In the figure, showing a square and an equilateral triangle, is $s$ larger or smaller than the radius of the circumscribed circle?

the circumscribed circle

  • $\begingroup$ You can convert the equilateral triangle+square into one big isosceles triangle using basic pythagorean theorem. The base would be the same as the square's base (s), while the diagonal can be calculated (Height is s+√3s/2, side is s/2, calculate the diagonal). Then i believe you can find the radius of a circle based on its inner triangle... I think. I'm too lazy to do the calculations though :P $\endgroup$
    – votbear
    Dec 26, 2017 at 7:30

3 Answers 3


The radius is S. Translate the top point down s units: the distance of this new point is s from all three vertices. Therefore the radius is s.

enter image description here

  • 2
    $\begingroup$ This is a great solution. $\endgroup$ Dec 26, 2017 at 18:57
  • $\begingroup$ "The distance of this new point is s from all three vertices." How did you know that this new point will be exactly "s" distance away from the bottom two points ? $\endgroup$ Jan 1, 2022 at 23:55

There's probably an elegant solution. Meanwhile, here is a bash:

As Votbear says, we should analyze the isosceles triangle formed by the bottom of the square and the tip of the triangle. The common side lengths will be $\sqrt{(s/2)^2+(\frac{2+\sqrt{3}}{2}s)^2}=s\sqrt{(2+\sqrt{3})}$.

Now let use use the formula $R=\frac{abc}{4K}$, where $R$ is the circumradius, $a,b,c$ are the sidelengths, and $K$ is the area. We get $R=\frac{(2+\sqrt{3})s^3}{4(\frac{2+\sqrt{3}}{4})}=s$. Wait what?

As for a possible elegant solution:

Starting with the top vertex and working counterclockwise, name the vertices of the pentagon ABCDE. Reflect B across AC to B'. This creates parallelogram ABCB', so B'A=B'C=s. Also note that B'A || BC || DE, so B'A || DE. But B'A = DE = s, so B'AED is also a parallelogram. So AE = B'D = s. Now we know that B'A = B'C = B'D, so B' is the circumcenter of the triangle ACD, and thus s is the circumradius of the given circle.


Another approach:


Reflection of the right leg of the triangle and right edge of the square about the orange line results in their common vertex being reflected to the centre of the circle, therefore the radius is $s$.


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