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Given a unit square (blue in the picture), pick a point on one edge and label it A. Label the distance from the nearest corner to A as x. Pick one of the corners opposite A and label it B. Call the remaining edge C. There is a unique square with one corner at A, one corner on edge C and with the remaining two corners forming an edge passing through B. What is the area of the new square in terms of x? The blue square is a unit square. What is the area of the orange square? (I haven't been able to work out a complete answer to this yet.)

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    $\begingroup$ pretty good question. thanks for it. I had a good time solving, putting into latex and so took more time to solve it :P $\endgroup$ – Oray Jul 18 at 12:30
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    $\begingroup$ It is more a mathematical question than a puzzle, unless there is a clever shortcut that solves it without much calculations. $\endgroup$ – Florian F Jul 18 at 14:00
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    $\begingroup$ @FlorianF Yes, and the solutions so far are all tedious algebraic calculations. $\endgroup$ – xnor Jul 18 at 15:04
  • $\begingroup$ Wow. Three very different and clever solutions. I'll wait one day to give others a chance, but it's going to be tough to pick an accepted answer. (I did consider posting this on the math exchange, but it seemed more like a puzzle than a simple question.) $\endgroup$ – Joshua Taylor Jul 18 at 17:43
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    $\begingroup$ @hexomino Not so much solving the quadratic as setting up the equation and doing the algebra. $\endgroup$ – xnor Jul 19 at 1:50
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Here is the answer:

$\frac{(2 - 2 x + x^2 + \sqrt{4x^3 - 3 x^4})}{2}$

More is coming!

Here is our original diagram completed with

the biggest square that I will show it will be square

for sure later;

enter image description here

the length $|DE|$ is our $x$ and let's put some specific angle that we are going to work with in our main square as below;

enter image description here

I call the length of side of the other square as $y$ and as you can see

from the $\alpha$ values, right triangles, and hypotenuse as $y$ all four right triangles in the biggest square is the same triangles. I do not want to get into much detail since it is kinda obvious. ($\Delta {GEI}$,$\Delta {EFC}$,$\Delta {FHK}$ and $\Delta {HGJ}$)

so we now know that;

$ICJK$ is a square and I will call the side length of that biggest square as $z$ from now on.

And Let's zoom where we want to focus and put our known equations;

enter image description here

Method 1

We know that from the figure above;

$tan(\alpha)=\frac{z-1}{x}=\frac{1-x}{z-1+x}$

from here we solve $z$ as;

$z=\frac{\sqrt{4 x - 3 x^2}-x}{2} + 1$

then using the equation below

$(z-1+x)^2+(1-x)^2 =y^2$

changing z value in terms of x later;

$(\frac{\sqrt{4 x - 3 x^2}+x}{2})^2+(1-x)^2 =y^2$

simply we find $y^2$ which is the area of the square we are looking for;

Method 2

$\cos{\alpha }=\frac{z+x-1}{y}$

$\sin{\alpha }=\frac{1-x}{y}$

$\cos{\beta }=\frac{1}{\sqrt{1+x^2}}$

$\sin{\beta }=\frac{x}{\sqrt{1+x^2}}$

and we know something else from sinus;

$\sin({\alpha+\beta})=\sin{\alpha}\cos{\beta}+\sin{\beta}\cos{\alpha}$

and we also know that;

5.

$\sin({\alpha+\beta})=\frac{y}{\sqrt{1+x^2}}$

if we combine these without using $1$ and notated that as $\cos{\alpha }$ alone we get;

$\sin({\alpha+\beta})=\frac{1-x}{y}\frac{1}{\sqrt{1+x^2}}+\cos{\alpha}\frac{x}{\sqrt{1+x^2}}=\frac{y}{\sqrt{1+x^2}}$

as a result;

6.

$\cos{\alpha }=\frac{y^2-1+x}{xy}$

and using 6. and 1. we are going to figure out what is $z$ in terms of $y$ and $x$ as below;

7.

$z=\frac{y^2-x^2-1}{x}+2$

so let's find our y value using these knowns;

$(z-1+x)^2+(1-x)^2 =y^2$

then put our new z value in terms of x and y as in 7;

$(\frac{y^2-x^2-1}{x}+x+1)^2+(1-x)^2 =y^2$

and solve for $y^2$ which is the area of the new square we are trying to find;

$Area=\frac{(2 - 2 x + x^2 + \sqrt{4x^3 - 3 x^4})}{2}$

and z in terms of only x becomes;

$z=\frac{\sqrt{4 x - 3 x^2}-x}{2} + 1$

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    $\begingroup$ Yessss I simplified my answer and it comes to the same as yours. Hurray, we got it! $\endgroup$ – Rand al'Thor Jul 18 at 12:21
  • $\begingroup$ I think you lost an x in your final Area equation, comparing it to your answer from Method 1. $\endgroup$ – Joshua Taylor Jul 18 at 17:53
  • $\begingroup$ @JoshuaTaylor nope, I dont think so, both equations must be equal to each other when simplified. you may try wolfram alpha ;) $\endgroup$ – Oray Jul 18 at 17:58
  • $\begingroup$ Oh, I see. The powers under the square root are different. $\endgroup$ – Joshua Taylor Jul 18 at 18:09
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As this tessellation of the tilted new square neatly matches an overlapped tiling of the unit square, the $w{\small\,\times\,}x$ overlap of unit squares equals the $1 \! - r^2$ difference in the areas of the two types of squares.

\begin{align} w \!\;x ~=~ & 1 - r^2 \\[.5ex] \Longrightarrow~~ x(1{-}w) ~=~ & r^2 \! - y \kern3em (~ y = 1-x ~) \\[-1ex] \\ r^2 ~=~ & y^2 + (1{-}w)^2 \\[.5ex] x^2 r^2 ~=~ & x^2 y^2 + (r^2\!{-}y)^2 \\[.5ex] \Longrightarrow~~ 0 ~=~ & (r^2)^2 - (x^2\!{+}2y)(r^2) + y^2 \!{+} x^2 y^2 \\[-1ex] \\ \textsf{area of new square} ~=~ & r^2 \\[.8ex] ~=~ & x^2\!{+}2y ~\pm\sqrt{ x^4 \!{+} 4 x^2 y {+} 4y^2 {-} 4y^2 \!{-} 4 x^2 y^2 } \over 2 \\[.8ex] ~=~ & x^2\!{-}2x{+}2 ~\pm\sqrt{ x^4 \!{+} 4 x^3 (1 \!{-} x) } \over 2 \\[1.5ex] ~=~ & 1 - x + {x^2 \over 2} \left( 1 + \sqrt{{4\over x}-3} ~ \right) \end{align}

(The “$ \small\pm $” was deduced to be “$ \small + $” by testing the formula on the case where  $ x = \large{1 \over 3} $$ \theta = 45^\circ $$ r = \large{2\sqrt2 \over 3} $  and  $ r^2 = \large{8 \over 9} $.)

Here are some tiling experiments, beginning with the easiest-by-hand 45° case, that led to selecting the straightforward version presented. The 45° case’s symmetry naturally created some fun red herrings.

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  • $\begingroup$ Now that's an answer! $\endgroup$ – Fattie Jul 20 at 13:21
  • $\begingroup$ Very cool answer! $\endgroup$ – justhalf Jul 23 at 1:15
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Firstly let's label everything:

labelled diagram

We have five similar right-angled triangles, which must be enough to get a lot of information about the interrelationships between the quantities $x,y,d,e,f,p,q,r$.

  1. The $p,d,x$ triangle is similar to the $1-d,y-p,r$ triangle. Compare the hypotenuses and longer adjacent sides to get $$d-d^2=py-p^2.$$

  2. Use Pythagoras's theorem in the $p,d,x$ triangle to get $x^2=p^2-d^2$. Combining this with the relation from point 1, we find $$x^2=py-d.$$

  3. The trapezium with parallel sides $d,e$ and perpendicular side $x+(1-x)=1$ has areas $\frac{d+e}{2}$. But it's also three right-angled triangles together, so it has area $\frac{py}{2}+\frac{dx}{2}+\frac{(1-x)e}{2}$. So we have $d+e = py + xd + (1-x)e$. Using $py=x^2+d$ from point 2, we find $d+e=x^2 + d + xd + e - ex$, which gives $0 = x+d-e$ therefore $$d = e-x.$$

  4. The $p,d,x$ triangle is similar to the $y,1-x,e$ triangle. Compare the non-hypotenuse sides to get $x(1-x) = de$. Using the relation from point 3, this means $x(1-x)= (e-x)e$, which gives a quadratic equation for $e$ in terms of $x$. Solving this, and using the fact that $e>x$ to know which root to sue, gives $$e = \frac{x+\sqrt{x^2-4(1)(x^2-x)}}{2}.$$

  5. Finally, use Pythagoras's theorem in the $y,1-x,e$ triangle to get $y^2=e^2+(1-x)^2$. Since $y^2$ is exactly the area of the new square, and using the relation from point 4, we get that the area of the new square is $$\left(\frac{x+\sqrt{4x-3x^2}}{2}\right)^2+(1-x)^2.$$

Simplifying this expression, the final answer is

$$\frac{x^2+(4x-3x^2)+2x\sqrt{4x-3x^2}}{4}+\frac{4-8x+4x^2}{4}$$ $$=\frac{x^2-2x+2+x\sqrt{4x-3x^2}}{2}$$

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I think this matches up with the other two solutions but uses coordinate geometry which is quicker here.

Let $B$ be the origin in the Cartesian plane $(0,0)$ then $A$ is located at the point $(x,1)$.
Let $C$ be located at the point $(1,y)$. I will designate the Cartesian coordinates as $(X,Y)$ so as not to confuse with $(x,y)$ above.
The line $AC$ is given by the equation $(y-1) X + (x-1)Y + (1- xy) = 0$.
This means that the perpendicular distance from $B$ to $AC$ is $\frac{1-xy}{\sqrt{(x-1)^2 + (y-1)^2}}$.
Also the distance between $A$ and $C$ is $\sqrt{(x-1)^2 + (y-1)^2}$.

In this instance these two quantities must be the same so we set them equal and solve for $y$. $$\frac{1-xy}{\sqrt{(x-1)^2 + (y-1)^2}} = \sqrt{(x-1)^2 + (y-1)^2} $$ $$\Rightarrow 1 -xy = (x-1)^2 + y^2 -2y + 1$$ $$\Rightarrow y^2 + (x-2)y +(x-1)^2= 0$$ and this means that $$ y = \frac{(2-x) \pm \sqrt{(x-2)^2 - 4(x-1)^2}}{2}$$ Note here that the plus sign puts $y>1$ so we take the minus sign. Then the area of the square is just $$(x-1)^2 + (y-1)^2 = (x-1)^2 + \left(\frac{x + \sqrt{(x-2)^2 - 4(x-1)^2}}{2}\right)^2 $$

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  • $\begingroup$ Your solution matches up with mine and Oray's, but I still need to read this and understand the method :-) $\endgroup$ – Rand al'Thor Jul 18 at 12:54
  • $\begingroup$ @Randal'Thor, this is the brute force tool in the geometry toolkit :) $\endgroup$ – justhalf Jul 23 at 1:16
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area function and square examples

area $F(x) = (1 - x)^2 + \left({x \over 2} + \sqrt{x - {3 \over 4} x^2} \right)^2$

${d \over dx} F(x) = 0 \implies x = {1 \over 3} \implies F( {1 \over 3} )$ is minimum.

Be $\vartheta$ the angle between AC and the horizontal, then $\vartheta(x) = \arctan \left( { {x \over 2} + \sqrt{x - {3 \over 4} x^2} } \over {1 - x} \right) \implies \vartheta({1 \over 3}) = 45°$

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I tried this problem for fun, but got quite a different answer than the others. Posting it for commentary, and educational purposes.

Definitions:

  • Points A, B and C and length x as described in the problem
  • Point D, diagonally across from B on the original square
  • Point E, top left of the original square
  • Length y (also: AC), side length of the created square
  1. First, I bisected the quadrangle ABCD along the line AC.
  2. The area of the triangle ABC is simple, as both the base and height are y. Therefore, the area is y^2 / 2
  3. The area of ACD is the same base (y), but the height is sqr(2)-y. The diagonal of the original square, minus the height of the other triangle. (y*sqrt(2) - y^2)/2
  4. This means that the area of ABCD is y*sqrt(2)/2
  5. The area of ABD is half the square, minus ADE. In other words, ABD = (1-x)/2, or AD/2
  6. By the same logic, we can say that the area of BCD is CD/2, we just don't quite know what CD is, yet.
  7. Knowing the area of ABCD and ABD, however, BCD = y*sqrt(2)/2 - (1-x)/2. This means that CD = y*sqrt(2) - (1-x)
  8. The Pythagorean triangle ACD is of course y^2 = (1-x)^2+(y*sqrt(2) - (1-x))^2
  9. That's a bit of a garbled mess, but if we work it out (it's just calculation by this point) we find that y = sqrt(2)(1-x)
  10. Finally, we go from a length to a square, so the area is 2(1-x)^2

Mind you, this is clearly wrong, as it implies AD = CD... which in many cases can't possibly be true... I fear I may have made a mistake in calculating the area of ACD, that's the only place where the error would be like this.

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