Key Concepts and Formulas

• Circumcenter, Orthocenter, and Centroid Relationship: For any triangle $\triangle ABC$, if $P$ is the circumcenter, $Q$ is the orthocenter, and $G$ is the centroid, then $P, G, Q$ are collinear (on the Euler line). The centroid $G$ divides the segment $PQ$ in the ratio $1:2$, i.e., $PG:GQ = 1:2$.
• Vector Representation: If $O$ is the origin, the position vector of the centroid $G$ is given by $\overrightarrow{OG} = \frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}}{3}$.
• Vector Identity (Circumcenter as Origin): If the circumcenter $P$ of $\triangle ABC$ is chosen as the origin, then the position vector of the orthocenter $Q$ is given by $\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$.

Step-by-Step Solution

Step 1: Identify the Given Information and the Goal

• We are given a $\triangle ABC$, where $P$ is the circumcenter and $Q$ is the orthocenter.
• We need to find the value of the vector sum $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$.

Step 2: Strategically Choose the Origin

• Why this step? The problem involves the circumcenter and orthocenter. A fundamental vector identity connects these points when the circumcenter is taken as the origin. Choosing $P$ as the origin simplifies the position vectors of $A, B, C$ to $\overrightarrow{PA}, \overrightarrow{PB}, \overrightarrow{PC}$ respectively, and the position vector of $Q$ becomes $\overrightarrow{PQ}$.
• Let $P$ be the origin. Thus, the position vector of $P$ is $\vec{0}$.
• The position vectors of the vertices $A, B, C$ with respect to $P$ are $\overrightarrow{PA}, \overrightarrow{PB}, \overrightarrow{PC}$.
• The position vector of the orthocenter $Q$ with respect to $P$ is $\overrightarrow{PQ}$.

Step 3: Apply the Key Vector Identity

• Why this step? There is a direct vector identity relating the circumcenter, orthocenter, and the vertices of a triangle when the circumcenter is the origin. This identity is crucial for efficiently solving this problem.
• The identity states that if $P$ is the circumcenter and $Q$ is the orthocenter of $\triangle ABC$, and $P$ is taken as the origin, then: $\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$

Step 4: Conclude the Result

• Why this step? We have found an expression for the required vector sum directly from the identity.
• The expression we need to evaluate is $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$.
• From Step 3, we see that this sum is exactly equal to $\overrightarrow{PQ}$.
• Therefore, $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{PQ}$

Step 5: Match with Options

• The result obtained is $\overrightarrow{PQ}$.
• Comparing this with the given options: (A) $\overrightarrow{QP}$ (B) $\overrightarrow{PQ}$ (C) $2\overrightarrow{PQ}$ (D) $2\overrightarrow{QP}$
• Our result matches option (B).

Common Mistakes & Tips

• Confusing Orthocenter and Centroid: While the centroid's position vector is the average of vertices' position vectors, the orthocenter's relationship with the circumcenter and vertices is more direct as stated in the key identity.
• Vector Direction: Be careful with vector directions. $\overrightarrow{PQ}$ is from $P$ to $Q$, while $\overrightarrow{QP}$ is from $Q$ to $P$. They are opposite vectors: $\overrightarrow{QP} = -\overrightarrow{PQ}$.
• Memorize the Identity: The identity $\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$ (with $P$ as origin) is a very powerful tool for problems involving circumcenter and orthocenter.

Summary

The problem asks for the vector sum $\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$, where $P$ is the circumcenter and $Q$ is the orthocenter of $\triangle ABC$. By choosing the circumcenter $P$ as the origin, we can directly apply a fundamental vector identity that states the position vector of the orthocenter $Q$ (relative to $P$) is equal to the sum of the position vectors of the vertices $A, B, C$ (relative to $P$). This identity gives $\overrightarrow{PQ} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}$, directly providing the answer.

The final answer is $\boxed{\overrightarrow{PQ}}$.