\section*{Problem 4 - 121024}

a) The students of a class are given the task to graphically determine \(\sqrt{12}\) and \(\sqrt{133}\).

For this, they should only use the altitude theorem or the leg theorem or both theorems (for each of the roots, one of these two theorems). A student solves both tasks using the same right-angled triangle.

What are all the possibilities for choosing suitable measures \(p\) and \(q\) of the lengths of the segments of the hypotenuse so that these measures \(p\) and \(q\) are also rational numbers?

b) Answer the same question for the case where \(\sqrt{11}\) and \(\sqrt{133}\) were to be determined.
}

Let \(a, b, c, q, p, h\) be the usual notations in a right-angled triangle. With \(p, q\), \(c\) is also rational. Therefore, the only possible lengths for the roots are \(a, b, h\). In particular, it holds that \(a^{2}+b^{2}=c^{2}, a^{2}-h^{2}=p^{2}\) and \(b^{2}-h^{2}=q^{2}\) are squares of rational numbers.

a) Since \(\sqrt{133}^{2}+\sqrt{12}^{2}=145\) is not a square number and \(\sqrt{12}<\sqrt{133}\) holds, the only possibility is \(a=\sqrt{133}, h=\sqrt{12}\) (or symmetrically \(b=\sqrt{133}\)). The system of equations from the altitude and cathetus theorems

\[
p q=12 \quad p(p+q)=133
\]

can be uniquely solved by \(p^{2}=121 \Rightarrow p=11\) and \(q=\frac{12}{11}\)

b) Since \(\sqrt{133}^{2}-\sqrt{11}^{2}=122\) is not a square, the only approach is \(a=\sqrt{133}, b=\sqrt{11}\). By adding the two cathetus equations

\[
p(p+q)=133 \quad q(p+q)=11
\]

we obtain \((p+q)^{2}=144 \Rightarrow p+q=12\). Substituting the sum yields \(p=\frac{133}{12}\) and \(q=\frac{11}{12}\).

Remark: The roots of two rational numbers are simultaneously constructible if the sum or difference is the square of a rational number.