Olympiad Maths Prep

Track / Stage 7 / 265 of 300 #1665 of 2000

Problem 1665

National olympiad second round; IMO P1/P4
Geometry Difficulty 7.7 Prove it

Each of two different lines parallel to the the axis OxOx have exactly two common points on the graph of the function f(x)=x3+ax2+bx+cf(x)=x^3+ax^2+bx+c. Let 1\ell_1 and 2\ell_2 be two lines parallel to OxOx axis which meet the graph of ff in points K1,K2K_1, K_2 and K3,K4K_3, K_4, respectively. Prove that the quadrilateral formed by K1,K2,K3K_1, K_2, K_3 and K4 K_4 is a rhombus if and only if its area is equal to 66 units.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

To prove that the quadrilateral formed by the points K1,K2,K3, K_1, K_2, K_3, and K4 K_4 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 1\ell_1 and 2\ell_2 be given by y=y1 y = y_1 and y=y2 y = y_2 respectively, where y1y2 y_1 \neq y_2 . These lines intersect the graph of f(x)=x3+ax2+bx+c f(x) = x^3 + ax^2 + bx + c at points K1,K2 K_1, K_2 and K3,K4 K_3, K_4 respectively.

2. Determine the x-coordinates of the intersection points:
Since 1 \ell_1 intersects the graph at y=y1 y = y_1 , we solve x3+ax2+bx+c=y1 x^3 + ax^2 + bx + c = y_1 . Let the roots be x1 x_1 and x2 x_2 . Similarly, for 2 \ell_2 , solve x3+ax2+bx+c=y2 x^3 + ax^2 + bx + c = y_2 . Let the roots be x3 x_3 and x4 x_4 .

3. **Form the points K1,K2,K3, K_1, K_2, K_3, and K4 K_4 :**
The points are K1=(x1,y1) K_1 = (x_1, y_1) , K2=(x2,y1) K_2 = (x_2, y_1) , K3=(x3,y2) K_3 = (x_3, y_2) , and K4=(x4,y2) K_4 = (x_4, y_2) .

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:
Distance between K1 and K2=x2x1 \text{Distance between } K_1 \text{ and } K_2 = |x_2 - x_1|
Distance between K3 and K4=x4x3 \text{Distance between } K_3 \text{ and } K_4 = |x_4 - x_3|
Distance between K1 and K3=(x3x1)2+(y2y1)2 \text{Distance between } K_1 \text{ and } K_3 = \sqrt{(x_3 - x_1)^2 + (y_2 - y_1)^2}
Distance between K2 and K4=(x4x2)2+(y2y1)2 \text{Distance between } K_2 \text{ and } K_4 = \sqrt{(x_4 - x_2)^2 + (y_2 - y_1)^2}
For the quadrilateral to be a rhombus, we need:
x2x1=x4x3and(x3x1)2+(y2y1)2=(x4x2)2+(y2y1)2 |x_2 - x_1| = |x_4 - x_3| \quad \text{and} \quad \sqrt{(x_3 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_4 - x_2)^2 + (y_2 - y_1)^2}

5. Calculate the area of the rhombus:
The area A A of a rhombus can be calculated using the formula:
A=12×d1×d2 A = \frac{1}{2} \times d_1 \times d_2
where d1 d_1 and d2 d_2 are the lengths of the diagonals. The diagonals can be calculated as:
d1=(x2x1)2+(y2y1)2 d_1 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
d2=(x4x3)2+(y2y1)2 d_2 = \sqrt{(x_4 - x_3)^2 + (y_2 - y_1)^2}
Given that the area is 6 units, we have:
12×d1×d2=6 \frac{1}{2} \times d_1 \times d_2 = 6
Simplifying, we get:
d1×d2=12 d_1 \times d_2 = 12

6. Conclusion:
The quadrilateral formed by K1,K2,K3, K_1, K_2, K_3, and K4 K_4 is a rhombus if and only if the area is equal to 6 units, as shown by the conditions derived above.

\blacksquare

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.