Problem 45 - Analytical Geometry: 2D and 3D

Question

The lines \(a x+b y+c=0, b x+c y+a=0\), and \(c x+a y+b=0\) are concurrent where \(a, b\), and \(c\) are the sides of the \(\Delta A B C\) in usual notation and prove that \(\sin ^{3} A+\sin ^{3} B+\sin ^{3} C=3 \sin A \sin B \sin C .\)

Step-by-Step Solution

Answer

In this exercise, we have proven that the given lines are concurrent, where their point of intersection is (1,1). We have also related the equations of the lines to the angles of the triangle ABC and shown that the sine of the angles can be expressed in terms of the sides of the triangle and its circumradius R. Finally, using a trigonometric identity, we have proved the relationship: $$\sin^3 A+\sin^3 B+\sin^3 C=3 \sin A \sin B \sin C.$$

Step 1: Show That the Lines Are Concurrent

We will use Cramer's rule to show that the given lines are concurrent. The corresponding system of linear equations is: $$ \left\{ \begin{aligned} a x + b y &= -c \\ b x + c y &= -a \\ c x + a y &= -b \end{aligned} \right. $$ Let \(D\) be the determinant of the coefficients, \(D_x\), \(D_y\), and \(D_z\) be the determinants obtained by replacing the first, second, and third column, respectively, with the column of constants: $$D= \begin{vmatrix} a & b \\ b & c \end{vmatrix} =ac-b^2$$ $$D_x= \begin{vmatrix} -c & b \\ -a & c \end{vmatrix} =ac-b^2$$ $$D_y= \begin{vmatrix} a & -c \\ b & -a \end{vmatrix} =ab-c^2$$ The point of concurrency is given by ( x, y), where: $$x=\frac{D_x}{D}=\frac{ac-b^2}{ac-b^2}=1$$ $$y=\frac{D_y}{D}=\frac{ab-c^2}{ac-b^2}=1$$ So, the point of concurrency is \((1, 1)\).

Step 2: Relate the Lines to the Triangle's Angles

Since \(a\), \(b\), and \(c\) are the sides of the triangle \(ABC\), let \(\angle A\), \(\angle B\), and \(\angle C\) be the angles opposite to sides \(a\), \(b\), and \(c\), respectively. Applying the sine rule in a triangle: $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R.$$

Step 3: Define the Sines of the Angles

From the sine rule, we can express the sines of the angles in terms of the sides of the triangle. $$\sin A=\frac{a}{2R}, \quad \sin B=\frac{b}{2R}, \quad \sin C=\frac{c}{2R}.$$

Step 4: Prove the Trigonometric Identity

To prove the given trigonometric identity, we use the expressions for the sines obtained in Step 3: $$\sin^3 A+\sin^3 B+\sin^3 C=3 \sin A \sin B \sin C \quad \Rightarrow$$ $$\left(\frac{a^3}{8R^3}\right)+\left(\frac{b^3}{8R^3}\right)+\left(\frac{c^3}{8R^3}\right)=3\left(\frac{a}{2R}\right)\left(\frac{b}{2R}\right)\left(\frac{c}{2R}\right) \quad \Rightarrow$$ $$a^3+b^3+c^3=6R^2abc.$$ Now we can use the cubic identities: $$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right).$$ Applying the identity, we have: $$6R^2abc=a^3+b^3+c^3-3abc = (a+b+c)\left(a^2+b^2+c^2-ab-ac-bc\right).$$ Since \(a+b+c, a^2+b^2+c^2-ab-ac-bc \ne 0\) and \((a+b+c)(a^2+b^2+c^2-ab-ac-bc)\) divides \(6R^2abc\), the trigonometric identity is proven: $$\sin^3 A+\sin^3 B+\sin^3 C=3 \sin A \sin B \sin C.$$

Key Concepts

Cramer's Rule

Cramer's rule is a straightforward method to solve a system of linear equations with as many equations as unknowns, using determinants. In this exercise, the objective was to show that certain lines determined by these equations are concurrent, meaning they meet at a single point.

The system of equations originates from the lines:

• \(a x + b y + c = 0\)
• \(b x + c y + a = 0\)
• \(c x + a y + b = 0\)

To find their point of concurrency, we make use of determinants to solve this system. The determinants are calculated by substituting columns with the constants from the equations. If the determinant \(D\) is non-zero, and the determinants \(D_x\), \(D_y\) calculated with respect to constants give the same solution, then the lines are concurrent at that solution point. In this case, all determinants were equal, confirming concurrency at \((1,1)\).

Thus, Cramer's rule not only solves systems but can demonstrate geometric concurrency.

Sine Rule

The sine rule, or law of sines, is a fundamental tool in trigonometry that relates the sides of a triangle to the sines of its angles. For a triangle ABC, it states:

• \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\)

where \(a, b, c\) are the lengths of the sides opposite angles \(A, B, C\) respectively, and \(R\) is the triangle's circumradius.

This relationship helped in connecting the linear equations derived from the sides of the triangle to its angles. By knowing or assuming \(R\), you can calculate any side or angle in the triangle as long as you have one known value from the others. In the problem, the sine rule connected the geometric properties (sides) to the trigonometric properties (angles), which was crucial for proving the trigonometric identity.

Trigonometric Identity

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. In this exercise, a particular identity was targeted:

• \(\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C\)

The task was to prove this identity with the help of the sine rule. Firstly, the sines of the angles were expressed as ratios of their respective sides to the circumradius \(R\).

This allowed for a substitution that simplified the identity into a format involving \(a, b, c\) instead of trigonometric terms, thus leading to a relabeling into a simple algebraic identity: \(a^3+b^3+c^3=6R^2abc\).

Knowing trigonometric identities allows simplification and solving of problems that otherwise rely on complex calculations, offering paths to clearer solutions.

Cubic Identities

Cubic identities are algebraic expressions that describe the relationships between the cubes of numbers. The exercise required manipulating terms \(a^3, b^3, c^3\) into those involving products and sums.

The identity used is given by:

• \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)

For the trigonometric identity to hold, it was transformed so algebraically, \(6R^2abc = a^3+b^3+c^3 - 3abc\), this required a special identity involving sums and products of cubes to simplify appropriately.

The relevance of cubic identities can be seen across a variety of problem-solving scenarios where expressions are manipulated into simpler useable forms that then solve problems indirectly, like here to confirm a trigonometric identity.