In the 10 by 10 grid of squares shown, point $P$ can be at any of the 41 points of intersection of pairs of gridlines inside (and not on) $\triangle F G H$. For each possible location of $P$, exactly three triangles are formed: $\triangle F P G$, $\triangle G P H, \triangle H P F$. How many of these 123 triangles have an area that is exactly half of the area of $\triangle F G H$ ?
(A) 5
(B) 7
(C) 3
(D) 11
(E) 9

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-018.jpg?height=415&width=422&top_left_y=1451&top_left_x=1277)
We begin by determining the area of $\triangle F G H$.

The base $G H$ has length 10, the perpendicular height of the triangle from $G H$ to $F$ is also 10, and so the area of $\triangle F G H$ is $\frac{1}{2} \times 10 \times 10=50$.

We wish to determine which of the 41 points is a possible location for $P$ so that $\triangle F P G$ or $\triangle G P H$ or $\triangle H P F$ has area $\frac{1}{2} \times 50=25$.

We begin by considering the possible locations for $P$ so that $\triangle G P H$ has area 25.

The base $G H$ has length 10, and so the perpendicular height from $G H$ to $P$ must be 5 (since $\frac{1}{2} \times 10 \times 5=25$ ).

Since the distance between two parallel lines remains constant, any point $P$ lying on a line that is parallel to $G H$ and that is 5 units from $G H$ will give a $\triangle G P H$ whose area is 25 .

The line labelled $\ell$ is parallel to $G H$ and lies 5 units above $G H$,

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-072.jpg?height=372&width=379&top_left_y=627&top_left_x=1559)
and so any point that lies on $\ell$ is a distance of 5 units from base $G H$. There are 5 points that lie on $\ell$ (that are at the intersection of gridlines) and that are inside $\triangle F G H$. These 5 points and one of the 5 possibilities for $\triangle G P H$ are shown.

Next, we consider the possible locations for $P$ so that $\triangle F P G$ has area 25.

Consider the point $X$ on $G H$ so that $F X$ is perpendicular to $G H$. Since $\triangle F G H$ is isosceles, then $F X$ divides the area of $\triangle F G H$ in half and so $\triangle F X G$ has area 25 .

However, if $X$ lies on $G H$, then $X$ does not lie inside $\triangle F G H$.

For this reason, $X$ is not a possible location for $P$, however it does provide some valuable information and insight.

If we consider the base of $\triangle F X G$ to be $F G$, then the perpendicular distance from $X$ to $F G$ is equal to the height required from base

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-072.jpg?height=387&width=377&top_left_y=1254&top_left_x=1557)
$F G$ so that $\triangle F X G$ has area 25 .

Any point $P$ that lies on a line that is parallel to $F G$ and that is the same distance from $F G$ as $X$ will give a $\triangle F P G$ whose area is also 25 .

(This is the same property that we saw previously for $\triangle G P H$, with base $G H$.)

How do we create a line that passes through $X$ and that is parallel to $F G$ ?

Beginning at $G$, if we move 5 units right and 10 units up we arrive at $F$.

Begin at $X$, move 5 units right and 10 units up, and call this point $Y$.

Can you explain why the line segment $X Y$ is parallel to $F G$ ?

Any point that lies on $X Y$ is a distance from base $F G$ equal to the required height of $\triangle F P G$.

There are 2 points that lie on $X Y$ (that are at the intersection of

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-072.jpg?height=385&width=379&top_left_y=1905&top_left_x=1559)
gridlines) and that are inside $\triangle F G H$.

(We note that we may move right 5 and up 10 by moving in 'steps' of right 1 and up 2 to arrive at each of these 2 points.)

These 2 points and one possibility for $\triangle F P G$ are shown.

Finally, we consider the possible locations for $P$ so that $\triangle H P F$ has area 25 .

As a result of symmetry, this case is identical to the previous case.

Thus, there are 2 possible locations for $P$ so that $\triangle H P F$ has area 25.

In total, there are $5+2+2=9$ triangles that have an area that is exactly half of the area of $\triangle F G H$.

ANswer: (E)