The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.

and no, i am not missing any values

260

are you missing any teeth?

pi

The answer of 260 given by adude23 in post #3 is correct, though I am amazed (and impressed) that anyone could have solved this so quickly, as it is quite a difficult problem overall. Here is an explanation for the mere mortals who may be slightly confused by such an abstract puzzle:

Suppose (for convenience) that the nine squares are numbered from 1 through 9 in order of size, with the smallest square being "1" and the largest square "9". Now, if we assume that square "1" has side length A and square "2" has side length B, it becomes possible to deduce that square "3" has side length A + B. It then follows that square "4" has side length (A + B) + A = 2A + B, square "5" has side length (2A + B) + (A + B) = 3A + 2B, square "6" has side length (3A + 2B) + (2A + B) = 5A + 3B, square "7" has side length (3A + 2B) + (A + B) + B = 4A + 4B, square "8" has side length (4A + 4B) + B = 4A + 5B, and square "9" has side length (5A + 3B) + (2A + B) + A = 8A + 4B.

Apart from the diagram, the only useful information given is that the length and width of the rectangle are natural numbers with no factors in common. Thus, it is very important that we find the rectangle's length-width ratio; if written in simplest form, this ratio will in fact tell us the actual dimensions, and then calculating the perimeter should be trivial.

The length of the rectangle can be looked at as the sum of the side lengths of squares "8" and "9", so in terms of A and B, it is (4A + 5B) + (8A + 4B) = 12A + 9B. The width can be viewed as the sum of the side lengths of squares "7" and "8", which in terms of A and B is (4A + 4B) + (4A + 5B) = 8A + 9B. So, the length-width ratio in terms of A and B is (12A + 9B)/(8A + 9B).

Unfortunately, this doesn't seem to help since there is no way to simplify the expression. However, the very handy property of substitution may just live up to its reputation here - can we express the rectangle's width in some other way? 

Indeed, we can. Suppose the width is looked at as the left dimension instead of the right. This reveals that it is also the sum of the side lengths of squares "6" and "9", or (5A + 3B) + (8A + 4B) = 13A + 7B. Since opposite sides of a rectangle are congruent, this value must equal the right-side width of 8A + 9B. So, 13A + 7B = 8A + 9B. Solving for B, we get B = 5A/2.  And NOW we can substitute 5A/2 in for B in the expression for length-width ratio. This gives [12A + 9(5A/2)]/[8A + 9(5A/2)] = (69A/2)/(61A/2) = 69/61.

At this point, it can be concluded from the given information that the rectangle has a length of 69 and a width of 61, because these are positive integers with no factors in common (meaning "relatively prime"). Thus, the perimeter is simply 2(69 + 61) = 260.   

 QED  

Lol... but seriously, I hope that helps.

I read on your profile that you are taking AP calculus as a freshmen. That is pretty impressive. If you are like me, then this problem may interest you. It is from this site: http://projecteuler.net/problem=317

The problems there are fun to solve but most require use of a computer and basic programming knowledge to solve. The problem below can be solved by hand with some knowlegde of calculus. A calculator still comes in handy.

A firecracker explodes at a height of 100 m above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of 20 m/s.

We assume that the fragments move without air resistance, in a uniform gravitational field with g=9.81 m/s².

Find the volume (in m³) of the region through which the fragments move before reaching the ground. Give your answer rounded to four decimal places.

^ That's a bit out of my league, but I've given it a shot anyway. I came up with 1,138,644.9758 cubic meters (hope I'm not TOO far off ).

That answer sounds bogus.

There is no way to assume that the squares that make up the rectangle are proportionatly related to each other.

 Yes, there is. I didn't make it altogether clear in my explanation, but the side lengths given in the first paragraph are based on adjacent squares. A chain of logic can be built starting with the 2 smallest squares, whose side lengths fit flush along a side of the 3rd-smallest square. And then, the sides of the 3rd smallest square and smallest square fit flush along a side of the 4th-smallest square... go back and read post #7 again, there's a reason why I numbered the squares and referred to them in that order.

cobra that was a great explanation

@browni, thanks for the problems, although I doubt I could solve them... my calculus skills are incredibly mediocre at best.

 The first digit is correct, and you have the correct number of digits.

Ringwraith: I bet you could solve it after you learn integration, unless you already have. You must have some real math skills to be able to take Calculus as a freshman. Actually can't remember exactly how to solve the problem, but I have the answer if anyone wants to try.

Well, let me try again, then.

Er, let's see here... d = (1/2)at^2, so 100m = (1/2)(9.81m/s/s)t^2  and t^2 = (100/4.905)s^2. This means t is approximately 4.51523641 seconds.

Putting that into v = d/t gives 20m/s = d/4.51523641s, and solving for d yields a distance of approximately 90.30472820 meters. Now, a 2-dimensional graph of the path of one of the particles, assuming we place the vertex at (0,100) to represent the starting point, will have a zero at (90.30472820, 0). Since ax^2 + 100 = y, a(90.30472820)^2 + 100 = 0  and a = -100/(90.30472820)^2 = -0.0122625.  Thus, the equation for the graph is -0.0122625x^2 + 100.

Integrating, we get (-0.0122625/3)x^3 + 100x. Substituting the value of 90.30472820 in for x yields 6,020.31521314 which should be the area under the curve. I'm not sure what to do from here, so I'll have to just wing it and see what happens.

If we place a cone inside the region so that the two figures have the same apex and base, that cone will have a volume of (1/3)(pi)(90.30472820m)^2 (100m) = 853,983.73186267m^3.  But if we take a triangular "slice" of the cone, it will have an area of (1/2)(90.30472820m)(100m) = 4,515.23640986m^2, which must be multiplied by 6,020.31521314/4,515.23640986 to match that of a 2-dimensional "slice" of the larger region. So, let's multiply the volume of the cone by that same ratio and see what happens: 853,983.73186267m^3 (6,020.31521314/4,515.23640986) = 1,138,644.9758m^3

No!! I got the same stupid thing again!  Oh well, I give up :(

And how exactly does this help me count the material to see if I am ahead?

I took advance calculus 30 yrs ago.  Can't apply it much anymore.

Didn't want to mess with kinematic equations, but this is how I think I'd do it anyway.

Kinematic eqs give you length of vertical axis and because you can get time also a vector distance from explosion to the region's surface.  Knowing these two distances the pythagorean then gives you an expression for the horizontal distance form vertical axis to surface.

Do triple integral for volume.  First is simply vertical axis length, 2nd is from 0 to the expression with the Pythagorean and third is from 0 to 2pi.

Does that sound right?

2x + 2(x+1)

After finding the envelope function y=f(x) with kinematics, I went with two single integrals, one of the envelope from 0 to the peak of the longest trajectory (the one associated with a 45-degree "launch" angle). Then integrate that trajectory from its peak to where it hits the ground, and multiply the lot by 2π. Using doubles might have been a little easier although it sounds like you would do the same amount of work I would.

 Hmm, the 45 degree angle would make all angles below it unnecessary to look at, but don't higher angles increase the volume due to height?  Or maybe I misunderstood.

Also I was thinking the volume moved though before the first particle (straight down) hits the ground.  I guess it does say all fragments.  In this case I suppose the shape is cone like.

With the cone model, simple geometry gives ~960,000 m3 and the answer is apparently over 1 million.  Interesting... I guess the correct shape is a "fat" cone.