Does the couch fit through the door?

Since the coach is longer than the height of the room door, you can not just get upright through the door. Instead, the coach has to fit through the door frame with its entire cross section of the L-shape. Therefore, we consider only the two-dimensional problem definition, whether the L-shape fits properly turned into the rectangular area of the door cross-section.

The cross section of the coach is a polygon $ABCDEF$ . We consider the line $g$ , which goes through the two points $C$ and $E$ . This line has a distance $d$ to the point $A$ . So if $d \leq 70$ , then the coach fits through the door frame.

We put the point $A$ in the origin of our coordinate system. The line $g$ is then spanned by the two position vectors $\vec C = (98,14)$ and $\vec E = (14,77)$ . This results in a straight line equation $g = \{\vec C + \lambda \cdot \vec V | \lambda \in \mathbb {R} \}$ with the difference vector $\vec V = \vec E - \vec C = (- 84, 63)$ , which is parallel to $g$ . The distance to point $A$ corresponds to the minimum length of all position vectors in $g$ . This minimum is determined by setting the derivative after the parameter $\lambda$ to zero: $\begin{aligned} & & d &= \min_{\lambda \in \mathbb{R}} |\vec C + \lambda \cdot \vec V| \\ \Rightarrow & & 0 &= \frac{d}{d \lambda} |\vec C + \lambda \cdot \vec V| \\ & & &= \frac{\frac{d}{d \lambda} (\vec C + \lambda \cdot \vec V^2)}{2 |\vec C + \lambda \cdot \vec V|} \\ & & &= \frac{\frac{d}{d \lambda} (\vec C^2 + \lambda^2 \cdot \vec V^2 + 2 \lambda \cdot \vec C \cdot \vec V)}{2 |\vec C + \lambda \cdot \vec V|} \\ & & &= \frac{ \lambda \cdot \vec V^2 + \vec C \cdot \vec V}{ |\vec C + \lambda \cdot \vec V|}\\ \Rightarrow & & \lambda &= - \frac{\vec C \cdot \vec V}{\vec V^2} \\ & & &= - \frac{\left( \begin{array}{c} 98 \\ 14 \end{array} \right) \cdot \left( \begin{array}{c} -84 \\ 63 \end{array} \right)}{ \left( \begin{array}{c} -84 \\ 63 \end{array} \right) \cdot \left( \begin{array}{c} -84 \\ 63 \end{array} \right) } \\ & & &= -\frac{-7350}{11025} \\ & & &= \frac{2}{3} \\ \Rightarrow & & d &= \left|\vec C + \frac{2}{3} \cdot \vec V\right| \\ & & &= \left| \left( \begin{array}{c} 98 \\ 14 \end{array} \right) + \frac{2}{3} \cdot \left( \begin{array}{c} -84 \\ 63 \end{array} \right) \right| \\ & & &= \left| \left( \begin{array}{c} 42\\ 56\end{array} \right) \right| \\ & & &= \sqrt{42^2 + 56^2} \\ & & &= 70 \end{aligned}$ Thus, the couch fits from the width exactly through the door.

As a second condition you would have to check if the height fits. However, this is not a problem since the diagonal of the L-shape has the length $\sqrt {98 ^ 2 + 77 ^ 2} \approx 125$ and is therefore much smaller than the door height.

The solution is illustrated here graphically:

The couch can fit if the L-shaped cross-section is rotated parallel to the wall with the door and then tipped diagonally, as shown:

If the sides of the couch are extended as in the diagram, then there are several similar right triangles we can use in our calculations to show that it will fit through the door.

Since $ID = 77 - 14 = 63$ , $IF = 98 - 14 = 84$ , $BD = 14$ , $FH = 14$ , and $\triangle IDF \sim \triangle BCD \sim \triangle HFG$ , $BC = 14 \cdot \frac{63}{84} = \frac{21}{2}$ and $HG = 14 \cdot \frac{84}{63} = \frac{56}{3}$ .

This means $AC = 77 + \frac{21}{2} = \frac{175}{2}$ and $AG = 98 + \frac{56}{3} = \frac{350}{3}$ . By Pythagorean's Theorem on $\triangle ACG$ , $CG = \sqrt{(\frac{175}{2})^2 + (\frac{350}{3})^2} = \frac{875}{6}$ . (Since this is less than the door's height of $180$ , we know that this orientation of the couch will fit through the door by height.)

Finally, since $\triangle ACE \sim \triangle ACG$ , $AE = \frac{350}{3} \cdot \frac{\frac{175}{2}}{\frac{875}{6}} = 70$ , and we know that this orientation of the couch will also fit (just barely) through the door by width.