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 \(x=1\).
Jun 02
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.
Jun 03
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: \(\sin(2x)=2\sin(x)\cos(x)\)
Jun 05
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 \(5\).
Jun 06
\[y=\sum_{n=0}^\infty{8\over36n^2+36n+5},\quad x=2\left({\pi\over y}\right)^2\]
First, we should focus on the summation to compute \(y\), then \(x\) follows
easily. Begin by decomposing the fraction. We can factor
\(36n^2+36n+5=(6n+1)(6n+5)\).
\[{A \over 6n+1}+{B \over 6n+5}={8 \over 36n^2+36n+5}\]
\[A(6n+5)+B(6n+1)=8 \Rightarrow n(6A+6B)+(5A+B)=8\]
\[6A+6B=0,5A+B=8 \Rightarrow A=2,B=-2\]
Then the summation is
\[2\sum_{n=0}^\infty\left({1\over6n+1}-{1\over6n+5}\right)\]
Now we can use an integration technique to evaluate two individual summations.
Consider the following integral for some parameter \(p\).
\[I(p)=\int_0^1{x^{p-1} dx\over 1-x^6}
=\int_0^1 x^{p-1}\left(1+x^6+x^{12}+x^{18}+\ldots\right)dx\]
\[=\int_0^1\sum_{n=0}^\infty x^{6n+p-1}dx=\sum_{n=0}^\infty\int_0^1 x^{6n+p-1}dx
=\sum_{n=0}^\infty{x^{6n+p}\over 6n+p}\Big|_0^1
=\sum_{n=0}^\infty{1\over 6n+p}\]
So this gives us a way to compute the summations with an integral. Using this,
the summation becomes the following integral:
\[2I(1)-2I(5)=2\int_0^1{dx\over1-x^6}-2\int_0^1{x^4dx\over1-x^6}
=2\int_0^1{(1-x^4)dx\over1-x^6}\]
Proceed by factoring
\[2\int_0^1{(1-x^2)(1+x^2)dx\over(1-x^2)(1+x^2+x^4)}
=2\int_0^1{(1+x^2)dx\over1+x^2+x^4}\]
To make progress, factor the denominator which we will do by using a little
thing that lets us express it as the difference of squares.
\[(1+x^2)^2=1+2x^2+x^4=(1+x^2+x^4)+x^2
\Rightarrow (1+x^2)^2-x^2=1+x^2+x^4\]
This gives us a \(a^2-b^2\) thing to factor the denominator.
\[2\int_0^1{(1+x^2)dx\over(x^2+x+1)(x^2-x+1)}\]
From here, we use partial fractions again. It can get complicated, but it turns
out we can do something a bit simpler that actually works in this case
(constant numerators instead of linear).
\[{A\over x^2+x+1}+{B\over x^2-x+1}={1+x^2\over(x^2+x+1)(x^2-x+1)}\]
\[A(x^2-x+1)+B(x^2+x+1)=1+x^2\]
\[x^2(A+B)+x(-A+B)+(A+B)=x^2+1\]
The conditions are \(-A+B=0\) for the linear part and \(A+B=1\) for the
quadratic and constant parts. So we end up with \(A=B={1\over2}\).
\[\int_0^1\left({1\over x^2+x+1}+{1\over x^2-x+1}\right)dx\]
Next split up the integral and complete squares, so we get \(\arctan\) in the
result.
\[\int_0^1{d(x+{1\over2})\over(x+{1\over2})^2+{3\over4}}
+\int_0^1{d(x-{1\over2})\over(x-{1\over2})^2+{3\over4}}\]
We can use the common \(\arctan\) integral for this with \(a^2={3\over4}\)
\[\int{du\over a^2+x^2}={1\over a}\arctan\left({x\over a}\right)\]
So now we have
\[{2\over\sqrt{3}}\arctan\left(2x+1\over\sqrt{3}\right)\Big|_0^1
+{2\over\sqrt{3}}\arctan\left({2x-1\over\sqrt{3}}\right)\Big|_0^1\]
\[={2\over\sqrt{3}}\left(
\arctan\left(\sqrt{3}\right)-\arctan\left({1\over\sqrt{3}}\right)
+\arctan\left({1\over\sqrt{3}}\right)
-\arctan\left({-1\over\sqrt{3}}\right)\right)\]
\[={2\over\sqrt{3}}\left({\pi\over3}+{\pi\over6}\right)
={2\over\sqrt{3}}{3\pi\over6}={\pi\over\sqrt{3}}\]
Finally, substitute to get \(x\)
\[x=2\left(\pi\over y\right)^2=2\left({\pi\sqrt{3}\over\pi}\right)^2=6\]
One other note about the integral we obtained is that there is a nice symmetry
property, which does not really make evaluating it any easier.
\[I=\int_0^1{(1+x^2)dx\over1+x^2+x^4}\]
Substitute \(u=x^{-1}\) and note the integral bounds swap cancels the negative
from computing \(dx\).
\[I=\int_1^\infty{1+u^{-2}\over1+u^{-2}+u^{-4}}{du\over u^2}
=\int_1^\infty{1+u^{-2}\over u^2+1+u^{-2}}du
=\int_1^\infty{u^2+1\over u^4+u^2+1}du\]
So we obtain the same integral but on the interval \([1,\infty)\) instead. So
\[2I=\int_0^\infty{(1+x^2)dx\over1+x^2+x^4}\]
In order to proceed, we still need to factor the denominator and do something
with completing the square to end up with \(\arctan\) like before, so the rest
of the process looks similar.
Jun 07
Find the smallest \(n\) such that a regular \(n\)-gon cannot be constructed
with a straightedge and compass.
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 \({2\pi\over n}\).
Jun 08
How many pairs of prime numbers sum to \(102\)?
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.
Jun 09
Find the tenths digit of \(\left(17+\sqrt{280}\right)^{17}\).
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.
Jun 10
The \(45^\circ\) angle subtends a circle arc of \(90^\circ\) so if we draw
another line from the center to the top left triangle point, we form a right
triangle. Both of the legs of the new triangle are radii of the circle so it is
a "45-45-90" triangle. From here we can determine that \(x=10\) since the
hypotenuse is \(10\sqrt{2}\).
Jun 11
How many proper divisors does \(265837\) have?
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 \(11\).
Jun 12
This is a quadrilateral so the angles sum to \(360^\circ\).
\[x+100+130+118=360\Rightarrow x=360-100-130-118=12\]
Jun 13
Find the number of Archimedean solids.
These solids are convex polyhedra which have regular polygon faces and are
vertex transitive (isogonal). The solids are listed below.
Truncated tetrahedron
Cuboctahedron
Truncated cube
Truncated octahedron
Rhombicuboctahedron
Truncated cuboctahedron
Snub cube
Icosidodecahedron
Truncated dodecahedron
Truncated icosahedron
Rhombicosidodecahedron
Truncated icosidodecahedron
Snub dodecahedron
There are 13 total. Sometimes it might be counted as 15 because the snub
cube and snub dodecahedron are chiral.