Understanding the Geometry of the Equilateral Triangle Problem

The problem involves an equilateral triangle Δ ABC, where the base BC is divided into three equal parts. We are given lines drawn parallel to the sides AB and AC, which intersect and form a parallelogram and a smaller triangle. We need to find the ratio of the sum of the areas of these two shapes to the area of the original triangle.

Setting up the Triangle and Trisection Points

Let the side length of the equilateral triangle Δ ABC be denoted by \(a\). The area of Δ ABC is given by the formula for the area of an equilateral triangle:

\( \text{Area}(\Delta \text{ABC}) = \frac{\sqrt{3}}{4} a^2 \)

The base BC is trisected at points D and E. This means the segment BC is divided into three equal parts: BD, DE, and EC.

• BD = DE = EC = \(\frac{BC}{3} = \frac{a}{3}\)

Analyzing the Parallel Lines and Formed Shapes

A line is drawn through D parallel to AB, meeting AC at F. Since DF is parallel to AB and D is on BC, Δ FDC is similar to Δ ABC. The ratio of corresponding sides in similar triangles is equal. The ratio of the base DC to BC is:

\( \frac{DC}{BC} = \frac{DE + EC}{BC} = \frac{a/3 + a/3}{a} = \frac{2a/3}{a} = \frac{2}{3} \)

Since Δ FDC \( \sim \) Δ ABC, the ratio of their sides is 2/3. Thus:

• FC = \(\frac{2}{3} AC = \frac{2}{3} a\)
• AF = AC - FC = \(a - \frac{2}{3} a = \frac{a}{3}\)
• DF = \(\frac{2}{3} AB = \frac{2}{3} a\)

Similarly, a line is drawn through E parallel to AC, meeting AB at G. Since EG is parallel to AC and E is on BC, Δ GBE is similar to Δ ABC. The ratio of the base BE to BC is:

\( \frac{BE}{BC} = \frac{BD + DE}{BC} = \frac{a/3 + a/3}{a} = \frac{2a/3}{a} = \frac{2}{3} \)

Since Δ GBE \( \sim \) Δ ABC, the ratio of their sides is 2/3. Thus:

• GB = \(\frac{2}{3} AB = \frac{2}{3} a\)
• AG = AB - GB = \(a - \frac{2}{3} a = \frac{a}{3}\)
• EG = \(\frac{2}{3} AC = \frac{2}{3} a\)

We observe that AG = \(\frac{a}{3}\) and AF = \(\frac{a}{3}\). Also, the line segment HG is part of EG (which is parallel to AC) and the line segment HF is part of DF (which is parallel to AB). Since AG lies on AB and AF lies on AC, and given the parallel constructions, the quadrilateral AGHF is a parallelogram.

Calculating the Area of Parallelogram AGHF

Parallelogram AGHF has adjacent sides AG and AF with lengths \(\frac{a}{3}\) each. The angle between AG (on AB) and AF (on AC) is the angle \(\angle BAC\), which is \(60^\circ\) because Δ ABC is equilateral.

The area of a parallelogram is given by the product of two adjacent sides and the sine of the angle between them.

\( \text{Area}(\text{AGHF}) = \text{AG} \times \text{AF} \times \sin(\angle \text{GAF}) \)

\( \text{Area}(\text{AGHF}) = \frac{a}{3} \times \frac{a}{3} \times \sin(60^\circ) \)

\( \text{Area}(\text{AGHF}) = \frac{a^2}{9} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3} a^2}{18} \)

Calculating the Area of Triangle DHE

H is the intersection of DF and EG. D and E are points on the base BC. The length of the base DE is \(\frac{a}{3}\).

To find the area of Δ DHE, we need the height of H above the base DE (which lies on BC). We can use coordinates to find the position of H.

Let B be the origin (0,0), C be (a,0), and A be (\(\frac{a}{2}, \frac{a\sqrt{3}}{2}\)).

The points D and E on BC are:

• D = (\(\frac{a}{3}\), 0)
• E = (\(\frac{2a}{3}\), 0)

The line DF passes through D(\(\frac{a}{3}\), 0) and is parallel to AB. The slope of AB is \(\frac{\frac{a\sqrt{3}}{2} - 0}{\frac{a}{2} - 0} = \sqrt{3}\).

Equation of line DF: \(y - 0 = \sqrt{3}(x - \frac{a}{3}) \implies y = \sqrt{3}x - \frac{a\sqrt{3}}{3}\).

The line EG passes through E(\(\frac{2a}{3}\), 0) and is parallel to AC. The slope of AC is \(\frac{\frac{a\sqrt{3}}{2} - 0}{\frac{a}{2} - a} = \frac{a\sqrt{3}/2}{-a/2} = -\sqrt{3}\).

Equation of line EG: \(y - 0 = -\sqrt{3}(x - \frac{2a}{3}) \implies y = -\sqrt{3}x + \frac{2a\sqrt{3}}{3}\).

The intersection point H is found by equating the y values:

\( \sqrt{3}x - \frac{a\sqrt{3}}{3} = -\sqrt{3}x + \frac{2a\sqrt{3}}{3} \)

\( 2\sqrt{3}x = \frac{2a\sqrt{3}}{3} + \frac{a\sqrt{3}}{3} = \frac{3a\sqrt{3}}{3} = a\sqrt{3} \)

\( 2x = a \implies x = \frac{a}{2} \)

Substitute x back into either equation to find y:

\( y = \sqrt{3}(\frac{a}{2}) - \frac{a\sqrt{3}}{3} = \frac{a\sqrt{3}}{2} - \frac{a\sqrt{3}}{3} = \frac{3a\sqrt{3} - 2a\sqrt{3}}{6} = \frac{a\sqrt{3}}{6} \)

The coordinates of H are \((\frac{a}{2}, \frac{a\sqrt{3}}{6})\). The height of H above BC (the x-axis) is its y-coordinate, \(\frac{a\sqrt{3}}{6}\).

The base of Δ DHE is DE, with length \(\frac{a}{3}\).

\( \text{Area}(\Delta \text{DHE}) = \frac{1}{2} \times \text{base} \times \text{height} \)

\( \text{Area}(\Delta \text{DHE}) = \frac{1}{2} \times \frac{a}{3} \times \frac{a\sqrt{3}}{6} = \frac{\sqrt{3} a^2}{36} \)

Calculating the Sum of Areas

The sum of the area of parallelogram AGHF and the area of Δ DHE is:

\( \text{Sum of Areas} = \text{Area}(\text{AGHF}) + \text{Area}(\Delta \text{DHE}) \)

\( \text{Sum of Areas} = \frac{\sqrt{3} a^2}{18} + \frac{\sqrt{3} a^2}{36} \)

\( \text{Sum of Areas} = \frac{2\sqrt{3} a^2}{36} + \frac{\sqrt{3} a^2}{36} = \frac{3\sqrt{3} a^2}{36} = \frac{\sqrt{3} a^2}{12} \)

Finding the Ratio of Areas

The ratio of the sum of the areas to the area of Δ ABC is:

\( \text{Ratio} = \frac{\text{Sum of Areas}}{\text{Area}(\Delta \text{ABC})} \)

\( \text{Ratio} = \frac{\frac{\sqrt{3} a^2}{12}}{\frac{\sqrt{3} a^2}{4}} \)

\( \text{Ratio} = \frac{\sqrt{3} a^2}{12} \times \frac{4}{\sqrt{3} a^2} \)

\( \text{Ratio} = \frac{4}{12} = \frac{1}{3} \)

Summary of Areas

The ratio of the sum of the area of parallelogram AGHF and the area of Δ DHE to the area of Δ ABC is \(\frac{1}{3}\).

Revision Table: Key Concepts Revisited

Additional Information: Further Exploration

This problem demonstrates how properties of equilateral triangles, similar triangles formed by parallel lines, and basic area formulas can be combined with coordinate geometry techniques to solve complex geometry problems. The specific ratio of 1/3 in this construction is a notable result.

The parallelogram AGHF, being formed by segments AG=\(\frac{a}{3}\) and AF=\(\frac{a}{3}\) with an angle of \(60^\circ\), is actually a rhombus made of two equilateral triangles of side \(\frac{a}{3}\). Its area could also be calculated as \(2 \times \frac{\sqrt{3}}{4} (\frac{a}{3})^2 = \frac{\sqrt{3} a^2}{18}\).

The point H(\(\frac{a}{2}, \frac{a\sqrt{3}}{6}\)) is on the median from A to BC of the triangle ABC. The median from A to BC meets BC at the midpoint M(\(\frac{a}{2}, 0\)). The equation of the median AM is the line connecting (\(\frac{a}{2}, 0\)) and (\(\frac{a}{2}, \frac{a\sqrt{3}}{2}\)), which is \(x = \frac{a}{2}\). The x-coordinate of H is \(\frac{a}{2}\), confirming it lies on the median from A. This gives an alternative way to find the x-coordinate of H.

Problems involving trisection points and parallel lines in triangles often lead to segments and areas with ratios involving 1/3, 2/3, 1/9, 4/9, etc., related to the square of the length ratios.