Here is an interesting problem..square inscribed in the square?

Question

I ran into this problem while tutoring one of my kids for the SAT math. Basically There is a large square with a smaller square inscribed into it. The problems provides the length of the sides of the larger square but asks for the area of the smaller square.

I solved the problem intuitively by assuming that the points where the smaller square inscribes the larger sqaure must be the midpoint of the larger side of the sqaure. Therefore the sides of the smaller sqaure can be solved using the 1:1:root 2 ratio and thus its area.

But my problem was finding a rigiourous proof using simple euclidean geometry to prove that infact the corners of the smaller square touch at the midpoint of the larger side of the sqaure as I do not have a geometry book handy and I could not find it on line.

Can you guys prove it? or suggest a proof? I hope my presentation of the question is not too confusing!

To CP and Pascal (I hope u would reply!)

U guys are right u can't prove it. I got this problem from SparksNotes SAT. And as far as I remember thats all they said. Even thoough my solution matched theirs, their. reasoning really bothered me. They said that their was a theorom that the smaller square touches the midpoints of the larger square side, which i did not recall.

I was bothered by this all nite and day. And the only thing i could derive, as CP noted, was pythagreans theorom and nothing else. I could not prove that it was the midpoint. .

Although the situation is possible to construct resulting in 4 isocolece right triangles. It is obviously not necessarily the case. Given only the deminisions of the larger square, different triangles and squres, and thus sq. areas can satisfy their conditions.

The book and I both had bad reasoning on that one. THanks guysI was going nuts trying to prove their mysterious theorom.

Answer 1

You can't prove it, because it isn't true. Let A=(0, 0), B=(4, 0), C=(4, 4), and D=(0, 4). Then ABCD is a square. Now let E=(1, 0), F=(4, 1), G=(3, 4), and H=(0, 3). Then EFGH is also a square inscribed in ABCD, but it does not touch the midpoints of the sides of ABCD.

The corollary of this is that it is impossible to uniquely determine the area of a square, knowing only the dimensions of a square in which it is inscribed. Are you sure that the problem did not provide additional information, such as an assertion that some pair of segments was congruent, or the length of one of the segments from a vertex of the smaller square to a vertex of the larger square?

Answer 2

a square inscribed inside a square is a proof of the Pythagorean Theorem. The inscribed square does not necessarily touch the midpoints. Your approach worked because you assumed that with this configuration that the Pythagorean Theorem was true (again it is a proof of it and so it did). You chose a specific configuration and then solved for the area analytically. At that moment, the generality of the squares lost meaning (it is does not have to be the midpoint) and the analyticity was correct because the theorem is correct.

Answer 3

problem-loose! The diameter of the circle could perhaps be the diagonal of the sq.. (Draw a diagram) you do now no longer even choose pi for this: If d = diagonal of a sqaure, then factor s = d/?2. So s = 30/?2 and perimeter P = 4s = 4(30/?2) = one hundred twenty/?2. the section is s² = (30/?2 )² = 900/2 = 450