Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Bulgaria

Problem:
Find the area of the triangle determined by the straight line with equation $x - y + 1 = 0$ and the tangent lines to the graph of the parabola $y = x^{2} - 4x + 5$ at its common points with the line.

Solution

Solution:
The common points of the graphs of the line and the parabola are $A(1, 2)$ and $B(4, 5)$ . The equations of the tangents to the graph of the parabola at $A$ and $B$ are $y = -2x + 11$ and $y = 4x - 11$ , respectively. The intersecting point of the two tangents is the point $C\left(\frac{5}{2}, -1\right)$ . The area of $\triangle ABC$ then equals $\frac{27}{4}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.