In this picture,

Rectangle triangle inside a equilateral one

$ABC$ is an equilateral triangle, whereas $ABP$ is a rectangle triangle. Let $P$ be inside the equilateral triangle, and $\alpha,\beta,\gamma$ three segments such that they sum up to the side of $ABC$ (and also to the hypotenuse of $ABP$, by construction).

Is it true that $P$ belongs to the red circle if and only $\gamma^2=2\alpha\beta$?

  • 2
    $\begingroup$ Consider ABP and apply Pythagoras' Theorem (and its converse). What can you get? $\endgroup$ – tonychow0929 Jun 5 '18 at 9:09
  • $\begingroup$ Hmmm Thanks Tony, but I don't get it :( $\endgroup$ – user559615 Jun 5 '18 at 9:11
  • 1
    $\begingroup$ You mean $(\alpha+\gamma)^2+(\beta+\gamma)^2=(\alpha+\beta+\gamma)^2$... $\endgroup$ – user559615 Jun 5 '18 at 9:12
  • $\begingroup$ Please use "right triangle", "right-angled triangle", or "rectangled triangle" (see e.g. this Wikipedia article). "rectangle triangle" is quite confusing. $\endgroup$ – Nominal Animal Jun 5 '18 at 15:38
  • $\begingroup$ Sorry, I correct it. $\endgroup$ – user559615 Jun 5 '18 at 15:39

$AB$ is the diameter of the red circle. If $P$ belongs to the red circle, $\angle APB=90^\circ$ (semi-circle). So by the Pythagorean theorem, $AB^2=AP^2+PB^2\implies (\alpha+\gamma+\beta)^2=(\alpha+\gamma)^2+(\beta+\gamma)^2\implies\gamma^2=2\alpha\beta.$

The converse of the Pythagorean theorem takes care of the "$\Leftarrow$" direction.

  • 1
    $\begingroup$ Yes, I see! Thanks a lot, Poyea (and also thanks to Tony, I got the same solution thanks to his suggestion). $\endgroup$ – user559615 Jun 5 '18 at 9:19

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