At temperature t, the lengths of the three sides are $\displaystyle (1+ \alpha_1t)L_1$, $\displaystyle (1+ \alpha_2t)L_2$, and $\displaystyle (1+ \alpha_3t)L_3$. In order that this be a right triangle for all t we must have (Pythagorean theorem)
$\displaystyle (1+ \alpha_1t)^2L_1^2+ (1+ \alpha_2t)^2L_2^2= (1+ \alpha_3t)^2L_3^2$
To solve that equation for $\displaystyle \alpha_3$, first divide both sides by $\displaystyle L_3^2$:
$\displaystyle (1+ \alpha_1t)^2\frac{L_1^2}{L_3^2}+ (1+ \alpha_2t)^2\frac{L_2^2}{L_3^2}= 1+ \alpha_3t$
now subtract 1 from both sides:
$\displaystyle (1+ \alpha_1t)^2\frac{L_1^2}{L_3^2}+ (1+ \alpha_2t)^2\frac{L_2^2}{L_3^2} 1= \alpha_3t$
and, finally, divide that by t:
$\displaystyle (1+ \alpha_1t)^2\frac{L_1^2}{tL_3^2}+ (1+ \alpha_2t)^2\frac{tL_2^2}{tL_3^2} \frac{1}{t}= \alpha_3$.
Since the original configuration was a right triangle we can replace [math]L_3^2[/math with $\displaystyle L_1^2+ L_2^2$:
$\displaystyle \alpha_3= (1+ \alpha_1t)^2\frac{L_1^2}{t(L_1^2+ L_2^2)}+ (1+ \alpha_2t)^2\frac{tL_2^2}{t(L_1^2+ L_2^2)} \frac{1}{t}$. If, for given $\displaystyle \alpha_1$, $\displaystyle \alpha_2$, $\displaystyle L_1$, and $\displaystyle L_2$, that is independent of t, then the answer is "yes", otherwise, "no".
