Puzzle: length of the diagonal
November 4, 2007 pm30 9:01 pm
A student walked into tutoring with a problem she picked up outside of school. The tutor had seen it before, and called me over. The answer wasn’t so tough – but we were looking for a cute solution. See what you find.
A rectangle is inscribed in a square. It is oriented 45∘in relation to the square, so that any two consecutive vertices of the rectangle, with the included vertex of the square, describe an isosceles triangle. The sum of the areas of these four isosceles triangles is 200. Find the length of the diagonal of the rectangle.
Can you find the diagonal? Can you find a cute method?
(The diagram is from a wonderful false proof. Click and play with it, if you have a moment)
7 Comments
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JD2718 
Hi,
I’ve solved this 2 ways but there’s not enough room in the margin of this page to include them :)
I get 20 as the diagonal (both ways!)
Let a=DO=DP.
Let b=OC=CN.
Let D=diagonal=PN
Let c=OP=sqrt(a^2+a^2)=sqrt(2a^2)
Let d=ON=sqrt(b^2+b^2)=sqrt(2b^2)
The area of the 4 isosceles triangles = a^2+b^2 = 200.
Then D^2=c^2+d^2=2a^2+2b^2=2(a^2+b^2)=2*200=400
So, D=20
The other way I’ll post in a little while as I’ve got to run right now.
That was pretty much my thinking.
POM is a right-angled triangle. Pythagoras’ Theorem states that the area of a square constructed on OM is equal to the sums of the areas of squares constructed on PO and PM.
The area of a square constructed on PO is 4 times the area of PDO, or 2 times the sum of the areas of PDO and MNB.
Similarly, the area of a square constructed on PM is 2 times the sum of the areas of PAM and ONC.
But the areas of PDO, MNB, PAM and ONC together is 200 square length units. Therefore the area of a square constructed on OM is twice this (400 square length units), so the length of OM is 20 length units.
The 2nd way, probably the one you thought of as “cute” is this:
Let a=DO=DP.
Let b=PA=AM.
Let D=diagonal=OM.
The trick is to notice that AM+MR=DO=a and since AM=b then MR=a-b.
So, a-b is one side of triangle OMR.
The second side is OR=DA=DP+DA=a+b.
The hypotenuse of the triangle, OM=D, is the side we’re trying to determine.
So, Pythagoras rescues us: MR^2+OR^2=OM^2, or
(a-b)^2+(a+b)^2=D^2.
Multiply things out:
a^2-2ab+b^2 + a^2+2ab+b^2 = D^2.
This simplifies to 2a^2+2b^2 = D^2 or 2(a^2+b^2) = D^2.
And, as in the first approach, a^2+b^2 is the area of the 4 triangles which is given as 200. So, we replace a^2+b^2 with 200 above and we get D^2 = 2(200) and D=20.
Sol
I also did it Sol’s first way. I don’t think it’s cute, but it’s straight-forward enough…
My cutesy way is to consider extreme cases.
Let the width and length of the rectangle become equal. Now we have a square inscribed in a square and the math is simple.
Consider the other extreme, when the length goes to 0. Now the “rectangle” (degenerate as it may be) is the diagonal of the square with area 200.
And the two answers are (of course) the same.
The cheat here is that I assumed that the answer was independent of the dimensions of the rectangle, but the problem is sort of phrased that way.
Jonathan
That is cute. And I think it is a reasonable assumption that the dimensions don’t impact the answer, since no dimensions are specified, and it is implied that there is a single answer.
It ir worth to think about it some time when it will be more free time… I hope I will not forget about this!