Penultimate POTW!

1. Call the length of the sides of the squares A.

2. Make a right triangle at the top (or bottom) of the parallelogram. One side will equal 6. Call the hypotenuse B and the third side C. We will also call the length of the long side of the full parallelogram D. Then we have these relationships:
Area of parallelogram = height x base = A*B. It will also equal 6*D, which we will use later.
Area of each triangle = 0.5*height x base = 0.5*A*(A-B)

3. Since we know the areas must be equal, A*B = 0.5A*(A-B), which means B = 0.5(A-B) which means 3B = A. This also means that D, the full length of the parallelogram, is SQRT [(2B)^2 + (3B)^2] or D = SQRT(13)*B
4. Since the area of the parallelogram = A*B and A = 3B, it equals 3B^2. It also equals 6*D = 6*SQRT(13)*B. Solving for B gives us B = 2 SQRT(13), and the area of the square is A*A = 3B*3B = 9B^2 = 468. The area of each piece is a third of this, or 156. Checking this against our area formulae in Step 2 shows this is indeed the case; A*B = 156 and 0.5A(A-B) = 156.

eric,

One of us has it wrong (probably me!) but I got a different answer:

Like you, let a be the side of the square, which is also one leg of the right triangle. Let b be the base of the parallelogram, which is also the hypotenuse of the right triangle. Now, the area of the parallelogram is 6b. The area of each of the triangles is equal to this area by construction, so the area of the square is 18b=a^2. The area of the right triangle is a*sqrt(b^2-a^2) = 6b since this area is equal to the area of the parallelogram. Square both sides of the second equation and you get a^2 * (b^2-a^2) = 36*b^2. Substitute 18b for a^2, and you get 18b * (b^2 - 18b) = 36b^2. Factor a "b" out of the left side to get 18b^2 * (b -18) = 36b^2. Divide both sides by b2 to get 18(b-18) = 36. Divide by 18 and get b-18 = 2 or b=20. The area of the parallelogram is thus 20*6 =120, and the area of the square is therefore 3 times this area or 360.

Sean,
I believe the area of the parallelogram is not 6B the way we defined B. Check on that. I could still be wrong about all my other stuff. :)

eric,

I think we are okay with our parallelogram areas. In your construction, you define D as the long side, so the area is 6D. The 6 is the distance between the parallels, which would imply that this is the distance along a line perpendicular to the two parallels. Since that line is perpendicular line connecting the opposite sides, it is an altitude of the parallelogram. The corresponding base would be the long side of the parallelogram. Therefore, in your solution, 6D is correct. In my solution, I define b as the long side of the parallelogram. 6b would then be the correct area.

I am not seeing any mistakes in either of our solutions. Do you suppose it's possible that Jason threw a problem with two valid solutions at us this week?

Perhaps the confusion is coming from the different right triangles we discuss in our solutions. I am adding no lines to the original problem diagram. The right triangles I talk about are the two right triangles that comprise the remaining area of the square that is not part of the parallelogram. These triangles are not only equal in area but congruent. For each, the side of the square is one leg; the long side of the parallelogram is the hypotenuse. If a and b are defined as the side of the square and the long side of the parallelogram respectively, then the other leg is sqrt(b^2-a^2). The area of each of these right triangles is then a*sqrt(b^2-a^2).

Okay, I am officially embarrassed. Eric's solution is right, and mine is not. My logic is sound, but my execution was off. I posted above that the area of each of the right triangles in the original diagram is a*sqrt(b^2-a^2) = 6b. That is, of course, wrong. The area is 1/2 * a * sqrt(b^2-a^2) = 6b which gives a*sqrt(b^2-a^2) = 12b. Proceeding then as before, square both sides to get a^2*(b^2-a^2) = 144b^2. Substitute 18b for a^2 to get 18b *(b^2 - 18b) = 144b^2. That then reduces to 18*(b-18)=144, giving b-18=8 or b = 26. From there, a^2 = 18b so a^2 = 18*26 = 468, exactly the same result as eric obtained.

Ah you are right, we defined B in different ways and your B is my D. Could your problem be here:

The area of the right triangle is a*sqrt(b^2-a^2)

I think its actually half that. A is one non-hypotenuse side of the triangle, Sqrt(B^2-A^2) is the other, so the area of the triangle is thus 0.5*A*Sqrt(B^2-A^2).

468 is the right answer. I arrived at it slightly differently. Divide one of the right triangles into two triangles by drawing a line segment from the right angle perpendicular to the hypotenuse. We know this distance must be 12: the hypotenuse is also the base of a parallelogram with height 6, and for the triangle to have the same area it would need twice the height.

If you call the side of the square M and the other leg of the big right triangles N, we know that MN/2=M(M-N) (area of the triangle on the left, area of the parallelogram on the right). Solve, and we get 3N=2M.

Go back to the first line we drew, which is length 12. Since the two triangles it made by dividing the big triangle are similar, with legs in the 3:2 ratio, it follows that the length of the hypotenuse of the big right triangle is 12*3/2+12*2/3=26. Then 26*6*3=468.

It just occurred to me that if you have two parallelograms of equal height, Q and R, and R has 1/nth the area of Q, then the base of R must be 1/nth the base of Q, too. That would have saved me a lot of effort...

I came at this problem from a different direction. Label the left side of the square A, then the left-hand line segment at the top of the square B, the hypotenuse of the left-side triangle C, and the right-hand line segment at the top of the square D.

Since the area of the parallelogram is (A^2)/3, then C = (A^2)/18 because its height (if C is its base) is 6. So the cosine of the bottom angle of the left-hand triangle is A/((A^2)/18), or 18/A.

If we construct the same little right triangle at the top of the parallelogram as others have, then D is its hypotenuse and one leg is 6, and because it's similar to the big left-hand triangle, then its most-acute angle must have the same cosine as we just calculated, so D = 6/(18/A), or D=A/3. (This is why the shortcut I described at the start of this comment would have been handy.)

Because B+D=A and D=A/3, then B=2A/3.

Since A^2+B^2=C^2, we can substitute and find A^2+(2A/3)^2=((A^2)/18)^2. So 13A^2/9=A^4/324... 13/9=A^2/324... so 468=A^2.

Hey Jason, where's POTW #10?
I have my answer all ready to go. :)