3.5: Triangle with the given sides

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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 $c$.

Theorem $\PageIndex{1}$

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.

$截屏2021-02-02 下午2.42.36.png$

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 $r$.

\[s(\beta, r):= AC.\]

Proposition $\PageIndex{1}$

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 $AB = r$.

$截屏2021-02-02 下午2.47.27.png$

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

\[\begin{array} {rcl} {s(r, \beta)} & = & {AC \le} \\ {} & \le & {AD + DC =} \\ {} & = & {2 \cdot AD <} \\ {} & < & {2 \cdot AY =} \\ {} & = & {\varepsilon} \end{array}\]

Corollary $\PageIndex{1}$

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.

$截屏2021-02-02 下午2.58.20.png$

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.

Theorem \(\PageIndex{1}\)

Proposition \(\PageIndex{1}\)

Corollary \(\PageIndex{1}\)