Question

Let \( ABC \) be an equilateral triangle with orthocenter at the origin and the side \( BC \) lying on the line \( x+2\sqrt{2}\,y=4 \). If the coordinates of the vertex \( A \) are \( (\alpha,\beta) \), then the greatest integer less than or equal to \( |\alpha+\sqrt{2}\beta| \) is:

• A. \(2\)
• B. \(4\)
• C. \(5\)
• D. \(3\)

Hint:

In equilateral triangle coordinate problems, placing the centroid or orthocenter at the origin greatly simplifies calculations using symmetry.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In an equilateral triangle, the orthocenter, centroid, and circumcenter coincide. The distance from the centroid to a side is \(1/3\) of the altitude.
Step 2: Detailed Explanation:
Let \(O(0, 0)\) be the centroid. The distance from \(O\) to side \(BC\) (\(x + 2\sqrt{2} y - 4 = 0\)) is:
\[ r = \frac{|0 + 0 - 4|}{\sqrt{1^2 + (2\sqrt{2})^2}} = \frac{4}{\sqrt{1 + 8}} = \frac{4}{3} \] The altitude of the triangle is \(h = 3r = 4\).
The distance from the centroid to vertex \(A\) is \(R = 2r = 8/3\).
Vertex \(A\) lies on the line passing through origin and perpendicular to \(BC\).
The equation of this line is \(2\sqrt{2}x - y = 0 \Rightarrow y = 2\sqrt{2}x\).
Also, \(A(\alpha, \beta)\) must be on the opposite side of \(BC\) relative to the origin.
The origin side is \(0 + 0 - 4 = -4\) (negative). Thus \(A\) must satisfy \(\alpha + 2\sqrt{2}\beta - 4>0\).
Since \(\alpha^2 + \beta^2 = R^2 = (8/3)^2 = 64/9\):
\[ \alpha^2 + (2\sqrt{2}\alpha)^2 = 64/9 \Rightarrow 9\alpha^2 = 64/9 \Rightarrow \alpha = \pm 8/9 \] If \(\alpha = -8/9\), then \(\beta = -16\sqrt{2}/9\). Testing sign: \(-8/9 - 64/9 - 4<0\) (same side).
So, we must have \(\alpha = -8/9, \beta = -16\sqrt{2}/9\) is not the case for vertex \(A\).
Wait, the orthocenter is at origin. The distance from orthocenter to vertex is \(R\). The vertex \(A\) is on the line segment such that \(O\) is between \(A\) and \(BC\).
Using vectors: \(\vec{OA} = -2\vec{OD}\) where \(D\) is the foot of perpendicular.
\(\vec{OD} = \frac{4}{9}(1, 2\sqrt{2})\). Then \(\vec{OA} = (-\frac{8}{9}, -\frac{16\sqrt{2}}{9})\).
\[ |\alpha + \sqrt{2}\beta| = \left| -\frac{8}{9} + \sqrt{2}\left(-\frac{16\sqrt{2}}{9}\right) \right| = \left| -\frac{8}{9} - \frac{32}{9} \right| = \left| -\frac{40}{9} \right| = \frac{40}{9} \approx 4.44 \] The greatest integer value is \(\lfloor 4.44 \rfloor = 4\).
Step 3: Final Answer:
The result is 4.