I want to calculate the area of the largest square which can be inscribed in a triangle of sides $a, b, c$ . The "square" which I will refer to, from now on, has all its four vertices on the sides of the triangle, and of course is completely inscribed within the triangle, and by "largest square", with the maximum area of such squares.

I did some research. I found that in the case of right triangles, there seemed to be only two kinds of squares being formed. One with a side of $\frac{ab}{a+b}$ and the other of $\frac{abc}{c^2 + ab}$, $c$ being the hypotenuse.

After some time, I realized that there was a common formula for all triangles. And that is:

$$side = \frac{ch}{c+h}$$ or $$\frac{1}{side} = \frac{1}{c} + \frac{1}{h}$$ where, $h$ is height on base $c$, or $$h = \frac{2\sqrt{(s(s-a)(s-b)(s-c))}}{c}$$

enter image description here

I achieved the above results using similar triangles, and have added a figure, to make the point clear. But that is all, I can prove. I tested for many triangles, and found that the smaller side always yields the larger square, but I am not able to prove it. For example, I can say that the largest square of a $4,7,10$ triangle, has area of approximately $\fraq{16}{3}$. I also want to make another conjecture (with an intuitive proof) that any triangle will give at most three squares.

So I want to ask you, are my calculations, proofs ... correct? Have I got them right? And those predictions? Are they right, and if so, proofs? Conclusively, do we have a sure-fire formula to get the area of the largest square?


2 Answers 2


There is a famous book of Polya ("How to solve it"), in which the problem of inscribing a square in a triangle is treated in a really interesting way, I strongly suggest the reading.

The inscribed square is clearly unique once we choose the triangle side where two vertices of the square lie. If we suppose that the square has two vertices on $AB$ and side $l$, then:

$$l+l\cot A + l\cot B = c,$$


$$ l = \frac{c}{1+\cot A+\cot B} = \frac{2R \sin A \sin B \sin C}{\sin C + \sin A\sin B}=\frac{abc}{2Rc+ab},$$

where $R$ is the circumradius of $ABC$. In order to maximize $l$, you only need to minimize $2Rc+ab = 2R\left(c+\frac{2\Delta}{c}\right)$, or "land" the square on the side whose length is as close as possible to $\sqrt{2\Delta}$, where $\Delta$ is the area of $ABC$.

  • 1
    $\begingroup$ Excellent! Its fantastic new way and quite easy to visualize. However I still have some doubts. How did you convert c/1+cotA+cotB to abc/2Rc+ab, I am not quite able to understand. And it does not completely answer my other points. $\endgroup$
    – Sawarnik
    Oct 19, 2013 at 19:25
  • 2
    $\begingroup$ In order to convert $\frac{c}{1+\cot A+\cot B}$, multiply the numerator and the denominator by $\sin A\sin B$, then use the relation $\sin A\cos B+\sin B\cos A=\sin(A+B)=\sin C$. After that, use the sine theorem: $a=2R\sin A$. For right triangles, $2R=c$, so your predictions for right triangles are right (no pun intended). $\endgroup$ Oct 19, 2013 at 20:20
  • 2
    $\begingroup$ Any triangle gives at most 3 squares since any triangle has three sides, and we have to "land" any inscribed square on a side. For any obtuse triangle there is only one inscribed triangle, "landed" on the longest side (have a look at Polya's construction). The $4-7-10$ triangle is obtuse since $4^2+7^2<10^2$, its area is $\frac{7\sqrt{39}}{4}\approx 11$ and its circumradius is $R=\frac{40}{\sqrt{39}}\approx 6.4$, so the side length of the (only) inscribed square is $\frac{280}{\frac{800}{\sqrt{39}}+28}\approx 1.8$ and the area of the inscribed square is $\approx 3.21733\ldots$ $\endgroup$ Oct 19, 2013 at 20:37
  • $\begingroup$ Its getting clearer, thanks! I have accepted it, although I still need your help. 1.8 does indeed correspond with my formula for the longest side. However, I don't get from where you mark this as the only square. Is there any other special condition on other triangles that I don't know of. $\endgroup$
    – Sawarnik
    Oct 20, 2013 at 12:22
  • 1
    $\begingroup$ Your formulas are related to right triangles, which have one inscribed square on the longest side (side length $\frac{abc}{c^2+ab}$) and one inscribed square sharing a right angle with the triangle (side length $\frac{ab}{a+b}$), so there is nothing strange that 1.8 does not fit your formulas, since the 4-7-10 triangle is not a right triangle, it is an obtuse one, and it is pretty clear that it is not possible to inscribe a square in an obtuse triangle, if we want that 2 vertices of the square lie on one of the shortest sides of the triangle. $\endgroup$ Oct 21, 2013 at 7:57

Say $x$ be the side of the largest square within a triangle with sides $a,b,c$. Say one side of the square is on the side $BC$. So we get a square and 3 triangles. Equating the area of the bigger triangle with the sum of area of the 3 small triangle and the square we get:

$$\frac12 ch = x^2\text{[area of the square]}+ \frac12 x(c - x) \text{[area of two small base triangle]} + \frac12 x( h - x)\text{[area of the upper triangle]}$$

Upon solving:

$$x= \frac{ch}{(c+h)}$$

  • 2
    $\begingroup$ @Sawarnik: you have asked if your calculations are correct. This is a second derivation (verification) of that result. $\endgroup$
    – robjohn
    Mar 11, 2014 at 17:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .