# Math Help - Construction Problem

1. ## Construction Problem

Given a river with parallel banks n and m, and two villages, A and B. Where should a bridge, CD, perpendicular to the banks, be constructed in order to minimize the total distance AC + CD + DB?

2. Pretend that the vertical segment, i.e., the bridge, is the last of the three segments. In other words, switch CD and DB.

3. I don't quite understand what you mean. I know that if the measure of angle ACE is congruent to the measure of angle BDF, then this creates the shortest distance. I just don't know why?

4. Hello, MATNTRNG1

$\text{Given a river with parallel banks }m\text{ and }n\text{, and two villages, }A\text{ and }B,$
$\text{where should a bridge }CD\text{, perpendicular to the banks, be constructed}$
$\text{in order to minimize the total distance }AC + CD + DB\,?$
Code:
            A
o
|   *
|       *   C
m - + - - + - - - - - ♥ - - - - - - - - -
:     |           |   *
w:     o           |       *   Q
:     P   *       |           o
:             *   |           |
n - + - - - - - - - - ♥ - - - - - + - - -
D   *       |
*   |
o
B

We can determine $\,w$, the width of the river.

From $\,A$ construct segment $AP$ so that: . $AP \perp m\,\text{ and }\,|AP| = w.$

Frim $\,B$ construct segment $BQ$ so that: . $BQ \perp n \,\text{ and }\,|BQ| = w.$

Draw $AQ$ intersecting $\,m$ at $\,C.$

Draw $PB$ intersecting $\,n$ at $\,D.$

$CD$ is the location of the bridge.

5. Soroban... You never cease to amaze! I know that you are correct but what would be the proof of this? Why does this create the shortest distance of AC + CD + DB?

6. I don't quite understand what you mean.
People traveling from A to B have to pass the bridge in any case. You can as well pretend that the bridge is the last segment of the journey. The rest of the way is the shortest when it is a straight line.

7. Hello, MATNTRNG1

This is my original diagram.

Code:
            A
o
|   *
|       *   C
m - + - - + - - - - - ♥ - - - - - - - - -
:     |           |   *
w:     o           |       *   Q
:     P   *       |           o
:             *   |           |
n - + - - - - - - - - ♥ - - - - - + - - -
D   *       |
*   |
o
B

Now reduce the river to a line (it has width 0).

Code:
            A
o
|   *
|       *   C
m - - - - + - - - - - ♥ - - - - - - - - -
D   *       |
*   |
o
B

Got it?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

emarakov has a truly elegant solution!

Code:
            A
o
*
*   C
m - - - - - - - - - - ♥ - - - - - - - - -
|
|
|
|
n - - - - - - - - - - ♥ - - - - - - - - -
D   *
*
o
B

The width of the river is constant.
We want to minimize the total diagonal distance: . $AC + DB.$

Switch $CD$ and $DB.$

Code:
            A
o
*
*   D
m - - - - - - - - - - ♥ - - - - - - - - -
*
*   B
o C
|
n -   - - - - - - - - - - - - - - + - - -
|
|
o
D

The total diagonal distance in a minimum when $AB$ is a straight line.

See?