Each of two different lines parallel to the the axis have exactly two common points on the graph of the function . Let and be two lines parallel to axis which meet the graph of in points and , respectively. Prove that the quadrilateral formed by and is a rhombus if and only if its area is equal to units.
Problem 1665
Official solution
To prove that the quadrilateral formed by the points and is a rhombus if and only if its area is equal to 6 units, we will follow these steps:
1. Identify the points of intersection:
Let the lines and be given by and respectively, where . These lines intersect the graph of at points and respectively.
2. Determine the x-coordinates of the intersection points:
Since intersects the graph at , we solve . Let the roots be and . Similarly, for , solve . Let the roots be and .
3. **Form the points and :**
The points are , , , and .
4. Prove the quadrilateral is a rhombus:
For the quadrilateral to be a rhombus, all sides must be of equal length. Calculate the distances between the points:
For the quadrilateral to be a rhombus, we need:
5. Calculate the area of the rhombus:
The area of a rhombus can be calculated using the formula:
where and are the lengths of the diagonals. The diagonals can be calculated as:
Given that the area is 6 units, we have:
Simplifying, we get:
6. Conclusion:
The quadrilateral formed by and is a rhombus if and only if the area is equal to 6 units, as shown by the conditions derived above.