You are supposed to solve this (part 3)

Source: Geogebra

Let $CE$ and $DF$ be 2 altitudes of the trapezoid $ACDB$ . Let $S$ be the area of the trapezoid $ACDB$ . c Let the side $AC$ be $a$ , the base $CB$ be $b$ , the altitude $CE$ be $h$ and $AE$ be $t$ . Notice that the total length of $AC$ , $CD$ , and $DB$ is equal to the length of XY. Hence, we have: $2a+b=1$ .

Consider the triangle $\triangle ACE$ , applying the Pythagoras theorem, we have:

${ AE }^{ 2 }={ AC }^{ 2 }-{ EC }^{ 2 }$

$\Rightarrow { h }^{ 2 }={ a }^{ 2 }-{ t }^{ 2 }$

$\Rightarrow { h }=\sqrt { { a }^{ 2 }-{ t }^{ 2 } }$

Notice that $CE // DF$ (since they are both the altitudes of the trapezoid ACDB) and $AB // CD$ (since they are both the bases of the trapezoid ACDB), hence $EFDC$ is a parallelogram $\Rightarrow \begin{cases} EF=CD=b \\ CE=DF \end{cases}$ .

Consider $\triangle ACE$ and $\triangle BDF$ , applying the Pythagoras theorem, we have:

$\begin{cases} { AE }^{ 2 }={ AC }^{ 2 }-{ EC }^{ 2 } \\ { BF }^{ 2 }={ BD }^{ 2 }-{ DF }^{ 2 } \end{cases}\\ \\ \\ \\$

Moreover, we had $\begin{cases} AC=BD \\ CE=DF \end{cases}\\ \\ \\ \\$ .

From that, we can conclude that $BF=AE=t$

The area of the trapezoid $ACDB$ is:

$S=\frac { (AB+CD)CE }{ 2 }$

$\Rightarrow S=\frac { (AE+EF+FB+b)h }{ 2 }$

$\Rightarrow S=(t+b)\sqrt { { a }^{ 2 }-t^{ 2 } }$

$\Rightarrow { S }^{ 2 }={ (t+b) }^{ 2 }({ a }^{ 2 }-{ t }^{ 2 })$

$\Rightarrow { S }^{ 2 }={ (t+b) }^{ 2 }({ a }-t)(a+t)$

$\Rightarrow { 3S }^{ 2 }={ (t+b) }(t+b)(3{ a }-3t)(a+t)$

Applying the $AM-GM$ inequality with 4 positive numbers: $t+b$ , $t+b$ , $3a-3t$ , $a+t$ , we have:

$\Rightarrow { 3S }^{ 2 }\le { (\frac { (t+b)+(t+b)+(3a-3t)+(a+t) }{ 4 } ) }^{ 4 }$

$\Rightarrow { 3S }^{ 2 }\le { (\frac { 2a+b }{ 2 } ) }^{ 4 }$

$\Rightarrow { 3S }^{ 2 }\le { (\frac { 1 }{ 2 } ) }^{ 4 }$

$\Rightarrow { S }^{ 2 }\le \frac { 1 }{ 48 }$

$\Rightarrow S\le \frac { \sqrt { 3 } }{ 12 }$ .

The equality holds if and only if $t+b= 3a-3t= a+t$

$\Rightarrow \begin{cases} a=b \\ 2a=4t \end{cases}$

$\Rightarrow \begin{cases} 3a=1 \\ \frac { t }{ a } =\frac { 1 }{ 2 } \end{cases}$

$\Rightarrow \begin{cases} a=\frac { 1 }{ 3 } \\ \frac { AE }{ AC } =\frac { 1 }{ 2 } \end{cases}$

$\Rightarrow \begin{cases} AC=\frac { 1 }{ 3 } \\ \sin { (\angle ACE) } =\frac { 1 }{ 2 } \end{cases}$

$\Rightarrow \begin{cases} \frac { x }{ y } =\frac { 1 }{ 3 } \\ \angle ACE=30° \end{cases}$

$\Rightarrow \begin{cases} x=1 \\ y=3 \\ \angle ACD=120° \end{cases}$

$\Rightarrow \begin{cases} x=1 \\ y=3 \\ z=120 \end{cases}$

$\Rightarrow xyz=360$ .

We have our final answer: $\boxed{360}$ .