Let's see how smart /sci/ really is. Information that

smart enough to not do your homework for you

homework board is that way
>>>/hm/

Hvad med du selv laver dine lektier, Rasmus?

>>8665270
Why would you bother bringing trig into it? It's a simple application of Pythagoras.

>>8665192
It looks like too much work, and I haven't been in a maths class since high school so this might be wrong, but basically here's what I'd do:
-Draw plan, O is the origin.
- Write the equation of the top semi-circle of the third circle (containing D) as a function f.
- Write the equation of the tangent line to f in any point.
- Solve for the tangent line should pass from the origin, therefore I'd have the actual equation of the tangent line (expressed by the radius)
-Write equation of second top semi-circle.
-Solve for when the tangent equation equals the equation of the second semi-circle, I'd have two solutions (the x coordinates of A and B).
-Find y coordinates of A and B (calculate their images with the equation of the second circle)
Now that I have A and B coordinates, I'd be able to calculate the distance between A and B (the answer to this riddlele).

>>8665283
Your mom is a simple application of Pythagoras.

>>8665283
elaborate

>>8665295
The (drawn) radius of the 3rd circle is perpendicular to the tangent line OD

>>8665300
Argh. Didn't read the question properly. You want AB, not OD.

Yeah, that might require a little trigonometry,

>>8665311
Your mom required a little trigonometry last night.

[math]\frac{8}{5}\,R[/math]

>>8665336
This is the right answer. But how did you get there? Another right answer:
|AB| = 2r cos(v) where v is the angle pictured by >>8665270 as 180° - [math]\alpha[/math]

Questions like this always make me want to generalise.

Given n touching circles on a line OE, and a line OD tangential to the nth circle, calculate the length of the intersection of that line with the kth circle (k<n).

(Assume unit circles.)

>>8665371
This guy's >>8665284 method works in this case I think.

>>8665371
Using this method: >>8665415
I got [math]\frac{4}{2n-1}\sqrt{n(n-1)-k(k-1)}[/math]

>>8665415

I don't understand what the fuck is going on.

>>8665636
OAC and OBC are both triangles with one side length 3R (OC), one side length R (AC or BC) and one angle [math]\theta[/math].

Using the cosine rule, you then get a quadratic equation, whose two solutions will be the lengths OA and OB.

Subtract one solution from the other to get AB.

>>8665270
I got as far as determining that OD = sqrt(24) and the angle is 11.537 deg. At least I think that's right.

I like that triangle thing in the middle - I'd never think of that.

>>8665576
The next obvious questions are:

For what pairs (n,k) is this a rational number?
(From the original question, clearly (3,2) is one such pair.)

Is it ever an integer?

i got 36.8 degrees for alpha. it shouldn't be an acute angle

>>8665725

Consider the surface n*(n-1) - k*(k-1) = y^2.
(n,k,y) = (0,0,0) is a point on it.
Draw a line through (0,0,0), and find
the other intersection of line and surface.
(n,k,y) = (a*t,b*t,c*t), with a,b,c and t rational.
(a*t)*((a*t)-1) - (b*t)*((b*t)-1) - (c*t)^2 = 0
t=(a-b)/(a^2-b^2-c^2);
n = a*(a-b)/(a^2-b^2-c^2);
k = b*(a-b)/(a^2-b^2-c^2);
y = c*(a-b)/(a^2-b^2-c^2);

If a,b,c are integers with a^2-b^2-c^2 = 1,
and a>b, n,k and y will be integers.

>>8665725
Nice family of solutions is (5n-2,3n-1) which always have length 8/5.
n=1 gives the original question (3,2)
n=2 gives (8,5), which I thought was neat.

>>8665883
This feels more like a moving of the goalposts than an actual solution, since you still have a constraint on the new parameters.

The question now becomes, for what values (a,b,c) does a^2-b^2-c^2 = 1?

Or to put it another way, a^2=b^2+c^2+1?

Like a shifted Pythagoras (>>8665283).

>>8665192
It's trivial.

>>8665914
>>8665725
here are the pairs less than 100
(3, 2)
(7, 3)
(8, 5)
(9, 8)
(13, 4)
(13, 8)
(15, 11)
(18, 11)
(19, 18)
(20, 8)
(21, 5)
(23, 14)
(26, 23)
(27, 23)
(28, 17)
(31, 6)
(33, 13)
(33, 17)
(33, 20)
(33, 32)
(37, 28)
(38, 11)
(38, 23)
(43, 7)
(43, 26)
(43, 38)
(43, 39)
(44, 32)
(45, 20)
(46, 18)
(48, 29)
(49, 33)
(51, 50)
(53, 32)
(55, 46)
(56, 53)
(57, 8)
(58, 35)
(59, 23)
(60, 53)
(62, 14)
(63, 18)
(63, 38)
(63, 59)
(68, 41)
(69, 53)
(72, 28)
(73, 9)
(73, 44)
(73, 53)
(73, 72)
(75, 26)
(77, 68)
(78, 47)
(79, 43)
(80, 68)
(83, 50)
(85, 33)
(85, 60)
(87, 83)
(88, 25)
(88, 53)
(91, 10)
(92, 17)
(93, 29)
(93, 56)
(93, 77)
(93, 88)
(94, 83)
(97, 48)
(98, 38)
(98, 50)
(98, 59)
(98, 95)
(99, 98)

>>8666114
From what I can see the first obvious family of solutions are:

[math](2n^2 + 1, 2n^2) [/math] giving rise to those solutions that are only 1 apart

>>8666114
>>8666121
and further the answer to the question I have posed in this thread

>>8666217

the solution to that gives rise to another family of solutions to the integral equation (generalising the (2n^2 + 1, 2n^2) solution), this is because, we need to find:

[math]m^2 = n(n-1) - k(k-1) = (n-k)(n+k - 1) [/math]

So, let [math]n - k = A^2 [/math], then you just need to make sure the other parenthesis is a square integer, and then you have another family of solutions (though potentially still not exhaustive)

>>8665914
>This feels more like a moving of the goalposts
well, it's a complete characterization of rational solutions, which was one of the questions asked.

Integer solutions are tougher, I think. Making a^2-b^2-c^2=1 would only be a possible way to find some.

>>8665883
plugging these into

A = 4/(2*n-1)*sqrt(n*(n-1)-k*(k-1))

we get

n=(b+1)/(2 b - c^2 + 1)
k=b/(2 b -c^2 + 1)
y=c/(2 b -c^2+1)
A = 4*c/(c^2+1).

A is a positive integer only when c=1, A=2.
Plugging c=1 into k=b/(2 b -c^2 + 1) gives k=1/2

There's no way to make n,k and A positive integers.

>>8666347
Here is an easier proof:

knowing the geometric background to this problem anyway, we know that if the length is going to be an integer, it will either be 1 or 2, since 2 is the maximum integer length going through a unit circle (The diameter), but since it cannot be the diameter, since if it was, it would go through the diameter of every circle, it must equal 1.

Then using some algebra and manipulating things we get the equation

[math]4n^2 - 4n + 1 = 16n(n-1) - 16k(k-1) [/math]

But then

[math]12n(n-1) - 16k(k-1) = 1 [/math]

so

[math]2(6n(n-1) - 8k(k-1)) = 1 [/math]

This is clearly impossible, as the LHS is even and the RHS is odd

Therefore no integral solutions

File: myproof.jpg (195KB, 1290x1600px) Image search: [Google]

195KB, 1290x1600px

>>8665367
Sorry can't bother to put it in LaTeX nor translate to english.

Good luck understanding my handwritting in spanish, I hope the drawings are good enough.

>>8665981
>the solution is trivial and is left to the examiner

man I wish I had the balls to do that on some final.

>>8665725

(n, k) yields a rational number exactly when 2k-1 and 2n-1 are part of a pythagorean triple with 2n-1 being the hypotenuse. The other number in the triple is divisible by 4, call it 4m. Then 2m is the root of (n-k)(n+k-1)

>>8666787

There's this one detail I don't understand:

¿Por qué ostias llamas al principal resultado "Lema 1"?

File: Circles puzzle.png (9KB, 720x540px) Image search: [Google]

9KB, 720x540px

Nightmare Mode

Find AB in terms of the radius r.

File: HELLMODE.png (55KB, 1038x916px) Image search: [Google]

55KB, 1038x916px

>>8667622
>find the sum of all chords

pro tip:
you can't

>>8665193
That's NOT the homework board

>>8667922
lol

ab=2R/29*Sqr( (19sqr(28)+4)^2-12*29^2)

1.9944*R

log2(π*r4)

>>8665192
Let us project D onto OE and call it D'. now create a right triangle with D, D' and the center of the third circle with I will call R. Then in terms of the angle D'DR (which happens to be equal to EDO):
DD'=rcos
OD'=5r-rsin
OD=r^2cos^2+25r^2-10r^2sin+r^2sin^2=r^2(26-10sin)
we also know that sin=rcos/r^2(26-10sin)
26rsin-10rsin^2=cos=sqrt(1-sin^2)
676r^2sin^2-520r^2sin^3+100r^2sin^4=1-sin^2
100r^2sin^4-520r^2sin^3+(676r^2+1)sin^2-1=0
I'm gonna solve this quartic for sin using this random website I found: http://equationsolver.intemodino.com/en/quartic-equation-calculator.html
Let sin=y-1.3
Then y^4+py^2+qy+s=0
p=1-338r^2
q=0
s=7.311616*10^10r^8-176640000r^6
y^4+(1-338r^2)y^2+7.311616*10^10r^8-176640000r^6=0
The solutions to this are:
[eqn] \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} [/eqn]
[eqn] - \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} [/eqn]
[eqn] \sqrt{ \frac{-1+338r^2+ \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} [/eqn]
[eqn] - \sqrt{ \frac{-1+338r^2+ \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} [/eqn]

>>8668700
For these to be real, it sets a general bound on r such that 1-676r^2+114244r^4+706560000r^6-2924640000r^8 must be greater than or equal to 0.
For the first two equations to work, 338r^2 must be greater than or equal to 1+sqrt(1-676r^2+114244r^4+706560000r^6-2924640000r^8). For the second two equations, 338r^2 must be greater than or equal to 1-sqrt(1-676r^2+114244r^4+706560000r^6-2924640000r^8). Plugging in y=sin+1.3 and solving for the angle, we can then find the equation of the line OD:
y=rsin
dy=sindr
x=rcos
dx=cosdr
dy/dx=tan
so the equation of the line is y=tan*x
And the equation of the top half of the second circle is y=sqrt(r^2-(x-3r)^2)=sqrt(-x^2+6rx-8r^2). Setting these two equations equal to each other, we can find the intersection points A and B:
tan*x=sqrt(-x^2+6rx-8r^2)
tan^2*x^2=-x^2+6rx-8r^2
(tan^2+1)x^2-6rx+8r^2=0
x=3r+/-sqrt(9r^2-8r^2(tan^2+1)))
x=r(3+/-sqrt(9-8sec^2))
A and B, in no particular order, are:
r(3+sqrt(9-8sec^2)), tan*r(3+sqrt(9-8sec^2))
r(3-sqrt(9-8sec^2)), tan*r(3-sqrt(9-8sec^2))
The difference between A and B is therefore
(2tan*r*sqrt(9-8sec^2))^2+(2r*sqrt(9-8sec^2)^2
4tan^2r^2(9-8sec^2)+4r^2(9-8sec^2)
4r^2(9-8sec^2)(tan^2+1)
4r^2(9-8sec^2)sec^2

>>8668702
Finally, we can write out the full answer. The distance will be:
[eqn] 4r^2 \Big(9-8sec^2 \Big( sin^{-1} \Big( \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) \Big) *sec^2 \Big( sin^{-1} \Big( \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) [/eqn]
and
[eqn] 4r^2 \Big(9-8sec^2 \Big( sin^{-1} \Big( - \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) \Big) *sec^2 \Big( sin^{-1} \Big( - \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) [/eqn]
For 338r^2 greater than or equal to 1+sqrt(1-676r^2+114244r^4+706560000r^6-2924640000r^8) and 1-676r^2+114244r^4+706560000r^6-2924640000r^8 greater than or equal to 0
and
[eqn] 4r^2 \Big(9-8sec^2 \Big( sin^{-1} \Big( \sqrt{ \frac{-1+338r^2+ \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) \Big) *sec^2 \Big( sin^{-1} \Big( \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) [/eqn]
and
[eqn] 4r^2 \Big(9-8sec^2 \Big( sin^{-1} \Big( - \sqrt{ \frac{-1+338r^2+ \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) \Big) *sec^2 \Big( sin^{-1} \Big( - \sqrt{ \frac{-1+338r^2- \sqrt{1-676r^2+114244r^4+706560000r^6-292464640000r^8}}{2}} -1.3 \Big) \Big) [/eqn]
For 338r^2 greater than or equal to 1-sqrt(1-676r^2+114244r^4+706560000r^6-2924640000r^8) and 1-676r^2+114244r^4+706560000r^6-2924640000r^8 greater than or equal to 0

>>8668682
Half marks, for not simplifying your answer as much as possible.

[math]AB=(\frac{8}{29}\sqrt{2+19\sqrt{7}})r[/math]

>>8668691
No marks. Approximation? Pfft. What is this, an engineering class?

(I'm assuming you were both answering >>8667622, otherwise it's an amazing coincidence.)

>>8665192
literally an old AIME problem, you can find all of these online

any 8th/9th grader who knows power of a point should solve this really quickly