtract away these 3 triangles areas from the big triangle to obtain \(x=48-8-8-18=14\).
Begin by using a common base for the logs with the change base formula. \[b={\log_2(6)\over\log_2(8)}={1\over3}\log_2(6)={1\over3}(1+\log_2(3))\] \[c={\log_2(12)\over\log_2(32)}={1\over5}(2+\log_2(3))\] Then substitute to solve for \(x\) \[x=\left({1+\log_2(3)-1\over3\log_2(3)}\cdot {2+\log_2(3)-2\over5\log_2(3)}\right)^{-1} =\left({1\over3}\cdot{1\over5}\right)^{-1}=15\]
Find the sum of the solutions of \[\left|{x\over2}-4\right|=2025\]
When \(|x/2-4|\geq0\) then solve \[{x\over2}-4=2025\Rightarrow{x\over2}=2029\Rightarrow x=4058\] When \(|x/2-4|<0\) then solve \[-\left({x\over2}-4\right)=2025\Rightarrow4-{x\over2}=2025\Rightarrow8-x=4050 \Rightarrow x=-4042\] Then the sum is \(4058-4042=16\).
Square the first equation to get something resembling the second. \[(y+z)^2=1^2\Rightarrow(y^2+z^2)+2yz=1\Rightarrow5+2yz=1\Rightarrow yz=-2\] Now square the second equation to get something resembling the third. \[(y^2+z^2)^2=5^2\Rightarrow(y^4+z^4)+2y^2z^2=25\] \[\Rightarrow x+2(yz)^2=25\Rightarrow x+2(-2)^2=25\Rightarrow x=17\] If we also wanted solutions for \(y,z\), then \(y+z=1\) and \(yz=-2\) give us two solutions which do satisfy the three equations: \(y=-1,z=2\) and \(y=2,z=-1\).
What runway number is due South?
We can find that runway numbers are from 01 to 36 (2 digits) indicating an azimuth angle to the nearest 10 degrees. Azimuth angles go clockwise from north so south would be \(180^\circ\), meaning runway number 18.
If we can get \(g(-5)\) to show up in the 2nd equation, we would be able to solve it. Notice how \(t=-2\) makes both instances of \(g\) have an argument of \(-5\). Then the 2nd equation can be used to find the solution: \[f(g(-5))=1-(-2+1)g(-5)\Rightarrow 2g(-5)-18=1+g(-5)\] \[\Rightarrow2x-18=1+x\Rightarrow x=19\]
First factor the quadratic equations. \[f(x)=\sqrt{x(58-x)}-\sqrt{(x-9)(58-x)}\] Notice how this restricts the domain to \(9\leq x\leq58\). A maximum occurs either on an endpoint or critical point since this is continuous. At \(x=58\) we get \(0\) and at \(x=9\) we get \(\sqrt{9(58-9)}=21\).
Next we need to check critical points. Start with the derivative of our objective function. \[f'(x)={58-2x\over2\sqrt{x(58-x)}}-{67-2x\over2\sqrt{(x-9)(58-x)}}=0\] The 2 fractions must be equal. Multiply each side by \(2\sqrt{x(x-9)(58-x)}\) which is valid for all \(x\) in the domain. \[(58-2x)\sqrt{x-9}=(67-2x)\sqrt{x}\] Now things get a bit messy with bigger numbers. We will omit some of the detail but square both sides to get polynomials and simplify, which ends up as a linear equation. \[963x-30276=0\Rightarrow x={30276\over963}={3364\over107}\] Since we squared things, check for extraneous solutions. We don't actually need to evaluate things exactly. Notice that our solution is \(x\approx31.4\) which makes \(58-2x<0\) and \(67-2x>0\). Since both denominators with the square roots are positive, \(f'(x)<0\), which means this solution is extraneous. Therefore the maximum is \(21\) at \(x=9\).
If we investigate further, we can see that \(f'(x)\) is finite and continuous between \(9<x<58\). It is nonzero on this interval so the sign is the same. Since the derivative must be the same sign on \((9,58)\), it must be negative since \(f(9)=21\) and \(f(58)=0\). \(f'(x)\) is undefined at both \(x=9\) and \(x=58\).
A clock shows time up to 12:00. What is the highest sum of the digits that the clock can show?
At 12:00, the sum is 3. At all other hours, :59 for minutes would maximize the sum. The highest sum for the hour we can have is 9: so the maximum is 9+5+9=23.
Find the smallest square that can be written as a sum of two integer squares.
This can be done with brute force. One common result is \(3^2+4^2=5^2=25\) so we have \(25\) as an upper bound. From there, we can try to form such sums for \(16,9,4,1\) and notice that it fails.
One subtle point is that the problem states integer squares which may be interpreted to include \(0^2\). That allows trivial solutions so we ignore it.
Find the nim sum of 42, 36, 12, 24.
The nim sum is defined as the binary exclusive or operation. First we convert the numbers to binary and compute exclusive or. \[\begin{align}&101010\\&100100\\&001100\\\oplus\quad&011000\\ \hline&011010\\\end{align}\] Converting back to decimal, the answer is \(26\).
Solve for \(y\) first. Notice that the integrand is periodic with period \(\pi\) so we can immediately simplify it a bit. Next apply the triple angle identity \(\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)\). Then substitute \(u=2t\). Also factor \(5\) out from the denominator which puts it in a form that we can use for geometric summation. \[y=4\int_0^\pi{4\cos^3(2t)-3\cos(2t)\over5-3\cos(2t)}dt ={2\over5}\int_0^{2\pi}{4\cos^3(u)-3\cos(u)\over1-{3\over5}\cos(u)}dt\] Next we turn the integrand into a geometric summation. Let \(x=\cos(u)\). Then \[{4x^3-3x\over1-{3\over5}x} =(4x^3-3x)\sum_{k=0}^\infty\left({3\over5}\right)^k x^k =4\sum_{k=3}^\infty\left({3\over5}\right)^{k-3}x^k -3\sum_{k=1}^\infty\left({3\over5}\right)^{k-1}x^k\] \[=4\sum_{k=3}^\infty\left({3\over5}\right)^{k-3}x^k -3\sum_{k=3}^\infty\left({3\over5}\right)^{k-1}x^k-3x-{9\over5}x^2\] \[=\sum_{k=3}^\infty\left({3\over5}\right)^k x^k\left(4\cdot{125\over27} -3\cdot{5\over5}\right)-3x-{9\over5}x^2 ={365\over27}\sum_{k=3}^\infty\left({3\over5}\right)^k x^k-3x-{9\over5}x^2\] Going back to the integral, we get \[y={2\over5}\int_0^{2\pi}\left[{365\over27} \sum_{k=3}^\infty\left({3\over5}\right)^k\cos^k(u)-3\cos(u)-{9\over5}\cos^2(u) \right]du\] \[={146\over27}\int_0^{2\pi}\sum_{k=3}^\infty \left({3\over5}\right)^k\cos^k(u)du -{6\over5}\int_0^{2\pi}\cos(u)du-{18\over25}\int_0^{2\pi}\cos^2(u)du\] Now the integral with \(\cos(u)\) can easily be found to be zero. The integral with \(\cos^2(u)\) can be evaluated with a double angle identity and leaves us with an extra \(18\pi/25\). The problem we have to solve now is \[y={146\over27}\sum_{k=3}^\infty \left({3\over5}\right)^k\int_0^{2\pi}\cos^k(u)du-{18\pi\over25}\] First we show that when \(k\) is even, the integral is zero, so we can eliminate those terms. Use the change of variables \(v=u-\pi\) \[I=\int_0^{2\pi}\cos^k(u)du=\int_{-\pi}^{\pi}\cos^k(v+\pi)dv\] \[=\int_{-\pi}^{\pi}\left(\cos(v)\cos(\pi)-\sin(v)\sin(\pi)\right)^kdv =-\int_{-\pi}^{\pi}\cos^k(v)dv\] But also since we are integrating one period, we also have \[I=\int_{-\pi}^{\pi}\cos^k(u)du\] Thus \(I=-I\Rightarrow I=0\). So we can eliminate the odd terms. We will reindex the series to start at \(k=0\) as well so there are 2 terms to subtract away. \[y={146\over27}\left[\sum_{k=0}^\infty\left({3\over5}\right)^{2k} \int_0^{2\pi}\cos^{2k}(u)du-\int_0^{2\pi}du-{9\over25}\int_0^{2\pi}\cos^2(u)du \right]-{18\pi\over25}\] \[={146\over27}\sum_{k=0}^\infty\left({3\over5}\right)^{2k} \int_0^{2\pi}\cos^{2k}(u)du-{146\over27}\left(2\pi+{9\pi\over25}\right) -{18\pi\over25}\] \[={146\over27}\sum_{k=0}^\infty\left({3\over5}\right)^{2k} \int_0^{2\pi}\cos^{2k}(u)du-{364\pi\over27}\] Next we need to find out what the cosine power integrals are for even powers. This is similar to the Wallis integrals but we will evaluate them here. Let \(I(2n)=\int_0^{2\pi}\cos^{2n}(t)dt\) with base case \(I(0)=2\pi\). To find a relationship, first use \(\cos^2(t)=1-\sin^2(t)\) to give us something for integration by parts. \[I(2n)=\int_0^{2\pi}\cos^{2n}(t)dt=\int_0^{2\pi}\cos^{2n-2}(t)dt -\int_0^{2\pi}\cos^{2n-2}(t)\sin^2(t)dt\] We get \(I(2n-2)\) and another integral which we will evaluate with the substitutions \(u=\sin(t)\) and \(dv=\cos^{2n-2}(t)\sin(t)dt \Rightarrow v=-{\cos^{2n-1}(t)\over2n-1}\). \[\int_0^{2\pi}\sin(t)\cos^{2n-2}\sin(t)dt =\left[\sin(t)\left(-{\cos^{2n-1}(t)\over2n-1}\right)\right]_0^{2\pi} +\int_0^{2\pi}{\cos^{2n}(t)dt\over2n-1}\] The \(uv\) part is zero so returning to before, we now have \[I(2n)=I(2n-2)-{1\over2n-1}I(2n)\Rightarrow I(2n)={2n-1\over2n}I(2n-2)\] This gives us a clear recurrence relation leading to the following \[I(2n)={2n-1\over2n}\cdot{2n-3\over2n-2}\cdot\ldots\cdot{1\over2}I(0) ={(2n-1)!!\over(2n)!!}\cdot2\pi\] \[={(2n)!\over((2n)!!)^2}\cdot2\pi={(2n)!\over(n!\cdot2^n)^2}\cdot2\pi ={2n\choose n}{2\pi\over4^n}\] Now we return to solving for \(y\) \[y={292\pi\over27}\sum_{k=0}^\infty \left(3\over5\right)^{2k}{2k\choose k}{1\over4^k}-{364\pi\over27}\] The \({2k\choose k}{1\over4^k}\) part of the summation makes this recognizable from the power series of \((1-x^2)^{-1/2}\) which is used with the generalized binomial theorem to derive the power series for \(\arcsin(x)\): \[{1\over\sqrt{1-x^2}}=\sum_{k=0}^\infty{2k\choose k}{x^{2k}\over4^k}\] So this whole summation is just \(\left(1-\left(3\over5\right)^2\right)^{-1/2}\) and we are almost done \[y={292\pi\over27}{1\over\sqrt{1-{9\over25}}}-{364\pi\over27} ={292\pi\over27}\cdot{5\over4}-{364\pi\over27}={365\pi\over27}-{364\pi\over27} =\boxed{\pi\over27}\] That was a lot of work. Now just a small evaluation to get \(x=27\).
First draw a radius and label an angle as \(\theta\).
The red line separates the triangle into 2. Right around the circle center, we have 2 triangles with 2 sides of length 7 meeting there, so the area is \[{1\over2}(7)(7)\sin(\theta)+{1\over2}(7)(7)\sin(\pi-\theta)=49\sin(\theta)\] Now we need \(\sin(\theta)\). We can first find \(\cos(\theta)\) with the law of cosines on the middle triangle with side lengths \(3,7,8\). \[8^2=3^2+7^2-2\cdot3\cdot7\cos(\theta)\Rightarrow\cos(\theta)={-1\over7}\] Now we can find \(\sin(\theta)\) and solve for the area \[49\sin(\theta)=49\sqrt{1-\cos^2(\theta)}=49\sqrt{1-{1\over49}} =49{\sqrt{48}\over\sqrt{49}}=49{4\sqrt{3}\over7}=28\sqrt{3}\] Therefore \(x=28\).
This is a linear system of equations. Rather than use the matrix method, we just need to cancel \(y\) so we can find \(x\). If we multiply the first by 3 and then add, we will cancel \(y\). \[\begin{align}30x+12y&=930\\7x-12y&=143\\\hline37x&=1073\end{align}\] Then \(x=29\). If we wanted, we can also find \(y=5\).
Jo travels 165 miles in 5 hours an arrives a half hour early. What speed should she travel to be right on time?
If 5 hours is half hour early then 5.5 hours is on time. So she should travel 165 miles in 5.5 hours. \[{165\text{mi}\over5.5\text{h}}=30\text{mi/h}\]