Length of the diagonal of a rectangle inscribed in a square

Question

I've seen this problem on a website, and have been struggling with it for the past week. Here's the problem

http://www.cemc.uwaterloo.ca/resources/potw/2016-17/English/POTWD-16-MT-NA-23-P.pdf

I tried to solve this problem by listing down all the equations including the pro-numerals that I could think of (e.g. $\overline{NK}^2 + \overline{MN}^2 = \overline{KM}^2$) but there always seems to be more pro-numerals than equations. Can anyone help me?
(P.S. Although this is a Year 9-10 problem, I'm only in Year 7)

Answer 1

Let $x = \overline{AM}$ and $y = \overline {MD}$. You are given $x^2 + y^2 = 50$. The white rectangle has sidelengths $\sqrt 2 x$ and $\sqrt 2 y$ so the length of the diagonal is given by $c^2 = (\sqrt 2 x)^2 + (\sqrt 2 y)^2 = 2 (x^2 + y^2) = 100$.

Thus $\overline{KM} = 10$.

Answer 2

Hint: The only pronumerals you need are the lengths of the sides of the isoceles right-triangles, i.e. the lengths of $AM$ and $MD$. Let $x = \overline{AM}$ and $y = \overline{MD}$. We deduce that $\overline{NM} = \sqrt{2}\,x$ and $\overline{ML} = \sqrt{2}\,y$. From the area given, we may deduce that $$ (x + y)^2 - (\sqrt{2}\,x)(\sqrt{2} \, y) = x^2 + y^2 = 50 $$ (Or, instead of subtracting the area of the rectangle from the area of the large square, you could note that the pairs of isosceles triangles can be combined into squares of side lengths $x$ and $y$ respectively).

Using this information, we are asked to deduce the value of $$ \overline{MK}^2 = (\sqrt{2}\,x)^2 + (\sqrt{2}\,y)^2 $$ Perhaps you could take it from there.

Note: There is not enough information to completely deduce every side-length. We have no way to solve for the value of $x$ in this situation.

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On a side note, I was thinking of the following setup for a "proof without words":

The full "big square" should have side length $2(x + y)$. The area of the red square is $\overline{KM}^2$. I wasn't able to complete the idea, but maybe someone will find this interesting.

Answer 3

By observing,

$BK=BN=DL=MD$

$AN=AM=CK=CL$

$KN=ML$ and $KL=MN$

Area of shaded region $= 50 \text{cm}^2$

$\text{Area of triangle BNK + Area of Triangle AMN + Area of triangle DML + Area of triangle CLK}=50\text{cm}^2$

$\dfrac{1}{2}(BN)(BK)+\dfrac 12 (AN)(AM) + \dfrac 12 (MD)(DL) + \dfrac 12 (CK)(CL)=50 \text{cm}^2$

$(BN)(BK)+(AN)(AM)=50 \text{cm}^2$

$BN^2 + AN^2=50 \text{cm}^2$

From Pythagoras theorem

$BK^2 + BN^2=NK^2$ and $AN^2 + AM^2 =MN^2$

Adding both

$BK^2+BN^2+AN^2+AM^2=NK^2+MN^2$

$50+50=KM^2$

$KM+±10$ (but length cannot be negative)

So $KM=10 \text{cm}$

Answer 4

Inspired by @Omnomnomnom's attempt, here's a picture-proof adapted from Böttcher's dissection proof (#36) of the Pythagorean Theorem:

(Every line shown is parallel or perpendicular to one of $\overline{CD}$, $\overline{CK}$, $\overline{KM}$, $\overline{KL}$, or $\overline{LM}$.)

The two pairs of right triangles from the problem combine to form the right-hand portion of the figure, and the square along $\overline{KM}$ (clearly?) has twice the area of that portion. So, $$|\overline{KM}|^2 = 2\cdot 50 = 100 \qquad\to\qquad |\overline{KM}| = 10$$

(Of course, it's not entirely obvious that the dissection is accurate. As Cut-the-Knot says of Böttcher's result: "I think cracking this proof without words is a good exercise for middle or high school geometry class." I suspect that there's a simpler diagram to make here, perhaps based on one of the many other proofs shown on CtK, but Böttcher's tends to be the one that comes first to my mind.)