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So here is an interesting but frustrating problem I have been working on: Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has one vertex on a vertex of the square. What is the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles?

T I had no idea how to solve it. Any help would be GREATLY appreciated. Thank you![enter image description here]1

TheLeogend
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  • Is your problem with identifying the largest square? Or with calculating its side-length? – Rob Arthan Apr 18 '18 at 23:46
  • I would say both. I have a hunch that the square would be tilted. But I can't figure out if that's true. – TheLeogend Apr 18 '18 at 23:50
  • If "inscribed" requires (as it usually does) that each vertex of the target square lie on an edge, then options would seem to be pretty limited. But maybe the question is abusing the term to mean "the largest square that fits in the space". Does the source of the exercise seem like it's looking for an elaborate optimization argument, or could this be merely a round-about way of asking for the distance between the vertex of the square and the endpoint of the common edge of the triangles? – Blue Apr 19 '18 at 00:05
  • If all the vertices have to be on an edge, wouldn't that make the problem impossible? So I'm guessing it has to be largest circle that fits in the space. – TheLeogend Apr 19 '18 at 00:32
  • Being on a vertex usually counts as being on an edge. So, the square whose diagonal joins the upper-right corner of your big square to the closest triangle vertex would be "inscribed". – Blue Apr 19 '18 at 00:40
  • I got the problem right! Thank you so much Blue! :) – TheLeogend Apr 19 '18 at 01:00
  • This is an AoPS question. Stop cheating. –  Aug 30 '20 at 03:03

1 Answers1

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Start by drawing it out and labeling it like so (sorry, it's not perfect, but good enough): A circle with two equilateral triangles inscribed

First find the side length of the equilateral triangle. $G$ is the midpoint of $\overline{FE}$. Let $\overline{GE} = x$. $\overline{DG} = \overline{GB}$, so using basic trigonometry, $\overline{DG} \land \overline{GB}=x\sqrt{3}$. Thus, $\overline{DB} = 2x\sqrt{3}$. Since $\overline{DB}$ is the diagonal of $ABCD$, it has length $10\sqrt{2}$. Then we can set up the equation $2x\sqrt{3}=10\sqrt{2}$. So the side length of the triangle is $2x = \frac{10\sqrt{2}}{\sqrt{3}}$.

Now look at the diagonal $\overline{AC}$, it is made up of twice the diagonal of the small square plus the side length of the triangle. Let the side length of the small square be $y$: $$\overline{AC} = \overline{AF}+\overline{FE}+\overline{EC}=y\sqrt{2}+\frac{10\sqrt{2}}{\sqrt{3}}+y\sqrt{2}=10\sqrt{2}.$$

Solve for y:

$$y\sqrt{2}+y\sqrt{2}+\frac{10\sqrt{2}\sqrt{3}}{3} = 10\sqrt{2} \\ y\sqrt{2}+y\sqrt{2}+\frac{10}{3}\left(\sqrt{6}\right) = 10\sqrt{2} \\ \frac{3y\sqrt{2}+3y\sqrt{2}+10\sqrt{6}}{3} = 10\sqrt{2} \\ \frac{6y\sqrt{2}+10\sqrt{6}}{3} = 10\sqrt{2} \\ 6y\sqrt{2}+10\sqrt{6} = 30\sqrt{2} \\ 6y\sqrt{2} = 30\sqrt{2}-10\sqrt{6} \\ y = \frac{30\sqrt{2}-10\sqrt{6}}{6\sqrt{2}} \\ y = \frac{\left(30\sqrt{2}-10\sqrt{6}\right)*\sqrt{2}}{12} \\ y = \frac{60-20\sqrt{3}}{12}.$$

Which simplifies to $\frac{5\left(3-\sqrt{3}\right)}{3}.$

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Joseph Eck
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  • Thank you so much! I check the answer and I was right. :) – TheLeogend Apr 19 '18 at 00:58
  • How do you know this is the largest square? – achille hui Apr 19 '18 at 03:33
  • @achillehui The lengths $\overline{AF}$ and $\overline{EC}$ are equal to each other, even though the picture is a bit off and it doesn’t appear so. A square is basically two equilateral triangles of equal size cut in half by the diagonal of the square. Those two spaces are where one can place the largest diagonal without overlapping the triangles when constructing the square – Joseph Eck Apr 19 '18 at 08:33
  • @JosephEck I also believe the circle you have is the largest one. However I cannot find any argument (even a heuristic one) that justify the claim and I wonder whether you have one. – achille hui Apr 19 '18 at 08:52
  • @achillehui This is the only square that fits the conditions in fact. – TheLeogend Apr 19 '18 at 13:47