The triangle $ABC$ below is equilateral, that is, it has all three sides of the same length and all its angles equal to $60^{\circ}$. Mr. Simas marks a point $H$ anywhere on the side $BC$ of the triangle. Then, he draws a segment parallel to the side $AC$, starting from $H$ and ending at point $I$ on the side $AB$. Next, he draws a segment parallel to the side $AB$, starting from $H$ and ending at point $J$ on the side $AC$, as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-05.jpg?height=623&width=711&top_left_y=702&top_left_x=775)

a) Knowing that the side $AB$ has a length equal to 1, calculate the perimeter of the quadrilateral $AIHJ$.

b) Mr. Simas continues drawing, as shown in the following figure, and draws the segments $LO$ and $MP$ perpendicular to the side $BC$.

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-05.jpg?height=625&width=708&top_left_y=1629&top_left_x=777)

Let $d$ be the length of the segment $IL$, let $f$ be the length of the segment $JM$, and let $x$ be the length of the segment $OP$. Show that

$$
x=\frac{1+d+f}{2}
$$
Solution

a) Since segment $I H$ is parallel to segment $A J$, and segment $A I$ is parallel to segment $J H$, we have that $\overline{I H}=\overline{A J}$ and $\overline{A I}=\overline{J H}$. Note: the bar above the segment denotes the length of the segment.

Since the side of triangle $A B C$ measures 1, we have that

$$
1=\overline{A B}=\overline{A I}+\overline{I B}=\overline{A I}+\overline{I H}
$$

Therefore, the perimeter of the quadrilateral is given by

$$
\overline{A I}+\overline{I H}+\overline{H J}+\overline{J A}=2(\overline{A I}+\overline{I H})=2
$$

b) Let's start by calling $\overline{B O}=m_{2}$ and $\overline{C P}=m_{1}$, as shown in the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-26.jpg?height=762&width=737&top_left_y=236&top_left_x=565)

The internal angles of an equilateral triangle are all equal to $60^{\circ}$. Therefore, $\measuredangle A C B=\measuredangle A B C=60^{\circ}$. Let's look at triangle $P C M$. Since $M P$ is perpendicular to $P C$, this triangle is a right triangle. Since the cosine of an angle (cos) is defined as the ratio of the adjacent side to that angle and the hypotenuse, we have that

$$
\cos 60^{\circ}=\frac{1}{2}=\frac{\overline{P C}}{\overline{M C}}=\frac{m_{1}}{\overline{M C}}
$$

From this, we conclude that $\overline{M C}=2 m_{1}$. By a similar argument, we also conclude that $\overline{L B}=2 m_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-26.jpg?height=760&width=737&top_left_y=1516&top_left_x=565)

Since triangle $H J C$ is equilateral, we have that $\overline{H J}=f+2 m_{1}$. Since $\overline{A I}=\overline{H J}$, we conclude that $\overline{A I}=f+2 m_{1}$. Since $\overline{A B}=1$, we have that $2 m_{1}+d+f+2 m_{2}=1$, and therefore, $m_{1}+m_{2}=\frac{1-d-f}{2}$. On the other hand, since $\overline{B C}=1$, we have that

$$
m_{1}+x+m_{2}=1
$$

Substituting one equation into the other, we get

$$
x=\frac{1+d+f}{2} .
$$