A large piece of canvas is in the shape of an isosceles triangle with $A C=B C=5 \mathrm{~m}$ and $A B=6 \mathrm{~m}$, as shown. Point $D$ is the midpoint of $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_30_d1e45f8438f865d01c92g-4.jpg?height=599&width=634&top_left_y=281&top_left_x=800)

The canvas is folded along median $C D$ to create two faces of a tent $(\triangle A D C$ and $\triangle B D C)$. The third face of the tent $(\triangle A B D)$ is made from a separate piece of canvas. The bottom of the tent $(\triangle A B C)$ is open to the ground.

![](https://cdn.mathpix.com/cropped/2024_04_30_d1e45f8438f865d01c92g-4.jpg?height=450&width=727&top_left_y=1141&top_left_x=756)

What is the height of the tent (the distance from $D$ to the base $\triangle A B C$ ) when the volume of the tent is as large as possible?
Since $\triangle A C B$ is isosceles with $A C=B C$ and $C D$ is a median, then $C D$ is perpendicular to $A B$.

Since $C D$ is perpendicular to $A B$, then when $\triangle A B C$ is folded along $C D, C D$ will be perpendicular to the plane of $\triangle A D B$.

Therefore, the volume of the tent can be calculated as $\frac{1}{3}$ times the area of $\triangle A D B$ times $C D$. Since $C D$ is fixed in length, then the volume is maximized when the area of $\triangle A D B$ is maximized.

Since $A D=D B$ are fixed in length, the area of $\triangle A D B$ is maximized when $\angle A D B=90^{\circ}$. (One way to see this is to note that we can calculate the area of $\triangle A D B$ as $\frac{1}{2}(A D)(D B) \sin (\angle A D B)$, which is maximized when $\sin (\angle A D B)=1$.)

Since $A B=6 \mathrm{~m}$, then $A D=D B=3 \mathrm{~m}$.

Since $B D=3 \mathrm{~m}$ and $B C=5 \mathrm{~m}$, then $C D=4 \mathrm{~m}$. ( $\triangle B D C$ is a $3-4-5$ triangle.)

Thus, the maximum volume of the tent is

$$
\frac{1}{3}\left(\frac{1}{2}(A D)(D B)\right)(D C)=\frac{1}{6}(A D)(D B)(D C)=\frac{1}{6}(3 \mathrm{~m})(3 \mathrm{~m})(4 \mathrm{~m})=6 \mathrm{~m}^{3}
$$

To find the height, $h$, of $D$ above the base $\triangle A B C$, we note that the volume equals $\frac{1}{3}$ times the area of $\triangle A B C$ times the height, $h$.

Since we know that the volume is $6 \mathrm{~m}^{3}$, then if we calculate the area of $\triangle A B C$, we can solve for $h$.

Since $A D=D B=3 \mathrm{~m}$ and $\angle A D B=90^{\circ}$, then $A B=3 \sqrt{2} \mathrm{~m}$ in the maximum volume configuration.

Therefore, $\triangle A B C$ has $A B=3 \sqrt{2} \mathrm{~m}$ and $A C=B C=5 \mathrm{~m}$.

Let $N$ be the midpoint of $A B$.

![](https://cdn.mathpix.com/cropped/2024_04_30_dfe99ade8da3ee5d11fdg-16.jpg?height=558&width=590&top_left_y=182&top_left_x=819)

Since $A C=B C$, then $C N$ is perpendicular to $A B$.

Since $B C=5 \mathrm{~m}$ and $N B=\frac{3 \sqrt{2}}{2} \mathrm{~m}$, then by the Pythagorean Theorem,

$$
C N^{2}=B C^{2}-N B^{2}=(5 \mathrm{~m})^{2}-\left(\frac{3 \sqrt{2}}{2} \mathrm{~m}\right)^{2}=25 \mathrm{~m}^{2}-\frac{18}{4} \mathrm{~m}^{2}=25 \mathrm{~m}^{2}-\frac{9}{2} \mathrm{~m}^{2}=\frac{41}{2} \mathrm{~m}^{2}
$$

and so $C N=\sqrt{\frac{41}{2}} \mathrm{~m}$.

Thus, the area of $\triangle A B C$ is $\frac{1}{2}(A B)(C N)=\frac{1}{2}(3 \sqrt{2} \mathrm{~m})\left(\frac{\sqrt{41}}{\sqrt{2}} \mathrm{~m}\right)=\frac{3 \sqrt{41}}{2} \mathrm{~m}^{2}$.

Finally, we have $\frac{1}{3}\left(\frac{3 \sqrt{41}}{2} \mathrm{~m}^{2}\right) h=6 \mathrm{~m}^{3}$ and so $h=\frac{12}{\sqrt{41}} \mathrm{~m}$.

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)