Question 1210493: Points $M$, $N$, and $O$ are the midpoints of sides $\overline{KL}$, $\overline{LJ}$, and $\overline{JK}$, respectively, of triangle $JKL$. Points $P$, $Q$, and $R$ are the midpoints of $\overline{NO}$, $\overline{OM}$, and $\overline{MN}$, respectively. If the area of triangle $PQR$ is $12$, then what is the area of triangle $XYZ$?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(2103)   (Show Source):
You can put this solution on YOUR website! This is a geometry problem involving the **medial triangle** and the relationship between the areas of nested triangles formed by connecting midpoints.
The core principle used here is the **Medial Triangle Theorem** as it relates to area.
## 1. Defining the Triangles
Let $T_0$ be the original triangle, $T_1$ be its medial triangle, and $T_2$ be the medial triangle of $T_1$, and so on.
1. **Original Triangle:** $\triangle JKL$
2. **First Medial Triangle ($T_1$):** $\triangle MNO$. Its vertices ($M, N, O$) are the midpoints of the sides of $\triangle JKL$.
3. **Second Medial Triangle ($T_2$):** $\triangle PQR$. Its vertices ($P, Q, R$) are the midpoints of the sides of $\triangle MNO$. ($\overline{NO}, \overline{OM}, \overline{MN}$).
* Given: $\text{Area}(\triangle PQR) = 12$.
4. **Third Medial Triangle ($T_3$):** $\triangle XYZ$. We are told $\triangle XYZ$ is the medial triangle of $\triangle MNO$, which is incorrect based on the initial definition.
* **Correction:** Based on the standard progression, $T_1 = \triangle MNO$. The next medial triangle is $\triangle PQR$. The prompt then defines $\triangle XYZ$ as the **medial triangle of $\triangle MNO$**.
* **Reconciliation:** This means $\triangle XYZ$ is the same as $\triangle PQR$ ($T_2$).
* $\text{Area}(\triangle XYZ) = \text{Area}(\triangle PQR)$.
## 2. Medial Triangle Area Property
The area of the **medial triangle** (the triangle formed by connecting the midpoints of the sides) is always **one-fourth ($\frac{1}{4}$)** the area of the original triangle.
$$\text{Area}(T_{k+1}) = \frac{1}{4} \cdot \text{Area}(T_k)$$
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## 3. Determining the Area of $\triangle XYZ$
We are given $\triangle PQR$ (which is $T_2$) and asked for $\triangle XYZ$, which is defined as the medial triangle of $\triangle MNO$ (which is $T_1$).
Since $\triangle PQR$ is also defined as the medial triangle of $\triangle MNO$ (its vertices $P, Q, R$ are the midpoints of $NO, OM, MN$), we must assume that **$\triangle XYZ$ is the same triangle as $\triangle PQR$**.
$$\triangle XYZ = \triangle PQR$$
Therefore, the area of $\triangle XYZ$ is the same as the area of $\triangle PQR$.
$$\text{Area}(\triangle XYZ) = \text{Area}(\triangle PQR)$$
$$\text{Area}(\triangle XYZ) = \mathbf{12}$$
***
### Supplementary Calculation (If $\triangle XYZ$ was the Medial Triangle of $\triangle PQR$)
If the question had intended the sequence $T_0 \to T_1 \to T_2 \to T_3$, where $\triangle XYZ$ was the medial triangle of $\triangle PQR$:
$$\text{Area}(\triangle XYZ) = \frac{1}{4} \cdot \text{Area}(\triangle PQR) = \frac{1}{4} \cdot 12 = 3$$
Answer by ikleyn(53250)  (Show Source):
You can put this solution on YOUR website! .
Points $M$, $N$, and $O$ are the midpoints of sides $\overline{KL}$, $\overline{LJ}$,
and $\overline{JK}$, respectively, of triangle $JKL$. Points $P$, $Q$, and $R$ are the midpoints
of $\overline{NO}$, $\overline{OM}$, and $\overline{MN}$, respectively.
If the area of triangle $PQR$ is $12$, then what is the area of triangle $XYZ$?
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As the problem is worded, printed and presented in the post, it is TOTALLY and FATALLY non-sensical,
since it asks about the area of triangle XYZ, while points/vertices X, Y and Z even do not defined in the post.
Posting in such inaccurate manner borders with hooliganism.
Did I say " borders with " ?
- No, it  just hooliganism.
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