3.5: Triangle with the given sides
Consider the triangle \(ABC\). Set
\(a = BC\), \(b = CA\), \(c = AB.\)
Without loss of generality, we may assume that
\(a \le b \le c.\)
Then all three triangle inequalities for \(\triangle ABC\) hold if and only if
\(c \le a + b\).
The following theorem states that this is the only restriction on \(a\), \(b\), and \(c\).
Assume that \(0 < a \le b \le c \le a + b\). Then there is a triangle with sides \(a, b\), and \(c\); that is, there is \(\triangle ABC\) such that \(a = BC\), \(b = CA\), and \(c = AB\).
- Proof
-
Fix the points \(A\) and \(B\) such that \(AB = c\). Given \(\beta \in [0, \pi]\), suppose that \(C_{\beta}\) denotes the point in the plane such that \(BC_{\beta} = a\) and \(\measuredangle ABC = \beta\).
According the Corollary \(\PageIndex{1}\), the map \(\beta \mapsto C_{\beta}\) is continuous. Therefore, the function \(b(\beta) = AC_{\beta}\)is continous (formally, it follows from Exercise 1.9.1 and Exercise 1.9.2 ).
Note that \(b(0) = c - a\) and \(b (\pi) = c + a\). Since \(c - a \le b \le c + a\), by the intermediate value theorem ( Theorem 3.2.1 ) there is \(\beta_0 \in [0, \pi]\) such that \(b(\beta_0) = b\), hence the result.
Assume \(r > 0\) and \(\pi > \beta > 0\). Consider the triangle \(ABC\) such that \(AB = BC = r\) and \(\measuredangle ABC = \beta\). The existence of such a triangle follows from Axiom IIIa and Proposition 2.2.2 .
Note that according to Axiom IV , the values \(\beta\) and \(r\) define the triangle \(ABC\) up to the congruence. In particular, the distance \(AC\) depends only on \(\beta\) and \(r\).
\[s(\beta, r):= AC.\]
Given \(r > 0\) and \(\varepsilon > 0\), there is \(\delta > 0\) such that
\(0 < \beta < \delta \Rightarrow s(r, \beta) < \varepsilon.\)
- Proof
-
Fix two points \(A\) and \(B\) such that \(AB = r\).
Choose a point \(X\) such that \(AY = \dfrac{\varepsilon}{2}\); it exists by Proposition 2.2.2 .
Note that \(X\) and \(Y\) lie on the same side of \((AB)\); therefore, \(\angle ABY\) is positive. Set \(\delta = \measuredangle ABY\).
Suppose \(0 < \beta < \delta\); by Axiom IIIa , we can choose \(B\) so that \(\measuredangle ABC = \beta\) and \(BC = r\). Further we can choose a half-line \([BZ)\) such that \(\measuredangle ABZ = \dfrac{1}{2} \cdot \beta\).
Note that \(A\) and \(Y\) lie on opposite sides of \((BZ)\) and moreover \(\measuredangle ABZ \equiv - \measuredangle CBZ\). In particular, \((BZ)\) intersects \([AY]\); denote by \(D\) the point of intersection.
Since \(D\) lies between \(A\) and \(Y\), we have that \(AD < AY\).
Since \(D\) lies on \((BZ)\) we have that \(\measuredangle ABD \equiv \measuredangle CBD\). By Axiom IV , \(\triangle ABD \cong \triangle CBD\). It follows that
\[\begin{array} {rcl} {s(r, \beta)} & = & {AC \le} \\ {} & \le & {AD + DC =} \\ {} & = & {2 \cdot AD <} \\ {} & < & {2 \cdot AY =} \\ {} & = & {\varepsilon} \end{array}\]
Fix a real number \(r > 0\) and two distinct points \(A\) and \(B\). Then for any real number \(\beta \in [0, \pi]\), there is a unique point \(C_{\beta}\) such that \(BC_{\beta} = r\) and \(\measuredangle ABC_{\beta} = \beta\). Moreover, \(\beta \mapsto C_{\beta}\) is a continuous map from \([0, \pi]\) to the plane.
- Proof
-
The existence and uniqueness of \(C_{\beta}\) follows from Axiom IIIa and Proposition 2.2.2 .
Note taht if \(\beta_1 \ne \beta_2\), then
\[C_{\beta_1} C_{\beta_2} = s(r, |\beta_1 - \beta_2|).\]
By Proposition \(\PageIndex{1}\), the map \(\beta \mapsto C_{\beta}\) is continuous.
Given a positive real number \(r\) and a pint \(O\), the set \(\Gamma\) of all points on distance \(r\) from \(O\) is called a circle with radius \(r\) and center \(O\).
Exercise \(\PageIndex{1}\)
Show that two circles intersect if and only if
\[|R - r| \le d \le R + r,\]
where \(R\) and \(r\) denote their radiuses, and \(d\) — the distance between their centers.
- Hint
-
The "only-if" part follows from the triangle inequality. To prove "if" part, observe that Theorem \(\PageIndex{1}\) implies existence of a triangle with sides \(r_1\), \(r_2\), and \(d\). Use this triangle to show that there is a point \(X\) such that \(O_1X = r_1\) and \(O_2X = r_2\), where \(O_1\) and \(O_2\) are the centers of the corresponding circles.