A Problem with the Year of Birth Inserted into It

              About Problem S:E24.345 from the SGMB magazine 12/2024, proposed for the 7th grade. Authors Doina STOICA and Mircea_Mario STOICA (the latter born in 1961; we were classmates).

          In translation:

              "Let $\;a, \;b\;$ and $\;c\;$ be the lengths of the sides of triangle $ABC$. We know that

$$\sqrt{a^2-20a+2061}+\sqrt{b^2-16b+100}+\sqrt{c^2-12c+52}=10+\sqrt{1961}.$$

                Prove that triangle $ABC$ is right-angled."

ANSWER CiP

$a=10,\;\;b=8,\;\;c=6\;\;$ and   $a^2=b^2+c^2$

                    Solution CiP

               The given equation can also be written as

$$\sqrt{(a-10)^2+1961}+\sqrt{(b-8)^2+36}+\sqrt{(c-6)^2+16}=10+\sqrt{1961} \tag{1}$$

We have that LHS $\geqslant \sqrt{1961}+\sqrt{36}+\sqrt{16}=\sqrt{1961}+6+4=10+\sqrt{1961}.$

It means that in (1) we have equality, therefore $a=10,\;b=8,\;c=6$ and the triangle is the Ancient Egyptian one.

$\blacksquare$