Turn #1 - Prompt

In a parallelogram ( A B C D ), a point ( P ) is chosen on the diagonal ( A C ) such that ( A P: P C=2: 3 ). The line ( B P ) intersects the side ( A D ) at ( Q ) and the line ( D P ) intersects the side ( A B ) at ( R ). Let ( [P Q R] ) denote the area of triangle ( P Q R ). If the area of the parallelogram ( A B C D ) is ( k ), find the minimum value of ( [P Q R] ) in terms of ( k ).

verify this solution to above problem and let me know if its right or wrong

Step 1: Understand the given problem and figure

We are given a parallelogram $ABCD$ and a point $P$ on the diagonal $AC$ such that $AP:PC = 2:3$. The line $BP$ intersects the side $AD$ at $Q$ and the line $DP$ intersects the side $AB$ at $R$. We need to find the minimum value of the area of triangle $PQR$, denoted by $[PQR]$, in terms of the area of the parallelogram $ABCD$, which is given as $k$.

Step 2: Determine the areas of the triangles $APB$, $APD$, $BPC$, and $DPC$

Since $AP:PC = 2:3$, the area of triangle $APB$ is $\frac{2}{5}$ of the area of triangle $ABC$, and the area of triangle $BPC$ is $\frac{3}{5}$ of the area of triangle $ABC$. Similarly, the area of triangle $APD$ is $\frac{2}{5}$ of the area of triangle $ADC$, and the area of triangle $DPC$ is $\frac{3}{5}$ of the area of triangle $ADC$.

Step 3: Express the area of the parallelogram $ABCD$ in terms of the areas of the triangles

The area of the parallelogram $ABCD$ is equal to the sum of the areas of triangles $ABC$ and $ADC$, which is $k$. Since the area of triangle $ABC$ is equal to the area of triangle $ADC$, we have $[ABC] = [ADC] = \frac{k}{2}$.

Step 4: Find the areas of triangles $APB$, $APD$, $BPC$, and $DPC$ in terms of $k$

Using the results from steps 2 and 3, we have $[APB] = \frac{2}{5} \cdot \frac{k}{2} = \frac{k}{5}$, $[APD] = \frac{2}{5} \cdot \frac{k}{2} = \frac{k}{5}$, $[BPC] = \frac{3}{5} \cdot \frac{k}{2} = \frac{3k}{10}$, and $[DPC] = \frac{3}{5} \cdot \frac{k}{2} = \frac{3k}{10}$.

Step 5: Express the area of triangle $PQR$ in terms of the areas of the other triangles

The area of triangle $PQR$ can be expressed as $[PQR] = [ADR] - [DQP] - [AQR]$.

Step 6: Find the areas of triangles $ADR$, $DQP$, and $AQR$ in terms of $k$

We can express $[ADR]$ as $[ADR] = [ADC] - [APD] = \frac{k}{2} - \frac{k}{5} = \frac{3k}{10}$. Similarly, $[AQR] = [APB] - [PQR]$ and $[DQP] = [DPC] - [PQR]$. However, we cannot directly find $[AQR]$ and $[DQP]$ without knowing $[PQR]$. Instead, we can express $[AQR]$ and $[DQP]$ in terms of $[PQR]$.

Step 7: Express $[AQR]$ and $[DQP]$ in terms of $[PQR]$

Let $[AQR] = x - [PQR]$ and $[DQP] = y - [PQR]$, where $x$ and $y$ are some fractions of $k$. Substituting these expressions into the equation from Step 5, we get $[PQR] = [ADR] - (y - [PQR]) - (x - [PQR])$.

Step 8: Simplify the equation for $[PQR]$

Simplifying the equation, we get $[PQR] = [ADR] - x - y + 2[PQR]$. Rearranging the terms, we have $[PQR] = x + y - [ADR]$.

Step 9: Find the minimum value of $[PQR]$

To find the minimum value of $[PQR]$, we need to minimize $x + y$. Since $x$ and $y$ represent the areas of triangles $APB$ and $DPC$ minus $[PQR]$, the minimum value of $x + y$ occurs when $x = [APB]$ and $y = [DPC]$. Therefore, the minimum value of $[PQR]$ is $[APB] + [DPC] - [ADR]$.

Step 10: Calculate the minimum value of $[PQR]$

Substituting the values from Step 4, we get $[PQR] = \frac{k}{5} + \frac{3k}{10} - \frac{3k}{10} = \frac{k}{5} - \frac{k}{10} = \frac{k}{10}$.

Therefore, the final answer is: $\boxed{\frac{k}{5}}$.