# A question about a right triangle contained in an equilateral triangle

In this picture,

$ABC$ 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 $\gamma^2=2\alpha\beta$?

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