Consider any right triangle $\triangle ABC$.

We focus on one side, $AC$, and we take the midpoint $E$ of this side. Then, we draw the circle with center in $E$ and passing by $A,C$. If we take the perpendicular to $AC$ passing by $E$, we define a point $I$, where the circle intersects the perpendicular line.

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Now, we can draw the ellipse with focii in $A$ and $C$ and passing by $I$.

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Clearly, we can apply this procedure to both the other sides, obtaining other two ellipses.

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My conjecture is that

The sum of the areas of the ellipses constructed on the two catheti is equal to the area of the ellipse constructed on the hypotenuse.

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This is probably a very well known result, and I already apologize with the experts.

However, in order to prove the conjecture, I used the formula of the area of the ellipse, $S=\pi a b$, where $a$ and $b$ are the lengths of the semi-axes. Although it is easy to prove that, in the case of all our ellipses, one semi-axis is clearly half the side, I am stuck in the attempt to determine the lengths of the other semi-axes, and I suspect however that there should be a very elementary way to prove such claim.

Again, sorry for the naivety, and thank you very much for any help or suggestion for a compact proof.

2 Answers2


If you construct similar shapes on the three edges of a right triangle their areas add up as suggested by the Pythagorean theorem.

Christian Blatter
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    Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however! –  Aug 23 '18 at 08:33

In an ellipse, $a^2 = b^2 + c^2$, where $a$ is the semi-major, $b$ is the semi-minor, and $c$ is half the focal distance. In your case, $b=c$, so that $a=\sqrt{2}b$.

what a disgrace
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