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In a triangle, three identical squares of side lengths 2.8 share a common vertex and are each touching two sides of the triangle. If one of the angles in the triangle is 75 degrees and is opposed to a side of 10.8, then what’s the area of the triangle?

  • 1
    $\begingroup$ To the close-voters: you might want to check how tricky this puzzle is, and how many neat "aha" steps are involved in the solution, before thinking it's a straightforward textbook problem rather than an olympiad-style geometry puzzle. $\endgroup$ Commented May 31, 2020 at 16:03
  • $\begingroup$ @Rand I'm not so familiar with PSE mores, but in general on SE sites closure of a question depends on the question and not on its answers. $\endgroup$
    – msh210
    Commented May 31, 2020 at 16:09
  • $\begingroup$ @msh210 You're right, but see the PSE policy on maths problems vs puzzles. To a SME it's quite obvious that this question is not a straightforward calculation problem and is going to involve some tricks, even if the exact details of the solution itself are far from obvious. $\endgroup$ Commented May 31, 2020 at 16:11

1 Answer 1


Let's label all the angles:

enter image description here

Note that we have: $$a+h+i=180,n+p+q=90,$$ $$a+b+c=90,f+h+j=90,g+i+k=90,$$ $$2d+p=180,2e+q=180,2m+n=180,$$ $$b+d=90,c+e=90,d+f=90,e+g=90,j+m=90,k+m=90,$$


$b=f$, $c=g$, $j=k$.

We're given that one of the angles of the big triangle is 75 degrees, so let's say $a=75$, which means

$b+c=15$, i.e. $f+g=15$, but also $h+i=105$, so $j+k=2(90)-(f+g)-(h+i)=60$. Also $j=k$, so these are $30$ giving $m=60$ and $n=60$. (This diagram is not to scale for sure!) Therefore we have an equilateral triangle, and the middle part of the $10.8$ side has length $2.8$. Also $p+q=30$ and $d+e=165$.

Now draw diagonals of the squares to make mini-triangles in the $h$ and $i$ corners. We have

$j=k=30$, so the angle at the $j$ / $k$ vertex of each mini-triangle is $75$ degrees. That means the two little triangles are similar to the big one, so $f+45=i$ and $g+45=h$. Equivalently $b+45=i$ and $c+45=h$, so we have another little similar triangle from drawing the diagonal of the third square to make a mini-triangle in the $a$ corner.

Now we have the following:

enter image description here

where $X+Y=8$ and (from comparing the two similar triangles) $XY=(2.8\sqrt{2})^2=15.68$. So the numbers $X$ and $Y$ are the roots of the quadratic equation $t^2-8t+15.68=0$, which means they are $$\frac{8\pm\sqrt{64-4(15.68)}}{2}=\frac{8\pm\sqrt{1.28}}{2}=4\pm\sqrt{0.32}=4\pm0.4\sqrt{2}.$$

Now we can use the SAS formula for area to find the area of one of those small triangles:

\begin{align*}\frac{1}{2}ab\sin C&=\frac{1}{2}(2.8\sqrt{2})(4\pm0.4\sqrt{2})\sin(75) \\ &=(5.6\sqrt{2}\pm1.12)(\frac{1+\sqrt{3}}{2\sqrt{2}}) \\ &=(2.8\pm0.28\sqrt{2})(1+\sqrt{3})\end{align*}

Now we can use the cosine rule to find the third side of one of those small triangles:

\begin{align*}&=\sqrt{(2.8\sqrt{2})^2+(4\pm0.4\sqrt{2})^2-2(2.8\sqrt{2})(4\pm0.4\sqrt{2})\cos(75)} \\ &=\sqrt{15.68+(16\pm3.2\sqrt{2}+0.32)-(22.4\sqrt{2}\pm4.48)(\frac{\sqrt{6}-\sqrt{2}}{4})} \\ &=\sqrt{32\pm3.2\sqrt{2}-(11.2\pm1.12\sqrt{2})(\sqrt{3}-1)} \\ &=\sqrt{(10\pm\sqrt{2})(3.2-1.12(\sqrt{3}-1))}\end{align*}

That third side is the one which corresponds to the side $10.8$ on the big triangle, so the ratio of their areas is


and the area of the big triangle

\begin{align*}&=\frac{10.8^2}{(10\pm\sqrt{2})(3.2-1.12(\sqrt{3}-1))}(2.8\pm0.28\sqrt{2})(1+\sqrt{3}) \\ &=\frac{116.64(0.28)(1+\sqrt{3})}{3.2-1.12(\sqrt{3}-1)} \\ &=\frac{32.6592(1+\sqrt{3})}{4.32-1.12\sqrt{3}} \\ &=\frac{32.6592(1+\sqrt{3})(4.32+1.12\sqrt{3})}{18.6624-3(1.2544)} \\ &=\frac{32.6592(7.68+5.44\sqrt{3})}{14.8992} \\ &=\frac{250.822656+177.666048\sqrt{3}}{14.8992} \\ &=\frac{31.352832+22.208256\sqrt{3}}{1.8624} \\ &=\frac{16.3296+11.5668\sqrt{3}}{0.97}.\end{align*}

I do wonder if I've missed a less calculation-heavy method though ... been doing all this by hand except for the last few steps where I reached for a calculator, although the above solution is still exact.

  • $\begingroup$ I think you should be able to use the sum of the squares of the two "third sides" and the sum of the areas of the two small triangles to get to the result with fewer roots $\endgroup$
    – user39583
    Commented May 31, 2020 at 16:59
  • $\begingroup$ @user39583 Well, I've now got an exact figure for the answer at least, but it's not especially pretty. $\endgroup$ Commented May 31, 2020 at 17:19
  • $\begingroup$ @user39583 Oops, good catch. I did some factorisation tricks to simplify the final answer a bit too. $\endgroup$ Commented May 31, 2020 at 17:58
  • $\begingroup$ Looks good now! $\endgroup$
    – user39583
    Commented May 31, 2020 at 18:02
  • $\begingroup$ The answer is around 37.49. $\endgroup$ Commented May 31, 2020 at 18:34

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