What is the area of the shaded part, compared to the entire circle?
Am I misunderstanding something or is there not enough information to solve this? Seems like there are too many degrees of freedom. For example, you can make the upper blue segment tiny and shift the bottom blue segment up so that it effectively occupies half of the outer circle (imagine the bottom blue segment is a semi circle with the same diameter as the outer circle). Alternatively, you could make the blue area almost zero by moving the two horizontal chords very close to each other (imagine the blue circles are huge and you're taking tiny segments of them).
The area of the circle is pi*r^2. The flat ends of the shaded circle-halves, if connected to each other, form a square. Half the diagonal of this square is r. The radius of the circle-halves is therefore r/sqrt(2). The shaded area then is pi*(r/sqrt(2))^2, or pi*½*r^2. Dividing the circle area by shaded area, we get that the shaded area is half that of the whole circle.
You are right. There isn't enough information. The two blue shapes should be semi-circles. (I'll change the picture to be more accurate.)
Given that they are semicircles, the answer is half the area of the circle.
You can arrive at this by using the diameter of the circle (2r) is equal to the height of the lower segment (ha) plus the radius of the lower semicircle (a) plus the radius of the upper circle (b) plus the height of the upper segment (hb)
ha = r - sqrt(r^2 - a^2)
hb = r - sqrt(r^2 - b^2)
So:
2r = r - sqrt(r^2 - a^2) + a + b + r - sqrt(r^2 - b^2)
a + b = sqrt(r^2 - a^2) + sqrt(r^2 - b^2)
Then do a lot of tedious algebraic manipulation to solve for r^2. And you arrive at
r^2 = a^2 + b^2
And the area of the two circles is 1/2 pi a^2 + 1/2 pi b^2 = pi/2 (a^2 + b^2) = pi/2 r^2
There is probably a more elegant approach, possibly involving the line connecting the left (or right) edge of both segments, which creates a segment that directly has length a^2 + b^2. But I didn't immediately see an easy way to directly relate this with the radius, and my cat sat down on my paper, so I couldn't continue working.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
r^2 = a^2 + b^2
...
There is probably a more elegant approach, possibly involving the line connecting the left (or right) edge of both segments, which creates a segment that directly has length a^2 + b^2. But I didn't immediately see an easy way to directly relate this with the radius, and my cat sat down on my paper, so I couldn't continue working.
There's a way to see it if you join the left corner of one with the right corner of the other (see diagram below). The line passes through the point of tangency of the two semicircles and is perfectly diagonal, forming a 45-degree angle with the bottom semicircle's flat side as an inscribed angle subtending the left arc (between the left corners). By a well-known inscribed-angle theorem, the central angle subtending the left arc is 90 degrees. Thus the two right triangles formed as below are congruent (one is a rotation of the other). Each triangle has leg lengths of a and b and hypotenuse r, and so r^2 = a^2 + b^2.
Hi. This isn't really a math "challenge", so much as a straightforward math question. I think the answer is some basic linear algebra, but I am ignorant of linear algebra.
https://i.imgur.com/yaFXZzw.png
If anyone can point me to the right vocabulary, I'd appreciate it. More specifically, I'm trying to implement this in Java. I already created the source to produce image #2, and now I am trying to produce image #3 from it.
Thanks.
Let R be a ring (possibly non-commutative and possibly without a multiplicative identity). Show that if x, y, and z are elements of R that satisfy x²+y-x²y = 0 and x² + z - zx² = 0, then y = z.
x2 having a square root in R is a red herring and does not factor into the solution.
zx2y = z(x2y) = (zx2)y = z(x2 + y) = (x2 + z)y = x2 + z + zy = x2 + y + zy. Therefore, y = z.
Hi. This isn't really a math "challenge", so much as a straightforward math question. I think the answer is some basic linear algebra, but I am ignorant of linear algebra.
https://i.imgur.com/yaFXZzw.png
If anyone can point me to the right vocabulary, I'd appreciate it.
I think it's referred to as perspective. I don't study graphics transformations a whole lot though.
Transforming 3D to 2D, in general, is just called projection.
Had to think about this for a while. Obviously, inf A can't be negative, by definition. Suppose inf A equals some positive number c. For any solution w, it is clear that -w is also a solution, so assume that w is positive. Let q = ceil(1/c). Then for any a ∈ ℤ, b ∈ ℕ, a + bw > 0 implies a + bw ≥ 1/q, in other words w > -a/b implies w ≥ (1 - qa)/qb.
Suppose we have that w > -a/b implies w ≥ (1 - ka)/kb for some k ∈ ℕ. If k = 1, then w is equal to (n - a)/b for some n ∈ ℕ. Otherwise, taking a = -(kn - 1)/(k - 1) - ukn and b = vkn, w > (kn - 1)/v(k - 1)kn + u/v implies w ≥ (kn+1 - 1)/v(k - 1)kn+1 + u/v for any n, u, v ∈ ℕ.
By induction, this means that w > -a/b implies that either w equals some ((1 - (k - 1)a)kn - 1)/(k - 1)knb, or w ≥ (1 - (k - 1)a)/(k - 1)b.
Since this is true for k = q, by further induction w must be rational, which is a contradiction.
I'm not sure that I follow your induction step. It looks way too convoluted, though. There are two proofs of this I like. The first is more direct, and the second invokes a topology flavor which is really cool.
A) The point to notice is that if x ∈ A and y ∈ A, with x > y, then x - y ∈ A. So, if the infimum is a positive number c, there are two possibilities. Either c is in A or c is not in A. If c is not in A, then there is an infinite number of elements of A, between c and 2c. Subtracting two of them we get a value smaller than c in A, contradiction.
Then, c is in A. But that means that the minimum distance between two elements of A is c, and that all numbers of the form nc, for n ∈ ℕ, are also in A. It's straightforward to see that the elements of the form nc are the only ones allowed, because if there were another one called x, sitting between nc and (n+1)c, then either x-nc or (n+1)c-x, would give a number smaller than c, contradicting the fact that c is the minimum.
Therefore, if the infinimum is positive, all elements of A are of the form nc, but that means that w is rational. So, inf A = 0
B) To prove it topologically, the key point to notice is that, for an irrational number w, the fractional parts of any multiple nw should be all different. If two of them were the same, we would have nw - mw = k, an integer, and w = k/(n-m) a rational number, which is absurd.
Therefore, the sequence nw-floor(nw) is a subset of A, has an infinite number of distinct elements, and is also bounded, since all elements are between 0 and 1. Therefore, we can apply the https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem and find a convergent subsequence. This means that, for any epsilon, we can find two elements of A x and y such that x - y < epsilon. Since x - y ∈ A, we just proved that there are elements of A which are arbitrarily small, which is enough to prove that inf A = 0.
Ah, should have thought of that! My idea was to apply the condition ¬(0 < a + wb < 1/k) repeatedly to get an ever increasing lower bound on w given an initial lower bound. By taking the limit of the resulting sequence, a new condition ¬(0 < a + wb < 1/(k - 1)) is constructed which is then applied in the same way, and so on, until we get to the point where the sequence diverges.
A) The point to notice is that if x ∈ A and y ∈ A, with x > y, then x - y ∈ A.
Forgot to reply until now, but that's actually a good way to look at this (I thought initially about doing it through continued fractions, but with this it's not necessary). Knowing that x ∈ A and y ∈ A implies |x-y| ∈ A, all we need to do now is use the Euclidean algorithm (least absolute remainder). Since both w (irrational) and 1 are in A, the algorithm never terminates, and you get a sequence of absolute remainders in A that converges to 0.
Anyway, here's a geometric problem that I saw somewhere on the internet:
Two purple squares and a green square are inscribed in a circle as shown in the image above. The top corner of the green square has the same y-coordinate as the top right corner of the upper purple square.
How many times more is the area of one of the purple squares compared to the area of the green square?
(IIRC it can be done geometrically, without algebra.)
I did it using algebra. I've never been good at geometric constructions. I didn't do the calculations myself, and used Wolfram Alpha instead, but the trigonometric manipulations seem reasonable.
First, transform the problem into triangles. Let's call the points where the green square meets the circle, taken clockwise, as A and B, they are followed by the two points where a purple square meets the circle, also clockwise, which we call C and D. We denote by O the center of the circumference.
Notice that the angle ABC has 135 degrees. That means the larger arc AC should be 270 degrees, which means that the angle AOC has 90 degrees. Besides that, since the angle BCD has 90 degrees, the points B, O and D are collinear.
From this, we call the angle AOB 2x (the factor of two will be useful). The angles BOC and COD are 90-2x and 90+2x, respectively. We are looking for a configuration where BC - AB/sqrt(2) = CD/2, which after trigonometry yields the equation:
2sin(pi/4-x) - sqrt(2)*sin(x) = sin(pi/4+x)
From here, we can probably simply open everything and solve an equation involving sin(x) and cos(x). There's a trick to solve it by forcing it to look like an equation of the form sin(x+phi) = something, which I don't remember the name. In my case, I simply plugged it on WA, which returns the answer x = arctan(1/5)
The area of the purple square over the green one is simply the square of the ratio of their sides, which is just sin(pi/4+x)/2*sin(x). Substituting x=arctan(1/5) in this, we get 3/sqrt(2). And squaring, we get that the purple square is 9/2 larger than the green one.
EDIT: In one of the math groups I participate, someone posted a Twitter post with the geometrical solution to this problem, which is pretty simple. If no one finds it, I can post it here.
The edge shared by the purple squares perpendicularly bisects a chord of the circle, and is thus a diameter.
Draw the perpendicular bisector of the northwest green edge, which is also a chord of the circle. This perpendicular bisector is also a diameter.
Note that the green and purple diameters meet at a 45° angle at the center of the circle.
There are any number of ways to proceed from here, but you might, for example, imagine translating the top purple square left until its bottom-right corner coincides with the center of the circle.
It is then easy to see that Green-East trisects its purple edge, and the desired result quickly follows.
That problem is from a Riddler post back on Mar 19. Their posted solution doesn't actually say too much. As pretty obviously there are a hundred ways to do this question.
I initially did it algebraically (as p4wn3r did, but without using trigonometry). After finding the answer, I tried to find a way to do it geometrically, and came up with the following:
The red line below is a diameter of the circle (since it's subtended by a 90° angle) and so the extension of the lower left edge of the green square is coincident with the bottom-right corner of the bottom square. So it goes through the top-left corner of the bottom square (bottom left corner of the top square), as shown below:
So the diameter of the green square is two-thirds the side length of the top purple square. To complete the problem geometrically, divide the green and purple square as follows:
This divides the purple square into 18 triangles and the green square into 4 triangles. So the ratio is 18/4 = 9/2.
Prove that no right triangle with integer side lengths can have a leg whose length is a multiple of the other.
Curiously, there is a proof involving only geometric construction.
HINT: Take the proposed triangle and fold the larger leg into the hypotenuse, one of the triangles created is similar to the larger one. What's going on with it?
Prove that no right triangle with integer side lengths can have a leg whose length is a multiple of the other.
Curiously, there is a proof involving only geometric construction.
HINT: Take the proposed triangle and fold the larger leg into the hypotenuse, one of the triangles created is similar to the larger one. What's going on with it?
I didn't expect something like this would have a geometric construction (it's obvious algebraically). But after seeing your hint, I can actually see now what it's like:
Here we have the triangle ABC with the hypotenuse horizontal. B is folded down onto the point D as shown, and it can be seen that ADE is similar to ABC.
Assume that ABC satisfies the property (AB, BC and AC are integers, with BC a multiple of AB).
BC=DC, so AD = BC-DC is an integer. By similarity, DE is a multiple of AD, and so DE is an integer. Also, BE=DE, so AE = AB-DE is an integer. So ADE is a smaller triangle satisfying the property. By infinite descent, this gives us an absurdity.
----
Here's a question that I was thinking some time ago but didn't post until now.
Let c be any real number greater than 1. Prove the following integral:
Ex: For c=2, the LHS is just the limit of tan-1(x) as x goes to infinity. That is pi/2, which is what the RHS works out to be. But how do we do the other values of c?
Here's a question that I was thinking some time ago but didn't post until now.
Let c be any real number greater than 1. Prove the following integral:
Ex: For c=2, the LHS is just the limit of tan-1(x) as x goes to infinity. That is pi/2, which is what the RHS works out to be. But how do we do the other values of c?
Contour integration works. Sorry for not drawing, I'm not that good at graphing software :P
First, make the substitution x=exp(z). The integrand becomes exp(z)dz/(1+exp(cz)), and the integration limits are from -infinity to +infinity.
Now, taken as a complex function, this has poles at exp(cz)=-1 => cz = (2n+1)*i*pi => z = (2n+1)*i*pi/c
Now we set up a contour to evaluate the integral. The contour goes from -infinity to +infinity at the line Im(z)=0 (A), and from +infinity to -infinity at the line Im(z)=2pi/c (B).
We call I the integral on segment (A). The integral on segment (B) is simply -I*exp(2*i*pi/c). Summing the two, we only have the contribution from the pole at z = i*pi/c. The residue is -exp(i*pi/c)/c.
Using the residue theorem, we have
I(1-exp(2*i*pi/c)) = -2pi*i*exp(i*pi/c)/c
I(exp(i*pi/c)-exp(-i*pi/c))/2i = pi/c
I*sin(pi/c) = pi/c => I = (pi/c)/(sin(pi/c))
EDIT: Another solution:
Set u=xc, dx = u^(1/c-1)*dx/c, we get the integral of u^(1/c-1)/c*(1+u) from 0 to infinity.
From beta function identities, notice that this is simply B(1/c,1-1/c)/c. In the wikipedia page, it can also be seen that B(1/c,1-1/c) = pi/sin(pi/c). Substituting this gives the answer.
Now we set up a contour to evaluate the integral. The contour goes from -infinity to +infinity at the line Im(z)=0 (A), and from +infinity to -infinity at the line Im(z)=2pi/c (B).
It sounds like the left and right sides of your contour are missing. I'm thinking of a rectangle like this, as t goes to infinity:
* On the left side, |ez/(1+ecz)| ≤ e-t/(1-e-ct) which approaches 0/(1-0) = 0, and so the integral (which is finite-length) approaches 0.
* On the right side, |ez/(1+ecz)| ≤ et/(ect-1) which, after using L'Hopital's Rule, approaches limit of e(1-c)t/c, which goes to 0 since c>1, and so the integral (which is finite-length) approaches 0.
The last point is important because, without it, nothing in the argument prevents it from working for (almost) all c, so you would end up with wrong conclusions like: when c=2/3, the integral is -3pi/2 (in fact, when c=2/3, the integral diverges to infinity).
I noticed that your substitution rewrites the function to avoid having any branch cuts. I did the contour integral differently, using the original function without substitution, but having a branch cut where the argument of the complex number is a negative number arbitrarily close to 0 (as shown in blue below). So it looks something like this, as r goes to 0 and R goes to infinity (here I used c=sqrt(21) to generate the poles, but it works in general for any c>1):
However, I am not aware of a way to do this without using contour integrals (and without using the beta function). I asked this question partly because I was wondering whether anyone had a way to evaluate this without using complex numbers or any of the non-elementary functions.
It sounds like the left and right sides of your contour are missing. I'm thinking of a rectangle like this, as t goes to infinity:
* On the left side, |ez/(1+ecz)| ≤ e-t/(1-e-ct) which approaches 0/(1-0) = 0, and so the integral (which is finite-length) approaches 0.
* On the right side, |ez/(1+ecz)| ≤ et/(ect-1) which, after using L'Hopital's Rule, approaches limit of e(1-c)t/c, which goes to 0 since c>1, and so the integral (which is finite-length) approaches 0.
There's another way to see this. Notice the asymptotic behavior of the integral as it goes to +infty and -infty on the real axis. At +infty, the exp(cz) term in the denominator dominates, and it falls off exponentially for c>1. At -infty, the 1 dominates, and the integral also falls off exponentially, because of the exp(z) term in the numerator.
If you have exponential decay, even if the contour length grows linearly, you can conclude it goes to zero. In a math test, where you need to be as rigorous as possible, you certainly have to mention this. However, I did mostly quantum field theory, which has LOTS of contour integrals, and it's almost impossible to be rigorous in all of them without wasting any time, so you end up doing these things intuitively :)
Two purple squares and a green square are inscribed in a circle as shown in the image above. The top corner of the green square has the same y-coordinate as the top right corner of the upper purple square.
How many times more is the area of one of the purple squares compared to the area of the green square?
(IIRC it can be done geometrically, without algebra.)
So the diameter of the green square is two-thirds the side length of the top purple square. To complete the problem geometrically, divide the green and purple square as follows:
This divides the purple square into 18 triangles and the green square into 4 triangles. So the ratio is 18/4 = 9/2.
Here we have the triangle ABC with the hypotenuse horizontal. B is folded down onto the point D as shown, and it can be seen that ADE is similar to ABC.
Assume that ABC satisfies the property (AB, BC and AC are integers, with BC a multiple of AB).
BC=DC, so AD = BC-DC is an integer. By similarity, DE is a multiple of AD, and so DE is an integer. Also, BE=DE, so AE = AB-DE is an integer. So ADE is a smaller triangle satisfying the property. By infinite descent, this gives us an absurdity.
----
Here's a question that I was thinking some time ago but didn't post until now.
Let c be any real number greater than 1. Prove the following integral:
Ex: For c=2, the LHS is just the limit of tan-1(x) as x goes to infinity. That is pi/2, which is what the RHS works out to be. But how do we do the other values of c?
* On the left side, |ez/(1+ecz)| ≤ e-t/(1-e-ct) which approaches 0/(1-0) = 0, and so the integral (which is finite-length) approaches 0.
* On the right side, |ez/(1+ecz)| ≤ et/(ect-1) which, after using L'Hopital's Rule, approaches limit of e(1-c)t/c, which goes to 0 since c>1, and so the integral (which is finite-length) approaches 0.
The last point is important because, without it, nothing in the argument prevents it from working for (almost) all c, so you would end up with wrong conclusions like: when c=2/3, the integral is -3pi/2 (in fact, when c=2/3, the integral diverges to infinity).
I noticed that your substitution rewrites the function to avoid having any branch cuts. I did the contour integral differently, using the original function without substitution, but having a branch cut where the argument of the complex number is a negative number arbitrarily close to 0 (as shown in blue below). So it looks something like this, as r goes to 0 and R goes to infinity (here I used c=sqrt(21) to generate the poles, but it works in general for any c>1):
However, I am not aware of a way to do this without using contour integrals (and without using the beta function). I asked this question partly because I was wondering whether anyone had a way to evaluate this without using complex numbers or any of the non-elementary functions.
