Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line x + 2√

Q: Wait, why is the orthocenter the same as the centroid? Is this always true?

That is a fantastic question! It is not true for all triangles, but it is a defining, beautiful property of the equilateral triangle. Because of the perfect symmetry, the altitude, median, angle bisector, and perpendicular bisector from any vertex all coincide. Since the orthocenter is the intersection of altitudes and the centroid is the intersection of medians, they must be the same point.

Q: Why did we have to check the 'same side' condition? Can't we just pick either coordinate?

Geometry is not just about solving algebraic equations; it is about spatial reality. When we solve $\alpha^2 = \frac{64}{81}$, we get two mathematical possibilities for the vertex $A$. However, only one of those points actually forms a triangle with the given base $BC$ such that the origin remains the centroid. The 'same side' check ensures that our vertex $A$ is positioned correctly relative to the base line $BC$.

Q: Why is the distance $AO = \frac{8}{3}$?

Remember the $2:1$ ratio property of the centroid. The centroid divides the median (which is also the altitude here) into two segments: the segment from the vertex to the centroid ($AO$) and the segment from the centroid to the base ($OD$). The ratio is $AO:OD = 2:1$. Since we calculated $OD = \frac{4}{3}$, it follows that $AO = 2 \times \frac{4}{3} = \frac{8}{3}$.

Question

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

Options:

(A) 2

(B) 5

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Symmetry

In an equilateral triangle $	riangle ABC$, the orthocenter, centroid, circumcenter, and incenter coincide at a single point. Given that the orthocenter is at the origin $O(0,0)$, we conclude that the centroid of the triangle is also at the origin.

This symmetry is our primary anchor. It allows us to treat the distance from the origin to any side as the distance from the centroid to that side.

### Calculating the Distance to the Base

The line $BC$ is defined by the equation $x + 2\sqrt{2}y - 4 = 0$. We calculate the perpendicular distance $OD$ from the origin $(0,0)$ to this line using the standard formula:

$$OD = \frac{|1(0) + 2\sqrt{2}(0) - 4|}{\sqrt{1^2 + (2\sqrt{2})^2}} = \frac{|-4|}{\sqrt{1 + 8}} = \frac{4}{3}$$

Since the centroid divides the median in a $2:1$ ratio, the distance from the vertex $A$ to the centroid $O$ is exactly twice the distance from $O$ to the base $BC$. Therefore, the length $AO$ is:

$$AO = 2 	imes \frac{4}{3} = \frac{8}{3}$$

### Determining the Coordinates of Vertex A

The slope of the line $BC$ is $m_{BC} = -\frac{1}{2\sqrt{2}}$. Because the altitude $AD$ is perpendicular to $BC$, its slope $m_{AD}$ must be the negative reciprocal:

$$m_{AD} = 2\sqrt{2}$$

Since the line $AD$ passes through the origin, its equation is $y = 2\sqrt{2}x$. Let the coordinates of vertex $A$ be $(\alpha, \beta)$, such that $\beta = 2\sqrt{2}\alpha$.

Given the distance $AO = \frac{8}{3}$, we apply the distance formula:

$$\alpha^2 + \beta^2 = \left(\frac{8}{3}ight)^2 = \frac{64}{9}$$

Substituting $\beta = 2\sqrt{2}\alpha$ into the equation:

$$\alpha^2 + (2\sqrt{2}\alpha)^2 = \frac{64}{9} \Rightarrow \alpha^2 + 8\alpha^2 = \frac{64}{9} \Rightarrow 9\alpha^2 = \frac{64}{9}$$

This yields $\alpha^2 = \frac{64}{81}$, or $\alpha = \pm \frac{8}{9}$.

### Final Calculation and Result

To ensure $A$ lies on the correct side of the line $BC$, we test the coordinates. We find that $\alpha = -\frac{8}{9}$ is the valid solution, resulting in $\beta = 2\sqrt{2}(-\frac{8}{9}) = -\frac{16\sqrt{2}}{9}$.

We now evaluate the expression $|\alpha + \sqrt{2}\beta|$:

$$|\alpha + \sqrt{2}\beta| = \left|-\frac{8}{9} + \sqrt{2}\left(-\frac{16\sqrt{2}}{9}ight)ight| = \left|-\frac{8}{9} - \frac{32}{9}ight| = \left|-\frac{40}{9}ight| = \frac{40}{9} \approx 4.44$$

The greatest integer less than or equal to $4.44$ is **4**.

Animated Solution for Mathematics - Straight Lines: Let $A B C$ be an equilateral triangle with orthocenter at the origin and the side $B C$ on the line $x + 22 y = 4$ . If the co-ordinates of the vertex A are $(α, β)$ , then the greatest integer less than or equal to $∣ α + 2 β ∣$ is

Select Answer:

Visualized Solution

The Equilateral Property

The Base Line $B C$

Distance to Side $B C$

Calculating $O D$

The Centroid Ratio

Slope of the Altitude

Equation of Altitude $A D$

The Distance Equation

Solving for $α$

Selecting the Correct Vertex

Testing the Values

Evaluating the Expression

The Final Answer

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Geometric Symmetry

Calculating the Distance to the Base

Determining the Coordinates of Vertex A

Final Calculation and Result

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