# Area of a part of a square

This question appears in the anime series Steins;Gate.

Is the value of $A$ uniquely determined by the three other areas, and the fact that the eight segments on the edges have the same length? If so, how can $A$ be determined?

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You should be able to solve this – Krazer May 16 '14 at 15:33
Unless I'm mistaken, all you can do is guess. If those marks are supposed to mean that all sides are the same length, then without actually giving a single length or angle all you can do is measure it if it's drawn to scale or guess that the sides are 5cm or so which makes A 32 cm squared. – Qiri May 16 '14 at 15:44
If we can assume that the angles are approximately right, we can see that $L^2 \in (16,32)$ (using Krazer's notation). If we assume only integer solutions to the lengths are permissible, then this means $L=5$. But since this isn't stated in the problem explicitly, it's hard to see if this is what is intended, and I'm not sure if A is unique or solvable without such assumptions. – Maroon May 16 '14 at 20:17
@Maroon We can do a little better than that, with $L^2\in (18,26)$. It seems to me that $A$ should be unique, since choosing the central point provides two degrees of freedom while we are given three values, but I can't prove it off the top of my head. – Alex Becker May 16 '14 at 20:20
@AlexBecker, it is unique, but my argument is hard to explain in words. Draw the inner square with edges at the midpoints of the original square. This new triangle is divided into four triangles with one common center point. Three of the areas must be known if we know the dimensions of the original square and given the three areas specified. Triangles of constant area require the vertex in the center to remain on a line parallel to the edge of the inner square. Thus two areas specified specify all areas. – nayrb May 16 '14 at 20:29

This problem can be solved by showing that the opposite areas in the diagram must add to the same value. This is proved below. From this we only need to solve $$20 \; + \;?\; = 32 + 16$$ and the missing area is $28$.

Proposition: $\boldsymbol{[HDEX] + [FBGX] = [EAFX] + [GCHX]}$.

Proof. It suffices to show that $$[HDEX] + [FBGX] = \frac12 [ABCD]$$ We reason as follows. \begin{align*} \frac12 [ABCD] &= [HDEY] + [FBGY] \\ &= [HDEX] + [EYX] + [XYH] + [FBGY] \\ \end{align*} Since $\triangle EYX$ and $\triangle GYX$ each have half of the base of $\triangle EGX$, they have the same area. Similarly $\triangle XYH$ and $\triangle XYF$ have the same area. So the above

\begin{align*} &= [HDEX] + [GYX] + [XYF] + [FBGY] \\ &= [HDEX] + [FBGX] \\ \end{align*}

as desired.

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Excellent! $-e^{i\pi}$. – evil999man May 17 '14 at 7:14
@Awesome I'm partial to $+e^{i \tau}$ – Michael Tong May 20 '14 at 23:39
I have no idea what $\tau$ means but my best bet is $\tau\in 2n\pi$ – evil999man May 21 '14 at 5:42

Let's assume that the figure is a square like this one:

Let $I$ and $H$ points of sides $AD$ and $AC$ such that $IKHA$ is a rectangle.

Let $IG =x$ and $HF=y$, we get three equations: $$w^2 - \frac{(x+y)}{2}w=16$$ $$w^2 + \frac{(x-y)}{2}w=20$$ $$w^2 + \frac{(x+y)}{2}w=32$$ Solving this system we get: $$w=2 \sqrt{6}$$ And the missing area is: $$A=28$$

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How do you get those equations? – Jack M May 17 '14 at 1:17
@JackM The area of each quadrilateral can be calculated as the area of a rectangle plus or minus the area of a triangle. For example the area of $GKFA$ is the area of $IKHA$ plus the area of $\triangle KHF$ plus the area of $\triangle KGI$. – RicardoCruz May 17 '14 at 2:47
Wait, am I misinterpreting $x$ and $y$? I thought they represented $IK$ and $HF$ respectively. – Jack M May 17 '14 at 4:42
@JackM I think $x = IG$ and $y = HF$. – 6005 May 17 '14 at 5:21
@JackM See my edit. I explained a bit more. – RicardoCruz May 17 '14 at 12:30

Let's name the areas $A_1, A_2, A_3, A_4$ in counterclockwise order from the upper right corner, so that $A_4$ is the unknown area, and let $d$ be the segment length, so the total area of the square is $(2d)^2 = 4d^2$. Let $P$ be the distinguished point in the middle of the square. We obviously have $$A_1+A_2+A_3+A_4=4d^2,$$ which doesn't help much. But we can do better than this. Connect the line segments from the midpoints of the square, making a smaller square with half the area of the original.

The resulting smaller square has sides of length $\sqrt 2 d$, and the four areas cut out by the same lines from $P$ are given by $A_i-\frac{d^2}{2}$. We can express these areas using the triangle area formula, with the bases given by the sides of the smaller square: Drop perpendiculars from $P$ to the sides of the smaller square, and denote the resulting heights by $h_1, h_2, h_3,$ and $h_4$ respectively. Then $A_i-\frac{d^2}{2} = \frac{1}{2}(\sqrt 2 d)(h_i)$. Now here's the key point: looking at the geometric picture, you can observe that $h_1+h_3 = h_2+h_4 = \sqrt 2 d$, so putting that together, we get $$A_1-\frac{d^2}{2} + A_3-\frac{d^2}{2} = \frac{1}{2}(\sqrt 2 d)(h_1+h_3) = \frac{1}{2}(\sqrt 2 d)(\sqrt 2 d) = d^2,$$ or $$A_1+A_3=2d^2.$$ Similarly we get $A_2+A_4=2d^2$. This allows us to solve for both $A_4$ and $d$.

Note that these equations still make sense if $P$ is not inside the interior square; we just have to give signed values for the $h_i$.

Here's an image to illustrate:

Either my image editing skills suck, or the original image was not to scale. Either way, hopefully this makes it clearer. The figure should be a square inside the larger square, so the sides should be perpendicular and equal in length to each other. What I've labeled is the four heights $h_1, h_2, h_3,$ and $h_4$. It would have gotten quite crowded if I had labeled the areas as well, but they're in the same respective regions. If someone else wants to offer a better image, please feel free!

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Would the downvoter please care to explain? – Dustan Levenstein May 16 '14 at 23:30

First, I assumed this was really a square, and not a square with rounded corners. I drew the diagram on a sheet of graph paper and labeled the || segment as having length of x, so the true center of the square is at (x, x).

Then I labeled the meeting point of the internal segments as (x-b, x-a), or a units below the center and b units to the left of center.

So we have $\tag{1} \label{1}$: \begin{align} \\ 20 & = (x-b)(x) + \frac12(x-b)(a) + \frac12(b)(x+a) \\ & = x^2 - bx + \frac12ax - \frac12ab + \frac12ab + \frac12bx \\ & = x^2 + \frac12ax - \frac12bx \end{align}

Similarly $\tag{2} \label{2}$: \begin{align} 32 & = (x+b)(x+a) - \frac12(b)(x+a) - \frac12(a)(x+b) \\ & = x^2 + ax + bx + ab - \frac12bx - \frac12ab - \frac12ax - \frac12ab \\ & = x^2 + \frac12ax + \frac12bx \end{align}

And $\tag{3} \label{3}$: \begin{align} 16 & = (x-a)(x-b) + \frac12(a)(x-b) + \frac12(b)(x-a) \\ & = x^2 - ax - bx + ab + \frac12ax - \frac12ab + \frac12bx - \frac12ab \\ & = x^2 - \frac12ax - \frac12bx \end{align}

Adding equations \ref{2} and \ref{3}, we find that $2x^2 = 48$, so $x = 2\sqrt6$. Subtracting equation \ref{3} from equation \ref{1}, we find that $ax = 4$, so $a = \frac13\sqrt6$. Subtracting equation \ref{1} from equation \ref{2}, we find that $bx = 12$, so $b = \sqrt6$.

Solving for A: \begin{align} A & = (x)(x-a) + \frac12(b)(x-a) + \frac12(a)(x+b) \\ & = x^2 - ax + \frac12bx - \frac12ab + \frac12ax + \frac12ab \\ & = x^2 - \frac12ax + \frac12bx \end{align}

Plugging in the values for $x$, $a$, and $b$ that we found above, we get: $$A = 28$$

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First name the unknown area A and half the length of the square's side n.

The total area is 2n X 2n = 4 (n^2)

The total area is also 16 + 20 + 32 + A

Setting those two equal and simplifying gives n^2 = 17 + (A/4)

Visually we know that 20 < A < 32

Also, A is divisible by 4, which gives two possibilities, 24 and 28

Using A = 24 gives n^2 = 23, which is possible but not likely in this case, as 23 does not have an integer as a square root

Using A = 28 gives n^2 = 25, and that makes everyone happy

As they say here in Maine, "Don't go past happy!"

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