Problem 20 - Complex numbers from A to...Z

Question

The diagonals \(A C\) and \(C E\) of a regular hexagon \(A B C D E F\) are divided by interior points \(M\) and \(N\), respectively, such that $$ \frac{A M}{A C}=\frac{C N}{C E}=r $$ Determine \(r\) knowing that points \(B, M\) and \(N\) are collinear.

Step-by-Step Solution

Answer

Answer: The value of r in terms of x is r = x/√3.

Step 1: Finding relationship between sides and segments

Since \(ABCDEF\) is a regular hexagon, all its sides are equal, and all its internal angles are equal to 120 degrees. Let the side length of the hexagon be \(s\). We can find the length of the segment \(AC\) using the law of cosines in triangle \(ABC\): $$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(120^{\circ})$$ $$AC^2 = s^2 + s^2 - 2(s)(s)\left(-\frac{1}{2}\right)$$ $$AC^2 = 3s^2$$ Similarly, we can find the length of the segment \(CE\) using the same method. Triangle \(CDE\) also has an internal angle of 120 degrees at vertex \(C\). $$CE^2 = CD^2 + DE^2 - 2(CD)(DE)\cos(120^{\circ})$$ $$CE^2 = s^2 + s^2 - 2(s)(s)\left(-\frac{1}{2}\right)$$ $$CE^2 = 3s^2$$ Let \(AM = xs\) and \(CN = ys\), so \(MC=(1-x)s\) and \(NE=(1-y)s\). By the given condition, we have: $$ \frac{AM}{AC}=\frac{xs}{\sqrt{3s^2}}=\frac{CN}{CE}=\frac{ys}{\sqrt{3s^2}} $$ We can cancel out the \(s\) and \(\sqrt{3}\) on both sides: $$ \frac{x}{\sqrt{3}}=\frac{y}{\sqrt{3}} $$ Therefore, we have: $$ x=y $$

Step 2: Using collinear condition

Since points \(B\), \(M\), and \(N\) are collinear, we know that triangles \(ABM\) and \(BMN\) are similar with the same ratio \(\frac{BM}{BN} = \frac{AM}{MN}\). Let \(BM=p\cdot s\) and \(BN=q\cdot s\), we have: $$ \frac{p}{q}=\frac{x}{1-x} $$ Now, we can use the trigonometry of the internal angle at \(B\) in both triangles \(ABM\) and \(BMN\) to eliminate variables and find the value of \(x\). Notice that \(\angle ABM = \angle MBC = 120^{\circ}\) and \(\angle MBN = 180^{\circ} - \angle MBA = 180^{\circ} - 120^{\circ} = 60^{\circ}\). Now, let's use the sine law in both triangles \(ABM\) and \(BMN\): In triangle \(ABM\): $$ \frac{AB}{\sin{\angle AMB}}=\frac{AM}{\sin{\angle MBA}} $$ $$ \frac{s}{\sin{60^{\circ}}}=\frac{xs}{\sin{120^{\circ}}} $$ In triangle \(BMN\): $$ \frac{BN}{\sin{\angle MNB}}=\frac{MN}{\sin{\angle NBM}} $$ $$ \frac{qs}{\sin{60^{\circ}}}=\frac{(1-x)s}{\sin{120^{\circ}}} $$ Now dividing these equations, we get: $$ \frac{p}{q}=\frac{x}{1-x} $$ From the previous steps, we already have that \(x=y\) and \(\frac{p}{q}=\frac{x}{1-x}\). Therefore, the given condition \(\frac{AM}{AC}=\frac{CN}{CE}=r\) becomes: $$ r=\frac{x}{\sqrt{3}}=\frac{y}{\sqrt{3}} $$ So we have: $$ r=\frac{x}{\sqrt{3}} = \frac{xs}{\sqrt{3s^2}} =\frac{AM}{AC} $$

Key Concepts

Regular hexagon

A regular hexagon is a six-sided polygon where all sides have equal length, and all internal angles are equal. Each angle in a regular hexagon measures 120 degrees. This symmetry means that any diagonal in the hexagon divides it into congruent triangles, which is helpful when solving geometric problems.
In geometry, regular hexagons are essential because of their perfect symmetry and recurring pattern, often appearing in tessellations and natural formations like honeycombs. Understanding regular hexagons involves recognizing these properties:

• Equal side lengths
• Each internal angle is 120 degrees
• Diagonals divide the hexagon into quadrilateral and triangular shapes

These properties provide the basis for analyzing relationships between segments, as seen in exercises involving regular hexagons.

Law of cosines

The Law of Cosines is a fundamental rule in trigonometry, especially useful in triangles that are not right-angled. This law helps calculate unknown sides or angles in a triangle when we know either:

• Two sides and their enclosed angle, or
• All three sides, to find any one of the angles

The formula states:
\[c^2 = a^2 + b^2 - 2ab \cos(C)\]
where \(c\) is the side opposite the angle \(C\), and \(a\) and \(b\) are the other two sides.

In a regular hexagon, the Law of Cosines is often used to determine the lengths of its diagonals. For instance, in our exercise, we use it in triangles like \(ABC\) and \(CDE\) to find the lengths of \(AC\) and \(CE\). Given that the internal angle at each vertex in the hexagon is 120 degrees, substituting \(120^{\circ}\) in the formula simplifies the calculations considerably.

Trigonometry in triangles

Trigonometry is a branch of mathematics that deals with relationships between the sides and angles of triangles. It's invaluable in solving problems involving unknown lengths or angles within triangles. Some foundational concepts include:

• Sine, cosine, and tangent functions, which relate angles to side lengths
• Trigonometric identities that simplify complex expressions
• The use of these functions within the Law of Sines and the Law of Cosines

These concepts are used to handle non-right angle triangles, common in geometry problems involving structures like hexagons.

In this exercise, trigonometry is employed using the Sine Law, which relates to the ratios of side lengths and sines of opposite angles in a triangle. This is crucial when determining segment proportions like \(AM\) and \(MN\), especially since points \(B\), \(M\), and \(N\) are collinear.

Similar triangles

Triangles are said to be similar when their corresponding angles are equal, and their corresponding sides are in proportion. Similar triangles share the same shape but may differ in size. This property is vital because it allows the equating of ratios of sides, leading to simplified calculations.
Key characteristics of similar triangles include:

• Corresponding angles are equal
• Corresponding sides are proportional in length

The similarity between triangles helps in solving problems where proportionality is involved, like finding specific segment lengths.

In the exercise, identifying the similarity of triangles \(ABM\) and \(BMN\) due to the collinearity of \(B\), \(M\), and \(N\) allows us to deduce relationships between segments like \(BM/p\) and \(BN/q\). This, in combination with given conditions, enables us to solve for \(r\), demonstrating the power of similarity in resolving complex geometric scenarios.