This was an old one. I think a solution was found but then lost. Yeah, it sounds stupid :)

Here, we received a claim that a solution was found again. It should not be terribly hard. Enjoy...

Here, we received a claim that a solution was found again. It should not be terribly hard. Enjoy...

Hey, thanks for reposting!

ReplyDeleteI bet the solution is lame now compared to some of the other problems posted, but here is what I found:

Let's call the tip of the angle O. The point where the upper segment meets the upper horizontal line is A and the one on the bottom is B.

Also, let's introduce the parameter Alpha as the angle between the center horizontal line and [OA]. This is the parameter that varies, and there is a specific value of Alpha that gives us the minimum value of x+y.

Now let's trace a vertical line passing by O.

If we take the symmetrical line of [OB] with respect to the vertical line, we get a segment [OB']. We can state that OB = OB', meaning y=OB'.

So, x+y = x+OB'.

when Alpha changes, the angle between [OA] and [OB'] changes. When that angle is 180 degrees - A, O, B' aligned - the distance x+OB' will be minimal (Straight path distance).

So the problem consists of finding the value of Alpha for which A, O and B' are aligned.

Under these conditions, by studying the angles, we get that Alpha = Theta/2.

We can also easily show that x=a/sin(Alpha) and y=b/sin(Alpha).

We get: min(x+y)=(a+b)/sin(Theta/2)

---------------------------

Using cosinus theorem:

ReplyDeletesqrt(x^2+y^2-2xycos(theta)) >= a+b

(x+y) >= sqrt((a+b)^2+2xy(cos(theta)+1))

If theta goes to 180, cos(theta)=-1, x+y get the minimum value which is a+b.

It is not the answer, but if we know while theta=180, x+y is equal to a+b, do we need write a function with parameter theta?

The solution of 'Anonimous' can't be right, because Fox 7 is a special case of this problem with a = 1, b = 2 and Theta = 90 degrees.

ReplyDeleteThe correct answer to Fox 7 is (1+2^(2/3))^(3/2).

And that's NOT the same result as I get with his formula...

:-(

Hey guys,

ReplyDeleteHenkie is right. I realize now where my (not-so-smart) mistake was: when Alpha changes, O is fixed but both A and B' change, so the straight line argument does not hold since the ends of the segments are moving.

I spent some time trying to transform the problem to a more approachable one geometrically, but so far, the best I got is prolonging the angle's legs to intercept the opposite horizonal lines. x+y will be proportional to the sum of the two full lines, which are now the diagonals of a trapeze.

After that, all of a sudden, I decided to switch for a frontal assault by simply writing the trigonometric equality then differentiating in order to find the value of Alpha. Except that the equation is too difficult to be solved for a general Theta...

Any ideas? Have I completely lost it?

Thanks!

PS. I did not quite understand what Contact said...

If the angle between x and the middle parallel line is A:

ReplyDeletex = a/sin(theta) and y = b/sin(theta-A)

x+y = asin(theta)cos(A) - asin(A)cos(theta) + bsin(A)

The minimum is a stationary point where d(x+y)/dA = 0.

d(x+y)/dA = -asin(theta)sin(A) - acos(A)cos(theta) + bcos(A) = 0

b - a(cos(theta)cos(A) + sin(theta)sin(A)) = 0

cos(A-theta ) = b/a

A = cos^-1(b/a) - theta

x + y = asin(theta)cos(cos^-1(b/a) - theta) - acos(theta)sin(cos^-1(b/a) - theta) + bsin(cos^-1(b/a) - theta)

3rd line is wrong.

ReplyDelete(6th line was also wrong by the way)

another anon

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ReplyDelete