Key Concepts and Formulas

• Orthocentre: The orthocentre of a triangle is the point where its three altitudes intersect. An altitude is a line segment from a vertex perpendicular to the opposite side.
• Slope of a Line: For a line in the form $Ax + By + C = 0$, the slope is $m = -\frac{A}{B}$. For a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$, the slope is $m = \frac{y_2 - y_1}{x_2 - x_1}$.
• Perpendicular Lines: Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 m_2 = -1$.

Step-by-Step Solution

Step 1: Identify Given Information

We are given the equations of the sides of triangle ABC:

• AB: $2x + y = 0$
• BC: $x + py = 15a$
• CA: $x - y = 3$ The orthocentre H is $(2, a)$, and $-\frac{1}{2} < a < 2$. We need to find the value of $p$.

Step 2: Calculate Slopes of the Sides

We find the slopes of the sides using the formula $m = -\frac{A}{B}$:

• Slope of AB ($m_{AB}$): $2x + y = 0 \implies m_{AB} = -\frac{2}{1} = -2$
• Slope of BC ($m_{BC}$): $x + py = 15a \implies m_{BC} = -\frac{1}{p}$
• Slope of CA ($m_{CA}$): $x - y = 3 \implies m_{CA} = -\frac{1}{-1} = 1$

Step 3: Find the Coordinates of Vertex A

Vertex A is the intersection of sides AB and CA. We solve the system of equations:

• $2x + y = 0 \implies y = -2x$
• $x - y = 3$ Substitute $y = -2x$ into the second equation: $x - (-2x) = 3 \implies 3x = 3 \implies x = 1$. Then $y = -2(1) = -2$. So, the coordinates of vertex A are $(1, -2)$.

Step 4: Apply Orthocentre Property (Altitude AH $\perp$ BC)

The altitude from A is perpendicular to BC. We find the slope of AH: $m_{AH} = \frac{a - (-2)}{2 - 1} = \frac{a + 2}{1} = a + 2$. Since AH $\perp$ BC, $m_{AH} \cdot m_{BC} = -1$. $(a + 2) \cdot \left(-\frac{1}{p}\right) = -1$ $\frac{a + 2}{p} = 1 \implies p = a + 2$ (Equation 1)

Step 5: Find the Coordinates of Vertex C

Vertex C is the intersection of sides BC and CA. We solve the system of equations:

• $x + py = 15a$
• $x - y = 3 \implies x = y + 3$ Substitute $x = y + 3$ into the first equation: $(y + 3) + py = 15a \implies y + py = 15a - 3 \implies y(1 + p) = 15a - 3 \implies y = \frac{15a - 3}{1 + p}$. Then $x = y + 3 = \frac{15a - 3}{1 + p} + 3 = \frac{15a - 3 + 3(1 + p)}{1 + p} = \frac{15a - 3 + 3 + 3p}{1 + p} = \frac{15a + 3p}{1 + p}$. So, the coordinates of vertex C are $\left(\frac{15a + 3p}{1 + p}, \frac{15a - 3}{1 + p}\right)$.

Step 6: Apply Orthocentre Property (Altitude CH $\perp$ AB)

The altitude from C is perpendicular to AB. We find the slope of CH: $m_{CH} = \frac{\frac{15a - 3}{1 + p} - a}{\frac{15a + 3p}{1 + p} - 2} = \frac{15a - 3 - a(1 + p)}{15a + 3p - 2(1 + p)} = \frac{15a - 3 - a - ap}{15a + 3p - 2 - 2p} = \frac{14a - 3 - ap}{15a + p - 2}$. Since CH $\perp$ AB, $m_{CH} \cdot m_{AB} = -1$. Thus, $m_{CH} = -\frac{1}{m_{AB}} = -\frac{1}{-2} = \frac{1}{2}$. $\frac{14a - 3 - ap}{15a + p - 2} = \frac{1}{2}$ $2(14a - 3 - ap) = 15a + p - 2$ $28a - 6 - 2ap = 15a + p - 2$ $13a - 4 = p + 2ap = p(1 + 2a)$ (Equation 2)

Step 7: Solve the System of Equations for 'a' and 'p'

We have:

1. $p = a + 2$
2. $13a - 4 = p(1 + 2a)$ Substitute Equation 1 into Equation 2: $13a - 4 = (a + 2)(1 + 2a)$ $13a - 4 = a + 2a^2 + 2 + 4a$ $13a - 4 = 2a^2 + 5a + 2$ $2a^2 - 8a + 6 = 0$ $a^2 - 4a + 3 = 0$ $(a - 1)(a - 3) = 0$ $a = 1$ or $a = 3$

Step 8: Check the Constraint for 'a' and Find 'p'

We are given $-\frac{1}{2} < a < 2$. If $a = 1$, then $-\frac{1}{2} < 1 < 2$, which is true. If $a = 3$, then $-\frac{1}{2} < 3 < 2$, which is false. Therefore, $a = 1$. Substitute $a = 1$ into Equation 1: $p = a + 2 = 1 + 2 = 3$

Common Mistakes & Tips

• Be careful with the signs when calculating slopes and applying the perpendicularity condition.
• Double-check your algebraic manipulations, especially when dealing with fractions.
• Remember to check the given constraint on the variable to eliminate extraneous solutions.

Summary

We used the properties of the orthocentre and altitudes to set up a system of equations involving the unknown parameters $a$ and $p$. Solving this system, and using the given constraint on $a$, allowed us to find the value of $p$.

The final answer is \boxed{3}.