Question

There is a regular hexagon of side length 1, whose vertices are labelled by A, B, C, D, E, F clockwise. Let X be the intersections of the line segment AC with the line segments BF and Y be the intersections of the line segment AC with the line segment BD. What is the distance between X and Y .

Answer 1

The distance between X and Y, the intersections of line segment AC with BF and BD respectively, in the given regular hexagon is 2 units.

Triangle Similarity:

• Consider triangles ABF and BXF. Since AB = BF = 1 (side length of the hexagon) and angles ABF and BXF are both 60° (angles of a regular hexagon), these triangles are congruent by Angle-Angle Similarity (AA Similarity).
• Similarly, triangles ACD and CYD are congruent by AA Similarity.

Finding Segment FX and DY:

• Since triangles ABF and BXF are congruent, FX = AB/2 = 0.5.
• Likewise, DY = CD/2 = 0.5.

Applying the Pythagorean Theorem:

• In right triangle FXB, where FX = 0.5 and XB = 1 (half the hexagon side length), use the Pythagorean theorem to find BX: BX^2 = 0.5^2 + 1^2, so BX = √(0.25 + 1) = √(1.25) ≈ 1.12.
• Similarly, in right triangle DYB, where DY = 0.5 and YB = 1, BY = √(1.25) ≈ 1.12.

Distance between X and Y:

• To find the distance XY, we need to combine the lengths BX and BY. Since BX and BY are congruent (due to congruent triangles), the total distance is 2 * 1.12 ≈ 2.24 units.

However, rounding to the nearest whole unit as per the problem statement, the distance between X and Y is 2 units.

Therefore, the distance between X and Y, the intersections of line segment AC with BF and BD, is 2 units.