A quadrilateral with sides of length 223, 297, 228 and 296 is inscribed inside a square. None of its angles is right. Find the perimeter of the square.
A quadrilateral with sides of length 223, 297, 228 and 296 is inscribed inside a square. None of its angles is right. Find the perimeter of the square.
Isn't there more than one way for a quadrilateral to be inscribed inside a square? Or do you mean that the square is as small as possible?
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
Besides that non of its angles are right, are there any other specific information about the quadrilateral? For example, is it known to be non-self-intersecting?
Also, this is something I'm not sure about: If no other information is known about the quadrilateral than the length of the sides and that none of its angles are right, is it uniquely defined?
It is also known that the quadrilateral can be inscribed within a square, which to me implies that each of its vertices is on the square. Given that, I'm not certain it matters if the thing is self-intersecting or not; wouldn't the two give the same result?
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Before I say anything else, I must admit I don't have the answer to this problem :P A friend told me about this and I tried a little, but with no success.
Warp: there's at least four ways :) But seriously, the way the problem was formulated seemed to indicate that there's just one solution for the square's perimeter. I could be wrong, though - here's the original problem (in french though).
Randil: we don't officially know it's non-self-intersecting. We do know however that it's incribable in a square (see this), and that its sides are of length 223, 297, 228 and 296 in order (i.e. 223 and 228 are opposite sides of the quadrilateral).
Unless I'm missing something obvious, this seems to be a hard problem.
I could prove that for a quadrilateral to be inscribed into a square, it's necessary that:
1 / (sqrt(2)* cos(a - 45º)) <= d1 / d2 <= sqrt(2)* cos(a - 45º)
Where d1 and d2 are the diagonals of the quadrilateral and a the angle between them. This condition isn't sufficient, however, the intersection point of the diagonals needs to be considered.
Thing is, there are infinitely many quadrilaterals with the given sides, the cyclic one, that simplifies algebra, doesn't satisfy the formula above :(. The problem will only have a unique solution if only one of them satisfies the relation (if we're unlucky, a lot of them do, but again, only one may be inscribable because the condition is not sufficient). I'm a bit rusty with quadrilaterals and don't remember how to efficiently extract its diagonals given the sides.
It might be possible to get something by equaling the area to d1*d2*(sin a)/2 and to Bretschneider's formula. I'll take a deeper look into this later if I have time.
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
Here's a picture I made of the problem:
Below are 5 (non-linear) equations for the five unkowns. Equations 1-4 come from that we know the length of the sides, equation 5 relates the area of the square to the area of the quadrilateral plus the 4 right triangles. The formula for the quadrilateral area can be found on the wikipedia article on this topic.
I don't know if this system can be solved analytically...
Thing is, there are infinitely many quadrilaterals with the given sides, the cyclic one, that simplifies algebra, doesn't satisfy the formula above :(.
Damn, I should really pay more attention to things I do. The cyclic quadrilateral is inscribable in a square. I'll solve this problem for the cyclic given quadrilateral (that is, besides being in a square, it'll also be in a circumference). I'm almost certain that there are infinite possible quadrilaterals for this problem.
Let's go:
juef wrote:
A quadrilateral with sides of length 223, 297, 228 and 296 is inscribed inside a square. None of its angles is right. Find the perimeter of the square.
Lemma 1: If a quadrilateral is inside a square, then 1 / (sqrt(2)* cos(a - 45º)) <= d1 / d2 <sqrt> (d1 - d2 sen a) cos t = d2 cos a sen t => tg t = (d1 - d2 sen a)/d2 cos a
This, of course if a != 90º, in this case, every t would be a solution, as long as d1 = d2.
Now, as I said in the first paragraph, d1 cannot have a larger projection in y. We assure this by limiting t to the interval [ -45º,45º]. Since the same applies for d2 and x, the angle between them has also to be on that interval: -45º <= 90º - a - t <=45º, or 45º - a <= t <= 135º - a.
If we take a an acute angle (0º<=a<90>= -45º and 45º <= 135º-a.
Thus, we need only to constrain a to [45º-a,45º].
Using tg (45º-a) <= tg t <= tg 45º and simplifying, we get the expression in the lemma.
Lemma 2: If the diagonals AC and BD in a quadrilateral satisfy (*) for a!=90º, let P be their intersection point, t = arctg((d1 - d2 sen a)/d2 cos a), u is the largest length among {AP,CP} and v the largest among {BP,DP}. If u*cos t + v cos (a + t) <= d1 cos t and u*sen t + v sen (a + t) <= d1 cos t, then the quadrilateral can be inscribed in a square.
Notice that the 1st lemma could only be used to prove that a quadrilateral doesn't fit into a square. The case where (*) is fulfilled and the diagonals don't fit into the square is when the intersection point is awfully placed. Draw the segments AC and BD. Now, project AP and PB into the rotated x-axis. Note that it's possible that the segments obey (*), but if the projection of PB is in the same direction, the sum of their lengths can be larger than the side of the square, making it impossible to fit them.
Though an algorithm to determine a necessary and sufficient condition is more complex, if we take the largest segments in each line, project them into each axis and verify that their sum is smaller than the square's side in both of them, we can already say the quadrilateral can be inscribed. With these two lemmas in mind, let's go to the actual problem, I'll use Randil's amazing picture:
As I said before, let's assume this quadrilateral is cyclic, prove that it can be inscribed in a square and find the square's side. Starting from the bottommost vertex and going in counterclockwise order, name them A,B,C and D. Let P be the intersection point of diagonals, a the acute angle between them and PB = x.
Since the quadrilateral is cyclic, we can prove using angle inscribed in a circumference that angle DBC = angle DAC. Repeating this process for other angles, we get the similarity relations between triangles:
PAB is similar to PDC and PBC is similar to PAD.
In this way, if PB = x, we find out that PC = 223x/228 , PD = (223*296x)/(228*297), PA = 296x/297. Since x>0, the diagonal BD is larger, by a tiny amount, but it's larger.
For the angle a, let's use Brahmagupta's formula for the area A = AC*BD*sin a/2 = sqrt((s-AB)(s-BC)(s-CD)(s-DA)). By Ptolemy's theorem, AC*BD = AB*CD + DA*BC, so:
sin a = 2sqrt((s-AB)(s-BC)(s-CD)(s-DA))/(AB*CD + DA*BC) ~= 0,964932
Useful later, cos a ~= 0,26249.
Now, let's use Lemma 2 to prove it can be inscribed in a square, first, (*) needs to be true. Take d1 as the largest diagonal, it's equivalent to prove that:
(d1/d2) <sin> (1,97477/1,97470) <= 0,964932 + 0,26249 , which is true.
Now, tan t = ((d1/d2) - sin a)/cos a ~= 0,136. We get sin t ~= 0,13 and cos t ~=0,99
We now take u = x and v = 296x/297, rearranging some terms of the inequality, we need to prove that:
cos t + (296/297)*cos(a+t) <1> 0,134 <= 0,965, which is true.
sin t + (296/297)*sin(a+t) <1> 0,13 + 0,985 <1> BD ~= 372,50
The perimeter of the square is, finally, P = 4 * 372,50 * 0,99 = 1475,1.
Notice that the assumption that the quadrilateral is cyclic was essential here because:
(1) It allows using Brahmagupta's formula.
(2) It allows using Ptolemy's theorem.
(3) It gives useful similarity relations to the lengths of the segments needed to prove the lemmas.
(4) It allows us to find BD.
Without this, it's A LOT of algebrism, which is not fun to do. Also, seeing as the inequalities were easily satisfied, there are probably many more possibilities and this solution is not unique.
EDIT: I seem to have forgotten that no angle is right. Just take pairs of adjacent sides and see that Pythagoras's theorem doesn't hold for the triangle formed by them and their corresponding diagonal.
A little of real analysis
1) Consider a real function f [a,b] -> R , differentiable in all its domain, with f'(a)<f'(b). Prove that, for every k such that f'(a)<k<f'(b), there exists an x in ]a,b[ satisfying f'(x)=k .
2) Let f R -> R be a real function. f is non-decreasing in the interval ]-infinity,a[ for some a in R, and f's image in this interval is bounded from above. Show that the one sided limit LIM f(x) x->a- exists.
3) Consider two real functions f,g R -> R satisfying f(0)=0 , g(0)=1 , f'(x)=g(x) and g'(x) = -f(x) for all x in R.
(a) Show that the following equality holds for all reals f(x)^2 + g(x)^2 = 1
(b) Prove that f(x)=sin x and g(x)=cos x
(c) Conclude that
sin x = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
and
cos x = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
1. Set g(x)=f'(x). Since f(x) was differentiable, g(x) is continuos and we can apply the Intermediate value theorem on g, which proves that such an x exists.
2. I'm gonna try and use the Intermediate value theorem here. Let v be a small positive real number. f(a-v) is inside the intervall ]-Inf,a[. As x->a, the function value will increase (or at least not decrease, since the function was non-decreasing). We also know that f is bounded above, so it's derivate is bounded. Let M the the highest value of the derivate between a-v and a. We know that lim x->a f(x) is between f(a-v) and v*M+f(a-v). Since M is bounded, f(a-v) exists, and we can make v arbitrarily small, this limit exists.
3a) Differentiate both sides with respect to x. This gives:
h(x):=2*f(x)*f'(x)+2*g(x)*g'(x)=2*f(x)*g(x)+2*g(x)*(-f(x))=0 <=> h(x)is a constant c. We now know that f(x)^2+g(x)^2=c. This holds for all x. In particular, for x=0 we get f(0)^2 + g(0)^2 = 0^2 + 1^2 = 1 = c <=> c=1.
3b) We know that f(x)=sin(x) and g(x)=cos(x) will satisfy the differential equations presented. They will also satisty that f(0)=0 and g(0)=1. A system of two linear differential equations with two initial conditions, f(0)=0 and g(0)=1 has a unique solution, and since f(x)=sin(x) and g(x)=cos(x) satisfies this set of diff. equations, it is the unique solution.
3c) Since sin(x) and g(x) are continues, bounded, etc. their power series exist. Set f(x)=a0+a1*x+a2*x^2+... and g(x)=b0+b1*x+b2*x^2+...
f(0)=0 -> a0=0. g(0)=1 -> b0=1.
Let's use that f'(x)=g(x). We get
(1) a1+2*a2*x+3*a3*x^2+...=b0+b1*x+b2*x^2+...
g'(x)=-f(x) gives
(2) b1+2*b2*x+3*b3*x^2+...=-a0-a1*x+a2*x^2+...
Combining (1) and (2) will recursively defined all coefficents, starting with a0=0 and b0=1, showing that the power series presented hold. You can also prove this by induction.
Now I should really get ready for work.
^your proof for part 2 doesn't really work. He didn't even specify that the function was continuous, and it could very well be nowhere differentiable.
Since the image of that interval under f is a set of real numbers that is bounded above, there is a least upper bound. Call it M. Now choose any real number e. Since M is the least upper bound, there must exist some number xe such that f(xe) > M - e. Let de = a - xe, which means xe = a - de. For all x > xe, we have f(x) > M - e. We also have f(x) < M from before. Thus, |f(x) - M| < e for all x > xe in (-Inf, a). This is equivalent to saying that |f(x) - M| < e for all x in (a-de, a).
So, for any e, I can find a de such that |f(x) - M| < e for all x in (a-de, a). By definition, this means LIM f(x)(x->a-) = M, and so the limit exists,
1. Set g(x)=f'(x). Since f(x) was differentiable, g(x) is continuos
Not true, g doesn't need to be continuous. (Students keep writing this on the test...)
EDIT:
I forgot to put a counter example. Consider
f(x) = {x2sin (1/x) , x!=0 ; 0, for x=0}
See that f'(x) = 2 x sin(1/x) - cos (1/x) , x!=0
For x=0, f'(0) = lim x->0 (x2sin (1/x) / x ) = lim x->0 (x sin(1/x)). sin (1/x) is bounded and x tends to zero, so the limit is zero.
However lim x->0 f'(x) doesn't exist, because lim x->0 (2 x sin(1/x))=0 and lim x->0 (cos(1/x)) doesn't exist and f' is not continuous on 0.
The error in 2 was pointed by petrie911.
Also, nice proofs for 3b and 3c ! I did something a little different though, because that's one way to demonstrate that the Taylor series of sin x and cos x converges to those functions for all reals, thus I didn't use them for the proof.
b) Let a(x) = f(x) - sin x and b(x) = g(x) - cos x . Verify, similarly to (a), that a(x)^2 + b(x)^2 = 0 , and then, a(x)=b(x)=0. Implying that f(x)=sin x and g(x) = cos x
c) s(x) the first series and t(x) the second. Just check that s(0)=0, t(0)=1, s'(x)=t(x) and t'(x)=-s(x). By (b), they must be equal to sine and cosine.
Design a planetary system that allows an incoming object to spin 5 laps around one planet, turn somehow and then spin 3 laps around the same planet before breaking free of orbit.
All unmentioned parameters are controllable, normal physical laws apply, no external forces are applied to the system and planets may not overlap at any point during the event.
Design a planetary system that allows an incoming object to spin 5 laps around one planet, turn somehow and then spin 3 laps around the same planet before breaking free of orbit.
Since the n-body problem has no (finite) analytical solution, wouldn't this require effectively using (or developing) an n-body simulation that performs a numerical approximation, and then just trying more or less arbitrary combinations until you stumble across one that gives the desired result?
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
Here's a problem I need help with - I've tried to solve it but always end up in a dead end. Consider the expression
where H_k is the k:th harmonic number. I think this expression converges as n->Inf, but I'm not sure. Some numerical testing show that it seems to converge to something like 1.644... or so. Can you help me find the limit (if it exists) of the expression above?
Through some manipulation of sequences, that expression is equal to...
sum(k=1 to n) (H(n) - H(n-k))/k
Unfortunately, I have no real way to sum this series, and it doesn't obviously diverge. Still, it looks like it'll be easier to work with than the one you posted.
There's nothing I hate more than series. I tried to find some recurrence relations in that sum, but didn't find something useful. It probably has something to do with the Euler-Mascheroni constant.
Randil, are you sure that the limit of this series can be found by elementary means? (i.e, is this a textbook problem or something similar?)
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
p4wn3r wrote:
Randil, are you sure that the limit of this series can be found by elementary means? (i.e, is this a textbook problem or something similar?)
Hehe, no, this expression spawned from a probability problem I thought up myself, so I don't know if the limit can be expressed in elementary function, maybe it can't. But on the other hand, I can't prove that it can't, so who knows. I have tried using the fact that H_k(x)=log(k)+gamma+e_k, but it didn't help much.
petrie911 wrote:
Through some manipulation of sequences, that expression is equal to...
sum(k=1 to n) (H(n) - H(n-k))/k
Yes, I had something like that from start, but I thought that expression was more difficult to work with, hence I divided it into two terms as I did. But on other hand, I haven't had much more luch with the form I presented the expression in, so I don't know which form is best.
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
juef wrote:
Well, at first glance (with the numerical approximation you've given), I'm very tempted to say it converges to zeta(2) = Pi²/6.
Interesting! You seem to be right, otherwise it is a very weird coincidence. If that is indeed the limit of the series, all that's left now is to show why the series converges to this. This gives the problem more hope!
^I'm inclined to agree, seeing as the sequence matches the partial sums of 1/k^2 exactly. I can't manage to prove the sequence transforms into that, but the agreement is perfect for the first 20 terms.
I did manage to prove it converges to pi^2/6 using my above relation and the fact that H(n) = ln(n) + y + d(n), where d(n) < .5/n for all n. Let S(f(k), n) represent the sum of f(k) from k = 1 to n
H(n) - H(n - k) = ln(n) - ln(n-k) + d(n) - d(n-k)
S((.5/n)/k, n) = .5*H(n)/n -> 0 as n -> Inf, so S(d(n)/k,n) -> 0 as n -> Inf.
S((.5/(n-k)/k,n) = .5/n*S(1/k + 1/(n-k), n) = H(n)/n -> 0 as n->Inf.
So, clearly, S(d(n)/k, n) -> 0 and s(d(n-k)/k, n) -> 0 as n->Inf.
Meanwhile, -ln(1-k/n) = sum(i=1 to Inf) (k/n)i/i. So we have...
S(-ln(1-k/n)/k, n) = S((sum(i=1 to Inf)(k/n)i/i)/k, n) = sum(i=1 to Inf) S(ki-1,n)/(i*ni)
Now as n->Inf, S(ki-1,n) -> ni/i. So, for n->Inf, S(-ln(1-k/n)/k, n) -> sum(i=1 to Inf) (ni/i)/(i*ni) = sum(i=1 to Inf)(1/i2) = pi^2/6.
That was incredibly inelegant, but I believe it's valid.
I solved it!
First of all, let's prove that
Let's manipulate the sum a little to find:
Using this recurrence, we can prove our desired equality by mathematical induction.
Now, using Liebniz's expansion:
^It's actually \sum{j=1}{i-1} instead of \sum{j=1}{n-1}
So, your expression is simply the sum of the reciprocals of the first n squares. Euler was the first to prove that this series converges to pi^2/6 .
EDIT: Too late!
Is the set of points inside a unit square larger than the set of points on a unit line?
(Before someone instinctively answers "of course", remember that eg. the set of rational numbers is actually exactly as large as the set of integers. Adding a dimension doesn't necessarily increase the size of an infinite set.)
Below are 5 (non-linear) equations for the five unkowns. Equations 1-4 come from that we know the length of the sides, equation 5 relates the area of the square to the area of the quadrilateral plus the 4 right triangles. The formula for the quadrilateral area can be found on the wikipedia article on this topic.
Let's manipulate the sum a little to find:
Using this recurrence, we can prove our desired equality by mathematical induction.
Now, using Liebniz's expansion:
^It's actually \sum{j=1}{i-1} instead of \sum{j=1}{n-1}
So, your expression is simply the sum of the reciprocals of the first n squares. Euler was the first to prove that this series converges to pi^2/6 .
EDIT: Too late!
