In equilateral triangle $A B C$, line segments are drawn from a point $P$ to the vertices $A, B$ and $C$ to form three identical triangles. The points $D, E$ and $F$ are the midpoints of the three sides and they are joined as shown in the diagram. What fraction of $\triangle A B C$ is shaded?
(A) $\frac{1}{5}$
(B) $\frac{5}{24}$
(C) $\frac{1}{4}$
(D) $\frac{2}{9}$
(E) $\frac{2}{7}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-097.jpg?height=366&width=374&top_left_y=1815&top_left_x=1361)
In equilateral triangle $A B C$, line segments are drawn from a point $P$ to the vertices $A, B$ and $C$ to form three identical triangles. The points $D, E$ and $F$ are the midpoints of the three sides and they are joined as shown in the diagram. What fraction of $\triangle A B C$ is shaded?
(A) $\frac{1}{5}$
(B) $\frac{5}{24}$
(C) $\frac{1}{4}$
(D) $\frac{2}{9}$
(E) $\frac{2}{7}$

## Solution 1

Since $P$ is a point of symmetry within $\triangle A B C$, the line segment $C P$ divides $\triangle E C F$ into 2 triangles of equal area. That is to say, the area of $\triangle E K C$ equals the area of $\triangle F K C$. Since the area of $\triangle E F C$ is $\frac{1}{4}$ the area of $\triangle A B C$, the area of $\triangle E K C=\left(\frac{1}{2} \times \frac{1}{4}\right)$ area of $\triangle A B C$ $=\frac{1}{8}($ area of $\triangle A B C)$.

Again since $P$ is a point of symmetry within $\triangle A B C$, the area of $\triangle A P C$ is $\frac{1}{3}$ the area of $\triangle A B C$. Since the shaded area is the area of $\triangle A P C-$ area of $\triangle K C E$, it represents $\left(\frac{1}{3}-\frac{1}{8}\right) \times$ area of $\triangle A B C=\frac{5}{24} \times$ area of $\triangle A B C$.

## Solution 2

Since $D, E$ and $F$ are the midpoints of the sides, we have four triangles of exactly the same area. That is to say, the areas of $\triangle A D E, \triangle D B F, \triangle D E F$, and $\triangle E F C$ are equal. Since $\triangle A M E$ equals half the area of $\triangle A D E$, it represents $\frac{1}{8}$ th the area of $\triangle A B C$.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-435.jpg?height=1592&width=388&top_left_y=838&top_left_x=1321)

Since the figure $M E N P$ is one of three identical shapes making up $\triangle D E F$ it is one third its area. Since $\triangle D E F$ itself is one quarter the area of $\triangle A B C$, the figure $M E N P$ is $\frac{1}{3} \times \frac{1}{4}$ or $\frac{1}{12}$ th the area of $\triangle A B C$. Overall, the shaded area is

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-436.jpg?height=219&width=219&top_left_y=324&top_left_x=1403)
$\frac{1}{8}+\frac{1}{12}=\frac{5}{24}$ th the area of $\triangle A B C$.

ANswer: (B)