Key Concepts and Formulas

• Centroid of a Triangle: The position vector of the centroid ($G$) of a triangle with vertices having position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3}$.
• Euler Line Property: For any non-equilateral triangle, the orthocenter ($H$), centroid ($G$), and circumcenter ($C_c$) are collinear. The centroid divides the segment $HC_c$ in the ratio $2:1$, i.e., $\vec{g} = \frac{1 \cdot \vec{c_c} + 2 \cdot \vec{h}}{1+2} = \frac{\vec{c_c} + 2\vec{h}}{3}$. This can be rearranged to $\vec{c_c} = 3\vec{g} - 2\vec{h}$.
• Vector Representation: Position vectors can be expressed as linear combinations of basis vectors.

Step-by-Step Solution

Step 1: Identify the position vectors of the vertices. Let the position vectors of the three vertices of the triangle be $\vec{a}$, $\vec{b}$, and $\vec{c}$. Given: $\vec{a} = 4 \vec{p}+\vec{q}-3 \vec{r}$ $\vec{b} = -5 \vec{p}+\vec{q}+2 \vec{r}$ $\vec{c} = 2 \vec{p}-\vec{q}+2 \vec{r}$

Step 2: Calculate the position vector of the centroid ($G$). The centroid's position vector is the average of the vertices' position vectors. $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3}$ Substitute the given vectors: $\vec{g} = \frac{(4 \vec{p}+\vec{q}-3 \vec{r}) + (-5 \vec{p}+\vec{q}+2 \vec{r}) + (2 \vec{p}-\vec{q}+2 \vec{r})}{3}$ Combine the coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$: $\vec{g} = \frac{(4-5+2)\vec{p} + (1+1-1)\vec{q} + (-3+2+2)\vec{r}}{3}$ $\vec{g} = \frac{1\vec{p} + 1\vec{q} + 1\vec{r}}{3}$ $\vec{g} = \frac{\vec{p}+\vec{q}+\vec{r}}{3}$

Step 3: Use the Euler line property to relate the orthocenter, centroid, and circumcenter. We are given the position vector of the orthocenter ($H$) as $\vec{h} = \frac{\vec{p}+\vec{q}+\vec{r}}{4}$. We are given the position vector of the circumcenter ($C_c$) as $\vec{c_c} = \alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$. The Euler line property states that $\vec{c_c} = 3\vec{g} - 2\vec{h}$. Substitute the calculated $\vec{g}$ and given $\vec{h}$: $\vec{c_c} = 3 \left(\frac{\vec{p}+\vec{q}+\vec{r}}{3}\right) - 2 \left(\frac{\vec{p}+\vec{q}+\vec{r}}{4}\right)$ Simplify the expression: $\vec{c_c} = (\vec{p}+\vec{q}+\vec{r}) - \left(\frac{\vec{p}+\vec{q}+\vec{r}}{2}\right)$ $\vec{c_c} = \frac{2(\vec{p}+\vec{q}+\vec{r}) - (\vec{p}+\vec{q}+\vec{r})}{2}$ $\vec{c_c} = \frac{2\vec{p}+2\vec{q}+2\vec{r} - \vec{p}-\vec{q}-\vec{r}}{2}$ $\vec{c_c} = \frac{(2-1)\vec{p} + (2-1)\vec{q} + (2-1)\vec{r}}{2}$ $\vec{c_c} = \frac{\vec{p}+\vec{q}+\vec{r}}{2}$

Step 4: Determine the values of $\alpha$, $\beta$, and $\gamma$. We have $\vec{c_c} = \alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ and we found $\vec{c_c} = \frac{\vec{p}+\vec{q}+\vec{r}}{2}$. By comparing the coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$: $\alpha = \frac{1}{2}$ $\beta = \frac{1}{2}$ $\gamma = \frac{1}{2}$

Step 5: Calculate the value of $\alpha+2 \beta+5 \gamma$. Substitute the determined values of $\alpha$, $\beta$, and $\gamma$: $\alpha+2 \beta+5 \gamma = \frac{1}{2} + 2\left(\frac{1}{2}\right) + 5\left(\frac{1}{2}\right)$ $\alpha+2 \beta+5 \gamma = \frac{1}{2} + 1 + \frac{5}{2}$ Combine the terms: $\alpha+2 \beta+5 \gamma = \frac{1}{2} + \frac{2}{2} + \frac{5}{2}$ $\alpha+2 \beta+5 \gamma = \frac{1+2+5}{2}$ $\alpha+2 \beta+5 \gamma = \frac{8}{2}$ $\alpha+2 \beta+5 \gamma = 4$

Common Mistakes & Tips

• Incorrect Euler Line Ratio: Ensure the correct ratio ($2:1$ for $H:C_c$ with $G$ in between) is used. The formula $\vec{g} = \frac{1 \cdot \vec{c_c} + 2 \cdot \vec{h}}{3}$ is crucial.
• Algebraic Errors: Be meticulous when combining coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$ during vector addition and subtraction.
• Misinterpreting the Question: Double-check that you are calculating the required expression ($\alpha+2 \beta+5 \gamma$) and not just finding $\alpha$, $\beta$, or $\gamma$ individually.

Summary

This problem leverages the fundamental relationship between the orthocenter, centroid, and circumcenter of a triangle, known as the Euler line property. First, we calculated the position vector of the centroid by averaging the position vectors of the vertices. Then, using the Euler line property ($\vec{c_c} = 3\vec{g} - 2\vec{h}$), we derived the position vector of the circumcenter in terms of $\vec{p}$, $\vec{q}$, and $\vec{r}$. By comparing this with the given form of the circumcenter's position vector, we determined the values of $\alpha$, $\beta$, and $\gamma$. Finally, we substituted these values into the expression $\alpha+2 \beta+5 \gamma$ to obtain the answer.

The final answer is \boxed{4} which corresponds to option (A).