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Problem:

A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.

Which area is greater?

a diagram showing overlapped squares, one forming a smaller blue square and the other an irregular red quadrilateral

Let the area of each large square be exactly $1$ unit squared. Then, the area of the blue square is exactly $1/4$ units squared. The same would apply to the red area if you were to rotate the square $k\cdot 45$ degrees for a natural number $k$.

Thus, I am assuming that no area is greater, and that it is a trick question $-$ although the red area might appear to be greater than the blue area, they are still the same: $1/4$.

But how can it be proven?

I know the area of a triangle with a base $b$ and a height $h\perp b$ is $bh\div 2$. Since the area of each square is exactly $1$ unit squared, then each side would also have a length of $1$.

Therefore, the height of the red triangle area is $1/2$, and so $$\text{Red Area} = \frac{b\left(\frac 12\right)}{2} = \frac{b}{4}.$$

According to the diagram, the square has not rotated a complete $45$ degrees, so $b < 1$. It follows, then, that $$\begin{align} \text{Red Area} &< \frac 14 \\ \Leftrightarrow \text{Red Area} &< \text{Blue Area}.\end{align}$$

Assertion:

To conclude, the $\color{blue}{\text{blue}}$ area is greater than the $\color{red}{\text{red}}$ area.

Is this true? If so, is there another way of proving the assertion?


Thanks to users who commented below, I did not take account of the fact that the red area is not a triangle $-$ it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.

This question is very similar to this post.


Source:

The Golden Ratio (why it is so irrational) $-$ Numberphile from $14$:$02$.

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    $\begingroup$ i think you can tile the red area 4 times to get the entire square $\endgroup$
    – gt6989b
    May 11, 2018 at 4:25
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    $\begingroup$ Hint: the sum of the two red sides that don't touch the center is $1$. $\endgroup$
    – dxiv
    May 11, 2018 at 4:29
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    $\begingroup$ @user477343 Glad the hint helped. You can make that into a full-fledged answer, and I'll +1 it. $\endgroup$
    – dxiv
    May 11, 2018 at 4:31
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    $\begingroup$ Is this a problem from "Brilliant" $\endgroup$ May 11, 2018 at 4:32
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    $\begingroup$ Note that purely from exam technique alone, the answer is likely to be "they are the same size". Indeed, the problem has not told you by how much the rotation occurs, and why privilege a rotation of $0$ over a rotation of some greater angle? This is not a proof; but the phrasing of the question has told you what answer to look for. (This is a more general point than your "it works this way for 45 degrees": this is a demonstration that no mathematical reasoning at all is required to exam-technique that the answer is "they're the same".) $\endgroup$ May 11, 2018 at 20:24

10 Answers 10

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some text

The four numbered areas are congruent.


[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.

enter image description here

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    $\begingroup$ This answer also deserves the tick for artistic reasons. $\endgroup$
    – BenM
    May 11, 2018 at 6:09
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    $\begingroup$ A great example of "proof by picture" that actually works. $\endgroup$
    – user133249
    May 11, 2018 at 14:46
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    $\begingroup$ This is not the same as the answer by Ross and Zoltan. I like this one better. Theirs was the first that came to my mind, too. $\endgroup$
    – Carsten S
    May 11, 2018 at 23:09
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    $\begingroup$ Can Wolfram Alpha draw that? $\endgroup$
    – Willtech
    May 12, 2018 at 2:44
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    $\begingroup$ @FrankShmrank The original poster asked within the question how it can be proven that the red area equals 1/4 (which would settle the title question). My answer makes it clear (without a formal proof, but in proof-by-picture, that’s par for the course) that the red area is one of four congruent areas that partition the unit square, so its area is 1/4. I agree my answer is less than a complete proof of the original question, but I think (and I guess many upvoters think) that it’s convincing. There are other excellent answers that are more traditionally proof-like, so upvote your favorite(s). $\endgroup$
    – Steve Kass
    May 19, 2018 at 17:14
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I think sketching the two identical triangles marked with green below makes this rather intuitive. This could also be turned into a formal proof quite easily.

Identical triangles

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    $\begingroup$ This method is similar to @RossMillikan 's answer above, but not quite the same :) I have to wait $9$ hours before I can upvote as I have reached my daily limit... but when I can, $$(+1)$$ $\endgroup$
    – Mr Pie
    May 11, 2018 at 15:00
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    $\begingroup$ It's not only similar, now that I read that solution, it's actually the exact same idea. Unfortunatly that answer didn't contain any images and I just looked at the images before posting my own answer. :) $\endgroup$
    – Zoltan
    May 11, 2018 at 15:05
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    $\begingroup$ Well congratulations on your first answer on the MSE! Yours is still a good answer :)) $\endgroup$
    – Mr Pie
    May 11, 2018 at 15:08
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    $\begingroup$ This is the clearest image to understand. +1 $\endgroup$
    – qwr
    May 12, 2018 at 20:36
  • $\begingroup$ @qwr Indeed! If only I could grab this answer and drag it below the accepted answer. That way, nobody would have to scroll all the way down to see this. It is my own answer that should probably be at the very bottom :) $\endgroup$
    – Mr Pie
    May 13, 2018 at 1:52
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enter image description here

Note that for equal angles $\angle A'OB' = \angle AOB = 90^\circ$, when we subtract a common part $\angle A'OB$ from both sides, we have $\angle AOA' = \angle BOB'$, so the red and cyan triangles are congruent: $\triangle AOA' \cong \triangle BOB'$.

That implies their areas are equal, and when we add a common part $\triangle A'OB$ we get area of the $AOB$ triangle equal to the area of the $A'OB'B$ quadrilateral. Finally, the area of the two squares' common part is constant, independent on the square's rotation angle.

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    $\begingroup$ Shouldn't be $\angle AOA' = \angle BOB'$? $\endgroup$
    – Pedro
    May 13, 2018 at 3:59
  • $\begingroup$ @Petro Right, thank you. $\endgroup$
    – CiaPan
    May 13, 2018 at 7:10
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    $\begingroup$ Do you mean to say that $\Delta AOA' \color{red}{\cong} \Delta BOB'$? $\endgroup$
    – Mr Pie
    May 14, 2018 at 3:08
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    $\begingroup$ This is the way I saw it $\endgroup$ May 15, 2018 at 2:59
  • $\begingroup$ @MrPie Yup, fixed. (Better late than later.) Thank you. $\endgroup$
    – CiaPan
    Jul 21, 2020 at 13:02
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The two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent.

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    $\begingroup$ I see what you mean. There was no need to describe the result when drawing the vertical and horizontal lines that define the blue area on the diagram of the red area $-$ it was clear as day that they would be equal after looking at the newly formed triangles! I like your method of showing they were equal :) $$(+1)$$ $\endgroup$
    – Mr Pie
    May 11, 2018 at 5:29
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By pinning a square's vertex to the center of the other, you guarantee a 90 degree slice outwards. This means we could tile 4 slices perfectly. A square has rotational symmetry of n=4. Since the rotation number is an integer multiple of the slice number, the area is invariant of rotation. You can apply this generally as well. A 120 degree slice of an equilateral triangle will be invariant. A 60 degree slice of a uniform hexagon will too. 120 degrees will work for the hexagon as well since that's 3 slices on a rotation number of 6.

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    $\begingroup$ FWIW I like this answer the best. It is a simple, brief proof that uses clear logic instead of math. $\endgroup$
    – Bohemian
    May 12, 2018 at 14:28
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    $\begingroup$ @Bohemian, the reasoning is of course maths. $\endgroup$
    – Carsten S
    May 12, 2018 at 23:24
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    $\begingroup$ @carsten but it’s basic geometry, without any calculations, arithmetic or formulae, such that someone without any mathematical know-how could follow. It’s only barely maths (and I’m not in the mood to play semantics) $\endgroup$
    – Bohemian
    May 13, 2018 at 8:37
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    $\begingroup$ @Bohemian: Whatever mood you are in, this is very much a mathematical answer. Looking for ideas like this will help you find solutions when manipulating formulæ gets you stuck. $\endgroup$
    – PJTraill
    May 13, 2018 at 21:39
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    $\begingroup$ @Bohemian, I may be a bit touchy on this subject, I hope I did not come across as rude. It is just that a recognize a misconception of what is mathematical in this, even though you may not hold it. It reminds me of beginners asking questions on how they can make their perfectly fine argument "more mathematical", by which they mean that they feel that they should use formulas. $\endgroup$
    – Carsten S
    May 14, 2018 at 15:37
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Let $f(\alpha)$ be the length of the segment from the center of the square to the outside of the square on the line at an angle of $\alpha$ degrees from the horizontal line pointing right.

Suppose that the first side of the square (in counterclockwise order) makes an angle of $\alpha$, then area you want is $\int\limits_{\alpha}^{\alpha+\frac{\pi}{2}} \frac{f(x)^2}{2} dx$ and since $f$ is periodic with period $\frac{\pi}{2}$ this is independent of $\alpha$.

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    $\begingroup$ This is much too advanced for my skill level. $$(+1)$$ $\endgroup$
    – Mr Pie
    May 11, 2018 at 4:35
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    $\begingroup$ in hindsight the other approach is better, but looking at it from the calculus point of view probably wont hurt :) $\endgroup$
    – Asinomás
    May 11, 2018 at 4:36
  • $\begingroup$ I am a high school student who is familiar with integrals and radians... but the statement, "$f$ is periodic," I don't know what that means. Is it ok if you could explain to me? Other than that, your answer is great! Thanks :) $\endgroup$
    – Mr Pie
    May 11, 2018 at 4:38
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    $\begingroup$ @AHB yeah I do. In my opinion, it is an act of kindness, especially when one has at least $-3$ downvotes or lower. However, I let the user know what I believe is (or might be) wrong with their question as if I did put a downvote. Also, I think there are some badges earnt when using all the upvotes in one day or something like that, idk for sure. I have only ever downvoted $1$ post, only to earn a badge of my first donwvote. $\endgroup$
    – Mr Pie
    May 12, 2018 at 10:03
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    $\begingroup$ @user477343 Yup, two actually, both bronze, 'Suffrage: 30 votes in a day', 'Vox Populi: all 40'. $\endgroup$
    – Artemis
    May 19, 2018 at 13:19
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$\hspace{5cm}$enter image description here

$$b^2+b^2=(a-c)^2+c^2 \Rightarrow \frac{b^2}{2}=\frac{(a-c)^2+c^2}{4}\\ S=\frac{b^2}{2}+\frac{(a-c)c}{2}=\frac{(a-c)^2+c^2+2(a-c)c}{4}=\frac{a^2}{4}.$$

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    $\begingroup$ The number of ways one can work this out is amazing!! Also, your answer is pure math(s)! However, I have to wait $1$ hour before I can upvote as I have reached my daily voting limit. $$(+1)$$ $\endgroup$
    – Mr Pie
    May 16, 2018 at 22:47
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    $\begingroup$ Could you elaborate on how you can assert that the 2 bs are actually equal to each other? $\endgroup$ May 19, 2018 at 14:52
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    $\begingroup$ @FrankShmrank, such problems help intuitive thinking and imagination. If the lower square turns clockwise, its top two sides will turn to the same angle with respect to their original positions and the sides $b$-$b$ will increase equally. $\endgroup$
    – farruhota
    May 19, 2018 at 16:21
  • $\begingroup$ @FrankShmrank I had the same problem, and then I deleted my comment and put up a new one (namely, my current one above) because I found out that by looking at the accepted answer, if the four triangles are congruent, then the two $b$s are equal to each other :) $\endgroup$
    – Mr Pie
    May 20, 2018 at 8:03
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Solution:

Although the red area is not a triangle, the sum of its sides that do not touch the centre is equal to $1$. This can only mean that no matter how many degrees the square is rotated, no area will be greater; the red area will always be equal to the blue area, i.e. $$\frac 14$$

Credit to @dxiv who pointed this out as a hint in a comment!

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    $\begingroup$ This is similar to Captain Morgan’s answer,but I find it less clearly expressed than that. $\endgroup$
    – PJTraill
    May 13, 2018 at 21:42
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    $\begingroup$ @PJTraill Yes, you are correct $-$ Captain Morgan has a much better answer :) $\endgroup$
    – Mr Pie
    May 13, 2018 at 21:49
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enter image description here

If we use $\overline{FB}$ for the base of $\triangle FEB$, then its altitude is $\frac 12s$. If we use $\overline{BG}$ for the length of the base of $\triangle BEG$, then its altitude is $\frac 12s$.

So the area of $\square FBGE$ is $\frac 12(\frac 12s)(s-x) + \frac 12(\frac 12s)(x) = \frac 14s^2$. Which is one-fourth of the area of the square.

The advantage of this method is that it allows you to break the square into $n$ pieces of equal area quite easily. In fact, the same method applies to an regular polygon.

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  • $\begingroup$ Thank you for letting me know. The diagram was very clear, and this method is very much applicable to the problem. $$(+1)$$ (in at least $19$ hours $-$ I have reached my daily voting limit). Also, how did you construct the picture? $\endgroup$
    – Mr Pie
    May 16, 2018 at 4:03
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    $\begingroup$ I used GeoGebra. $\endgroup$ May 16, 2018 at 4:48
  • $\begingroup$ Thank you, again, for telling me :) $\endgroup$
    – Mr Pie
    May 16, 2018 at 6:55
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    $\begingroup$ This is by far the best answer as it doesn't assert anything that isn't given. All the other answers assert things that aren't necessarily known. It's sad that the chosen answer was chosen because that user had the lowest rep. $\endgroup$ May 19, 2018 at 14:55
  • $\begingroup$ @FrankShmrank all the answers are great, imho. $\endgroup$
    – Mr Pie
    May 20, 2018 at 8:02
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The two areas are equal. On the diagram with the red area draw the vertical and horizontal lines that define the blue area. The red area has a triangular region added to the left of the blue area and a triangular region above and to the right removed from the blue area. Those two triangles are congruent.

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