Solving the Overlapping Squares Puzzle

[Dafoe]

I'm kind of lost on this one. Could someone please help me.



Two equally big squares with the sides 12 cm partly covers each other as the figure shows. One of the squares corner is in the other squares center. Decide the area of the shadowed part.

http://g.imagehost.org/view/0352/Jobbigare_3

 

[CompuChip]

First let's assume that the rectangles have the same orientation, i.e. the sides of the second one intersect the sides of the first one at right angles. Can you find the answer then?

Once you have that, you can consider the case as drawn above. I suggest adding two auxiliary triangles, by drawing lines perpendicular to the sides of the "first" rectangles sides and going through the top two vertices of the other one.

 

[Dafoe]

Any other way to solve the problem?

 

[HallsofIvy]

You want something harder? Why? I can't imagine anything simpler than the way CompuChip suggested.

 

[DaveC426913]

You want something harder? Why? I can't imagine anything simpler than the way CompuChip suggested.

I haven't seen this problem before but following CompuChip's idea, the answer practically jumps off the page just by imagining the lines.

 

[CompuChip]

But if you insist on something harder, you could try parametrizing the squares in Cartesian coordinates (hint: if you want it really hard, use polar coordinates) calculate the overlap and integrate.

 

[Dafoe]

But if you insist on something harder, you could try parametrizing the squares in Cartesian coordinates (hint: if you want it really hard, use polar coordinates) calculate the overlap and integrate.

how do you do that ?

 

[HallsofIvy]

Apparently, you have not understood anything any of the four responses to your question have said.

Or are you honestly saying, "Yes, I could do it with simple geometry, but I would much rather use calculus"?

Okay, choose your coordinate system so that the center of the first square is (0,0), the x-axis is horizontal and the y-axis is vertical. Taking 1 cm as the unit, the sides of the first square are given by x= 6, x= -6, y= 6, and y= -6. If the angle that side of the second square, going up and to the right from (0,0), is \theta, then the equation of that line is y= tan(\theta)x, the equation of the side opposite it is y= tan(\theta)x- 12 and the eqations of the other two sides are y= -cot(\theta)x and y= -cot(\theta)(x-12). You could find the area of that by dividing it into two parts with a vertical line where the side of the second square cuts the base of the first and integrating each of those.<br /> <br /> Okay? That&#039;s about the hardest way I can think of doing that.<br /> <br /> If you really did not understand CompuChip&#039;s first response, draw vertical and horizontal lines from the center of square one to the right and bottom sides. What can you say about the two right triangles formed?

 

[Capt. McCoy]

Do you still want a more difficult solution?go back to compuchip's first solution...for most of us here, we'd consider that a solution on how to get to the other side of the road...where as for you it's similar to an average person trying to understand the concept of relativity...

compuchip's second solution for you is like an average person comprehending quantum physics...

hallsofivy's solution four you though is like an average person looking for an equation on the randomness of the universe...

so one suggestion...

TRY TO UNDERSTAND COMPUCHIP'S FIRST SOLUTION...

 

[matheinste]

Would it be correct to say that if we assume, as I suppose we must,that there is enough information in the question to arrive at an answer, the orientation of the second square is arbitrary and so we can take any orientation we want. In that case the answer is immediately obvious but may be too easy.

Matheinste.

 

[DaveC426913]

Would it be correct to say that if we assume, as I suppose we must,that there is enough information in the question to arrive at an answer, the orientation of the second square is arbitrary and so we can take any orientation we want. In that case the answer is immediately obvious but may be too easy.

Matheinste.

We don't have to assume. There is enough information. Compuchip's solution is about as simple as it gets.

I'm not sure what the OP dislikes about it.

 

[matheinste]

We don't have to assume. There is enough information. Compuchip's solution is about as simple as it gets.

Of course you are correct. I just like the idea that lack of given specific values in a problem is often in itself suggestive of an answer. Such as in this case you are free to orient the second square as suggested by Compuchip because if the orientation had a bearing on the answer you would need more information.

Matheinste

 

[HallsofIvy]

It is clear that the orientation does NOT matter because if you drop a perpendicular from the center of the first square (corner of the second square) to the bottom edge and draw a perpendicular from the center of the first square to the right edge, you get two congruent triangles. One is inside the "overlap" and the other is outside it. Whatever area you lose one one side by rotating, you gain on the other- the area of the "overlap" is the same no matter what the angle is.

(I confess I didn't see that until after I read Compuchip's post.)

Another problem that uses the concept you mention, "that lack of given specific values in a problem is often in itself suggestive of an answer" is this:

You drill a cylindrical hole of length L all the way through a sphere. What is the volume of the material left?

This problem does not give the radius of either the sphere or the cylindrical hole. If you assume those radii are not relevant, then the problem is easy: imagine a hole of infinitesmal radius. Now we must have L= 2R so R= L/2 and the volume of the original sphere is (4/3)\pi R^3= (1/2)\pi L^3. And, of course, NO material is lost so the answer to "What is the volume of the material left?" is (1/2)\pi L^3 no matter what the radii are.

It is an interesting exercise in integration to show that answer is correct. (The integration comes in calculating the volume of the "caps" that are removed at each end of the cylindrical hole.)

 

[DaveC426913]

It is clear that the orientation does NOT matter because if you drop a perpendicular from the center of the first square (corner of the second square) to the bottom edge and draw a perpendicular from the center of the first square to the right edge, you get two congruent triangles. One is inside the "overlap" and the other is outside it. Whatever area you lose one one side by rotating, you gain on the other- the area of the "overlap" is the same no matter what the angle is.

(I confess I didn't see that until after I read Compuchip's post.)

Another problem that uses the concept you mention, "that lack of given specific values in a problem is often in itself suggestive of an answer" is this:

You drill a cylindrical hole of length L all the way through a sphere. What is the volume of the material left?

This problem does not give the radius of either the sphere or the cylindrical hole. If you assume those radii are not relevant, then the problem is easy: imagine a hole of infinitesmal radius. Now we must have L= 2R so R= L/2 and the volume of the original sphere is (4/3)\pi R^3= (1/2)\pi L^3. And, of course, NO material is lost so the answer to "What is the volume of the material left?" is (1/2)\pi L^3 no matter what the radii are.

It is an interesting exercise in integration to show that answer is correct. (The integration comes in calculating the volume of the "caps" that are removed at each end of the cylindrical hole.)

It has been 28 years since I was posed this very question by a romantic rival. I never got the answer (though I did get the girl.)

...

But now that I look at your explanation, I see that it is not the answer I was expecting. You have simply provided the algebraic formula to find the answer. The answer is still dependent on plugging in real variables.

 

[DrGreg]

Prof ofIvy's post reminds me of a similar one about an annulus, i.e. the region between two concentric circles. If the longest straight line you can draw within an annulus has length L, what's the area of the annulus? If you assume there is a unique answer, you can just make the inner radius zero.

However in this case you don't need to have studied calculus to verify the answer really is unique.